text:
# How to apply hermite InterpolatingPolynomial?

I found that to do hermite InterpolatingPolynomial
I need to write:

InterpolatingPolynomial[{{{1}, 2, 3}, {{2}, 6, 7, 8}}, x]

however,i need to apply it on a table:

 Table [{xi = a + (b - a)/n i ,  N[Round[10^7  f [xi]]/10^7],
N[ N[Round[10^7  f2 [xi]]/10^7]]}, {i, 0, n}]  ]

so how can i do that?

thanks.

you have it close, just need an extra {} on the x term: ( making up a simple example..)

n = 4; a = 0; b = 1; f[x_] := 10^12 Sin[2 x] ; f2[x_] := 10^12 Cos[x] ;
poly = InterpolatingPolynomial[
Table[{{xi = a + (b - a)/n i}, N@Round[ f[xi], 10^7],
N@Round[f2[xi], 10^7]}, {i, 0, n}], x] // Simplify

0.+ 1.*10^12 x + 6.76392*10^13 x^2 - 7.54631*10^14 x^3 + 3.6613*10^15 x^4 - 9.58558*10^15 x^5 + 1.44937*10^16 x^6 - 1.26549*10^16 x^7 + 5.91826*10^15 x^8 - 1.14596*10^15 x^9

Plot[{f[x], poly}, {x, 0, 1},
PlotStyle -> {Thickness[.01], {Thin, Red}}]

I cleaned things up using the 2 arg form of Round.

You should notice I deliberately fed it the wrong derivative just for illustration, otherwise you get a perfect fit for this example.

Edit: if you already have the table of the form

 table={{x1,f[x1],fp[x1]},{x2,f[x2],fp[x2]},...}

you do like this to add in the { }:

 InterpolatingPolynomial[{List@First@#, Sequence @@ Rest@#} & /@ table, x]


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# Algebra Word Problem Solver

Algebra problems tend to be worded a little differently from the typical 'Ann had a dozen apples' kind of arithmetic questions. Algebra deals with variables and numbers which are arranged in equations to form generalizations. Hence the terms used are understandably a bit different in order to convey the relation between different elements. Algebra word problems require students to decipher the meaning and form the correct equations. Algebra word problem solver is an online tool which will help users understand how algebra word problems are stated. With the word problem solver for algebra, students learn what the different terms mean and what kind of relation they imply.

## Word Problem Solver Algebra

Online algebra word problem solvers are available for all grades and levels of algebra. Algebra 1 word problem solvers and algebra 2 word problem solvers cover the beginner and intermediate levels of algebra. Free algebra word problem solvers also give you tips and hints on which terms or phrases to look out for. Learning to identify these will enable you to understand what you are expected to find out much quicker. College algebra word problem solvers are designed to handle the advanced course load and are very popular among college students who turn to the internet for a substantial amount of course help and research.

## Solved Examples

Question 1: The base of a triangle is 2 m more than twice its altitude. If the area of the triangle is 30 square m. Find its altitude.

Solution:
Let the altitude of the triangle = x

therefore, base of the triangle = 2 + 2x

Given, area of the triangle = 30

Step 1:

Area of the triangle = $\frac{1}{2}$ Base * Height

=> 30 = $\frac{1}{2}$ * (2 + 2x) * x

=> 30 * 2 = (2 + 2x) * x

=> 60 = 2x + 2x2

=> 2x2 + 2x - 60 = 0

Step 2:

Solve for x, 2x2 + 2x - 60 = 0

=> 2x2 + 12x - 10x - 60 = 0

=> 2x(x + 6) - 10(x + 6) = 0

=> (2x - 10)(x + 6) = 0

Step 3:
either 2x - 10 = 0    or    x + 6 = 0

=> 2x = 10     or     x = - 6

Length must be positive, so neglect x = - 6.

=> 2x = 10

=> x = 5

Hence altitude of the triangle is 5 m. answer

Question 2: The sum of the ages of a father and his son is 45. 5 years ago, father's age is three years more than thrice the sons age. Find the age of the son 4 years hence.

Solution:
Let the age of the son = x years

Therefore the age of the father = 45 - x
(Sum of their ages is 45)

Step 1:
5 years ago:

Son's age = x - 5

and father's age  = 45 - x - 5 = 40 - x

Step 2:
The problem states:

Father's age = 3 + 3(Son's age)

=> 40 - x = 3 + 3(x - 5)

=> 40 - x = 3 + 3x - 15

=> 40 - x = 3x - 12

=> 3x + x = 40 + 12

=> 4x = 52

Divide each side by 4

=> x = $\frac{52}{4}$

=> x = 13

Son's present age = 13 years

=> After 4 years son will be of = 13 + 4 = 17 years.

format: plain


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Are Newton's gravity waves detectable by a laser interferometer?

Newton's theory of gravity supports "gravity waves" in that moving objects cause changing gravitational fields. For example, two bodies rotating around their center of mass will have a stronger gravitational field when they are longitudinally oriented than when they are transverse oriented. Given two masses of mass $M$ orbiting on a circle of radius $r$, at a distance $d$ from an observer, the strength of the attraction is: $$\begin{matrix} F_{min} &=& \frac{2GM}{d^2},\\ F_{max} &=& \frac{2GM}{d^2}\frac{1+r^2/d^2}{(1-r^2/d^2)^2} \end{matrix}$$

This should be detectable at long distance.

My question is this: With that sort of gravity, would the laser interferometer based gravity wave detectors be able to detect the gravity wave? An example is the LIGO, Laster Interferometer Gravitational-wave Observatory.

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You're asking purely about a Newtonian treatment, not the full GR analysis, right? –  David Z Apr 7 '11 at 2:55
Dear Carl, there are no "waves" in Newtonian gravity. A "wave" is something that has shape - minima and maxima - that depends on space. There are no oscillations like that, as a function of space, in Newtonian gravity. The gravitational force is uniformly decreasing away from the source. This totally changes the situation. You apparently want to measure the time depedendence of the quadrupole moment - but it's very tiny, a hopeless power law. You can't measure "tides" from distant galaxies or binary stars. Gravity waves, which only exist in GR, have effects that drop slower with the distance. –  Luboš Motl Apr 7 '11 at 5:32
@Lubos; Instead of "waves" call it what you like. My question is whether it's detectable at LIGO. –  Carl Brannen Apr 8 '11 at 20:46

This is exactly the approach taken in Bernard Shutz's note "Gravitational waves on the back of an envelope" (Am. J. Phys. 50 vol 5 pp 412). The abstract reads:

Using only Newtonian gravity and a little special relativity we calculate most of the important effects of gravitational radiation, with results very close to the predictions of full general relativity theory. Used with care, this approach gives helpful back‐of‐the‐envelope derivations of important equations and estimates, and it can help to teach gravitational wave phenomena to undergraduates and others not expert in general relativity. We use it to derive the following: the quadrupole approximation for the amplitude h of gravitational waves; a simple upper bound on h in terms of the Newtonian gravitational field of the source; the energy flux in the waves, the luminosity of the source (called the ‘‘quadrupole formula’’), and the radiation reaction in the source; order‐of‐magnitude estimates for radiation from supernovae and binary star systems; and the rate of change of the orbital period of the binary pulsar system. Where our simple results differ from those of general relativity we quote the relativistic ones as well. We finish with a derivation of the principles of detecting gravitational waves, and we discuss the principal types of detectors under construction and the major limitations on their sensitivity.

(If you don't have access to Am J Phys, this talk seems to recapitulate the details.)

A major difference in this Newtonian scalar theory from the real GR theory of gravitational waves is in the effect of the waves on an inertial test particle. This Newtonian theory predicts that the waves would appear as an oscillating force along the direction between the source and the test particle. By contrast, GR predicts an oscillating differential tidal effect in the plane perpendicular to the line connecting the source and the test mass.

As a result, while I think LIGO would still detect the "Newtonian" form of gravitational waves, the antenna pattern of the detector would be different. The L-shaped LIGO detector has optimal sensitivity to a source located directly overhead in the GR case (allowing the gravitational wave to stretch one arm while it is compressing the other). There would be no sensitivity to a "Newtonian" source directly overhead. However, you could detect it if the "Newtonian" source were aligned with either arm.

By the way, "Newtonian noise" (the near-field action of Newtonian gravity arising from density waves in the material near the detector) is a real concern for terrestrial gravitational wave detectors!

P.S. To be pedantic, it is best to avoid the term "gravity wave" (as opposed to "gravitational wave"), since a "gravity wave" ("Newtonian gravity wave" even!) is something completely different.

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Thanks for posting something referenced! –  Sklivvz Apr 7 '11 at 18:10

I don't think so. To first order, the maximum variation in your Newton's gravity force is $$F_{min}-F_{max}=\frac{6GMmr^2}{d^4}$$

If we assume that the effects of gravity propagate at light speed, then in order for one mirror to experience $F_{max}$ while the other experiences $F_{min}$, the objects must rotate 90 degrees in the time $l/c$ where $l$ is the path length of the interferometer (4 kilometers for LIGO).

If we choose the speed of light as an upper bound for the speed of the rotating objects, this means that they can be separated by no more than about 5 kilometers ($r=2.5$).

A lower bound on $d$ can be 4 lightyears, the distance to the nearest star.

Let's suppose the mass of a mirror is $m = 100 kg$, and the mass of the rotating objects is 20 billion solar masses, the largest known mass in the universe.

Punching this into google calculator gives a force difference of about $10^{-28}$N, which is 4 orders of magnitude smaller than any force yet measured - and of course most of these assumptions are unphysically generous.

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It's newtonian gravity, so there is no propagation speed. The effect is instantaneous. That and the falling off as $1/d^4$ like you mention, makes this not very "gravitational radiation" like in my opinion. –  John Apr 7 '11 at 3:06
yeah, I wasn't sure about that.. I was thinking you could maybe treat it in a semi-relativistic manner by assuming a propagation speed of c, kind of like basic relativistic E&M and retarded potentials. There's probably lots of problems with this approach, but I think it's less wrong than assuming an instantaneous propagation which clearly violates special relativity. –  user2963 Apr 7 '11 at 3:24
The main problem is that there's no wave equation for the waves to satisfy. You can do something ad-hoc like applying retarded potentials, but that makes sense in Maxwell theory because retarded and advanced potentials solve Maxwell's equations. What is the "Newtonian" wave equation? –  Jerry Schirmer Apr 7 '11 at 4:59
The operation of LIGO is actually more easily understood in the long wavelength approximation, in which the gravitational wave strain changes very slowly compared to the time needed for light to circulate in the interferometer arms. It is not necessary to arrange for one mirror (of the two in an arm) to see a maximum of the wave while the other sees a minimum, since we are looking for a tidal effect. –  nibot Apr 7 '11 at 16:17

This does not imply gravity waves. This can be seen by the following calculation. The mass configuration is that of two masses in a circular orbit around each other. This is a dipole term, which defines ${\vec p}~=~m{\vec d}$, for $\vec d$ the distance between the two masses. So this dipole moment projected along the field determines a dipole potential $$\phi_d(r)~=~{\vec p}\cdot\nabla\frac{GMm}{r}$$ If this generates power in waves that escape the system, then a time derivative is dependent on $md{\vec d}/dt$. However, momentum conservation tells us this must be zero in the initial frame of the two masses.

