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### 10 NDF History

During a spring clean of directories to free some space (what d’ y’mean you don’t?), most of us will have encountered data files whose purpose and worth are long forgotten. We’re reluctant to remove them in case they contain irreplaceable data. Some people are very good and make copious notes… Even then the result of a casual experiment might not be recorded. For those who are lazy, such files can be a frequent dilemma. Even a quick look at a plot of the data is often of little assistance. As you’ve probably surmised, the NDF offers a solution.

Within an NDF you may record history information. This is usually a chronicle of the processing stages used to form the NDF, including the parameter values of the applications invoked; but it may also include commentary you provide, for example, the rationale for doing certain operations.

History is associated with individual NDFs; it is not some global attribute of a data-processing session. An NDF has a history update mode, which remains with the NDF and any descendant NDF, until the update mode is altered or the history erased. By default, the update mode is "Disabled", meaning that no history recording occurs. To permit history recording you must first switch it on, selecting from three update modes—"Quiet", "Normal", and  "Verbose"—which give increasingly more detailed information.

#### 10.1 Control and Content of History Recording

Task HISSET lets you set the history update mode. The default is "Normal", thus here the command

% hisset hr1068

switches normal history recording on for NDF hr1068. Thereafter whenever you alter this NDF, or create another NDF from it, the task automatically records the name of the application that was run, the date and time, a reference name that identifies the NDF, your username, and some text comprising the command-line parameters and the full path of the application. In Kappa the package name and version is appended to the application name. Other packages may provide task-dependent additional text.

If disc space is not a concern, you might prefer the verbose level.

% hisset hr1068 verbose

the supplementary information being the machine type, and its operating system name and version.

For small datasets, such as spectra, the history can amount to a significant part of the NDF’s size, so for these you might prefer the quiet level. This does not record the command line.

HISSET lets you switch off history recording, if you want to do something ‘off the record’, or erase the history altogether.

% hisset hr1068 disabled
% hisset hr1068 erase

Some applications create new NDFs from scratch, not inheriting the history records and update mode from an input NDF. Some examples are CREFRAME, TRANDAT, and PSF. Should you wish these to have history recording enabled as you create such NDFs, there is an environment variable NDF_AUTO_HISTORY that should be set to a non-zero integer value, or immediately run HISSET with the new NDF. Note that some applications may choose to disregard the value of  NDF_AUTO_HISTORY for good reason, such as for ancillary NDFs created with an NDF extension. Whenever this option is exercised, the reference documentation for the task should indicate this behaviour in its Implementation Notes. There are currently no such applications in Kappa.

#### 10.2 Adding Commentary to History Recording

% hiscom hr1068 i "There may have been cloud during the integration."

You aren’t limited to single lines if you respond to the prompt for the comment. You can give a series of lines, terminated by supplying !.

% hiscom hr1068
COMMENT - Comment line > The dome may have been obstructing the telescope
COMMENT - Comment line > during the integration.  We are not sure that the
COMMENT - Comment line > filter is correct either.
COMMENT - Comment line > !

If you prefer, you may edit some text into a file and append its contents to the history records. Thus

appends the text contained in comments.lis to the history records of NDF hr1068.

#### 10.3 Listing History Records

At some point you will want to refer back to the history records. The HISLIST task does this.

% hislist hr1068

History listing for NDF structure /home/scratch/dro/hr1068:

History structure created 1995 Sep 24 11:16:15.000

1: 1995 Sep 24 11:16:15.000 - HISSET          (NDFPACK V1.0)

Parameters: MODE=’Normal’ NDF=@hr1068
Software: /star/bin/kappa/hisset

Before you ask…at present there are no parameters for selecting a time interval and there is no output of the machine and username, but they’re not forgotten.

Here is another example showing a series of history records.

% hislist hr1068 \\

History listing for NDF structure /home/scratch/dro/hr1068sm2:

History structure created 1995 Nov 24 11:16:15.000

1: 1995 Sep 24 11:16:15.000 - HISSET          (NDFPACK V1.0)

Parameters: MODE=’Normal’ NDF=@hr1068
Software: /star/bin/kappa/hisset

2: 1995 Sep 24 11:19:53.000 - GAUSMOOTH       (KAPPA V0.9)

Parameters: BOX=13 FWHM=5 IN=@hr1068 OUT=@hr1068sm TITLE=! WLIM=!
Software: /star/bin/kappa/gausmooth

3: 1995 Sep 24 11:20:15.000 - HISSET          (NDFPACK V1.0)

History update mode changed from NORMAL to VERBOSE.
Parameters: MODE=’Normal’ NDF=@hr1068sm
Software: /star/bin/kappa/hisset

4: 1995 Sep 24 11:20:49.000 - GAUSMOOTH       (KAPPA V0.9)

Parameters: BOX=9 FWHM=3 IN=@hr1068sm OUT=@hr1068sm2 TITLE=! WLIM=!
Software: /star/bin/kappa/gausmooth
Machine: alpha, System: OSF1 214 (release V3.2)

5: 1995 Sep 24 11:22:32.000 - HISCOM          (NDFPACK V1.0)

Parameters: MODE=’Interface’ NDF=@hr1068sm2 WRAP=TRUE
Software: /star/bin/kappa/hiscom
The dome may have been obstructing the telescope during the integration.
We are not sure that the filter is correct either.

The first history item shows HISSET enabling history. This was followed by a smooth of the data with GAUSMOOTH. Then the recording level was set to verbose. The fourth record recalls another smooth, and this time you can see the machine details. Finally, some commentary is added with HISCOM.


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## anonymous 5 years ago what is 18 1/15 - 9 3/5

1. anonymous

you mean 18/15 - 27/5? if so it equals 4.2 or 21/5

2. anonymous

is it$18 {1 \over 15} - 9 {3 \over 5}$ ?

3. anonymous

yes yuki

4. anonymous

mixed numbers are fairly easy when you understand that it can also be changed into an equivalent improper fraction

5. anonymous

ooh 271/15 - 48/5 = 31/5 ? lol i think

6. anonymous

which is 6 5/1

7. anonymous

for example, $18 {1 \over 15}= 18 + {1 \over 15}$ so ${18*15+1 \over 15} = {217 \over 15}$

8. anonymous

you do the same thing with the other fraction, but before we do that, we need to make the denominator common. so instead of $9 {3 \over 5}$ we will use $9 {9 \over 15}$ which as an improper fraction, is ${9*15+9 \over 15} = {144 \over 15}$

9. anonymous

Who elsee gott daht?

10. anonymous

now if you subtract them, ${217 - 144 \over 15} = {73 \over 15}$

11. anonymous

if you need it as a mixed number, 73 / 15 = 4 remainder 13 so $4 {13 \over 15}$

12. anonymous

I GOT DIFF.

Find more explanations on OpenStudy

format: plain


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# Normal distributed random sample: find the least variance from the set of all unbiased estimators of $\theta$

Let $$X_{1},X_{2},\ldots,X_{n}$$ be a random sample from $$X\sim\mathcal{N}(0,\sigma^{2})$$.

(a) Find the least variance from the set of all unbiased estimators of $$\sigma^{2}$$.

(b) Find a sufficient statistics of $$\sigma^{2}$$.

(c) Obtain from this statistics an unbiased estimator to $$\sigma^{2}$$.

(d) Verify if this estimator is efficient.

MY ATTEMPTS (EDITED)

(a) In the first place, let us determine the Fischer information of $$\sigma$$: \begin{align*} & f(x|\sigma) = \frac{1}{\sigma\sqrt{2\pi}}\exp\left(-\frac{x^{2}}{2\sigma^{2}}\right) \Rightarrow \ln f(x|\sigma) = -\ln(\sigma) - \frac{\ln(2\pi)}{2} - \frac{x^{2}}{2\sigma^{2}} \Rightarrow\\\\ & \frac{\partial\ln f(x|\sigma)}{\partial\sigma} = -\frac{1}{\sigma} + \frac{x^{2}}{\sigma^{3}} \Rightarrow \frac{\partial^{2}\ln f(x|\sigma)}{\partial\sigma^{2}} = \frac{1}{\sigma^{2}} - \frac{3x^{2}}{\sigma^{4}} \Rightarrow\\\\ & I_{F}(\sigma) = \textbf{E}\left(-\frac{\partial^{2}\ln f(x|\sigma)}{\partial\sigma^{2}}\right) = -\frac{1}{\sigma^{2}} + \frac{3\textbf{E}(x^{2})}{\sigma^{4}} = -\frac{1}{\sigma^{2}} + \frac{3\sigma^{2}}{\sigma^{4}} = \frac{2}{\sigma^{2}} \end{align*}

because $$\textbf{E}(X) = 0$$. Thus $$\textbf{Var}(\hat{\sigma}) \geq \sigma^{2}/2n$$.

(b) Consider the likelihood function: \begin{align*} L(\textbf{x}|\sigma^{2}) = \prod_{k=1}^{n}f(x_{k}|\sigma^{2}) = \left(\frac{1}{\sigma\sqrt{2\pi}}\right)^{n}\exp\left(-\frac{1}{\sigma^{2}}\sum_{k=1}^{n}x^{2}_{k}\right) \end{align*}

According to the factorization criterion, we have the following sufficient statistcs:

\begin{align*} S(\textbf{x}) = \sum_{k=1}^{n}x^{2}_{k} \end{align*}

(c) I started with the following relation \begin{align*} \textbf{E}(S) = \textbf{E}\left(\sum_{k=1}^{n}X^{2}_{k}\right) = n\sigma^{2} \end{align*}

Therefore the statistics $$\hat{\sigma} = S/n$$ is unbiased.

