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Rasek [7]
1 year ago
12

A chemist combined 0.440 L of an unknown calcium solution with an excess of ammonium chromate. This resulted in the precipitatio

n of calcium chromate. The mass of the precipitate was 346.7 mg. What was the molar concentration of Ca2 in the original sample
Chemistry
1 answer:
Rina8888 [55]1 year ago
6 0

The concentration of the original calcium ions is 0.005 M

<h3>What is concentration?</h3>

The term concentration has to do with the amount of substance in solution. We know that the concentration can be measured in a lot of units such as mole/litre, grams per litre, percentage and so on.

As such we have the equation;

Ca^2+(aq) + (NH4)2CrO4(aq) --------> CaCrO4(s) + 2NH4^+(aq)

Number of moles of the precipitate = 346.7 * 10^-3 g/156 g/mol

= 0.0022 moles

Now;

1 mole of Ca^2+ produces 1 mole of CaCrO4 hence 0.0022 moles of CaCrO4 was produced by  0.0022 moles of CaCrO4.

Given that the volume of the solution is  0.440 L, the concentration of the solution is;   0.0022 moles/0.440 L

= 0.005 M

Learn more about molarity:brainly.com/question/8732513

#SPJ1

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The ideal gas heat capacity of nitrogen varies with temperature. It is given by:
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Answer:

A)  1059 J/mol

B)  17,920 J/mol

Explanation:

Given that:

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We can determine C_v from above if we make C_v the subject of the formula as:

C_v=C_p-R

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C_V = 21.106-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4

A).

The formula for calculating change in internal energy is given as:

dU=C_vdT

If we integrate above data into the equation; it implies that:

U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4\,) du

U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T/1+(5.82*10^{-7})T2/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)

U2-U1= 1059J/mol

Hence, the internal energy that must be added to nitrogen in order to increase its temperature from 450 to 500 K = 1059 J/mol.

B).

If we repeat part A for an initial temperature of 273 K and final temperature of 1073 K.

then T = 273 K & T2 = 1073 K

∴

U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T/1+(5.82*10^{-7})T2/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)

U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})273/1+(5.82*10^{-7})1073/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)

U2-U1= 17,920 J/mol

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Can you be more specific

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