So in order for gravity waves to exist one must have quadrupole moments and higher. However, this contributes to gravity waves in the post Newtonian expansion to second order. The

A gravity wave is a perturbation on a background metric $\eta_{ab}$ with the total metric $$g_{ab}~=~\eta_{ab}~+~h_{ab}.$$ The flat background metric has zero Ricci curvature so that to first order in the perturbation expansion $R_{ab}~=~\delta R_{ab}$, which enters into the Einstein field equation $R_{ab}~-~\frac{1}{2}Rg_{ab}~=~\kappa T_{ab}$, where $\kappa~=~8\pi G/c^4$ is the very small coupling constant between the momentum-energy source and the spacetime configuration or field. The Ricci curvature to first order is then $$R_{ab}~=~{1\over 2}\Big(\partial_c\partial_a{h^c}_b~+~\partial_c\partial_b{h^c}_a~-~\partial_a\partial_bh)~-~\partial_c\partial^ch_{ab}\Big).$$ The harmonic gauge $g^{bc}\Gamma^a_{bc}~=~0$, to first order as $\partial_c{h^c}_a~=~\frac{1}{2}\partial_\mu h$ the Einstein field equation gives $$\partial^c\partial_ch_{ab}~-~\frac{1}{2}\eta_{ab}\partial^c\partial_ch~=~{{16\pi G}\over {c^4}}T_{ab},$$ which is well defined for the traceless metric term ${\bar h}_{ab}~=~h_{ab}~-~\frac{1}{2}\eta_{ab}h$ with the simple wave equation $$\partial^c\partial_c{\bar h}_{ab}~=~{{16\pi G}\over {c^4}}T_{ab}.$$ This is then a basic wave equation for a tensor field with two independent components $h_{\times\times}$ and $h_{++}$ for polarization directions.

What is of interest is the source of the gravity waves $T_{ab}$. The metric perturbation far from the source $r~=~|{\bf x}|~>>~|{\bf y}|$ is approximated by, $${\bar h}_{ab}~=~{{4G}\over{rc^4}}\int_V d^3yT_{ab}(t~-~r/c,~{\bf y}).$$ The covariant constancy of the momentum-energy tensor $\nabla\cdot {\bf T}~=~0$ results in a number of relationships. The energy $T_{tt}~=~\rho U_0U_0$ component of with these relationships gives that for the quadrupole moment $I_{ij}~=~\int_Vx_ix_j\rho(t,~{\bf x})$ the metric wave is given by $${\bar h}_{ij}~=~{{2G}\over {rc^4}}{{d^2}\over{dt^2}}I_{ij}(t,~{\bf x}).$$ This term is different from a naïve idea of gravity waves suggested here. The coupling constant is $~\sim~G/c^4$ which comes from the Einstein field equation and is extremely small. The post Newtonian terms to $c^{-4}$ are second order, and this is a departure from a Newtonian gravity.

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Isn't that one has to content with quadrupole moment, because mass dipoles don't exist? (Whether antimatter/matter might make up such a mass dipole I dont know.) Black holes circling each other are surmised to emit gravity waves. What is the difference to Carls masses? –  Georg Apr 7 '11 at 13:56


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# Freezing water in a closed container [duplicate]

We know that density of ice reduces by about 8% during freezing, this means it expands to have little higher volume. But if I fill water in a container (entire volume) which has very low coefficient of thermal expansion at 0 $^\circ$C, will the space constraints stop water from freezing?

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## marked as duplicate by nemu, Brandon Enright, John Rennie, jinawee, Waffle's Crazy PeanutDec 18 '13 at 17:29

Does the space constraint of a cast-iron engine block stop a pure water coolant from freezing? –  User58220 Dec 19 '13 at 0:12
I worked with a professor who used to do the exploding cast iron ball full of water demo. Energetic and surprising even though you know what is coming. –  dmckee Dec 19 '13 at 4:40

If you force your liquid in a container you will increase the pressure during the freezing process. This will lower the freezing temperature and thus will indeed "stop the water from freezing". However, if you cool down further, probably at some point a different crystal structure (or the same with lower inter-atomic distances) will form and thus the water will still freeze, just with higher density.

While I do not know this effect from water I know for sure this happens to other materials if the volume is limited.

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For thermo with no stress/strain you look at the free energy: $$F=U-TS$$ but when you incorperate stress/strain there is an additional term like: $$F=U-TS-\sigma d\epsilon$$ Which incorperates the energy of compressing the solid.


text:
# Homework Help: Change in speed of the bullet after striking an object

1. Dec 17, 2012

### omc1

1. The problem statement, all variables and given/known data
A gangster fires a bullet (m1 = 30 g, v1 = 190 m/s) at Stupendous Man (m2 = 31 kg), but it simply bounces away elastically. If Stupendous man was standing on a frictionless surface, what is the change in speed of the bullet after striking stupendous man? Be careful with signs and units!

2. Relevant equations conservation of momentum, mv1+mv2=mv1+mv2

3. The attempt at a solution the man is not moving initially, and final, i got 0.183 for the final bullet speed but thats not right, Iam confused about how to find the final for the bullet?

2. Dec 17, 2012

### SammyS

Staff Emeritus
You haven't explained how you got that result, so it's hard to say where you went wrong.

The problem doesn't state the direction in which the bullet bounces away, so I would assume that the bullet goes back toward the location it was fired from.

Did you use the fact that it's an elastic collision ?

Such a problem can often be solved most easily in the center of mass reference frame. If you do, don't round-off anything until your final answer.

3. Dec 17, 2012

### omc1

i thought i was using the formula for elastic collision, .03*190=.03*V2+31*V2, but I though the man did not move so I am confused about that, how would I apply the cm to this?

4. Dec 17, 2012

### SammyS

Staff Emeritus
That is for an inelastic collision. That's evident because you have both the bullet and the man traveling at the same velocity.

5. Dec 17, 2012

### omc1

but i dont know either final velocity...

6. Dec 17, 2012

### Vineeth T

Is the answer 0.367ms-1.

Soln:
We have two equations (one from momentum conservation and the other from kinetic
energy conservation) with two variables (final velocities of the bullet and the man).Solve these
two equations and get the final velocity of the man in terms of initial velocity of both and their masses.

The final equation will be:

V=M2U2/(M1+M2)

where M1 is the mass of the man,
M2 is the mass of the bullet,
U2 is the velocity of the bullet.

7. Dec 18, 2012

### omc1

yes, that is the answer, why is the mass combined because the bullet bounces off.

8. Dec 18, 2012

### SammyS

Staff Emeritus
In your above equation (from your Original Post), you have two unknowns. You need another equation in order to get a solution.

Let's call the final velocity of the man, V2, and the final velocity of the bullet, u2.

The initial velocities being V1 = 0 m/s and u1 = 190 m/s .

Similarly, let M be the mass of the man, M = 31 kg. Let m be the mass of the bullet, m = 30 grams = 0.03 kg.

(I took it upon myself to change some variable names.)

Conservation of momentum gives $\ mu_1+MV_1=mu_2+MV_2\$ where V1=0, for the (initially) stationary man.
$\ mu_1=mu_2+MV_2\$

The other equation you need is from conservation of kinetic energy (It's an elastic collision.)
$\ (1/2)m{u_1}^2=(1/2)m{u_2}^2+(1/2)M{V_2}^2\$

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text:
# A limit without Taylor series or l'Hôpital's rule $\lim_{n\to\infty}\prod_{k=1}^{n}\cos \frac{k}{n\sqrt{n}}$

Computing without Taylor series or l'Hôpital's rule

$$\lim_{n\to\infty}\prod_{k=1}^{n}\cos \frac{k}{n\sqrt{n}}$$

What options would I have here? Thanks!

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The best option is to prove Taylor series formula and use this answer –  no identity Feb 24 '13 at 22:41
@Norbert: Teacher says that we need no Taylor here. –  Chris's sis Feb 24 '13 at 22:47
Did the teacher suggest anything, apart from saying what not to do? –  Pedro Tamaroff Feb 24 '13 at 22:56
@PeterTamaroff: no clue from his side. All I know is that he is always right. –  Chris's sis Feb 24 '13 at 22:58
Is using the fact that $\sin(x) \geq x-\frac{x^3}{6}$and $\log(1+x)\geq x-\frac{x^2}{2}$ legal? –  Ishan Banerjee Feb 25 '13 at 13:32

The following tries to use only "basic" inequalities, avoiding L'Hopital and Taylor.

The inequality $$\tag1 e^x\ge 1+x\qquad x\in\mathbb R$$ should be well-known and after taking logarithms immediately leads to $$\tag2 \ln(1+x)\le x\qquad x>-1$$ and after taking reciprocals $$\tag3 e^{-x}\le \frac1{1+x}\qquad x>-1$$ Substituting $-\frac x{1+x}$ for $x$ in $(3)$ and taking logarithms, we have $$\tag4 \frac x{1+x}\le \ln(1+x)\qquad x>-1.$$ For $0<x<\frac\pi2$ we have $$\ln\cos x=\ln\sqrt{1-\sin^2x}=\frac12\ln(1-\sin^2x)$$ and by $(2)$ and $(4)$ $$\tag 5 -\frac12\tan^2x=\frac12 \frac{-\sin^2x}{1-\sin^2x}\le \ln\cos x\le -\frac12\sin^2x.$$ Since $0\le \sin x \le x\le \tan x$ for $0\le x<\frac\pi2$ and the quotient of the upper and the lower bound in $(5)$ is just $\cos^2x$ we find for $0<x\le y<\frac\pi2$ (using $\cos^2y\le \cos^2x$) $$-\frac12 x^2\cdot\frac1{\cos^2y}\le \ln\cos x\le -\frac12 x^2\cdot \cos^2y.$$ Letting (for $n\ge1$) $y=\frac1{\sqrt n}$, $x=\frac k{n\sqrt n}$ with $1\le k\le n$, this gives $$-\frac{k^2}{2n^3}\cdot\frac{1}{\cos^2\frac1{\sqrt n}}\le \ln\cos \frac k{n\sqrt n}\le -\frac{k^2}{2n^3}\cdot\cos^2\frac1{\sqrt n}.$$ Using $1^2+2^2+\ldots + n^2=\frac{n(n+1)(2n+1)}{6}$ we find by summation $$-\frac{n(n+1)(2n+1)}{12n^3}\cdot\frac{1}{\cos^2\frac1{\sqrt n}}\le\sum_{k=1}^n\ln\cos\frac k{n\sqrt n}\le-\frac{n(n+1)(2n+1)}{12n^3}\cdot\cos^2\frac1{\sqrt n}$$ and by taking the limit $$\sum_{k=1}^\infty\ln\cos\frac k{n\sqrt n}=-\frac16$$ and ultimately $$\prod_{k=1}^\infty\cos\frac k{n\sqrt n}=e^{-\frac16}.$$

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I think it is $\exp(-1/6)$, but this makes sense. –  Pedro Tamaroff Feb 24 '13 at 23:37
Oops, I see I have $h^2$ in $(5)$ where there should be $\sqrt h$. Not only does this change th elimit to invoke l'Hopital, but it also makes the error estimate $(2)$ vs. $(1)$ wrong (though that could be mended). I'll have to rethink. –  Hagen von Eitzen Feb 25 '13 at 11:21
@HagenvonEitzen: your style of solving problems is somewhat different from the "usual" styles. I like to see that. Consequently, I definitely choose your answer. –  Chris's sis Feb 25 '13 at 16:48

The best way (I think) to solve this problem is to use $\ln(1-x)=-x+O(x^2)$ and $\sin x=x+O(x^3)$. In fact, \begin{eqnarray*} \lim_{n\to\infty}\ln\prod_{k=1}^n\cos\frac{k}{n\sqrt{n}}&=&\lim_{n\to\infty}\sum_{k=1}^n\ln\cos\frac{k}{n\sqrt{n}}=\lim_{n\to\infty}\frac{1}{2}\sum_{k=1}^n\ln\cos^2\frac{k}{n\sqrt{n}}\\ &=&\lim_{n\to\infty}\frac{1}{2}\sum_{k=1}^n\ln(1-\sin^2\frac{k}{n\sqrt{n}})\\ &=&-\lim_{n\to\infty}\frac{1}{2}\sum_{k=1}^n\left[\left(\frac{k}{n\sqrt{n}}\right)^2+O\left(\left(\frac{k}{n\sqrt{n}}\right)^4\right)\right]\\ &=&-\lim_{n\to\infty}\frac{1}{2}\sum_{k=1}^n\frac{k^2}{n^3}=-\frac{1}{6}. \end{eqnarray*} Hence $$\lim_{n\to\infty}\prod_{k=1}^n\cos\frac{k}{n\sqrt{n}}=e^{-\frac{1}{6}}.$$

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I am just going to prove that the product converges. To get the exact value, you need more advanced stuff.