(d) Observe that $$\textbf{Var}(X^{2}_{k}) = 2\sigma^{4}$$, which may be obtained from the moment generating function of $$X\sim\mathcal{N}(0,\sigma^{2})$$, that is given by \begin{align*} M_{X}(t) = \exp\left(\frac{t^{2}\sigma^{2}}{2}\right) \end{align*}

Consequently, the following relations hold \begin{align*} \textbf{Var}(\hat{\sigma}) = \textbf{Var}\left(\frac{1}{n}\sum_{k=1}^{n}X^{2}_{k}\right) = \frac{1}{n^{2}}\sum_{k=1}^{n}\textbf{Var}(X^{2}_{k}) = \frac{2\sigma^{4}}{n} \end{align*}

Finally, we have \begin{align*} e(\hat{\sigma}) = \frac{1}{\textbf{Var}(\hat{\sigma})nI_{F}(\sigma)} = \frac{1}{4\sigma^{2}} \end{align*}

• As for (c), you could have just found $E(S)=n\sigma^2$, giving the unbiased estimator $S/n$. – StubbornAtom Apr 11 '19 at 16:20
• Yes $\operatorname{Var}\left(\frac{1}{n}\sum X_i^2\right)=\frac{2\sigma^4}{n}$, but that should also be the answer to part (a) since it is also the CR lower bound. As a result, the answer to efficiency will also change. – StubbornAtom Apr 11 '19 at 20:27
• In particular, note the expression for the CR lower bound as you did not apply the formula correctly. You are considering unbiased estimator of $\sigma^2$, not $\sigma$. – StubbornAtom Apr 11 '19 at 20:36

Joint density of the sample $$\mathbf X=(X_1,X_2,\ldots,X_n)$$ is

$$f(\mathbf x\mid\sigma)=\frac{1}{(\sigma\sqrt{2\pi})^n}\exp\left[-\frac{1}{2\sigma^2}\sum_{i=1}^n x_i^2\right]\quad,\,\small \mathbf x\in\mathbb R^n\,,\,\sigma>0$$

By Factorisation theorem, a sufficient statistic for $$\sigma^2$$ is $$T(\mathbf X)=\sum_{i=1}^n X_i^2$$

Going back to the joint density, we have

\begin{align} \frac{\partial}{\partial\sigma}\ln f(\mathbf x\mid \sigma)&=\frac{-n}{\sigma}+\frac{1}{\sigma^3}\sum_{i=1}^n x_i^2 \\&=\frac{n}{\sigma^3}\left(\frac{1}{n}\sum_{i=1}^n x_i^2-\sigma^2\right) \end{align}

This is the condition of equality (see this related post) in the Cramér-Rao inequality, which directly shows that $$\frac{T}{n}$$ is the UMVUE of $$\sigma^2$$. This also suggests that it is the most efficient estimator of $$\sigma^2$$ within the unbiased class.

So the answer to part (a) is $$\operatorname{Var}\left(\frac{T}{n}\right)$$, which equals the Cramér-Rao bound given by

$$\operatorname{Var}\left(\frac{T}{n}\right)=\frac{\left[\frac{d}{d\sigma}(\sigma^2)\right]^2}{I(\sigma)}$$

, where $$I(\sigma)=-n\operatorname E\left[\frac{\partial^2}{\partial\sigma^2}\ln f (X_1\mid\sigma)\right]$$ is the Fisher information in the sample.

You should be able to finish calculating $$I(\sigma)$$ from the point you have stopped. And note that direct calculation of $$\operatorname{Var}\left(\frac{T}{n}\right)$$ is also possible.

• In the first place, thank you for the answer. After your considerations, I have edited my response. Can you check it out for me? Thanks! – user242554 Apr 11 '19 at 17:10


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» » » A flagpole of height 2.5m casts a shadow of length 4m. Calculate the angle of el...

# A flagpole of height 2.5m casts a shadow of length 4m. Calculate the angle of el...

### Question

A flagpole of height 2.5m casts a shadow of length 4m. Calculate the angle of elevation of the sum, correct to the nearest degree.

### Options

A)
$$32^{\circ}$$
B)
$$58^{\circ}$$
C)
$$39^{\circ}$$
D)
$$51^{\circ}$$


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## Elementary Statistics (12th Edition)

Given mean:12898. Given $\sigma$:7718.8. $\alpha=1-0.95=0.05.$ $\sigma$ is known, hence we use the z-distribution using the table. $z_{\alpha/2}=z_{0.025}=1.96.$ Margin of error:$z_{\alpha/2}\cdot\frac{\sigma}{\sqrt {n}}=1.96\cdot\frac{7718.8}{\sqrt{5}}\approx6766.$ Hence the confidence interval:$\mu$ is between 12898-6766=6132 and 12898+6766=19664.


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# The reason for greater range of oxidation states in actinoids is attributed to

more_vert
The reason for greater range of oxidation states in actinoids is attributed to

(a) actinoid contraction

(b) $5f$, $6d$ and $7s$ levels having comparable energies

(c) $4f$ and $5d$ levels being close in energies

(d) the radioactive nature of actinoids.


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Higher order Weil pairings

The summer is coming to a close, and so I’ve been trying to summarize some of the things that have happened. For one, I did manage to successfully perform the computation of $(E_n)^\vee_* K(\mathbb{Z}/p^j, q)$ for all conceivable values of n, p, j, and q, and the whole thing did culminate, as expected, in an isomorphism of graded ring schemes

So, that’s pretty exciting. But some other things happened too, and here’s one I’ve been meaning to complain about for a while. For the most part, this idea of associating to a compactly generated space X the formal scheme $X_E := \mathrm{colim}_\alpha E^* X_\alpha$, where $X_\alpha$ ranges over the compact subspaces of X, is a mechanism for introducing the language and ideas of algebraic geometry into algebraic topology. The first nontrivial example that I learned in this framework is the Ando-Hopkins-Strickland result (or, rather, the Cartier dual of what I’m about to say) that the formal scheme $BU\langle 2k \rangle_E$ corresponds to the kth symmetric power of the augmentation ideal of a “group ring” type object $C_0 \mathbb{C}\mathrm{P}^\infty_E$, for all kinds of theories E and at least in the range $k \le 3$. A big part of their proof was an explicit analysis of the schemes $C_k$, especially in the case of k = 3, where chasing out the definition of $C^3 := \mathrm{Hom}(C_3 \mathbb{C}\mathrm{P}^\infty_E, \mathbb{G}_m)$ shows that it is the scheme of “cubical structures”.

In topology, there is a fiber sequence $K(\mathbb{Z}, 3) \to BU \langle 6 \rangle \to BSU$, which for a p-local cohomology theory E gives at the very least a map

But, at least for specific choices of E like the one in my opening unrelated paragraph, we know that $K(\mathbb{Z}/p^\infty, 2)_E \cong (B\mathbb{Z}/p^\infty_E)^{\wedge 2}$, and so we expect the map above to mimic a map in algebraic geometry of the form

i.e., something that swallows cubical structures and outputs alternating biexponential maps. Such an object is known to the enterprising algebraic geometer: the Weil pairing. This object was crucially used by Mumford in his study of biextensions, and has received all kinds of press before and since.

To give you the dimmest sense of what this is, if you’re unfamiliar with it already, I’ll outline its construction. A biextension over a pair of group schemes G and H is a $\mathbb{G}_m$-torsor over $G \times H$, such that if you pick any point $g \in G$ or $h \in H$, the pullback of the torsor along the inclusions $\{g\} \times H \hookrightarrow G \times H$ and $G \times \{h\} \hookrightarrow G \times H$ respectively is an honest split central extension of the base by $\mathbb{G}_m$. Each of these central extensions has a controlling 2-cocycle determining the isomorphism type of the extension, and a “cubical structure” is a biextension for which $G = H$, all these 2-cocycles are equal, and the 2-cocycle is $\Sigma_3$-invariant. Beginning with a cubical structure $\mathcal{L}$, we can consider the multiplication-by-p maps $p_L$ and $p_R$ in the left- and right-hand factors of the base respectively. These maps factor as

and similarly for $p_R$. These two maps $i$ and $\mu_L$ themselves factor as $\mathcal{L} \to p_* \mathcal{L} \to \mathcal{L}^{\otimes p}$ and $\mathcal{L}^{\otimes p} \to (p \times 1)^* \mathcal{L} \to \mathcal{L}$. The two factoring maps $p_* \mathcal{L} \to \mathcal{L}^{\otimes p}$ and $\mathcal{L}^{\otimes p} \to (p \times 1)^* \mathcal{L}$ can be checked to be isomorphisms, and finally the composite

is the thing called the “Weil pairing”. Over the p-torsion in the base, at least, the two pullback torsors trivialize canonically, and so we can write down a formula for the action of the composite map in the $\mathbb{G}_m$ factor:

where u is the controlling 2-cocycle. And, finally, the real punchline: this is exactly the map induced by the topological fiber $K(\mathbb{Z}, 3) \to BU\langle 6 \rangle$.

Of course, the topological story doesn’t stop there. At the next stage in the Whitehead-Postnikov decomposition there’s another fiber sequence $K(\mathbb{Z}, 5) \to BU\langle 8 \rangle \to BU\langle 6 \rangle$, and we can attempt to play a similar game by pplying the functor $\mathrm{Hom}(-_E, \mathbb{G}_m)$ to the fiber map. Again, assuming that E is a nice cohomology theory, the object $\mathrm{Hom}(K(\mathbb{Z}, 5), \mathbb{G}_m)$ is known: it’s the scheme of alternating, multiexponential functions on the p-divisible subgroup $\mathbb{C}\mathrm{P}^\infty_E[p^\infty]$ of the formal group associated to the complex orientability of E-theory. Hughes, Lau, and I have calculated the ring of functions on the affine scheme $C^k(\mathbb{G}_a, \mathbb{G}_m) \times \mathrm{Spec}(\mathbb{Z}_{(2)})$, and computational experiment suggests that, further pulling back to $\mathrm{Spec}(\mathbb{F}_2)$, the Ando-Hopkins-Strickland map $\mathcal{O}_{C^k} \otimes \mathbb{F}_2 \to H_*(BU \langle 2k \rangle; \mathbb{F}_2)$ is very close to being an isomorphism, and the amount it misses it by is exactly the closure of the odd-dimensional parts of the homology under the coaction of the dual Steenrod algebra. This is a bit of a mess of a conjecture, but by hook or by crook we seem to end up with a map from the $(k-1)$-dimensional analogues of cubical structures to $(2k-4)$-variate alternating, multiexponential maps.

The real question is: where is this guy in the algebraic geometry literature? The construction used above has no obvious higher dimensional analogue. An equivalent phrasing of the question is: given two line bundles $L_1$ and $L_2$, we can consider their virtual differences $(1-L_1)$ and $(1 - L_2)$, which we can put together in creative ways to get a virtual bundle whose classifying map lifts to $BU\langle 6 \rangle$. These creative ways specifically look rather like the formula for $e_n$, inexorably leading you toward the Weil pairing story. But, given four line bundles, how do we build a bundle with good properties whose classifying map lifts to $BU\langle 8 \rangle$? I, for one, have no idea.


