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dedylja [7]
3 years ago
7

Calculate the mass percent of a solution that is prepared by adding 68.1 g of NaOH to 341 g of H20.

Chemistry
1 answer:
Sati [7]3 years ago
5 0

mass percent of the solution = 16.64%

Explanation:

We calculate the mass percent of a solution using the following formula:

mass percent = (mass of the solute / mass of the solution) × 100

mass of the solute = 68.1 g of NaOH

mass of the solution = mass of the solute + mass of water (H₂O)

mass of the solution = 68.1 + 341 = 409.1 g

mass percent = (68.1 / 409.1) × 100

mass percent = 16.64%

Learn more about:

mass percent

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The first step in industrial nitric acid production is the catalyzed oxidation of ammonia. Without a catalyst, a different react
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<u>Answer:</u> The value of K_{eq} is 4.84\times 10^{-5}

<u>Explanation:</u>

We are given:

Initial moles of ammonia = 0.0280 moles

Initial moles of oxygen gas = 0.0120 moles

Volume of the container = 1.00 L

Concentration of a substance is calculated by:

\text{Concentration}=\frac{\text{Number of moles}}{\text{Volume}}

So, concentration of ammonia = \frac{0.0280}{1.00}=0.0280M

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The given chemical equation follows:

                  4NH_3(g)+3O_2(g)\rightleftharpoons 2N_2(g)+6H_2O(g)

<u>Initial:</u>        0.0280        0.0120

<u>At eqllm:</u>    0.0280-4x   0.0120-3x   2x       6x

We are given:

Equilibrium concentration of nitrogen gas = 3.00\times 10^{-3}M=0.003

Evaluating the value of 'x', we get:

\Rightarrow 2x=0.003\\\\\Rightarrow x=0.0015M

Now, equilibrium concentration of ammonia = 0.0280-4x=[0.0280-(4\times 0.0015)]=0.022M

Equilibrium concentration of oxygen gas = 0.0120-3x=[0.0120-(3\times 0.0015)]=0.0075M

Equilibrium concentration of water = 6x=(6\times 0.0015)]=0.009M

The expression of K_{eq} for the above reaction follows:

K_{eq}=\frac{[H_2O]^6\times [N_2]^2}{[NH_3]^4\times [O_2]^3}

Putting values in above expression, we get:

K_{eq}=\frac{(0.009)^6\times (0.003)^2}{(0.022)^4\times (0.0075)^3}\\\\K_{eq}=4.84\times 10^{-5}

Hence, the value of K_{eq} is 4.84\times 10^{-5}

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