Start with $|\cos(x)-1| \le x^2/2$. This can be proved from $\cos(2x) = \cos^2(x)-\sin^2(x) = 1 - 2 \sin^2(x)$, so $|\cos(2x)-1| = 2|\sin^2(x)|$. Since $|\sin(x)| \le x$ for all $x$, $|\cos(2x)-1| \le 2x^2$ or $|\cos(x)-1| \le x^2/2$.

Since $|\cos(x)| \le 1 + x^2/2$, $|\cos(\frac{k}{n\sqrt{n}})| \le 1 + \frac{k^2}{2n^3}$. Since $\sum_{k=1}^n \frac{k^2}{2n^3} = \frac1{2n^3} \sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{12n^3} < 1$ for $n > 1$, the product converges (since $\prod (1+a_n) < \exp(\sum a_n)$ if $a_n > 0$).

-


text:
# A Narrow-stencil finite difference method for approximating viscosity solutions of fully nonlinear elliptic partial differential equations with applications to Hamilton-Jacobi-

This paper presents a new narrow-stencil finite difference method for approximating the viscosity solution of second order fully nonlinear elliptic partial differential equations including Hamilton-Jacobi-Bellman equations. The proposed finite difference method naturally extends the Lax-Friedrichs method for first order problems to second order problems by introducing a key stabilization and guiding term called a "numerical moment". The numerical moment uses the difference of two (central) Hessian approximations to resolve the potential low-regularity of viscosity solutions. It is proved that the proposed scheme is well posed (i.e, it has a unique solution) and stable in both the l-2 norm and the l-infinity norm. The highlight of the paper is to prove the convergence of the proposed scheme to the viscosity solution of the underlying fully nonlinear second order problem using a novel discrete comparison argument. This paper extends the one-dimensional analogous method of Feng, Kao, and Lewis to the higher-dimensional setting. Numerical tests are presented to gauge the performance of the proposed finite difference methods and to validate the convergence result of the paper.

## Authors

• 11 publications
• 1 publication
03/12/2019

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## 1 Introduction

This paper develops and analyzes a narrow-stencil finite difference approximation of the viscosity solution to the following fully nonlinear second order Dirichlet boundary value problem:

 (1a) F[u](x)≡F(D2u,∇u,u,x) =0, ∀x∈Ω, (1b) u(x) =g(x), ∀x∈∂Ω,

where () is a bounded domain and denotes the Hessian matrix of at . The PDE operator is a fully nonlinear second order differential operator in the sense that is nonlinear in (or in at least one of its components). We assume is continuous and is Lipschitz in its first three arguments. Moreover, is assumed to be elliptic and satisfy a comparison principle (defined in Section 2.3). The paper focusses on a particular class of fully nonlinear PDEs called Hamilton-Jacobi-Bellman (HJB) equations which correspond to , where

 (2a) H[u] ≡minθ∈Θ(Lθu−fθ), (2b) Lθu =Aθ(x):D2u+bθ(x)⋅∇u+cθ(x)u.

HJB equations arise in various applications including stochastic optimal control, game theory, and mathematical finance (cf.

[15]).

Numerical fully nonlinear PDEs has experienced some rapid developments in recent years. Various numerical methods have been proposed and tested (cf. [12, 23] and the references therein). In order to guarantee convergence to the viscosity solution of problem (1), most of these works follow the framework laid out by Barles-Souganidis in [1] which requires that a numerical scheme be monotone, consistent, and stable. It follows from a well-known result of [22] that, in general, monotone schemes are necessarily wide-stencil schemes as seen in [9, 10, 26, 29]. Consequently, to design convergent narrow-stencil schemes, one has to abandon the standard monotonicity requirement (in the sense of Barles-Souganidis [1]) and, instead, deal with non-monotone schemes. This in turn prohibits one to directly use the powerful and handy Barles-Souganidis (analysis) framework. On the other hand, it should be pointed out that monotone narrow-stencil schemes could be constructed for problem (1) in special cases. For example, when the matrix is diagonally dominant for all , the methods proposed in [20, 2] become narrow-stencil schemes even though they require wide-stencils for general negative semi-definite matrices . For isotropic coefficient matrices (i.e., diagonal), a monotone linear finite element method was proposed in [18]. Finally, we also mention that when satisfies a Cordes condition, a least square-like interior penalty discontinuous Galerkin (IP-DG) method was proposed in [30]. This method is obviously a narrow-stencil scheme, and it was proved to be convergent even though it is not monotone.

The primary goal of this paper is to develop a convergent narrow-stencil finite difference method for problem (1) in the general case when is uniformly elliptic. Thus, we only require to be uniformly negative definite as long as the underlying PDE satisfies the comparison principle. The method to be introduced in this paper can be regarded as a high dimensional extension of the one-dimensional finite difference method developed and analyzed by the authors in [14]. We note that this extension is far from trivial in terms of both method design and convergence analysis because of the additional difficulties caused by the second order mixed derivatives on the off-diagonal entries of the Hessian matrix for . Indeed, constructing “correct” finite difference discretizations for those mixed derivatives and controlling their “bad” effect in the convergence analysis poses some significant new challenges. Recall that the main reason to use wide-stencils in the works cited above is exactly to “correctly” approximate the second order mixed derivatives. Hence, the new challenges which must be dealt with in this paper can be regarded as the price we need to pay for not using wide-stencils.

To achieve our goal, several key ideas are introduced and utilized in this paper. First, we introduce multiple discrete approximations of and to locally capture the behavior of the underlying function when it has low regularity. The reason for using such multiple approximations is to compensate for the low resolution of a fixed grid, especially when using a narrow-stencil finite difference approximation. Second, as explained above, using narrow-stencils forces us to abandon the hope of constructing monotone finite difference schemes (again, in the sense of Barles-Souganidis [1]). However, in order to obtain a convergent scheme, the anticipated finite difference method must have some “good” structures. Our key motivating idea is to impose such a structure, called generalized monotonicity, on our finite difference scheme. We shall demonstrate that besides allowing for the use of narrow-stencils, generalized monotonicity is a more natural concept/structure than the usual monotonicity concept/structure, and it is much easier to verify. Third, to obtain practical generalized monotone finite difference methods, two important questions to answer are: how to build such a scheme and what is the primary mechanism that makes the method work? Inspired by the vanishing moment method developed in [11, 13], we propose a higher order stabilization term/mechanism, called a numerical moment, as a means to address these questions. We will see that the stabilization term/mechanism is the key for us to overcome the lack of monotonicity in the admissibility, stability, and convergence analysis.

The remainder of this paper is organized as follows. In Section 2 we introduce the basics of viscosity solution theory and elliptic operators as well as the corresponding notation. We also define our primary finite difference operators that will be used throughout the paper and list some identities for those difference operators. In Section 3 we first formulate our finite difference method and then prove several properties of the scheme. Section 4 is devoted to studying the well-posedness of the scheme. A contraction mapping technique based on a pseudo-time explicit Euler scheme is employed to prove the existence and uniqueness for the proposed finite difference scheme and to derive an

-norm stability estimate. In Section

5, an -norm stability estimate for the second-order central difference quotients of the numerical solution is first derived using a Sobolev iteration technique. Then an -norm stability estimate is derived based on combining both estimates. The results in Sections 4 and 5 rely upon some auxiliary linear algebra results that can be found in Appendix A. In Section 6 we present a detailed convergence analysis for the proposed finite difference method. In Section 7 we provide some numerical experiments that validate and complement the theory. The paper is concluded in Section 8 with a brief summary and some final remarks.

## 2 Preliminaries

We begin by recalling the viscosity solution concept for problem (1) as documented in [6, 8, 5] and defining basic difference operators that will be used to construct our finite difference method. We define the relevant function space notation in Section 2.1, state the definition of viscosity solutions in Section 2.2, and recall the definition of ellipticity and the comparison principle in Section 2.3. The comparison principle ensures uniqueness of solutions to (1). The specific assumptions for HJB equations will be given in Section 2.4. Lastly, in Section 2.5, we define our basic difference operators which will serve as the building blocks for constructing various discrete gradient and Hessian operators. The discrete gradient and Hessian operators will be used to define our narrow-stencil finite difference method in Section 3.

### 2.1 Function spaces and notation

Standard function and space notation as in [4] and [16] will be adopted in this paper. For example, for a bounded open domain , , , and are used to denote, respectively, the spaces of bounded, upper semicontinuous, and lower semicontinuous functions on . For any , we define

 v∗(x)≡limsupy→xv(y)andv∗(x)≡liminfy→xv(y).

Then, and , and they are called the upper and lower semicontinuous envelopes of , respectively. We also let denote the set of symmetric real-valued matrices.

In presenting the relevant viscosity solution concept for fully nonlinear second order PDEs, we shall use to denote a general fully nonlinear second order operator. More precisely, . The general second order fully nonlinear PDE problem is to seek a locally bounded function such that is a viscosity solution of

 (3) G(D2u,∇u,u,x) =0in ¯¯¯¯Ω,

where we have adopted the convention of writing the boundary condition as a discontinuity of the PDE (cf. [1, p.274]). We shall treat the Dirichlet boundary condition explicitly in the formulation of our finite difference method for (3).

### 2.2 Viscosity solutions

Due to the full nonlinearity of the differential operator in (3), the standard weak solution concept based upon lowering the order of derivatives on the solution function using integration by parts is not applicable. Thus, a different solution concept is necessary for fully nonlinear PDEs. Among several available weak solution concepts for fully nonlinear PDEs, the best known one is the viscosity solution notion whose definition is stated below.

Let denote the second order operator in (3).

(i) A locally bounded function is called a viscosity subsolution of (3) if , when has a local maximum at ,

(ii) A locally bounded function is called a viscosity supersolution of (3) if , when has a local minimum at ,

(iii) A locally bounded function is called a viscosity solution of (3) if is both a viscosity subsolution and a viscosity supersolution of (3).

###### Remark 2.1

(a) Applying the above definition to the Dirichlet boundary value problem (1), we require the viscosity solution satisfies the Dirichlet boundary condition (1b) in the viscosity sense. Using the notation of [5, Section 7], this can be described as follows. First, define the boundary operator by

 (4) B(A,q,v,x)≡v(x)−g(x)

based on the Dirichlet boundary condition (1b), and then define the differential operator by

 G(D2u,∇u,u,x)≡{F[u](x),if x∈Ω,B[u](x),if x∈∂Ω.

Then we have

 G∗(A,q,v,x)={F(A,q,v,x),if x∈Ω,min{F(A,q,v,x),B(A,q,v,x)},if x∈∂Ω

and

 G∗(A,q,v,x)={F(A,q,v,x),if x∈Ω,max{F(A,q,v,x),B(A,q,v,x)},if x∈∂Ω.

Using the boundary operator , we have explicitly written the boundary condition as a discontinuity of the PDE operator .

(b) If is continuous, then another way to enforce the Dirichlet boundary condition is in the pointwise sense. As explained in [5], the enforcing of the Dirichlet boundary condition in the viscosity sense and in the pointwise sense may not be equivalent in general.