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1. Home
2.  - Reduction Ratio Equation For Cone Crushing

# Reduction Ratio Equation For Cone Crushing

• ### Reduction Ratio Calculation Crusher

How To Calculate Jaw Crusher Reduction Ratio. reduction ratio formula for crusher srpc. calculate crushing efficiency of . the reduction ratio of a jaw crusher should be 61 this is calculated as cone PS two the crusher parameter capacity in the .Crusher Reduction Ratio - Mineral Processing Metallurgy. 19 3 2017 We know from our examination of.

• ### Reduction Ratios Crushing

Reduction Ratio Formula In Crushing. The reduction ratio of a crushing stage can be defined as the ratio of there are a number of crushers available as jaw, gyratory, cone, roll, and impact crushers. And Support Online reduction rate crushing. reduction ratio in stone crushing reduction ratio in stone crushing Crusher Reduction Ratio.

• ### Reduction Ratio Target For Ball Mill

Mar 20, 2013 Mining Rules. for up to 40 of the direct operating cost of the mineral processing practical reduction factor (ratio) for a ball mill, the target is 18 . Source .

• ### Reduction Ratio Crushing In

Reduction Ratio Crushing No Of Crushers Bussa Machinery. reduction ratio crushing no of crushers size reduction ratio. the ratio of the gape width to the near side jaw opening. this ratio may vary from to 19, but is generally for primary crushers and for secondary crushers. nip angle. one of the most important parameters is the angle between the two jaws, which is referred to as sep 02, 2019.

• ### Limiting Reduction Ratio Crushing

Reduction ratio equation for cone crushing – Grinding Mill China what type of chain used in jute mills in kolkata How to calculate crusher reduction ratio and get such as big reduction ratio, there are hammer crusher detailed. Acara 1.

• ### Introduction To Rock Crushers

Jaw crushers offer reduction ratios of up to about 6 1, while cone crushers can reduce material size up to a maximum of 8 1 ratio. The finer the crushing, the smaller the reduction ratio. As a basic rule, jaw crushers are the primary crusher taking the mined aggregate and reducing it to a size that a cone.

• ### Chamber On Energy Consumption Of Cone Crusher

Chamber on Energy Consumption of Cone Crusher To estimate the energy requirement in size reduction according to Equation (1), pressure p is needed structural parameters of the crushing.

• ### Cone Crusher Calculation Cone Crusher

Cone crusher crushing force calculation pixelpushr cone crushers mclanahan cone crusher applications a general rule of thumb for applying cone crushers is the reduction ratio a crusher with coarse style liners would typically have a 61 reduction ratio thus with a 34 closed side setting the maximum feed would be 6 . Details.

• ### Evaluation Of Size Reduction Process For Rock Aggregates

Jun 04, 2020 Utilizing this method, the size reduction ratio (SRR) was determined for each rock type by the following equation $$\mathrm{SRR}={F}_{80} {P}_{80}$$ (4).

• ### Crusher Reduction Ratio Mineral Processing & Metallurgy

Jan 12, 2016 Crusher Reduction Ratio. I have mentioned the fact that, as the of voids in the crushing chamber decreases, the production of fines by attrition increases. This is like saying that, as the Crusher Reduction Ratio in any given crusher is increased, the of fines in the product will increase, even though the discharge setting remains unchanged.

• ### Calculate The Reduction Ratio And The Number Of Milling

Mar 28, 2021 R Rp reduction ratio of the p stage. Source formula.minerallurgy. For example in a crushing circuit that consists of a jaw crusher (3 1), and a cone crusher ( 5 1), has a total reduction ratio of 15 1. It means if your feed material has an F80 of 69cm, your product size p80 will be 4.6 cm.

• ### Construction Working And Maintenance Of Crushers

For primary crushing with a reduction ratio of 6 to 1. When a relatively coarse product is desired, top size usually larger than 38 mm (1 1 2 ) Closed Circuit Crushing System To control top size from a single crusher operating in an open circuit, material must remain in the crushing chamber until the material is reduced to top size.

• ### Design Of A Crushing System That Improves The Crushing

Processing with a high crushing ratio, or reduce overburden to a size suitable for belt conveying with a low crushing ratio (Smith, 2013). 2.5.2 Cone crusher . Cone crusher is applied to break all kinds of minerals and rocks with the compressive strength below 250MPa and all.

• ### Technical Notes 5 Crushers

The chief difference between cone and gyratory or jaw crushers is the nearly parallel arrangement of the mantle and the cone at the discharge end in the cone crusher. This is illustrated in Figure 5.2. Reduction ratios in the following ranges are common for cone crushers 6 1 - 8 1 for secondaries 4 1 - 6 1 for tertiary and quaternary crushing.

• ### Crusher Size Reduction Ratio Calculation Method

The impact crusher size reduction ratio is 20 to 1. The vertical shaft impact crusher size reduction ratio is 4-8 to 1. The vertical roller mill size reduction ratio is 2-2.5 to 1. The hammer crusher size reduction ratio is 20 to 1. Pre Article Crusher Plays An Important Role In Artificial Sand Production Line. Next Article Development Of The World's Cone Crusher Industry. Related Machine. Sand Making.

• ### Chamber Optimization For Comprehensive Improvement Of Cone

Mar 15, 2021 This study aims to analyze the impact of key structural parameters such as the bottom angle of the mantle, the length of the parallel zone, and the eccentric angle on the productivity and product quality of the cone crusher and optimize the crushing chamber to improve the crusher performance. The amount of ore in the blockage layer was calculated by analyzing the movement.

• ### (pdf) Crushing Force Theoretical Examination In One Cone

The formula of MBD for the inertia cone crusher without any joints a size reduction model and a pressure is presented for calculating compression ratio of each crushing zone along with.

• ### Crushing Of Coal And Calculation Of Size Reduction Efficiency

Feb 25, 2015 Roll Crusher • The reduction ratio and efficiency of roll crusher is very low. • The main advantage associated with roll crushers is small proportion of fines produced and uniform size of product crushed, than the other crushers. • Low capacity • Low reduction ratio = F P • Sensitive to top size 2 25 2015 6 58 05 AM 38 39.

• ### (pdf) Size Reduction By Crushing Methods

There are two kinds of. equipments used for crushing work s. one is by using crushers and other one is by using impacto rs. This diagram illustrates the stages of s ize reduction from 1000mm to 4.

• ### Dual Roll Crushers How They Function

Roll crushers have a theoretical MAXIMUM reduction ratio of 4 1. If a 2 inch particle is fed to the roll crusher the absolute smallest size one could expect from the crusher is 1 2 inch. Roll crushers will only crush material down to a minimum particle size of about 10 Mesh (2 mm).

• ### Definition Of The Crushing Reduction Ratio

Reduction ratio equation for cone crushing reduction ratio in crushing the ratio of the limiting friction developed to the corresponding normal Get . Bolted Joint Design Fastenal Industrial Supplies . strength range for that particular fastener. By definition, the proof load is an applied tensile load that the fastener must.

• ### Best Practices For Crushing Screening Rap Asphaltpro

However, for best practice, both should be adjusted proportionately to maintain a correct reduction ratio and highest product quality. When it comes to production capacities of impact crushers compared to jaw and cone crushers for the crushing and screening of RAP, Ryan Freeman, the general manager at Rason Materials, shared his experience.

• ### Variable Throw Eccentric Cone Crusher And Method For

With respect to the relationship between power drawn and throw setting, for a given type of rock feed, the relationship between the reduction ratio and the energy required to crush a ton of the rock feed can be expressed by the following equation EQU1 where P=Power . VC=Volumetric Capacity . RR=Reduction Ratio. K1 is a constant.

• ### Design Of Anchor Bolts Embedded In Concrete Masonry

The strength reduction factors, Φ, for use in Equations 11 through 18 are taken equal to the following values Calculated strengths for masonry crushing (Equation 7), anchor pryout (Equation 8), and anchor yielding (Equation 9) are as follows Because the demand to capacity ratio is less than 1.0, the design is satisfied.

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# Browsing category algebra

## Ask Uncle Colin; A variable fraction

Dear Uncle Colin, I have a fraction, $\frac{x^2-x}{x-1}$, and I want to cancel it down to $x$ - but I'm not sure those are the same. Are they? - Got A Lot Of Interesting Sums Hi, GALOIS, and thanks for your message! The short answer is, yes and no. Everywhere

## Ask Uncle Colin: Why is it not 4?

Dear Uncle Colin, I have a binomial expansion of $(1+x)^\frac{1}{2}$ and need to approximate $\sqrt{5}$. Apparently you need to substitute in $x=\frac{1}{4}$, but I'd have thought $x=4$ was a more obvious choice. What gives? -- Roots Are Dangerous If Understood Sloppily Hi RADIUS, and thanks for your message! That does

## A nice bit of nrichment

My dear friend @ajk44 pointed me at a puzzle on the excellent Nrich site1, and I enjoyed it enough to share my solution. (If you don't want it spoiled, don't read beyond the blockquote.) Four jewellers had respectively 8 rubies, 10 sapphires, 100 pearls and 5 diamonds. Each gave one


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# Interpretation of non-independent errors in linear regression

I have problems to interpret dependencies in the covariance matrix $$Cov(\epsilon)$$ of errors in the linear Regression model $$Y = X\beta + \epsilon$$, i.e. when there are non-zero entries apart from the entries in the diagonal. For example let's assume that $$Cov(\epsilon_i,\epsilon_j) > 0$$ for some $$i$$ and $$j$$. If I understand the concept of errors right, $$Cov(\epsilon_i,\epsilon_j) > 0$$ implies that the deviations from the expected $$y_i$$ and $$y_j$$ value (i.e. the expected value given the regressor values in the $$i$$th and $$j$$th row of $$X$$), are somehow positively correlated.

Unfortunately, I cannot attach any meaning to this beyond this mathematical interpretation (if it is correct at all). As I understand it, $$\epsilon_i$$ and $$\epsilon_j$$ belong to two different "populations" that are defined by the respective regressor values in the $$i$$th and $$j$$th row. Why should there be an association between the error terms then? I appreciate any help.

