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Crazy boy [7]
2 years ago
9

Calorimetric studies show that the reaction is exothermic. 2NO2(g) N2O4(g) + 14.1 kcal. Based on this information, which one--if

any--of the following additional changes would increase the molar concentration at equilibrium of N2O4(g)?
decrease the pressure

increase the temperature

decrease the concentration of NO2(g)

stir the reaction mixture

none of the above
Chemistry
1 answer:
FinnZ [79.3K]2 years ago
5 0
In order for you to get the answer, please have in mind the following situation: To increase the molar concentration of N2O4(g), 2NO2(g) should also increase for equilibrium to occur. Now, this equation is exothermic. By <span>Le Chatelier's principle, equilibrium constant and reaction constants also come into play in terms of increasing or decreasing the temperature. After that I know you can find the answer. </span>
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you find that 7.36g of a compound has decomposed to give 6.93g of oxygen. the only other element in the compound is hydrogen. if
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In order to prepare very dilute solutions, a lab technician chooses to perform a series of dilutions instead of measuring a very
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<u>Answer:</u> The final concentration of potassium nitrate is 5.70\times 10^{-6}M

<u>Explanation:</u>

To calculate the molecular mass of solute, we use the equation used to calculate the molarity of solution:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

We are given:

Mass of potassium nitrate (solute) = 0.360 g

Molar mass of potassium nitrate = 101.1 g/mol

Volume of solution = 500.0 mL

Putting values in above equation, we get:

\text{Molarity of }KNO_3=\frac{0.360\times 1000}{101.1\times 500.0}\\\\\text{Molarity of }KNO_3=7.12\times 10^{-3}M

To calculate the molarity of the diluted solution, we use the equation:

M_1V_1=M_2V_2          .......(1)

  • <u>Calculating for first dilution:</u>

M_1\text{ and }V_1 are the molarity and volume of the concentrated KNO_3 solution

M_2\text{ and }V_2 are the molarity and volume of diluted KNO_3 solution

We are given:

M_1=7.12\times 10^{-3}M\\V_1=10mL\\M_2=?M\\V_2=500.0mL

Putting values in equation 1, we get:

7.12\times 10^{-3}\times 10=M_2\times 500\\\\M_2=\frac{7.12\times 10^{-3}\times 10}{500}=1.424\times 10^{-4}M

  • <u>Calculating for second dilution:</u>

M_2\text{ and }V_2 are the molarity and volume of the concentrated KNO_3 solution

M_3\text{ and }V_3 are the molarity and volume of diluted KNO_3 solution

We are given:

M_2=1.424\times 10^{-4}M\\V_2=10mL\\M_3=?M\\V_3=250.0mL

Putting values in equation 1, we get:

1.424\times 10^{-4}\times 10=M_3\times 250\\\\M_3=\frac{1.424\times 10^{-4}\times 10}{250}=5.70\times 10^{-6}M

Hence, the final concentration of potassium nitrate is 5.70\times 10^{-6}M

8 0
2 years ago
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