The above definitions can be informally interpreted as follows. Without a loss of generality, we may assume whenever has a local maximum at or whenever has a local minimum at . Then, is a viscosity solution of (3) if for all smooth functions such that “touches” the graph of from above at , we have , and for all smooth functions such that “touches” the graph of from below at , we have . A geometric interpretation of the definition of viscosity solutions can be found in Figure 1. Since the definition is based on locally moving derivatives from the viscosity solution to a smooth test function, the concept of viscosity solutions relies upon a “differentiation-by-parts” approach. This is in contrast to weak solution theory where derivatives are moved to a test function using an “integration-by-parts” approach, which, in general, is not a local operation.

### 2.3 Ellipticity and the comparison principle

An advantage of the viscosity solution concept for fully nonlinear problems is that, due to its entirely local structure, it is flexible enough to seek solutions in the space of bounded functions. To ensure the existence and uniqueness of viscosity solutions, additional structure conditions on the differential operator are often necessary. The most common such conditions are an ellipticity requirement and a comparison principle. The following definitions of ellipticity and the comparison principle are standard (cf. [5]).

(i) Equation (3) is said to be uniformly elliptic if, for all , there holds

 λtr (A−B)≤G(B,p,v,x)−G(A,p,w,x)≤Λtr (A−B)

whenever , where and means that is a nonnegative definite matrix.

(ii) Equation (3) is said to be proper elliptic if, for all , there holds

 G(A,p,v,x)≤G(B,p,w,x)

whenever and , where and .

(iii) Equation (3) is said to be degenerate elliptic if, for all , there holds

 G(A,p,v,x)≤G(B,p,v,x)

whenever , where .

Problem (3) is said to satisfy a comparison principle if the following statement holds. For any upper semicontinuous function and lower semicontinuous function on , if is a viscosity subsolution and is a viscosity supersolution of (3), then on .

The comparison principle is sometimes called a strong uniqueness property due to the fact it immediately yields the uniqueness of viscosity solutions for problem (3) (cf. [1]). When is differentiable, the degenerate ellipticity condition can also be defined by requiring that the matrix is negative semi-definite while the proper ellipticity condition additionally requires the number is nonnegative (cf. [16, p. 441]). We then clearly see that the ellipticity concept for nonlinear problems generalizes the notion for linear elliptic equations.

###### Remark 2.2

Since we have assumed that satisfies the comparison principle, it follows that that problem (3) has a unique viscosity solution .

### 2.4 Hamilton-Jacobi-Bellman equations

As mentioned in Section 1, a specific class of fully nonlinear PDEs that fits the structure requirements in this paper are (stationary) Hamilton-Jacobi-Bellman (HJB) equations described by (2).

We shall assume that is uniformly (in ) negative definite in the sense that there exists constants such that

 (5) −Λ|ξ|2≤Aθ(x)ξ⋅ξ≤−λ|ξ|2∀ξ∈Rd,x∈Ω,θ∈Θ.

We also assume there exists a constant such that

 ∥Aθ∥L∞(Ω),∥bθ∥L∞(Ω),∥cθ∥L∞(Ω)≤K

and for some constant for all and . Thus, the equation is proper elliptic and satisfies the comparison principle. We note that for stochastic optimal control applications, we have (cf. [15]). This case will also be considered. However, such problems may not satisfy a comparison principle as discussed in [24] and [28].

Under the above assumptions on the coefficients and the control set, problem (1)–(2) has a unique solution and is globally Lipschitz continuous. Thus, is differentiable almost everywhere and the weak derivatives are bounded by the constant . Since is closed and bounded, there actually exists a function such that

 H[u]=minθ∈Θ(Lθu−fθ)=Aθ∗:D2u+bθ∗⋅∇u+cθ∗u=0in Ω.

Thus, , , and , and we have is differentiable everywhere with bounded first order derivatives.

The analysis assumes a global Lipschitz condition for ease of presentation; however, locally Lipschitz should be sufficient. The global Lipschitz assumption excludes Monge-Ampère-type equations which are only conditionally degenerate elliptic [4] and locally Lipschitz. In light of the recent work [10], we know that the Monge-Ampère equation has an equivalent HJB reformulation. Much of the work of this paper can be extended to the Monge-Ampère equation via its HJB reformulation.

### 2.5 Difference operators

We introduce several difference operators for approximating first and second order partial derivatives. The multiple difference operators will be used to help resolve the low regularity of viscosity solutions and will play a key role in motivating and defining our new narrow-stencil FD methods.

Assume is a -rectangle, i.e., . We shall only consider grids that are uniform in each coordinate , . Let be an integer and for . Define , , , and . Then, . We partition into sub--rectangles with grid points for each multi-index . We call a mesh (set of nodes) for . We also introduce an extended mesh which extends by a collection of ghost grid points that are at most one layer exterior to in each coordinate direction. In particular, we choose ghost grid points such that for some , . We set and is defined by replacing by in the definition of and then removing the extra multi-indices that would correspond to ghost grid points that are not in the set to ensure .

#### 2.5.1 First order difference operators

We define the standard forward and backward difference operators as well as the central difference operator for approximating first order derivatives including the gradient operator. The forward and backward difference operators will serve as the building blocks for constructing all of the first and second order difference operators in this paper.

Let

denote the canonical basis vectors for

. Define the (first order) forward and backward difference operators by

 (6) δ+xi,hiv(x)≡v(x+hiei)−v(x)hi,δ−xi,hiv(x)≡v(x)−v(x−hiei)hi

for a function defined on and

 δ+xi,hiVα≡Vα+ei−Vαhi,δ−xi,hiVα≡Vα−Vα−eihi

for a grid function defined on the grid . Note that “ghost-values” may need to be introduced in order for the above difference operators to be well-defined on the boundary of . We also define the following central difference operator:

 (7) δxi,hi≡12(δ+xi,hi+δ−xi,hi)for i=1,2,⋯,d.

Lastly, we define the “sided” and central gradient operators , , and by

 (8) ∇±h≡[δ±x1,h1,δ±x2,h2,⋯,δ±xd,hd]T,∇h≡[δx1,h1,δx2,h2,⋯,δxd,hd]T.

#### 2.5.2 Second order difference operators

We now introduce a number of difference operators that approximate second order derivatives including the Hessian operator. Using the forward and backward difference operators introduced in the previous subsection, we have the following four possible approximations of the second order differential operator :

 Dμνh,ij≡δνxj,hjδμxi,hifor μ,ν∈{+,−},

which in turn leads to the definition of the following four approximations of the Hessian operator :

 (9) Dμνh≡[Dμνh,ij]di,j=1for μ,ν∈{+,−}.

To construct our numerical methods in the next section, we also need to introduce the following three sets of averaged second order difference operators:

 (10a) ˆδ2xi,xj;hi,hj ≡12(D+−h,ij+D−+h,ij)=12(δ−xj,hjδ+xi,hi+δ+xj,hjδ−xi,hi), (10b) ˜δ2xi,xj;hi,hj ≡12(D−−h,ij+D++h,ij)=12(δ+xj,hjδ+xi,hi+δ−xj,hjδ−xi,hi), (10c) ¯¯¯δ2xi,xj;hi,hj ≡12(ˆδ2xi,xj;hi,hj+˜δ2xi,xj;hi,hj)=14(D++h,ij+D+−h,ij+D−+h,ij+D−−h,ij) =14(δ+xj,hjδ+xi,hi+δ−xj,hjδ+xi,hi+δ+xj,hjδ−xi,hi+δ−xj,hjδ−xi,hi)

for all . For notation brievity, we also set

 δ2xi,hi≡ˆδ2xi,hi≡ˆδ2xi,xi;hi,hi,˜δ2xi,hi≡˜δ2xi,xi;hi,hi,¯¯¯δ2xi,hi≡¯¯¯δ2xi,xi;hi,hi.

Using the above difference operators, we define the following three “centered” approximations of the Hessian operator :

 (11) ˆD2h=[ˆδ2xi,xj;hi,hj]di,j=1,˜D2h=[˜δ2xi,xj;hi,hj]di,j=1,¯¯¯¯¯D2h=[¯¯¯δ2xi,xj;hi,hj]di,j=1.

We will also consider the (standard) central approximation of the Hessian operator defined by

 (12) [D2h]ij≡⎧⎨⎩δ2xi,hiif i=j,¯¯¯δ2xi,xj;hi,hj,if i≠j.

We now record some facts that can be easily verified for the various second order operators. First,

 2ˆD2h=D+−h+D−+h,2˜D2h=D−−h+D++h,2¯¯¯¯¯D2h=ˆD2h+˜D2h.

Second, a simple computation reveals

 (13a) ˆδ2xi,xj;hi,hjv(x) =−v(x+hiei−hjej)−v(x+hiei)−[v(x−hjej)−v(x)]2hihj −v(x−hiei+hjej)−v(x−hiei)−[v(x+hjej)−v(x)]2hihj, (13b) ˜δ2xi,xj;hi,hjv(x) =v(x+hiei+hjej)−v(x+hiei)−[v(x+hjej)−v(x)]2hihj +v(x−hiei−hjej)−v(x−hiei)−[v(x−hjej)−v(x)]2hihj, (13c) ¯¯¯δ2xi,xj;hi,hjv(x) =v(x+hiei+hjej)+v(x−hiei−hjej)4hihj −v(x−hiei+hjej)+v(x+hiei−hjej)4hihj.

Lastly, there holds the relationships

 (14a) δ2xi,hi =δ±xi,hiδ∓xi,hi, (14b) δ2xi,2hi =¯¯¯δ2xi,hi=δxi,hiδxi,hi.

We have is the standard (second order) three-point central difference operator for approximating second order derivatives in one-dimension, is a lower-resolution (second-order) three-point central difference operator for approximating second order derivatives in one-dimension, is a (second order) five-point central difference operator for approximating second order derivatives in one-dimension, is the standard (second order) central difference operator for approximating second order mixed derivatives, and and are two alternative (second order) central difference operators for approximating second order mixed derivatives. Observe that in two dimensions, corresponds to a 7-point stencil, and correspond to 9-point stencils, and corresponds to an 11-point stencil, as seen in Figure 2. The operators and use nearest neighbors in the stencil (including diagonal directions) while and use neighbors two steps away in each Cartesian direction.

We lastly introduce the (diagonal) variables and so that we can define the central difference operators and by

 (15a) δ2ξji,hv(x) ≡v(x−hiei+hjej)−2v(x)+v(x+hiei−hjej)h2i+h2j, (15b) δ2ηji,hv(x) ≡v(x+hiei+hjej)−2v(x)+v(x−hiei−hjej)h2i+h2j

for all with . Then, a direct computation yields

 (16a) [¯¯¯¯¯D2hv]ij =[D2hv]ij=h2i+h2j4hihj(δ2ηji,hv−δ2ξji,hv), (16b) [ˆD2hv]ij =hi2hjδ2xi,hiv+hj2hiδ2xj,hjv−h2i+h2j2hihjδ2ξji,hv, (16c) [˜D2hv]ij =−hi2hjδ2xi,hiv−hj2hiδ2xj,hjv+h2i+h2j2hihjδ2ηji,hv,

relationships that are illustrated in Figure 3. Lastly, there holds

 (17) [˜D2hVα]ii−[ˆD2hVα]ii =˜δ2xi,hiVα−δ2xi,hiVα =2(δ2xi,2hiVα−δ2xi,hiVα) =12(δ2xi,hiVα−ei−2δ2xi,hiVα+δ2xi,hiVα+ei)

for all .

## 3 A narrow-stencil finite difference method

We now propose a new finite difference method for approximating viscosity solutions of the fully nonlinear second order problem (1). In the construction of such a narrow-stencil scheme, our main idea is to design a numerical operator that utilizes the various numerical Hessians defined in the previous section so that a relaxed (or generalized) monotonicity condition is satisfied. The crux of the formulation is to introduce the concept of numerical moments which can be considered analogous to the concept of numerical viscosities used in the Crandall and Lions’ finite difference framework for (first order) Hamilton-Jacobi equations. Below, we first define our narrow-stencil finite difference method, and we then describe the idea and details of numerical moments. We also prove a generalized monotonicity result for our proposed numerical operator.