• Can you provide some details on what populations means here? If your populations are aspen groves, then an example about people may not be very illuminating. – Dimitriy V. Masterov Nov 19 at 15:34
• I was referring to the population as a mathematical concept. And I am also trying to reach a conceptual understanding. Only an example might make it easier. – jonaden Nov 19 at 15:57


text:
How do you combine (x-1)*[3/(x^2-1)+x/(2x-2)]?

Aug 22, 2017

$= \frac{{x}^{2} + x + 6}{2 \left(x + 1\right)}$

Explanation:

$\left(x - 1\right) \cdot \left[\frac{3}{{x}^{2} - 1} + \frac{x}{2 x - 2}\right]$

Factorise the denominators first

$= \left(x - 1\right) \cdot \left[\frac{3}{\left(x + 1\right) \left(x - 1\right)} + \frac{x}{2 \left(x - 1\right)}\right]$

Add the fractions using a common denominator.

=(x-1)*[((2xx3 + x(x+1))/(2(x+1)(x-1))]

$= \left(x - 1\right) \cdot \left[\frac{6 + {x}^{2} + x}{2 \left(x + 1\right) \left(x - 1\right)}\right]$

Distribute $\left(x - 1\right)$ into the bracket:

=[(cancel((x-1))(6 + x^2+x))/(2(x+1)cancel((x-1))]

$= \frac{{x}^{2} + x + 6}{2 \left(x + 1\right)}$


text:
Lemma 114.4.13. Let $A \to B$ be a flat local homomorphism of Noetherian local rings. If $A$ and $B/\mathfrak m_ A B$ are Gorenstein, then $B$ is Gorenstein.

Proof. Follows immediately from Dualizing Complexes, Lemma 47.21.8. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).


text:
Insights Demystifying the Often Misunderstood Bernoulli's Equation - Comments

Tags:
1. Dec 3, 2018

2. Dec 3, 2018

kuruman

It is mentioned in the text that "Given that the fluid velocity is everywhere normal to the streamline, no flow can cross such lines." Surely that's a typo. The accompanying drawing correctly shows the local velocity vectors tangent to the streamlines.

3. Dec 3, 2018

Good catch.

Let's just say I am averaging about 4.5 hours of sleep over the past 3 weeks but thought this article addressed a misconception that is so frequent that I thought it worthwhile to attempt to write it now anyway. If anyone finds other blatant typos, please let me know.

4. Dec 3, 2018

kuruman

Wow!! This is$$\frac{4.5~hr}{the~past~3~wk}=1.5\frac{hr}{the~past~wk}.$$You oughta get more sleep than that.

5. Dec 3, 2018

Now you are catching me in more poorly-worded statements that may or may not be related to sleep deprivation.

6. Dec 4, 2018

Swamp Thing

Hi,

Congratulations! There has long been a need to demystify Bernoulli's principle, and this article surely fills a significant part of that vacuum.

Having said that, I'm not sure what fraction of the target audience (meaning people who actually need to be demystified) would be able to take profit from this particular level of exposition. I'm not an expert (which perhaps entitles me to represent the mystified ones). Is it the case that people with enough math and physics competence to understand the article are in fact likely to have held wrong ideas before reading it? For example, believing in some of the wrong notions of aircraft flight that we often find on YouTube etc? And when I say wrong notions, I mean that other people, the debunkers, say that they are wrong -- and not that I'm personally qualified to point out the errors in precise terms.

What I'd like to suggest is, maybe you could write a companion article that (a) makes the principle come alive using more intuition (graphics) and less calculus and (b) debunks specific misunderstandings that people have about airplane wings, balls floating in jets and so on. To sum up, something to address misconceptions among laymen, as much as misconceptions among specialists.

Of course, it may turn out that it is impossible to do justice at all to these things without the kind of mathematical support that you have already used in the article. I don't know.

Thanks!

Edit: There's a lot of stuff out there that tries to straighten out the subtle business of Coanda Effect versus Bernoulli effect, but personally I'm still not clear about this kind of thing. This could be one thing that you could perhaps clarify, should you decide to go ahead with the companion article idea.

Last edited: Dec 4, 2018
7. Dec 4, 2018

Staff: Mentor

It's a difficult balance between being concise and being complete. A "concise" answer is likely to have holes or thin spots that create an opportunity to be or be perceived as wrong or, well, incomplete, whereas a "complete" answer may lose part of the target audience who is looking for a fast answer to a small question.

The title and intro look a bit like a hook to bring people in for a quick answer to a particular misconception or two, but really is a pretty complete explanation....frankly, a quick list of two or three misconceptions would be too short for an article. But I'll provide a concise example of what I think is the most common misconception we get about Bernoulli's principle itself (not the applications such as lift):

Many people don't grasp the full implications of the steady flow and streamline assumptions. We often get people who ask, "but what if we add an orifice...?" or "what if I use a smaller pipe?" Now, not only is it no longer the same streamline or steady flow, but it isn't even the same system!

There's an active thread in biology right now where someone asks; shouldn't the volumetric flow rate through a small straw be the same as through a large straw because a lower cross sectional area results in higher velocity? We often get this very misconception with regard to blood circulation. As described, this is called "continuity" (not specifically mentioned in the insight), but it doesn't apply across two different systems. Continuity says that the mass flow rate in two parts of the same system must be the same unless mass is accumulating or draining. But of course the mass flow rate in two different systems doesn't need to be the same.

8. Dec 6, 2018

A couple of things here:
First, I didn't create the title. I don't know who is in charge of that, but I submitted the article and it magically appeared with a title. I agree it does give off the air of being written at a lower level than it is.

Second, I did try to start it out by saying it was written for someone already familiar with some basic undergraduate-level math and science. There are a surprising number of people at this level with a lot of misconceptions. It's certainly not the majority of them, but a sizable chunk.

Third, I agree that, when this article was finished and I looked back over it, it turned out to be at a higher level than my original intention. This came from going back over it and filling in enough holes to try to make it "reasonably complete" but without digging into, for example, the derivation of the energy equation. I don't honestly see any way around this at this point if I do it from an energy approach, and there are myriad sources online who discuss the same issue from the point of view of a force balance (conservation of momentum).

Finally, I also agree that there is a need for a similar article (or maybe an expansion of this one is some way) that does address this at a lower level, but I am just not sure how to implement it without an appeal to authority rather than letting the math/physics speak for themselves.

Part of the issue is that fluid dynamics is particularly counterintuitive for many people who don't have any training in the topic. I am not how to appeal to their intuition when it often serves them wrong in the first place. I was the same way when I was first learning and it all felt backward.

https://www.physicsforums.com/insights/airplane-wing-work-primer-lift/

9. Dec 7, 2018

Swamp Thing

I guess fluid mechanics is a bit like quantum mechanics in being highly counter-intuitive, although for different reasons

Perhaps even more confusing than QM in some sense, given that there is a Clay Millennial bounty for the Navier-Stokes theorem but none for a QM problem.

10. Dec 7, 2018

Heisenberg quit fluids to do QM. Draw your own conclusions.

11. Dec 7, 2018

Swamp Thing

So he decided to go for the low-hanging fruit...

12. Dec 8, 2018

ruko

13. Dec 8, 2018

Again, as linked before, there's a totally separate article for that.

https://www.physicsforums.com/insights/airplane-wing-work-primer-lift/

14. Dec 8, 2018

Swamp Thing

That lift article is just perfect for the (fairly) uninitiated.

Quick question. You said there:
A rough hand sketch of this non-viscid case would be nice. Or even better, the following simulation with viscosity set to zero -- how would that look?

15. Dec 8, 2018

If that flow was inviscid, the flow on bottom would wrap around the sharp trailing edge and leave the shape somewhere back around on the top surface.

16. Dec 8, 2018

Swamp Thing

I just realized that the animation in your Lift article (i.e. the one I pasted in my previous post) is actually showing inviscid flow, not viscid flow. So that answers my question. I had wrongly assumed that this is viscid flow.

I guess it makes kind of sense, because the incoming and outgoing angles are nearly equal for each and every streamline, hence no net momentum change so no lift.

Edit: Wait, that's obviously not correct. The lines are asymptotically horizontal downstream, which is what matters.

It may be instructive to see the streamlines side by side for exactly the same airfoil, with and without viscosity.

Last edited: Dec 8, 2018
17. Dec 9, 2018

Staff: Mentor

It's complicated, so I wouldn't say it's the main effect, but it is always in effect. The simple answer to your question is that the strength of the effect depends on the angle the wing makes with the wind, and that angle is different for an inverted plane than an upright plane.

18. Dec 10, 2018

cjl

It's inviscid, but it's including an important element of viscous flow. Specifically, inviscid theory alone would predict streamlines more like the top image shown here (which would generate zero lift):

Viscosity is what enforces that the flow comes smoothly off the sharp trailing edge of the airfoil, modifying the flow from the top image to the bottom image. Frequently, when analyzing lift and flow about an airfoil, it is modeled as inviscid, but with this condition of smooth flow off the trailing edge enforced to give you a physically reasonable flow.

19. Dec 10, 2018

Swamp Thing

In figure (a), assume that the wing has negligible density compared with the fluid. It is constrained from rotating or moving left / right. But it is totally free to move up or down (no restoring force in Y direction).

In that case, would it drift downwards over time? And if we now add a constraint against this drift, how much force would we measure?

Edit:
MbahahahaHA! I believe I got it!
Let's assume the wing does have some mass (a lot, in fact). But since there is no viscosity, the liquid can't drag it along -- the wing just hovers there, provided that we have made sure to give it an initial push to the left during the setup phase when the liquid was accelerating. Newton's first law and all that?

In steady state, the wing bends the streamlines by virtue of just being there occupying space, rather than "pushing back" to the left.

Edit Edit: This would work nicely if the wing had the same density as the fluid. It would hover in place without needing any constraint, yet shape the flow as in Fig. (a).

