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love history [14]
3 years ago
5

Write a balanced chemical equation for the standard formation reaction of solid chromium(iii) nitrate crno33 .

Chemistry
1 answer:
Drupady [299]3 years ago
7 0

Explanation:

Standard formation of an substance is defined as formation of one mole of substance from its constituting elements.

Chromium(III) nitrate is composed of chromium, nitrogen and three oxygen atoms.

2Cr(s)+3N_2(g)+9O_2\rightarrow 2Cr(NO_3)_3(s)

A balanced chemical equation for the standard formation reaction of solid chromium(III) nitrate

Cr(s)+\frac{3}{2}N_2(g)+\frac{9}{2}O_2\rightarrow Cr(NO_3)_3(s)

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Inevitable manufacturing imperfections in optical fiber leads to birefringence, which is one cause of pulse broadening in fiber-optic communications. Such imperfections can be geometrical (lack of circular symmetry), due to stress applied to the optical fiber and/or due to bending of the fiber.

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What will be the new volume if 125 mL of He gas at 100 degree Celsius and .979 atm is cooled to 26 degree Celsius and the pressu
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r-ruslan [8.4K]

The molar concentration will be greater than 0.01 M KIO_{3}.


Since more of the compound was measured out than what was calculated, you can think of the solution as being 'stronger' than what it was calculated to be. Since a 'stronger' concentration results in a number that is higher, the molarity of this solution is going to be greater than 0.01 M.

7 0
3 years ago
If there are 3.10 moles of O, how many moles of each of the compounds are present?
Burka [1]

Explanation:

The question pretty  much requires us to find the amount of moles of each compounds based on the number of moles of O given.

H2SO4

1 mol of H2SO4 contains 4 mol of O

x mol of H2SO4 would contain 3.10 mol of O

x = 3.10 * 1 / 4 = 0.775 mol of H2SO4

C2H4O2

1 mol of C2H4O2 contains 2 mol of O

x mol of C2H4O2 would contain 3.10 mol of O

x = 3.10 * 1 / 2 = 1.55 mol of C2H4O2

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1 mol of NaOH contains 1 mol of O

x mol of NaOH would contain 3.10 mol of O

x = 3.10 * 1 / 1 = 3.10 mol of NaOH

3 0
3 years ago
A 10.5 mL sample of vinegar, containing acetic acid, was titrated using 0.460 M NaOH solution. The titration required 19.13 mL o
laila [671]

Explanation:

Step 1:

A good first step for a problem like this is to write down the chemical formula and balance it.

It appears here that we have 10.5 mL of vinegar, which IS acetic acid, and 19.13 mL of 0.460 M NaOH. That will give us the following balanced chemical equation:

CH3COOH + NaOH ------> NaCH3COO + H2O

All of the constituents come out to a value of 1, conveniently.

Step 2:

Since all of our stoichiometric coefficients are one, we can use a shortcut to answer this equation. I don't know if it has a name, but I just call it the titration formula. It goes something like this:

M1 * V1 = M2 * V2

M stands for Molarity and V stands for volume. 1 and 2 being the before the reaction and after the reaction.

So, our M1 for this is going to be what the question says was used for this titration. That's 0.460M NaOH.

Our V1 is going to be the initial volume of the sample, which was 10.5 mL

Our V2 is going to be 19.13, which is the volume when we're finished.

It's clear that we don't know M2, so let's find it.

Keep in mind that it's easier to convert to liters pretty much always, so I've done that by dividing the mL values each by 1000.

Using some algebra, we can see that we now have:

0.460 M * 0.0105 L = x M * 0.01913 L

Which goes to:

\frac{0.00483mol}{0.01913L} = 0.252 M

<h3>So our M2, the molar concentration of acetic acid in this vinegar, is equal to 0.252 M. </h3>
3 0
3 years ago
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