### 3.1 Formulation of the narrow-stencil finite difference method

In Section 2.5.2 we defined four basic sided discrete Hessian operators for . These four operators are the building blocks for constructing our numerical operator that approximates the differential operator . However, inspired by our earlier 1-D work [14], we construct the numerical operator so that it only depends explicitly upon the central difference operators and with regard to the Hessian argument in . Recall that denotes the extended mesh of that includes ghost mesh points and is the corresponding extended index set of (see Section 2.5).

Our specific narrow-stencil finite difference method is defined by seeking a grid function such that for all

 (18a) ˆF[Uα,xα] =0 for xα∈Th∩Ω, (18b) Uα =g(xα) for xα∈Th∩∂Ω,

where

 (19) ˆF[Uα,xα] ≡˜F(D−−hUα,D−+hUα,D+−hUα,D++hUα,∇+hUα,∇−hUα,Uα,xα) ≡ˆF(ˆD2hUα,˜D2hUα,∇+hUα,∇−hUα,Uα,xα) ≡F(¯¯¯¯¯D2hUα,∇hUα,Uα,


text:
Enter the total torque (N-m) and the change in angular position (radians) into the calculator to determine the Work Done By Torque.

## Work Done By Torque Formula

The following formula is used to calculate the Work Done By Torque.

Wt = t * d(θ)

• Where Wt is the Work Done By Torque (Joules)
• t is the total torque (N-m)
• d(θ) is the change in angular position (radians)

## How to Calculate Work Done By Torque?

The following two example problems outline how to calculate the Work Done By Torque.

Example Problem #1:

1. First, determine the total torque (N-m). In this example, the total torque (N-m) is given as 40.
2. Next, determine the change in angular position (radians). For this problem, the change in angular position (radians) is given as 100.
3. Finally, calculate the Work Done By Torque using the equation above:

Wt = t * d(θ)

Inserting the values from above and solving the equation with the imputed values gives:

Wt = 40 * 100 = 4000 (Joules)

Example Problem #2:

Using the same process as above, first define the variables required by the formula. In this case, these values are:

total torque (N-m) = 15

change in angular position (radians) = 30

Entering these values yields:

Wt = 15 * 30 = 450 (Joules)