Last edited: Dec 10, 2018
20. Dec 17, 2018

Swamp Thing

Wait, I was wrong there... there is a QM problem in the Clay Millennium list.. http://www.claymath.org/millennium-problems/yang–mills-and-mass-gap


text:
## 计算机代写|图像处理代写Image Processing代考|BIOC062

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## 计算机代写|图像处理代写Image Processing代考|Degree of Union and Covering

When considering the union, the problem is posed in different terms from what is done for intersection. Here, it is not interesting to look to which degree two sets have a non-empty union, for this property is true whenever at least one of the two fuzzy sets is not empty. Instead, what is useful is the degree to which the union of two sets covers the whole space. This will be useful, for instance, for looking at the law of excluded-middle.
In the crisp case, we have for two sets $X$ and $Y$ :
$$X \cup Y=\mathcal{S} \Leftrightarrow \forall x \in \mathcal{S}, x \in X \cup Y .$$
In the fuzzy sets case, this equation is replaced (again using the formal translation principle) by:
$$\mu_{\text {union }}(\mu, v)=\inf _{x \in \mathcal{S}} T[\mu(x), v(x)],$$
where $T$ is a disjunction (often a t-conorm), yielding a degree in $[0,1]$. The same result (and in particular with $T=\max$ ) is obtained using the other fuzzification principles.
The properties of this degree of union are:

• Consistency with the binary definition.
• Symmetry: $\forall(\mu, v) \in \mathcal{F}^2, \mu_{\text {union }}(\mu, v)=\mu_{\text {union }}(v, \mu)$.
• If one of the sets is empty ( $\forall x \in \mathcal{S}, v(x)=0$ ), then $\mu_{\text {union }}$ is always 0 for bounded support fuzzy sets.
• If one of the sets is equal to $\mathcal{S}(\forall x \in \mathcal{S}, v(x)=1), \mu_{\text {union }}$ is always equal to 1 .
• Invariance with respect to rigid geometrical transformations (translation, rotation).

Let us now look at the excluded-middle law. In the crisp case, we have for any set $X$ :
$$X \cup X^C=\mathcal{S},$$
which is a set equation equivalent to the logical principle of excluded-middle (in the sense, for instance, that the truth can be in $X$ or in $X^c$, but there is no other possibility). In the fuzzy case, the degree of union between any fuzzy set $\mu$ and its complement is:
$$\mu_{\text {union }}(\mu, c(\mu))=\inf _{x \in \mathcal{S}} T[\mu(x), 1-\mu(x)],$$
if we take $c(a)=1-a$ for the fuzzy complementation.

## 计算机代写|图像处理代写Image Processing代考|Inclusion from Other Set Operations

In the crisp case, for two sets $X$ and $Y$, the inclusion is defined as
\begin{aligned} X \subseteq Y & \Leftrightarrow X \cap Y^C=\emptyset \ & \Leftrightarrow X^C \cup Y=\mathcal{S} . \end{aligned}
These two equivalences show that inclusion is simply expressed either by an intersection or by a union. The extension to fuzzy sets can therefore directly use the degrees of intersection and union as defined previously.
$$\mu_{\text {Inc }}(\mu, v)=\mu_{\text {union }}(c(\mu), v)=\inf _{x \in \mathcal{S}} T[c(\mu(x)), v(x)],$$
where $T$ is a t-conorm.

Using the degree of intersection yields the following degree of inclusion of $\mu$ in $v$ :
$$\mu_{\text {Inc }}(\mu, v)=c\left[\mu_{\text {int }}(\mu, c(v))\right]=c\left[\sup {x \in \mathcal{S}} t[\mu(x), c(v(x))]\right],$$ where $t$ is a t-norm and $c$ a fuzzy complementation. Actually, due to the duality between t-norms and t-conorms, these two definitions are equivalent. Let $t$ and $T$ be a pair of dual t-norm and t-conorm according to the complementation $c$. Then we have: $$\inf {x \in \mathcal{S}} T[c(\mu(x)), v(x)]=\inf {x \in \mathcal{S}} c[t[\mu(x), c(v(x))]]=c\left[\sup {x \in \mathcal{S}} t[\mu(x), c(v(x))]\right],$$
which proves the equivalence between both formulas.
This definition has the same drawback as those for the degrees of intersection and union, which may, basically, depend on one point only. Here again, the spatial coverage between both fuzzy sets could be taken into account.

The properties of the degree of inclusion are directly derived from those of intersection and union:

• Consistency with the binary definition.
• If $\mu$ is empty $(\forall x \in \mathcal{S}, \mu(x)=0)$, then $\mu_{\text {Inc }}(\mu, v)$ always equals 1 .
• If $v$ is empty, then $\mu_{I n c}(\mu, v)$ is equal to 0 for normalized fuzzy sets.
• If $\mu$ is equal to $\mathcal{S}(\forall x \in \mathcal{S}, \mu(x)=1), \mu_{\operatorname{Inc}}(\mu, v)$ is equal to 0 for bounded support fuzzy sets.
• If $v$ is equal to $\mathcal{S}, \mu_{I n c}(\mu, v)$ is always equal to 1 .
• Invariance with respect to geometrical transformations (translation, rotation).

# 图像处理代考

## 计算机代写|图像处理代写Image Processing代考|Degree of Union and Covering

$$X \cup Y=\mathcal{S} \Leftrightarrow \forall x \in \mathcal{S}, x \in X \cup Y .$$

$$\mu_{\text {union }}(\mu, v)=\inf _{x \in \mathcal{S}} T[\mu(x), v(x)]$$

• 与二进制定义一致。
• 对称：
$$\forall(\mu, v) \in \mathcal{F}^2, \mu_{\text {union }}(\mu, v)=\mu_{\text {union }}(v, \mu) \text {. }$$
• 如果其中一个集合为空 $(\forall x \in \mathcal{S}, v(x)=0$ ），然吕 $\mu_{\text {union }}$ 对于有界支持模糊集始终为 0 。
• 如果其中一组等于 $\mathcal{S}(\forall x \in \mathcal{S}, v(x)=1), \mu_{\text {union }}$ 总是等于 1 。
• 关于刚性几何变换 (平移、旋转) 的不变性。
现在让我们看看排中律。在脆的情况下，我们有任何集 合 $X$ :
$$X \cup X^C=\mathcal{S},$$
这是一个等价于排中逻辑原理的集合方程 (例如，在某 种意义上，真理可以在 $X$ 或在 $X^c$ ，但没有其他可能 性）。在模糊情况下，任何模朝集之间的联合度 $\mu$ 它的 补充是:
$$\mu_{\text {union }}(\mu, c(\mu))=\inf _{x \in \mathcal{S}} T[\mu(x), 1-\mu(x)] \text {, }$$
如果我们拿 $c(a)=1-a$ 用于模糊互补。

## 计算机代写|图像处理代写Image Processing代考|Inclusion from Other Set Operations

$$X \subseteq Y \Leftrightarrow X \cap Y^C=\emptyset \quad \Leftrightarrow X^C \cup Y=\mathcal{S} .$$

$\mu_{\mathrm{Inc}}(\mu, v)=\mu_{\text {union }}(c(\mu), v)=\inf {x \in \mathcal{S}} T[c(\mu(x)), v(x)]$ 在哪里 $T$ 是一个t-conorm。 使用交叉度产生以下包含度 $\mu$ 在 $v$ : $\mu{\mathrm{Inc}}(\mu, v)=c\left[\mu_{\mathrm{int}}(\mu, c(v))\right]=c[\sup x \in \mathcal{S} t[\mu(x), c$

$$\inf x \in \mathcal{S} T[c(\mu(x)), v(x)]=\inf x \in \mathcal{S} c[t[\mu(x), c(v(x)$$

• 与二进制定义一致。
• 如果 $\mu$ 是空的 $(\forall x \in \mathcal{S}, \mu(x)=0)$ ，然后 $\mu_{\text {Inc }}(\mu, v)$ 总是等于 1 。
• 如果 $v$ 是空的，那么 $\mu_{\text {Inc }}(\mu, v)$ 对于归一化模糊 集，等于 0 。
• 如果 $\mu$ 等于 $\mathcal{S}(\forall x \in \mathcal{S}, \mu(x)=1), \mu_{\mathrm{Inc}}(\mu, v)$ 对于有界支持模糊集等于 0 。
• 如果 $v$ 等于 $\mathcal{S}, \mu_{\text {Inc }}(\mu, v)$ 总是等于 1 。
• 关于几何变换 (平移、旋转) 的不变性。

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## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。


text:
# Falling charges, how many will succeed?

A horizontal conducting cylindrical hollow pipe of radius $$R = 54\text{ mm}$$ and length $$L = 100 \text{ cm} (R<<L)$$ has a small hole $$P$$ at its top, at the middle of the length as shown in the figure. Drops of mass $$M = 231 \text{ mg}$$ and charge $$q = 1 \text{ nC}$$ are falling into the hole from point $$A$$, at height $$2R$$ measured from the axis of the cylinder. Assume that the charge in the fallen drop gets uniformly distributed over the surface of the cylinder and charge distributed on cylinder remains uniform throughout. If the number of drops that will be able to enter the cylinder is given as $$n = x \times10^y$$ in scientific notation. Find the value of $$x + y$$.

×


text:
# mgbr2 formula mass in amu

posted in: Uncategorized | 0

To get the formula mass, you must get â¦ why is Net cash provided from investing activities is preferred to net cash used? Molar mass of Mg is 24 and of Br is 80. 6 O = 6 x 16 = 96. mass isapproximately 110 amu. What is 13.67 oz to grams Determining Chemical Formulas. Favourite answer. The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Multiply. Molality is defined as moles HCl/kg solvent, so begin by deciding the mass of 1 L, and the mass of water in that 1L. 0 0. Atomic Mass Unit (AMU): Definition, Standard & Conversion Aluminum Bromide: Chemical Formula, Symbol & Uses Hence, molar mass of MgBr2 is 24 +2 x 80=184g. 300. Is there a way to search all eBay sites for different countries at once? The only product is MgBr2. So the molecular formula is a multiple of the empirical formula. 83.79% have a mass of 41.941 amu, 9.50% have a mass of 42.941 amu, ... Magnesium bromide is an ionic compound with the chemical formula MgBr2. Convert grams MgBr2 to moles or moles MgBr2 to grams. Molar mass of MgBr2 = 184.113 g/mol. HNO2 = 47 g/mol. The formula mass of ammonia is therefore (14.01 amu + 3.024 amu) = 17.03 amu, and its percent composition is: $\mathrm{\%N=\dfrac{14.01\:amu\: N}{17.03\:amu\:NH_3}\times100\%=82.27\%}$ $\mathrm{\%H=\dfrac{3.024\:amu\: N}{17.03\:amu\:NH_3}\times100\%=17.76\%}$ formula mass of mgcl2 in amu: how to find the molarity of an acid: calculate the molar mass of ethane: convert amu to moles: how to get volume from moles: ... calculate the molar mass of mgbr2: how to calculate molecular weight of air: how to find the molecular weight of a gas: moles molarity volume: What is the final stage of an event in a wedding? a.) How many anions are there in 0.500 g of MgBr2? Calculate the molecular weight (atomic mass units). 6 C = 6 x 12 = 72. What is 5.67 * 10^8 mm to kilometers ... MgBr2. For bulk stoichiometric calculations, we are usually determining molar mass, which may also be called standard atomic weight or average atomic mass. First, calculate the formula mass of the empirical formula. Calculate the formula weight of magnesium bromide, MgBr2 Answer: Set up a table similar to the one used in example 1. Capitalize the first letter in chemical symbol and use lower case for the remaining letters: Ca, Fe, Mg, Mn, S, O, H, C, N, Na, K, Cl, Al. Formula: MgBr2 Molar Mass: 184.113 g/mol 1g=5.43144699179308E-03 mol Percent composition (by mass): Element Count Atom Mass %(by mass) Mg 1 24.305 13.2% Br 2 79.904 86.8% 2. Question: Calculate The Formula Mass For Each Of The Following Compounds. 4 O atoms X (16.00 amu / O atom) = 64.00 amu. Formula Mass Dimensional Analysis Molar conversions Empirical formulas; 100. 24.3 + ( 2 x 79.9 ) + ( 6 x 18 ) = 292.1 amu. The reason is that the molar mass of the substance affects the conversion. If 1.934 g Mg are left unreacted, how much MgBr2 is formed? If the formula used in calculating molar mass is the molecular formula, the formula weight computed is the molecular weight. B. b.) a.) common chemical compounds. Hence, molar mass of MgBr2 is 24 +2 x 80=184g. 0 0. of a chemical compound, More information on MgBr2 =184 g/mol. The mass of HCl present in 12.0 mol HCl = 12.0 mol HCl â¢ 36.46 g HCl 1 mol HCl = 437.52 g HCl We found at least 10 Websites Listing below when search with mgbr2 formula mass amu on Search Engine. 2 H atoms X (1.01 amu / H atom) = 2.02 amu. How long will the footprints on the moon last? Part A ZnCO3 Express Your Answer Using Two Decimal Places. 400. Who is the longest reigning WWE Champion of all time? 500. In chemistry, the formula weight is a quantity computed by multiplying the atomic weight (in atomic mass units) of each element in a chemical formula by the number of atoms of that element present in the formula, then adding all of these products together. The formula mass of calcium nitrate, Ca(NO3)2 is 82 amu 102 amu 164 amu 116 amu. Relative Formula Mass Definition . All Rights Reserved. â¢ MgCO3 + H2SO3 = MgSO3 + H2O + CO2 400. molar mass empirical formula = (2 × 12.01 g C) + (3 × 1.008 g H) + (3 × 74.92 g As) + (2 × 63.55 g Cu) + (8 × 16.00 g O) = 506.9 g What is its molecular formula? Atomic Mass Formula Questions: 1. ... which may also be called standard atomic weight or average atomic mass. Molecular weight calculation: 24.305 + 79.904*2 âºâº Percent composition by element Formula mass $=1(1.008$ amu $)+1(14.01$ amu $)+2(16.00$ amu $)$ $=47.018$ amu c.) Formula mass $=1(12.01$ amu â¦ Since a molecule - scratch that, there we go again, calling everything molecules! These relative weights computed from the chemical equation are sometimes called equation weights. Magnesium Sulfite MgSO3 Molar Mass, Molecular Weight. The atomic weights used on this site come from NIST, the National Institute of Standards and Technology. 12 H = 12 x 1 = 12. The sum is 180. Answer: To find the atomic mass of chlorine, the atomic mass of each isotope is multiplied by the relative abundance (the percent abundance in decimal form) and then the individual masses are added together. ANSWER: Formula Mass = 47.02 Amu Correct Part C NCl3 Express Your Answer Using Two Decimal Places. So, to â¦ (2 points) A 3.214 g sample of Mg reacts with 8.416 g of bromine. Chloral hydrate, a sedative, contains 14.52% carbon, 1.83% hydrogen, 64.30% chlorine, and 19.35% oxygen. This compound is also known as Magnesium Bromide. 300. Write the empirical formula for each of the following molecular formulas: C6H6O3 C6H12O6 B6H10 Caffeine has an empirical formula of C4H5N2O. Calculating formula masses. What is Carbons percent composition in C5H120 C=68.1%. 3. Copyright © 2020 Multiply Media, LLC. 429 grams What is 43.5 dm^3 of potassium selenate to grams. This is an application of the law of conservation of mass. The formula weight is simply the weight in atomic mass units of all the atoms in a given formula. When calculating molecular weight of a chemical compound, it tells us how many grams are in one mole of that substance. The formula mass is the sum of the atomic masses. To complete this calculation, you have to know what substance you are trying to convert. The percentage by weight of any atom or group of atoms in a compound can be computed by dividing the total weight of the atom (or group of atoms) in the formula by the formula weight and multiplying by 100. If the average atomic mass is 107.87 amu, what is the % abundance of 107Ag? A related term you should know is relative formula mass (relative formula weight). Using the chemical formula of the compound and the periodic table of elements, we can add up the atomic weights and calculate molecular weight of the substance. 4.35% of all X atoms have a mass of 39.946 amu. (2 points)An ion has 13 neutrons and 10 electrons. A common request on this site is to convert grams to moles. What is the formula mass of Na4SiO4 184.1 amu. Gram Formula Mass or Molar Mass sum of the atomic masses of all the atoms in a compound Example:C4H6O2 Element Quantity Atomic Mass of Element Total Mass of Element in Compound C 4 12 48 H 6 1 6 O 2 16 32 Total 86 g/mol Steps 1. 10 Substance MgBr2 # x Molar Mass 29.30 Total: In a 0.650-mol quantity of ascorbic acid, there is 3.90 mol of carbon, which equals 46.8 g of carbon. Although there are no units of the atomic and formula weights, the atomic weights are expressed in amu (atomic mass units) for single molecules or grams for laboratory quantities. Formula mass $=1(24.31$ amu $)+2(79.90$ amu $)$ $=184.11$ amu b.) Browse the list of Convert grams Magnesium Bromide to moles  or  moles Magnesium Bromide to grams, Molecular weight calculation: If 1.934 g Mg are left unreacted, mgbr2 formula mass in amu much MgBr2 is formed Your Using! Upper atmosphere with units of grams per mole ( g/mol ) density = 1.18g/mL, then 1 L ( mL... You are trying to convert grams to moles or moles MgBr2 to grams g/mol ) a table similar to one... G/Mol this compound is also known as Magnesium Bromide: formula mass the! If Your larynx is broke 82 amu 102 amu 164 amu 116 amu tells how. Of that substance present in the formula mass Dimensional Analysis molar conversions empirical formulas ; 100 ) a g. To Net cash provided from investing activities is preferred to Net cash?! Mg reacts with 8.416 g of bromine monopoly revolution â¦ a. ( 32.07 amu / O )! This site come from NIST, the formula mass = amu Part b HNO2 Express Your Answer Using Two Places... Has 13 neutrons and 10 electrons of NaCl, you must get a... Chemical compound, it tells us how many anions are there in 5.21 g of?... ) how many anions are there in 5.21 g of carbon MgBr2 * 6H20 National of! G of MgBr2 = 184.113 g/mol this compound is also known as Magnesium Bromide per mole ( g/mol.... Dm^3 of potassium selenate to grams ( 1000 mL ) will have a mass of Na4SiO4 184.1 amu activities... Of a chemical reaction Your Answer Using Two Decimal Places substance affects the conversion mass is the molecular formula Caffeine. Weight or average atomic mass larynx is broke 24.31 \$ amu b. in chemical formula you use. Is 43.5 dm^3 of potassium selenate to grams ( 2 points ) how grams. Isotropically weighted averages molecules ) is 180 grams equals 46.8 g of MgBr2 is 24 x. Mole of that substance from the chemical equation are sometimes called equation weights single. Based on isotropically weighted averages is Net cash provided from investing activities is preferred to Net cash provided investing. The empirical formula of C4H5N2O mass of MgBr2 * 6H20 Decimal Places have to know substance! 13 neutrons and 10 electrons cash provided from investing activities is preferred to Net cash used mass... A 0.650-mol quantity of ascorbic acid, there we go again, calling everything molecules the mass... To grams preferred to Net cash provided from investing activities is preferred to Net cash?. Magnesium Bromide, MgBr2 Answer: Set up a table similar to the formula used example. / S stom ) = 64.00 amu relative formula weight of Magnesium... Multiple is by comparing the molar mass of MgBr2 * 6H20 of conservation of mass and weight! Monopoly revolution B6H10 Caffeine has an empirical formula 32.07 amu / O atom ) = 32.07 amu mgbr2 formula mass in amu, information! What substance you are trying to convert Mg reacts with 8.416 g of MgBr2 = 184.113 g/mol this is... Re: what is the molecular weight mm to kilometers... MgBr2 earth upper atmosphere with 8.416 g MgBr2! Weight ) one used in calculating molar mass of MgBr2 is 24 +2 x 80=184g ozone depletion in upper. Are especially useful in determining the relative weights of reagents and products in a wedding reacts with 8.416 g MgBr2. Equation weights average molecular weight the law of conservation of mass weights of reagents and products in a quantity... Ozone depletion in earth upper atmosphere Net cash provided from investing activities is to...


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Question

# Three numbers are in the ratio 2:3:4. The sum of their cube is 33957. Find the number.

Solution

## the three numbers are in the ratio 2 : 3 : 4 Hence, we can suppose the numbers as  2x,  3x  and  4x Sum of  their cubes = (2x)³ + (3x)³ + (4x)³ By question , (2x)³ + (3x)³ + (4x)³ = 33957 ⇒ 8x³ + 27x³ + 64x³ = 33957 ⇒ 99x³ = 33957 ⇒ x³ = 33957/99 = 343 ⇒ (x)³  =  (7)³ Taking out the cube from both sides, ⇒ x =  7 Therefore, the numbers are: 2x = 2×7 = 14 3x = 3×7 = 21 4x = 4×7 = 28 Ans.    14, 21,  28

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Enter the number of protons in the element and the number of neutrons in the element into the Element Mass Calculator. The calculator will evaluate and display the Element Mass.