text:
10 What Is The Conjugate Base Of H2po4

Asked on December 20, 2019 by Kanti Thangavel. The conjugate base is: _____ 9. (HINTbases are proton acceptors). the more concentrated the conjugate base. We want to find the conjugal base form of the substance. Which of the following will give a buffer with a pH near 4. The conjugate base of a strong acid is a very weak base, and, conversely, the conjugate acid of a strong base is a very weak acid. 8 times 10 to the negative 5, 1. Weak acids have varying strengths dependent on their pKas, so you have acids that are weak acids, but strong relative to other weaker acids, and vice versa. Acid Base. Use The “The Weaker The Acid, The Stronger The Conjugate Base” Principle To Obtain The Strengths Of Bases From A pKa Table. H2PO4– + H2O mc030-1. A buffer frequently used in the laboratory contains varying proportions of the conjugate acid–base pair H2PO4-1 and HPO 4-2. The conjugate base of an acid, even a polyprotic one, always has one less H than the acid. What is the acid that reacts with this base when ammonia is. F- and HF is a conjugate acid base pair. Acid Ka1 Ka2 Ka3 H2CO3 4. H2O gained a proton to form H3O^+; therefore, H2O is the base and H3O^+ is its conjugate acid. What is the conjugate base of H2PO4–(aq)?Select one:a. E) none of these. When a Brønsted-Lowry acid donates its proton during a chemical reaction, for example, the new product is the conjugate base. 8x 10-4) has a pH of 3. 095L of water. H2PO4(−) Identify as an acid or a base: A solution that is 0. H3PO4/H2PO4^- is the acid/conjugate base pair. And what is the conjugate base of this beasty?. 3 x 10 -5 Kb = 1. 00 × 10-13 mole of strong base NaOH to 1. Acid–Base Equilibrium Constants: K a , K b , p K a , and p K b The magnitude of the equilibrium constant for an ionization reaction can be used to determine the relative strengths of acids and bases. HF is an acid and HPO42-is its conjugate base. The stronger a base is: the the its Ka value its conjugate base its Kb value K a) its conjugate acid Conjugate Base? Example 1: Consider the following acids and their Ka values. Acid strength decreases and base strength increases down the table. which is the reverse of the equilibrium H 2 O (acid) + B (base) ⇌ OH − (conjugate base) + BH + (conjugate acid). K a K b = K w = 1 x 10-14. Here’s a nifty way to look at it: Suppose we have the following reaction: NH3 ←> NH4+ + OH- We know that this is a shorter way to write: NH3 + H2O ←> NH4+ + OH- Which is basically telling us that NH3 takes away a hydrogen proton from H2O to become. A buffer frequently used in the laboratory contains varying proportions of the conjugate acid–base pair H2PO4-1 and HPO 4-2. What is the conjugate base for each of the following? HSO4 HNO3 H2PO4 Soya HCI H30* H3PO4 H2SO4 H2O H2CO3 HCO3. Be able to identify or provide the conjugate acids and conjugate bases of any acid or base. The conjugate base of a weak acid is. Miller , a Sean T. 12 × 10-11. The term conjugate comes from the Latin stems meaning "joined together" and refers to things that are joined, particularly in pairs, such as Brnsted acids and bases. NH3, NH4+ C. Conjugate Acid-Base Pairs - Duration: 4:36. Conjugate acids and conjugate bases are the acids and bases that lose or gain protons. stoichiometry of acid-base reaction 2. Calculate the volume of each stock solution required to make the buffer. HCO3- -----CO3^2- + H+ H20 -----H+ +OH-Thus the conjugate base of acid HCO3- and H2O are CO3^2- and OH-. Step 3 To get the conjugate base of H2PO4-, it must act as Bronsted acid and should give H+ ion. guided reading activity world history, i wish that had duck feet guided reading level,. ⚛ The weaker the base, the stronger its conjugate acid. What is the conjugate acid for each of the following? HSOHasy sulfuric acid NH NH4N HPO- Hot SO - HS 04- NO, NH+ HO_HSO+ нсо" На со 3 1 CI HPO42- Ha P04 10. 8 x 10-11 10. And z can be written z=x+iy where x, y ∈ ℝ. 8x 10^-23 3. Login to reply the answers Post; crete. 21 HCO3- CO32- 4. A conjugate base as one less proton than the acid. Hence is the Bronsted acid (Answer) Acid loses H+ (proton) to forms conjugate Base. It is what is left over after an acid has donated a proton during a chemical reaction. See full answer below. (HINTbases are proton acceptors). HA, an additional 5. What is the conjugate base in the following equation: HBr + H2 O \longrightarrow H3 O+ + Br-. Provide details on what you need help with along with a. This is a two-step problem 1. 1 Select the pair of substances in which an acid is listed followed by its conjugate base. Similar Questions. ⚛ The stronger the base, the weaker its conjugate acid. The diprotic acid -- H2CO3. Significance The primary purpose and importance of buffers in pharmacy is to protect pharmaceutical preparations from any sudden change in pH even when limited concentrations of acid or base are added. Toenjes a and Jeffery L. Miller , a Sean T. What is the Bronsted base in the following equation: HBr + H2 O \longrightarrow H3 O+ + Br-23. 00 × 10-13 mole of strong base NaOH to 1. General solution when you know the concentrations of conjugate acid and base. The hydroxide ion OH −, a well known base, is here acting as the conjugate base of the acid water. Identify the formula for the conjugate acid of H2PO4−. So what we need to look at is what's gonna happen when we remove or give up and h plus I on from each of these comments Right now it's important to remember the charge here, not just the hydrogen, because that's going to affect the charge of our contra get base. This principle is discussed in detail in the article on acid-base reaction theories. Identify the formula for the conjugate base of HPO42−. NH3 –> NH2(-) or CH4 –> CH3 (-). 8 times 10 to the negative 5, 1. A solution containing appreciable amounts of a weak conjugate acid-base pair is called a buffer solution, or a buffer. H2PO4- is anacid and F- is its conjugatebase. What is the conjugate base of HSO4- A. The following table gives the acid-dissociation constant values, Ka, for carbonic acid, H2CO3, and phosphoric acid, H3PO4. What is the Henderson-Hasselbalch. of the conjugate base Bz-in the solution AFTER the titration, then calculate pH. Imagine a generic acid. Ammonia (NH 3) acts as a weak base in aqueous solution. HF is an acid and HPO42-is its conjugate base. HCO3- -----CO3^2- + H+ H20 -----H+ +OH-Thus the conjugate base of acid HCO3- and H2O are CO3^2- and OH-. The conjugate can be very useful because when we multiply something by its conjugate we get squares like this:. H3PO4 is a tribasic acid, thus ionising in three steps. 8 × 10-4 CH3CH2NH3+ (from ethylamine) 2. So when the weak acid (phosphoric acid) loses a hydrogen ion, it will form on the other side the weak conjugate base ie. What is complex conjugate pair? The set of 2 elements: a complex number $z$ and its conjugate $\overline {z}$, form a pair of conjugates. Conjugate base (CB) – remaining part of an acid after it has 5. Chemistry 12 Unit 4 - Acids, Bases and Salts Tutorial 14 - Solutions Page 5 Answer to Question 8 on page 17 of Tutorial 14. H2O gained a proton to form H3O^+; therefore, H2O is the base and H3O^+ is its conjugate acid. Examples of Use. Likewise, when a base accepts a proton during a chemical reaction, the new product is the conjugate acid. Acid Ka1 Ka2 Ka3 H2CO3 4. The conjugate base of a weak acid is a strong base, and the conjugate base of a strong acid is a weak base. a) Write the two steps in the dissociation of carbonic acid (H2CO3). The stronger a base is: the the its Ka value its conjugate base its Kb value K a) its conjugate acid Conjugate Base? Example 1: Consider the following acids and their Ka values. 90 x 10–2 [H. See full answer below. What is the conjugate acid for each of the following? HSOHasy sulfuric acid NH NH4N HPO- Hot SO - HS 04- NO, NH+ HO_HSO+ нсо" На со 3 1 CI HPO42- Ha P04 10. 1 M ephedrine (weak base Kb = 2. " But if one is strong, the other must be weak. TABLE OF CONJUGATE ACID-BASE PAIRS Acid Base K a (25 oC) HClO 4 ClO 4 – H 2 SO 4 HSO 4 – HCl Cl– HNO 3 NO 3 – H 3 O + H 2 O H 2 CrO 4 HCrO 4 – 1. 2– (aq) + H. Here’s a nifty way to look at it: Suppose we have the following reaction: NH3 ←> NH4+ + OH- We know that this is a shorter way to write: NH3 + H2O ←> NH4+ + OH- Which is basically telling us that NH3 takes away a hydrogen proton from H2O to become. K a K b = K w = 1 x 10-14. Here’s a nifty way to look at it: Suppose we have the following reaction: NH3 ←> NH4+ + OH- We know that this is a shorter way to write: NH3 + H2O ←> NH4+ + OH- Which is basically telling us that NH3 takes away a hydrogen proton from H2O to become. So when the weak acid (phosphoric acid) loses a hydrogen ion, it will form on the other side the weak conjugate base ie. Acid Base Conjugate Conjugate. In the reaction of ammonia, water is an acid because it donates a hydrogen ion to ammonia. 12 × 10-11. Because the product of K a times K b is a relatively small number, either the acid or its conjugate base can be "strong. H2PO4(−) Identify as an acid or a base: A solution that is 0. The more stable/weaker the conjugate acid,. Toenjes a and Jeffery L. 35E-10, and Cb = 0. A pair of molecules or ions related to one another by the gain or loss of a single hydrogen ion is called a conjugate acid-base pair. The idea of conjugate acids and bases is determined by the reactants in a chemical reaction. Conjugate base is allways ONE proton H+ less than its acid which it is conjugated to. For example, the label on a package of artificial fruit-juice mix. Ammonia is defined as a base under this theory. The conjugate base of an acid is formed when the acid donates a proton. Jul 21,2020 - Which of the following statement(s) is/are correct?a)The pH of 1. D) The Weaker The Conjugate Acid. What is the acid that reacts with this base when ammonia is. B) The Stronger The Conjugate Base. 1 M then we get pH= 10. H 2CO 3 is the stronger acid; NH 4 + has the stronger conjugate base. Nalbandian , a Eric M. Correct answers: 1, question: An acid has an acid dissociation constant of 2. ) After an acid has given up its proton, it is capable of getting back that proton and acting as a base. 35E-10) then by applying the above equations with Ca = 0. The proper way to say it is that “weak acids have strong CONJUGATE bases“. 00 L of aqueous solution. HCO3- is a conjugate acid, H 2 CO 3. Kb is equal to 1. 2pH = pK1 + pK2. A solution of acetic acid and sodium acetate (CH 3 COOH + CH 3. A conjugate base is what is left after an acid gives up a proton. Ionization Constants for Conjugate Pairs. The H2PO4^- ion is named dihydrogen phosphate ion. 35E-10, and Cb = 0. Likewise, when a base accepts a proton during a chemical reaction, the new product is the conjugate acid. HA, an additional 5. HPO42-, H2PO4- B. It is what is left over after an acid has donated a proton during a chemical reaction. The conjugate base is formed when an acid loses its proton. deviation: noun aberrance , aberrancy , aberration , alteration , anomalousness, anomaly , antipathy , antithesis , branching off , breach of practice , change of. 8x 10-4) has a pH of 3. 10 molar solution of each of 2nd reaction H2O OH-the salts above is acidic, neutral or basic. This is what happens when you reply to spam email | James Veitch - Duration: 9:49. D) F– > CN– > H2O. The conjugate base of an acid, even a polyprotic one, always has one less H than the acid. Be able to identify or provide the conjugate acids and conjugate bases of any acid or base. of the conjugate base Bz-in the solution AFTER the titration, then calculate pH. Strong acids, such as HNO3 have weak conjugate bases, so NO3-is a weak base. H2PO4– + H2O mc030-1. The Conjugate Method is a term and methodology made most popular by Louie Simmons, owner of the Westside Barbell Club and the creator of what has become known as the “Westside” method of training. State the formula of the conjugate base of H2PO4-HPO4^2-ii) State the formula of the conjugate acid of H2PO4-H3PO4 iii) A solution of calcium dihydrogenphosphate Ca(H2PO4)2 in water acts as a buffer solution. 2pH = pK1 + pK2. F- is the conjugate base and HF is the conjugate acid. Since Louie started publishing his articles on his methodologies in Power Lifting USA back in the 1990’s, the methods grew wildly in popularity …. H3PO4 is a tribasic acid, thus ionising in three steps. 8x 10-5 and for hydoylamine is 1. If so, write the formulas of its conjugate acid and its conjugate base. What is the Bronsted base in the following equation: HBr + H2 O \longrightarrow H3 O+ + Br-23. 8 x 10-5 3-5 x 10-4 6. In the reaction of ammonia, water is an acid because it donates a hydrogen ion to ammonia. The conjugate base of H 2. 4 has the stronger conjugate base. Because the product of K a times K b is a relatively small number, either the acid or its conjugate base can be "strong. HCl is the stronger acid; HF has the stronger conjugate base. The conjugate acid of HPO42 Concepts and reason The concept used to solve this problem is based on the conjugate acid and conjugate base. so : HPO4 2-and the conjugate base of that is. 69 × 10-10 HC2H3O2 (acetic acid) 1. Correct answers: 1, question: An acid has an acid dissociation constant of 2. which is the reverse of the equilibrium H 2 O (acid) + B (base) ⇌ OH − (conjugate base) + BH + (conjugate acid). Let v be the harmonic conjugate of u=Im(e^(z²)), then: uᵪ=vᵧ, uᵧ=-vᵪ. Check the amphoteric species. 4 has the stronger conjugate base. The stronger the acid, the weaker the conjugate base. 8 times 10 to the negative 5, 1. Understanding the hydration of these anions at the molecular level represents a long-standing goal in chemistry. NH3 –> NH2(-) or CH4 –> CH3 (-). What is the conjugate base of H2PO4–(aq)?Select one:a. H3PO4 is a tribasic acid, thus ionising in three steps. 69 × 10-10 HC2H3O2 (acetic acid) 1. Back to top. A conjugate base as one less proton than the acid. The dihydrogenphosphate ion, H2PO4 ^–, has both a conjugate acid and a conjugate base. guided reading activity world history, i wish that had duck feet guided reading level,. H2PO4(−) Identify as an acid or a base: A solution that is 0. Acids and bases exist as conjugate acid-base pairs. which is the reverse of the equilibrium H 2 O (acid) + B (base) ⇌ OH − (conjugate base) + BH + (conjugate acid). H2PO4(−) Identify as an acid or a base: A solution that is 0. The equilibrium constants (Ka) for HCN and HF in H2O at 25°C are 6. 8x 10-5 and for hydoylamine is 1. HPO42- is the conjugate base and. The conjugate base of bicarbonate, HCO 3- is carbonate, CO3 2-. Nalbandian , a Eric M. HPO42-, H2PO4- B. ⚛ The stronger the base, the weaker its conjugate acid. Acids and bases are thus regarded. Conjugate Acid-Base Pairs - Duration: 4:36. Because the product of K a times K b is a relatively small number, either the acid or its conjugate base can be "strong. Toenjes a and Jeffery L. Login to reply the answers Post; crete. E) None Of The Above. 5x10-12 Answer: Q > Ksp, therefore, La(OH)3 precipitates (a) acid = proton donor; base = proton acceptor (b) Acid Conjugate base 1981 D 1 reaction st NH4 + NH3 Al(NO3)3 K2CO3 NaHSO4 NH4Cl H2O OH-(a) Predict whether a 0. Similarly, a base (B) can gain a proton from. Post a Question. This is what happens when you reply to spam email | James Veitch - Duration: 9:49. CH3COOH, CH3COOH2+ D. Let's do that math. A conjugate base as one less proton than the acid. , July 23, 2020 /PRNewswire/ — Geisinger researchers have concluded there is sufficient evidence to consider including DNA changes that cause neurodevelopmental and psychiatric disorders in genomic screening programs. a conjugate base is a species which is formed when acid loses a proton. Identify the formula for the conjugate acid of H2PO4−. In the reaction below identify the acid, base, conjugate acid and conjugate base: HCl + H2O --> H3O+ + Cl-10. Significance The primary purpose and importance of buffers in pharmacy is to protect pharmaceutical preparations from any sudden change in pH even when limited concentrations of acid or base are added. Every time a Brnsted acid acts as an H +-ion donor, it forms a conjugate base. The conjugate acid of HPO42 Concepts and reason The concept used to solve this problem is based on the conjugate acid and conjugate base. So the conjugate base of CH 4, CH 3 (-) is an extremely strong base. 0 10-8M solution of HCl is 8b)The conjugate base of H2PO4- is HPO42-c)Autoprotolysis constant of water increases with temperatured)When a soltion of a weak monoprotic acid is titrated again a strong base, at half - neutralization point pH = (1/2) pKa. Ah, hydrogen ion. Which conjugate acid is stronger? base * c. Gustafson * a. 6 times 10 to the negative 10. 8x 10-4) has a pH of 3. Preview this quiz on Quizizz. This relationship is known as the Henderson-Hasselbalch equation. This is the base ionization reaction. If so, write the formulas of its conjugate acid and its conjugate base. Calculate the volume of each stock solution required to make the buffer. The hydroxide ion OH −, a well known base, is here acting as the conjugate base of the acid water. A pair of molecules or ions related to one another by the gain or loss of a single hydrogen ion is called a conjugate acid-base pair. HCO3- is a conjugate acid, H 2 CO 3. This is a two-step problem 1. The idea of conjugate acids and bases is determined by the reactants in a chemical reaction. D daughter isotope In a nuclear equation the compound remaining after the parent isotope (the original isotope) has undergone decay. which is the reverse of the equilibrium H 2 O (acid) + B (base) ⇌ OH − (conjugate base) + BH + (conjugate acid). The H2PO4^- ion is named dihydrogen phosphate ion. Ah, hydrogen ion. CH3COOH and CH3COO- is a conjugate acid base pair. right for each of these acids. The conjugate base of a weak acid is a strong base, and the conjugate base of a strong acid is a weak base. Define conjugate. Preview this quiz on Quizizz. HNO3, asked by Juan on March 10, 2013. 1 M then we get pH= 10. deviation: noun aberrance , aberrancy , aberration , alteration , anomalousness, anomaly , antipathy , antithesis , branching off , breach of practice , change of. Weak Acid Conjugate Base Ka pKa HC2O4- C2O42- 6. Question: 1) What Is The Conjugate Base Of H2PO4⁻ ?A) HPO42-B) PO43-C) H3PO4D) H3O+E) OH⁻ 2) The Stronger The Acid, Then Which Of The Following Is True?A) The Stronger The Conjugate Acid. O) is dissolved in water, the solution turns basic from the reaction of the oxide ion (O. a conjugate base is a species which is formed when acid loses a proton. Since Louie started publishing his articles on his methodologies in Power Lifting USA back in the 1990’s, the methods grew wildly in popularity …. Methane (CH 4) is a weak acid, but it can’t act as a base – it doesn’t have a lone pair. Solution: (a) The proton is transferred from the sulfate to the cyanide so: HSO4-(aq)/SO42-(aq) and CN-(aq)/HCN(aq ) are the two acid-base pairs. If so, write the formulas of its conjugate acid and its conjugate base. Be able to identify or provide the conjugate acids and conjugate bases of any acid or base. Conjugate acids and conjugate bases are the acids and bases that lose or gain protons. ) {eq}\mathbf{OH^-}{/eq} is the Bronsted base. TABLE OF CONJUGATE ACID-BASE PAIRS Acid Base K a (25 oC) HClO 4 ClO 4 – H 2 SO 4 HSO 4 – HCl Cl– HNO 3 NO 3 – H 3 O + H 2 O H 2 CrO 4 HCrO 4 – 1. (b) Give the conjugate acidof the following Brønsted–Lowry…. , July 23, 2020 /PRNewswire/ — Geisinger researchers have concluded there is sufficient evidence to consider including DNA changes that cause neurodevelopmental and psychiatric disorders in genomic screening programs. A base is a substance which can accept one or more protons (Bronsted Acid) + (Bronsted Base) ----> Conjugate Base + Conjugate acid Here we see donates or loses ion to form. conjugate base of an acid has one less proton than the acid. We want to find the conjugal base form of the substance. O(l) Æ NO3-(aq) + H3O + (aq) H2O is the strongest base. 1 M then we get pH= 10. Draw out the conjugate bases of the acids on your pka table by removing a proton. H2O gained a proton to form H3O^+; therefore, H2O is the base and H3O^+ is its conjugate acid. F- and HF is a conjugate acid base pair. The conjugate base of a weak acid is a strong base, and the conjugate base of a strong acid is a weak base. ⚛ A weak Brønsted-Lowry acid has a strong conjugate base. 2×10−13 1) At a pH of 7. Which answer corectly lists the Bronsted-Lowry conjugate of propylamine, C 3H7NH2? a. Methane (CH 4) is a weak acid, but it can’t act as a base – it doesn’t have a lone pair. And what is the conjugate base of this beasty?. (HINTbases are proton acceptors). Nalbandian , a Eric M. In the following reactionHF(aq) + HPO42-(aq) DF-(aq) +H2PO4-(aq)a. Imagine a generic acid. Conjugate Acid-Base Pairs - Duration: 4:36. O) is dissolved in water, the solution turns basic from the reaction of the oxide ion (O. What is the conjugate acid in the following equation: HBr + H2 O \longrightarrow H3 O+ + Br-A. F- is the conjugate base and HF is the conjugate acid. 00 L of aqueous solution. The conjugate base of H2SO4 is ___. H3PO4 H+ + H2PO4- pK1 H2PO4- H+ + HPO42- pK2 HPO41- H+ + PO43- pK3 BUFFERING When the pH is held ‘steady’ because of the presence of a conjugate acid/base pair, the system is said to be buffered In the environment, we must think about more than just one conjugate acid/base pairings in solution Many different acid/base pairs in solution. 15 x 10-5moles is needed to generate the conjugate base in situ. Post a Question. Another example, if we have a solution with 0. a conjugate base is a species which is formed when acid loses a proton. A buffer frequently used in the laboratory contains varying proportions of the conjugate acid–base pair H2PO4-1 and HPO 4-2. In H2O the conjugate base is H2PO4-, being conjugated to the acid H3PO4. H2O gained a proton to form H3O^+; therefore, H2O is the base and H3O^+ is its conjugate acid. the stronger its conjugate base. It is what is left over after an acid has donated a proton during a chemical reaction. A conjugate base as one less proton than the acid. H2PO4 HSO4 H2SO4 NH HCO, HO HO 9. Similar Questions. Conjugate Acid (CA) – acid formed when a base accepts a proton. Miller , a Sean T. Similarly, a base (B) can gain a proton from. B) a weak base. The stronger a base is: the the its Ka value its conjugate base its Kb value K a) its conjugate acid Conjugate Base? Example 1: Consider the following acids and their Ka values. The conjugate base of a weak acid is. Ammonia (NH 3) acts as a weak base in aqueous solution. The conjugate base of H2SO4 is ___. The conjugate acid BH + of a base B dissociates according to BH + + OH − ⇌ B + H 2 O. Conjugate base is allways ONE proton H+ less than its acid which it is conjugated to. 1 Select the pair of substances in which an acid is listed followed by its conjugate base. HA + H2O H3O+ + A- /OH- Acid Base Conjugate Conjugate Acid Base or HA A- + H+ Acid Conjugate Conjugate Base Acid Acid strength is specified by its dissociation constant Molar concentration for: HA + H2O H3O+ + A- reactants products HA H3O+ A- H2O a measure of relative proton affinities for each conjugate acid base pair. The prefix ''di'' indicates the presence of two hydrogen atoms. jpg H3O+ + HPO42– Which of the following is a base–conjugate… Get the answers you need, now! 1. When hydrobromic acid (HBr) reacts with. The dihydrogenphosphate ion, H2PO4 ^–, has both a conjugate acid and a conjugate base. Nalbandian , a Eric M. Fundamentals According to the Bronsted-lowery theory, a conjugate acid is a species which is formed by reaction of proton with a base. Merely said, the Conjugate Acid And Base. CA = H2PO4 and CB = PO4-2. (b) Give the conjugate acidof the following Brønsted–Lowry…. O(l) Æ NO3-(aq) + H3O + (aq) H2O is the strongest base. E) None Of The Above. Every time a Brnsted acid acts as an H +-ion donor, it forms a conjugate base. 2– (aq) + H. -1, as well as other conjugate acid–base pairs. Thus conjugate base of H3PO4 is H2PO4. The conjugate acid BH + of a base B dissociates according to BH + + OH − ⇌ B + H 2 O. Use The “The Weaker The Acid, The Stronger The Conjugate Base” Principle To Obtain The Strengths Of Bases From A pKa Table. In H2O the conjugate base is H2PO4-, being conjugated to the acid H3PO4. The conjugate acid of a weak base will be a weak acid. Phosphoric acid, H_3PO_4, is the parent acid. What is complex conjugate pair? The set of 2 elements: a complex number $z$ and its conjugate $\overline {z}$, form a pair of conjugates. Thus, a strong acid must have a weak conjugate base. Miller , a Sean T. 35E-10) then by applying the above equations with Ca = 0. Log in or register to post comments. what the conjugate base of CH3NH3+ CH3NH2, methyl amine. Note: when calculating the equilibrium for a base, consider the water reaction to realize these two useful relations: pH + pOH = 14 [H+][OH-] = 1. So what we need to look at is what's gonna happen when we remove or give up and h plus I on from each of these comments Right now it's important to remember the charge here, not just the hydrogen, because that's going to affect the charge of our contra get base. Which conjugate acid is stronger? base * c. Fundamentals According to the Bronsted-lowery theory, a conjugate acid is a species which is formed by reaction of proton with a base. Significance The primary purpose and importance of buffers in pharmacy is to protect pharmaceutical preparations from any sudden change in pH even when limited concentrations of acid or base are added. A) a strong base. H3PO4 H+ + H2PO4- pK1 H2PO4- H+ + HPO42- pK2 HPO41- H+ + PO43- pK3 BUFFERING When the pH is held ‘steady’ because of the presence of a conjugate acid/base pair, the system is said to be buffered In the environment, we must think about more than just one conjugate acid/base pairings in solution Many different acid/base pairs in solution. H2PO4 HSO4 H2SO4 NH HCO, HO HO 9. 2k points). Fundamentals According to the Bronsted-lowery theory, a conjugate acid is a species which is formed by reaction of proton with a base. It allows one to determine the pH of a solution from knowing the pK a of the buffer and the concentration of the acid and conjugate base components. equilibrium calculation Strategy —find the conc. Correct answers: 1, question: An acid has an acid dissociation constant of 2. The pH is: 1. Likewise, when a base accepts a proton during a chemical reaction, the new product is the conjugate acid. In the reaction of ammonia, water is an acid because it donates a hydrogen ion to ammonia. Thus, HPO42- is the conjugate base of H2PO4- (which is an acid in step II, but is the conjugate base of H3PO4 in step I). So when the weak acid (phosphoric acid) loses a hydrogen ion, it will form on the other side the weak conjugate base ie. DANVILLE, Pa. 0 10-8M solution of HCl is 8b)The conjugate base of H2PO4- is HPO42-c)Autoprotolysis constant of water increases with temperatured)When a soltion of a weak monoprotic acid is titrated again a strong base, at half - neutralization point pH = (1/2) pKa. HA + H2O H3O+ + A- /OH- Acid Base Conjugate Conjugate Acid Base or HA A- + H+ Acid Conjugate Conjugate Base Acid Acid strength is specified by its dissociation constant Molar concentration for: HA + H2O H3O+ + A- reactants products HA H3O+ A- H2O a measure of relative proton affinities for each conjugate acid base pair. Calculate the Ka for the conjugate acids (HON 16. Asked on December 20, 2019 by Kanti Thangavel. Step 3 To get the conjugate base of H2PO4-, it must act as Bronsted acid and should give H+ ion. The stronger the base, the weaker its conjugate acid. If so, write the formulas of its conjugate acid and its conjugate base. Phosphoric acid, H_3PO_4, is the parent acid. Buffers also have commercial applications. H 2O acquires the H + ions most. Ah, hydrogen ion. Toenjes a and Jeffery L. ⚛ The stronger the base, the weaker its conjugate acid. HA + H2O H3O+ + A- /OH- Acid Base Conjugate Conjugate Acid Base or HA A- + H+ Acid Conjugate Conjugate Base Acid Acid strength is specified by its dissociation constant Molar concentration for: HA + H2O H3O+ + A- reactants products HA H3O+ A- H2O a measure of relative proton affinities for each conjugate acid base pair. 10 Draw the conjugate acid of each species as in Example 9. Preview this quiz on Quizizz. 8 x 10-4 Rank these acids in order of increasing strength: so. 65 calculate the concentration of hydrofluoric acid. Because the product of K a times K b is a relatively small number, either the acid or its conjugate base can be "strong. The conjugate acid base pair differs by one proton. Significance The primary purpose and importance of buffers in pharmacy is to protect pharmaceutical preparations from any sudden change in pH even when limited concentrations of acid or base are added. In the reaction of ammonia, water is an acid because it donates a hydrogen ion to ammonia. The conjugate base is: _____ 9. General solution when you know the concentrations of conjugate acid and base. conjugate base: substance formed when an acid loses a hydrogen ion. HS-4-,SO4^2- B. Another example, if we have a solution with 0. 