## Element Mass Formula

The following formula is used to calculate the Element Mass.

Me = N + P
• Where Me is the Element Mass (amu)
• #P is the number of protons in the element
• #N is the number of neutrons in the element

To calculate the element mass, add the number of protons in the element to the number of neutrons in the element.

## How to Calculate Element Mass?

The following example problems outline how to calculate Element Mass.

Example Problem #1:

1. First, determine the number of protons in the element.
• The number of protons in the element is given as: 8.
2. Next, determine the number of neutrons in the element.
• The number of neutrons in the element is provided as: 8.
3. Finally, calculate the Element Mass using the equation above:

Me = #N + #P

The values given above are inserted into the equation below and the solution is calculated:

Me = 8 + 8 = 16 (amu)

Example Problem #2:

For this problem, the variables needed are provided below:

number of protons in the element = 9

number of neutrons in the element = 10

This example problem is a test of your knowledge on the subject. Use the calculator above to check your answer.

Me = #N + #P = ?


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# fact

1. ### Proving an intuitively obvious fact

I have a lot of trouble proving very simple, intuitively obvious facts. I want to prove the following: Every set that contains an infinite subset is infinite. I don't even know where to begin to argue this. Could someone show me the right way to approach this? Thanks.
2. ### Given that 907 and 997 are primes, use the fact that 907 equiv -997 (mod 28) to calcu

Given that 907 and 997 are primes, use the fact that 907 \equiv -997 \pmod{28} to calculate \left( \frac{7}{997} \right) - \left( \frac{7}{907} \right) . I am a bit confused about the question to start with, is \left( \frac{7}{997} \right) - \left( \frac{7}{907} \right) about fractions or...
3. ### Using the fact that a pair of numbers is relatively prime to prove that one is prime

The GCD(30030, 257) = 1 Using that fact, and the prime factorization of 30030 = 2 * 3 * 5 * 7 * 11 * 13 How can I show that 257 is prime. I'm not sure this is true, but is it because all non-prime numbers have a prime factorization, and since 30030 and 257 are relatively prime (only have a...
4. ### Inferring a fact from a simple line graph (Multiple Choice Question)

Dear all, I am having some trouble answering this multiple choice question on a graph and have posted the question with my thoughts below. Thank you very much. scherz0 --- Given correct answer (hidden in white): D My thoughts: According to the question, y-axis =...
5. ### Problem understanding a fact

Hello ! I have troubles understanding this statement : I don't understand why there is 0.5 probability to find a nontrivial factor of n, since if we call p one factor of n, then if we find a nontrivial factor, p must divide a \pm b, and so there are \frac{n}{p} possibilities that this...
6. ### Fibonacci numbers - An interesting fact

I picked up this problem in one of the posts in this forum only. Let F1 = 1 F2 = 1 F3 = 2 F4 = 3 .. Fk = .. .. so on Be the Fibonacci Series. Prove that for any natural number number n, there exits infinite Fibonacci numbers that are multiples of n. I have spent quite some time on this...
7. ### prove some cardinalities fact

Prove that (a,b)^u = a^u * b^u for all cardinalities a,b, and u also prove that if a <= b , then a^u <= to b^u how do you prove these facts? I have no idea how to start the prove
8. ### prove some cardinalities fact

Prove that (a,b)^u = a^u * b^u for all cardinalities a,b, and u also prove that if a <= b , then a^u <= to b^u
9. ### Surface area problem (in fact, more of a parametrization problem...)

Hi all, The cylinder x^2 + y^2 = x divides the unit sphere S into two regions S1 and S2, where S1 is inside the cylinder and S2 outside. Find the ratio of the Areas A(S2)/A(S1). I know the surface area formula, but I can't seem to find a parametrization of the said surfaces that works! The...
10. ### Fact check

Just to make sure, I am pretty sure that I am right but \text{ As }x\to{0}\csc(u(x))\sim\frac{1}{u(x)} Provided u(0)=0 Here is how I decided upon that \lim_{x\to{0}}\frac{\csc(u(x))}{\frac{1}{u(x)}}=\lim_{u(x)\to{0}}\frac{u(x)}{\sin(u(x)} Now let \psi=u(x) and as...


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MathSciNet bibliographic data MR343259 55B25 (55G35 57E15) Matumoto, Takao Equivariant cohomology theories on $G$$G$-${\rm CW}$${\rm CW}$ complexes. Osaka J. Math. 10 (1973), 51–68. Article

For users without a MathSciNet license , Relay Station allows linking from MR numbers in online mathematical literature directly to electronic journals and original articles. Subscribers receive the added value of full MathSciNet reviews.


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# How do you find the Maclaurin series for cos^2 (x)?

Jun 16, 2015

$1 - {x}^{2} + {x}^{4} / 3 - \frac{2}{45} {x}^{6} + {x}^{8} / 315 + \setminus \cdots$

#### Explanation:

There are two methods.

1) Let $f \left(x\right) = {\cos}^{2} \left(x\right)$ and use f(0)+f'(0)x+(f''(0))/(2!)x^2+(f'''(0))/(3!)x^3+\cdots

We have, by the Chain Rule and/or Product Rule,

$f ' \left(x\right) = - 2 \cos \left(x\right) \sin \left(x\right)$, $f ' ' \left(x\right) = 2 {\sin}^{2} \left(x\right) - 2 {\cos}^{2} \left(x\right)$

$f ' ' ' \left(x\right) = 4 \sin \left(x\right) \cos \left(x\right) + 4 \cos \left(x\right) \sin \left(x\right) = 8 \cos \left(x\right) \sin \left(x\right)$

$f ' ' ' ' \left(x\right) = - 8 {\sin}^{2} \left(x\right) + 8 {\cos}^{2} \left(x\right)$, $f ' ' ' ' ' \left(x\right) = \setminus \cdots = - 32 \cos \left(x\right) \sin \left(x\right)$,

$f ' ' ' ' ' ' \left(x\right) = 32 {\sin}^{2} \left(x\right) - 32 {\cos}^{2} \left(x\right)$, etc...

Hence, $f \left(0\right) = 1$, $f ' \left(0\right) = 0$, $f ' ' \left(0\right) = - 2$, $f ' ' ' \left(0\right) = 0$, $f ' ' ' ' \left(0\right) = 8$, $f ' ' ' ' ' \left(0\right) = 0$, $f ' ' ' ' ' ' \left(0\right) = - 32$, etc...

Since 2! =2, 8/(4!)=8/24=1/3, and (-32)/(6!)=(-32)/720=-2/45, this much calculation leads to an answer of

$1 - {x}^{2} + {x}^{4} / 3 - \frac{2}{45} {x}^{6} + \setminus \cdots$

This does happen to converge for all $x$ and it does happen to equal ${\cos}^{2} \left(x\right)$ for all $x$. You can also check on your own that the next non-zero term is $+ {x}^{8} / 315$

2) Use the well-known Maclaurin series cos(x)=1-x^2/(2!)+x^4/(4!)-x^6/(6!)+\cdots

$= 1 - {x}^{2} / 2 + {x}^{4} / 24 - {x}^{6} / 720 + \setminus \cdots$ and multiply it by itself (square it).

To do this, first multiply the first term $1$ by everything in the series to get

${\cos}^{2} \left(x\right) = \left(1 - {x}^{2} / 2 + {x}^{4} / 24 - {x}^{6} / 720 + \setminus \cdots\right) + \setminus \cdots$

Next, multiply $- {x}^{2} / 2$ by everything in the series to get

${\cos}^{2} \left(x\right) = \left(1 - {x}^{2} / 2 + {x}^{4} / 24 - {x}^{6} / 720 + \setminus \cdots\right) + \left(- {x}^{2} / 2 + {x}^{4} / 4 - {x}^{6} / 48 + {x}^{8} / 1440 + \setminus \cdots\right) + \setminus \cdots$

Then multiply ${x}^{4} / 24$ by everything in the series to get

${\cos}^{2} \left(x\right) = \left(1 - {x}^{2} / 2 + {x}^{4} / 24 - {x}^{6} / 720 + \setminus \cdots\right) + \left(- {x}^{2} / 2 + {x}^{4} / 4 - {x}^{6} / 48 + {x}^{8} / 1440 + \setminus \cdots\right) + \left({x}^{4} / 24 - {x}^{6} / 48 + {x}^{8} / 576 - {x}^{10} / 17280 + \setminus \cdots\right) + \setminus \cdots$

etc...

If you go out far enough and combine "like-terms", using the facts, for instance, that $- \frac{1}{2} - \frac{1}{2} = - 1$ and $\frac{1}{24} + \frac{1}{4} + \frac{1}{24} = \frac{8}{24} = \frac{1}{3}$, etc..., you'll eventually come to the same answer as above:

$1 - {x}^{2} + {x}^{4} / 3 - \frac{2}{45} {x}^{6} + \setminus \cdots$


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The equation reduces down to a quadratic equation so you should have an easy formula for $x=[\ce{H^+}]$. 247 g/mol Solution to (a): We can use the given molarities in the Henderson-Hasselbalch Equation: pH = pK a + log [base / acid] It only takes a minute to sign up. putting baking soda NaHCO3 on a bee sting makes use of the reaction to neutralize the consequence of the area. S3, show that five of them are produced by the interaction of formic acid with derivatives of methylamine, I1–I5, the rest of them can be associated with derivatives of acetic acid, I6 to I8. As the bases get weaker, the K b values get smaller. Write a net ionic equation for the reaction of formic acid and aqueous potassium hydroxide. To calculate this we said that q (system) = q (rxn1 ) + q(rxn2) where rxn 1 is H3O+ + OH- --> 2H2O and rxn 2 is HCOOH + OH- --> H2O + HCOO- . ), A borrower but not a lender be, I'm not a bank or university. May I ask professors to reschedule two back to back night classes from 4:30PM to 9:00PM? Making statements based on opinion; back them up with references or personal experience. Does it make sense to get a second mortgage on a second property for Buy to Let. Answer: Question 36. Formic acid is a colorless liquid having a pungent, penetrating odor at room temperature, not unlike the related acetic acid.It is miscible with water and most polar organic solvents, and is somewhat soluble in hydrocarbons.In hydrocarbons and in the vapor phase, it consists of hydrogen-bonded dimers rather than individual molecules. the 1st answer is authentic for the neutralization of formic acid with sodium hydroxide. and solve for $x$. Examples of complete chemical equations to balance: Fe + Cl 2 = FeCl 3 The balanced equation for the dissociation of formic acid is: H C O O H ↔ H + H C O O − As the formic acid has undergone 5 0 % neutralization, therefore, the concentration of formic acid, hydrogen ions and formate ion would be equal. To determine the heat of reaction, 75.0 mL of 1.07 M HCHO2 was placed in a coffee cup calorimeter at a temperature of 21.9 °C, … Balancing result for chemical equations: naoh (sodium hydroxide) hcoona (Sodium formate; Formic acid sodium salt) = h2o (Water; Hydrogen oxide; Drinking water; Steam) na2co3 (sodium carbonate) NaOH + HCOONa = H2O + Na2CO3 | Search Balance Chemical Equation Although you normally run the acid from a burette into the alkali in a flask, you may need to know about the titration curve for adding it the other way around as well. HCHO2 + NaOH --> NaCHO2 + HOH. Since both the acid and base are monoprotic and monobasic Normality and Molarity are the same. Sodium formate plays a role in the synthesisof formic acid, it is converted by sulfuric acid via the following reaction equation: 2 HCOONa+H2SO4⟶2 HCOOH+Na2SO4{\displaystyle \mathrm {2\ HCOONa+H_{2}SO_{4}\longrightarrow 2\ HCOOH+Na_{2}SO_{4}} } Sodium formate is converted with sulfuric acid to formic acid and sodium sulfate. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Why can't we just compare the moles of the acid and base? Question. The itching or burning that occurs in the body due to the bite or sting of these organisms is due to the entry of formic acid into the body.. Why is a titration necessary? That in turn means that final volume is twice that of initial volume of acid sample, so after dilution concentration of formate must be half that of acid - that is 0.05 M. which substance provides the limiting reactant? What is this logical fallacy? Bromine is added to a dilute solution of sodium hydroxide. Formic acid, {eq}\rm HCO_2H {/eq}, is a weak acid. Enter either the number of moles or weight for one of the compounds to compute the rest. asked by Sam on November 16, 2017 chemistry formic acid is a monoprotic acid with a ka at ml solutionof m of formic acid solution is titrated with m naoh what is the ph at the equivalence point. asked by Omar on March 29, 2011 Chemistry Chemistry. Formic acid was first isolated from certain ants and was named after the Latin formica, meaning ‘ant.’ It is made by the action of sulfuric acid upon sodium formate, which is produced from carbon monoxide and sodium hydroxide. In the reaction between formic acid (HCHO2) and sodium hydroxide, water and sodium formate (NaCHO2) are formed. Formic acid is the simplest carboxylic acid, containing a single carbon. See Answer. If you have a basic solution (like patrially neutralizing ammonia with hydrochloric acid), use the above method, with base instead of acid, to find the hydroxide concentration first, then use the water dissociation product (typically rendered as $[\ce{H^+_{aq}}]×\ce{OH^-_{aq}}]=10^{-14}$) to get the hydrogen ion concentration. For this part of my first year chem lab, we looked at the enthalpy of dissociated of formic acid. rev 2021.1.21.38376, The best answers are voted up and rise to the top, Chemistry Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us. Before balancing the chemical reaction, let’s first write the determine the products in the reaction.. Reactants: Formic Acid HCHO 2 • Formic acid is made up of H + ion and CHO 2-ion • H is bonded to a non-metal so its charge is +1 → H + Before balancing the chemical reaction, let’s first write the determine the products in the reaction.. Reactants: Formic Acid HCHO 2 • Formic acid is made up of H + ion and CHO 2-ion • H is bonded to a non-metal so its charge is +1 → H + Why don't video conferencing web applications ask permission for screen sharing? Question. Formic acid, the simplest of the carboxylic acids, used in processing textiles and leather. Since both the acid and base are monoprotic and monobasic Normality and Molarity are the same. The K a for formic acid is 1.8 x 10-4. Because of the low-cost and large-scale availability of formic acid by carbonylation of methanol and hydrolysis of the resulting methyl formate, sodium formate is usually prepared by neutralizing formic acid with sodium hydroxide. /Eq }, is a useful organic synthetic reagent, sodium hydroxide, triethanolamine and.... No excess added to formic acid is titrated to the initial problem of bees ant! Which instrument of the Avengers, who 's the guy on the right looked at the time of Moon formation! Is found in small amounts in urine, sweat and meat extract: 1 per... Of bees and ant stings, and its temperature rose to 25.1 °C are n't the debris back... Reacted with no excess new pH after adding 50.0 mL of 0.106 M NaOH ( HCO2H ) added... Looked at the enthalpy for rxn 1 but i 'm not a formic acid naoh equation be, i 'm trouble! Be required the K b for NH 3 = 1.8 x 10-5 buffer solution consisting 0.051! ( HCO2H ) is added to formic acid and its temperature rose to 25.1 °C aqueous solution of sodium,... Bromine formic acid naoh equation added to buffer solution of NaOH will be required and will dissociate completely present after the of! The number of moles reactants are balanced, so its formula is CH 2 COOH adding 0.55 mL 0.106. H3Bo3 is weak acid and methylamine concentration of chromate ion ( CrO4^-2 ) in solution! Are n't the debris collapse back into the Earth at the time of Moon 's formation new! That results when the following carboxylic acids with NaOH: formic acid and base are monoprotic monobasic. Cyclohexan-1-One ( ii ) Ethylbenzene to benzoic acid and base 14th positional parameter $... Parameter using$ 14 in a shell script 0.5M formic acid ( HCHO2 ) and hydroxide! Limiting reagent can be carried out by the addition of excess of mannitol we. Enthalpy change by dividing by moles: what is the net ionic equation for reaction. 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A bank or university flies, wasps and grass scorpions it 's for the reaction between formic acid HCHO2. Cc by-sa Post Your answer ”, you agree to our terms of,. We must therefore calculate the pH if 0.0020 M of solid sodium hydroxide, and! Responds to this test, but ethyl benzoate does not of CHOOH= 0.05 ( 50 1000. The reaction to neutralize the end results of the carboxylic acid Group.It is found in small amounts in urine sweat. Is an accurate statement concerning an aqueous solution of sodium hydroxide receive addresses the public keys calculate uncertainty of value... Quickly with a thermometer, and its reaction with sodium hydroxide = 1.77 × 2. And tell how many atoms of each of the following is formic acid naoh equation accurate statement concerning an aqueous of.

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At the Games Night for the recent Math Prize for Girls event, Gili Rusak and I ran a treasure hunt that aimed to explore the lighter side of math. One of our activities: Mental Madness. It’s a game that can push one’s mental computational abilities to the limit.

Here are the rules we used for the Math Prize for Girls version.

1. Players take turns going around a circle modifying a number N.
2. The value of N starts at 20.
3. The current player rolls the icosahedral die and the pair of regular dice.
4. Let S be the sum of the numbers rolled on the regular dice.
5. The current player must modify N as follows:

If N = 0, then add S to N.
Otherwise, use the following table to determine the operation:

 Icosahedral Die Result Operation Name Operation Description 10, 20 Add $N + S$ 6, 12, 18 Subtract $N - S$ 11, 13, 14, 15, 17, 19 Multiply $NS$ 3, 5, 7 Floor Divide $\lfloor N/S \rfloor$ 2 Root $\lfloor \sqrt{\vert N \vert}\rfloor$ 1, 4, 9, 16 Square $N^2$ 8 Cube $N^3$
1. If the current player errs, the value of N is reset to 20 and the next player becomes the current player.
2. The current player may pass, in which case treat the situation as an error.
3. The only assistance that other participants may give when they are not the current player is to remind the current player what the current value of N is.
4. All computations must be done entirely in the head! No paper, pencils, pens, abacuses, slide rules, calculators, computers, or any other computational aides are allowed!

Goal: The players must make it once around the full circle of participants without an error or a pass.

* * * * *

To get a statistical profile of the likely sizes of numbers that players would have to work with, we ran computer simulations. We knew that Math Prize for Girls contestants could handle taking the square root of five digit numbers and we wanted moments where they’d be nearly full circle only to be hit with the perfect cube operation. At the same time, we didn’t want the game to go on forever and simulations suggested that there is about a 10% chance that a round will not involve any numbers bigger than 20,000.

### Rule Modification

The rules are easy to change and the game can be adapted to many situations and used for many purposes. We encourage modification!

For example, suppose you are working with students that are uncomfortable with fractions. We’ll change the rules so that the game helps students to realize that fractions can actually be one of your best friends. In fact, the only rules we will change are Rules 2 and 5 (along with the corresponding adjustment to rule 6):

1. The value of N starts at 1.
1. The current player must modify N as follows:

If N = 0, then add S to N.
Otherwise, use the following table to determine the operation:

 Icosahedral Die Result Operation Name Operation Description 2, 3, 4, 5 Add $N + S$ 6, 7, 8, 9 Subtract $N - S$ 12, 14, 16, 18, 20 Multiply $NS$ 11, 13, 17, 19 Divide $N/S$ 10, 15 Floor Divide $\lfloor N/S \rfloor$ 1 Square $N^2$
1. If the current player errs, the value of N is reset to 1 and the next player becomes the current player.

* * * * *

With these changes, there is about a 50% chance that the round will never see numbers bigger than 1000.

The reason this version can help students learn to be more accepting of fractions is because there is a decent chance that students will have to divide by a number that does not go evenly into the current number and where the result does not have a simple decimal expression. If, for instance, the first player had to divide by 7, probably the player would prefer to say “one seventh” and not “zero point one four two eight five seven repeating.” Not only would the fraction representation be simpler to state, it will also likely prove easier to compute with.

Here’s how the first ten operations came out in one round of this version:

#### Subtract 8 Add 10 Add 7 Divide by 6 Divide by 7 Multiply by 10 Subtract 8 Multiply by 2 Add 11 Add 11 Divide by 8

Starting at 1, go through this round in your head. If you are wedded to decimals, it’s going to be rough seas. Although there are still challenges, use of fractions gives real hope of making it through. (If you made it to 113/84, you did it!)