10 M HC2H3O2 and 0. HPO42- is an acid andH2PO4- is itsconjugatebase. HPO4 2- (aq) + H2O (l) --> H2PO4-(aq) + OH-(aq) HPO42- and H2PO4- is a conjugate acid base pair. A conjugate base as one less proton than the acid. Thus conjugate base of H3PO4 is H2PO4. " But if one is strong, the other must be weak. 2×10−13 1) At a pH of 7. Phosphoric acid, H_3PO_4, is the parent acid. It turns out that you multiply them, Ka times Kb for a conjugate acid-base pair is equal to Kw. The fact that HCl is a strong acid implies that its conjugate base Cl – is too weak a base to hold onto the proton in competition with either H 2 O or H 3 O +. If it loses a proton, H^+, we conserve both mass and charge, and H_2PO_4^- results. so : HPO4 2-and the conjugate base of that is. The K a of hypoiodous acid (HOI) is 2. 6 x 10-10 33 Acid -Base Reactions Strategy —find the conc. Conjugate Acid-Base Pairs - Duration: 4:36. CH3COOH is the conjugate acid and CH3COO- is the conjugate base. Which answer corectly lists the Bronsted-Lowry conjugate of propylamine, C 3H7NH2? a. Ah, hydrogen ion. 0 10-8M solution of HCl is 8b)The conjugate base of H2PO4- is HPO42-c)Autoprotolysis constant of water increases with temperatured)When a soltion of a weak monoprotic acid is titrated again a strong base, at half - neutralization point pH = (1/2) pKa. ) If a solution of EIF (Ka= 6. Log in or register to post comments. Conjugate-base anions are ubiquitous in aqueous solution. Which conjugate acid is stronger? base * c. The conjugate base of any comppound is the compound formed after the removal of H+ from them respectively. And z can be written z=x+iy where x, y ∈ ℝ. O(l) Æ NO3-(aq) + H3O + (aq) H2O is the strongest base. 15 x 10-5moles is needed to generate the conjugate base in situ. 2×10−13 1) At a pH of 7. Which answer corectly lists the Bronsted-Lowry conjugate of propylamine, C 3H7NH2? a. D) F– > CN– > H2O. Similar Questions. ) Given the Kb for ammonia is 1. In both cases identify the conjugate acid– base pairs. 76 when the acid and conjugate base are mixed in equimolar proportions? Acid Ka NH4+ (from ammonia) 5. These are, respectively: asked Jul 30, 2019 in Chemistry by Satkriti ( 69. H3PO4/H2PO4^- is the acid/conjugate base pair. H3PO4 + H2O ==> H3O^+ + H2PO4^-So H3PO4, phosphoric acid, lost a proton to form H2PO4^- so H2PO4^- is the conjugate base of H3PO4, the acid. F- is the conjugate base and HF is the conjugate acid. Calculate the Ka for the conjugate acids (HON 16. 12 × 10-11. DANVILLE, Pa. what the conjugate solution bottom regarding CH3NH3+ "CH3NH2, methyl amine. 65 calculate the concentration of hydrofluoric acid. Conjugate Acid-Base Pairs - Duration: 4:36. Weak acids have varying strengths dependent on their pKas, so you have acids that are weak acids, but strong relative to other weaker acids, and vice versa. H3PO4 + H2O <---> H2PO4- + H3O+ H2PO4- + H20 <---> HPO4^-2 +H3O+ HPO4^-2 + H20 <---> PO4^3- +H3O+. Because the product of K a times K b is a relatively small number, either the acid or its conjugate base can be "strong. ) If a solution of EIF (Ka= 6. This is just a re-statement of what is implicit in what has been said above about the distinction between strong acids and weak acids. The conjugate base of H2PO-4 is - 3137012 yl and cyclopentyl. The idea of conjugate acids and bases is determined by the reactants in a chemical reaction. stoichiometry of acid-base reaction 2. 10 molar solution of each of 2nd reaction H2O OH-the salts above is acidic, neutral or basic. 5x10-12 Answer: Q > Ksp, therefore, La(OH)3 precipitates (a) acid = proton donor; base = proton acceptor (b) Acid Conjugate base 1981 D 1 reaction st NH4 + NH3 Al(NO3)3 K2CO3 NaHSO4 NH4Cl H2O OH-(a) Predict whether a 0. Ah, hydrogen ion. Weak Acid Conjugate Base Ka pKa HC2O4- C2O42- 6. 2k points). H2CO3 after donating a H+ forms HCO3-. 3E-5 ) and 0. The more stable/weaker the conjugate base, the stronger the acid. Write the reaction that occurs, and identify the conjugate acid– base pairs. Methane (CH 4) is a weak acid, but it can’t act as a base – it doesn’t have a lone pair. Acid Ka1 Ka2 Ka3 H2CO3 4. Correct answer is option 'B,C'. Toenjes a and Jeffery L. Likewise, when a base accepts a proton during a chemical reaction, the new product is the conjugate acid. Ammonia (NH 3) acts as a weak base in aqueous solution. 01 M ephedrine hydrochloride (conjugate acid Ka = 1E-14 / 2. com A conjugate base is a species formed by the removal of proton from an acid. E) None Of The Above. Identify the formula for the conjugate acid of H2PO4−. Check the amphoteric species. H2CO3 after donating a H+ forms HCO3-. General solution when you know the concentrations of conjugate acid and base. The conjugate acid of HPO42 Concepts and reason The concept used to solve this problem is based on the conjugate acid and conjugate base. These are, respectively: asked Jul 30, 2019 in Chemistry by Satkriti ( 69. F- is the conjugate base and HF is the conjugate acid. The stronger a base is: the the its Ka value its conjugate base its Kb value K a) its conjugate acid Conjugate Base? Example 1: Consider the following acids and their Ka values. The conjugate acid BH + of a base B dissociates according to BH + + OH − ⇌ B + H 2 O. HA + H2O H3O+ + A- /OH- Acid Base Conjugate Conjugate Acid Base or HA A- + H+ Acid Conjugate Conjugate Base Acid Acid strength is specified by its dissociation constant Molar concentration for: HA + H2O H3O+ + A- reactants products HA H3O+ A- H2O a measure of relative proton affinities for each conjugate acid base pair. CH3COOH is the conjugate acid and CH3COO- is the conjugate base. Tabulated below are several examples of conjugate acid-base pairs. 6 times 10 to the negative 10. Conjugate-base anions are ubiquitous in aqueous solution. For example, H 2 S O 4 → H S O 4 − + H +. accept (gain) a proton. HF is an acid and HPO42-is its conjugate base. Define conjugate. H2CO3 after donating a H+ forms HCO3-. A) a strong base. The prefix ''di'' indicates the presence of two hydrogen atoms. HPO4 (2-) is the conjugate base of H2PO4-. Conjugate-base anions are ubiquitous in aqueous solution. con·ju·gat·ed , con·ju·gat·ing , con·ju·gates v. The hydroxide ion OH −, a well known base, is here acting as the conjugate base of the acid water. Keep in mind that having conjugate acid - base pair in solution makes. what the conjugate base of CH3NH3+ CH3NH2, methyl amine. Thus conjugate base of H3PO4 is H2PO4. Acid–Base Equilibrium Constants: K a , K b , p K a , and p K b The magnitude of the equilibrium constant for an ionization reaction can be used to determine the relative strengths of acids and bases. Complete the table below. When a Brønsted-Lowry acid donates its proton during a chemical reaction, for example, the new product is the conjugate base. of the conjugate base Bz-in the solution AFTER the titration, then calculate pH. General solution when you know the concentrations of conjugate acid and base. CH3COOH, CH3COOH2+ D. 8x 10-4) has a pH of 3. The conjugate can be very useful because when we multiply something by its conjugate we get squares like this:. Another example, if we have a solution with 0. A conjugate Lewis base-Brønsted acid catalyst for the sulfenylation of nitrogen containing heterocycles under mild conditions† Christopher J. ) After an acid has given up its proton, it is capable of getting back that proton and acting as a base. , July 23, 2020 /PRNewswire/ — Geisinger researchers have concluded there is sufficient evidence to consider including DNA changes that cause neurodevelopmental and psychiatric disorders in genomic screening programs. The prefix ''di'' indicates the presence of two hydrogen atoms. 01 M, Ka = 4. Acid Base Conjugate Conjugate. General solution when you know the concentrations of conjugate acid and base. 6 times 10 to the negative 10. As well: H3PO4 is conjugated acid to the base H2PO4-. HPO42- is an acid andH2PO4- is itsconjugatebase. A strong acid loses a proton easily, and hence the conjugate base of a strong acid is a weak base. Conjugate Acid And Base Worksheet With Answers is available in our digital library an online access to it is set as public so you can download it instantly. The H2PO4^- ion is named dihydrogen phosphate ion. H3PO4/H2PO4^- is the acid/conjugate base pair. B) The Stronger The Conjugate Base. 1 xl 0-8 , which is the stronger base? a. 8 times 10 to the negative 5, and let's get out the calculator and let's go ahead and do that math. Because the product of K a times K b is a relatively small number, either the acid or its conjugate base can be "strong. Correct answers: 1, question: An acid has an acid dissociation constant of 2. 1 M ephedrine (weak base Kb = 2. A strong acid loses a proton easily, and hence the conjugate base of a strong acid is a weak base. Acid HC H O HOCN 1. H2PO4– + H2O mc030-1. O) is dissolved in water, the solution turns basic from the reaction of the oxide ion (O. HPO4 2- (aq) + H2O (l) --> H2PO4-(aq) + OH-(aq) HPO42- and H2PO4- is a conjugate acid base pair. A strong base, on the other hand, must have a weak conjugate acid. The conjugate base is: _____ 9. These are, respectively: asked Jul 30, 2019 in Chemistry by Satkriti ( 69. E) none of these. The Conjugate Method is a term and methodology made most popular by Louie Simmons, owner of the Westside Barbell Club and the creator of what has become known as the “Westside” method of training. What is the conjugate base in the following equation: HBr + H2 O \longrightarrow H3 O+ + Br-. So what we need to look at is what's gonna happen when we remove or give up and h plus I on from each of these comments Right now it's important to remember the charge here, not just the hydrogen, because that's going to affect the charge of our contra get base. The acid with the strongest conjugate base -- HCN. The more stable/weaker the conjugate acid,. Conjugate Acid-Base Pairs - Duration: 4:36. Gustafson * a. The conjugate acid BH + of a base B dissociates according to BH + + OH − ⇌ B + H 2 O. " But if one is strong, the other must be weak. Post a Question. 5x10-12 Answer: Q > Ksp, therefore, La(OH)3 precipitates (a) acid = proton donor; base = proton acceptor (b) Acid Conjugate base 1981 D 1 reaction st NH4 + NH3 Al(NO3)3 K2CO3 NaHSO4 NH4Cl H2O OH-(a) Predict whether a 0. 0 0 posted by DrBob222 Jun 10, 2010. The following table gives the acid-dissociation constant values, Ka, for carbonic acid, H2CO3, and phosphoric acid, H3PO4. F- is the conjugate base and HF is the conjugate acid. 65 calculate the concentration of hydrofluoric acid. The more stable/weaker the conjugate base, the stronger the acid. 8 times 10 to the negative 5, and let's get out the calculator and let's go ahead and do that math. com A conjugate base is a species formed by the removal of proton from an acid. 19 H2PO4- HPO42- 6. H2O gained a proton to form H3O^+; therefore, H2O is the base and H3O^+ is its conjugate acid. The hydroxide ion OH −, a well known base, is here acting as the conjugate base of the acid water. HCO3- -----CO3^2- + H+ H20 -----H+ +OH-Thus the conjugate base of acid HCO3- and H2O are CO3^2- and OH-. 0 * 10^-14 Examples of calculating pKa for acids and bases: 1) A 4. Acid Ka1 Ka2 Ka3 H2CO3 4. The fact that HCl is a strong acid implies that its conjugate base Cl – is too weak a base to hold onto the proton in competition with either H 2 O or H 3 O +. In the following reactionHF(aq) + HPO42-(aq) DF-(aq) +H2PO4-(aq)a. H3PO4 + H2O ==> H3O^+ + H2PO4^-So H3PO4, phosphoric acid, lost a proton to form H2PO4^- so H2PO4^- is the conjugate base of H3PO4, the acid. In solution, these two species (the acid and its conjugate base) exist in equilibrium as illustrated by Eq. Recall that every acid has a conjugate base: The K a of hypochlorous acid (HOCl) 3. -1, as well as other conjugate acid–base pairs. Identify the formula for the conjugate acid of H2PO4−. Thus, a strong acid must have a weak conjugate base. H2PO4- is anacid and F- is its conjugatebase. 40, what is the ratio of the molar concentrations of HPO42− to H2PO4−?. Post a Question. equilibrium calculation Strategy —find the conc. 35E-10, and Cb = 0. 2 –) with water. H3PO4/H2PO4^- is the acid/conjugate base pair. 8x 10-4) has a pH of 3. In this particular case, the water acts as the base (the proton acceptor) and is amphiprotic meaning it can act as a base or acid and it can both donate and accept protons. So what we need to look at is what's gonna happen when we remove or give up and h plus I on from each of these comments Right now it's important to remember the charge here, not just the hydrogen, because that's going to affect the charge of our contra get base. Weak acids have varying strengths dependent on their pKas, so you have acids that are weak acids, but strong relative to other weaker acids, and vice versa. The conjugate base of HNO 2 is NO 2- HNO 2 + H 2 O ↔ H 3 O + + NO 2-The conjugate acid of CN-is HCN. Conjugate acids and conjugate bases are the acids and bases that lose or gain protons. What is the conjugate base for each of the following? HSO4 HNO3 H2PO4 Soya HCI H30* H3PO4 H2SO4 H2O H2CO3 HCO3.

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# Camera Control¶

Panda3D’s camera is considered a PandaNode. It can therefore be manipulated as any other node.

The actual camera is defined in ShowBase as a NodePath named base.cam. There is also a plain node above the camera, which is a NodePath called base.camera. Generally you want to control the latter with your code.


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# [Tex/LaTex] How to remove figure label

captionsfloatrowfloats

I'm using floatrow package to organise my figures, but then would like to add an image in the preliminary chapters which should NOT contain figure label e.g. Figure 1, but only the caption To my family. I also want to eliminate the underline of caption which floatrow does. Is there a way to get around this? The code is as below:

\begin{figure}[!h]
\ffigbox[\FBwidth]
{\color{blue}\includegraphics[width=6cm,height=6cm]{./Figures/flower.png}}
{\caption{To my family}}
\end{figure}

You can use \caption* command instead of \caption provided by caption package.

\documentclass{article}
\usepackage{floatrow}
\usepackage{caption}

\begin{document}

\begin{figure}[h]
\ffigbox[\FBwidth]
{Your picture is here.}
{\caption*{To my family}}
\end{figure}

\end{document}