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dsp73
3 years ago
5

A chemist measures the amount of hydrogen gas produced during an experiment. She finds that 264. g of hydrogen gas is produced.

Calculate the number of moles of hydrogen gas produced. Round your answer to 3 significant digits. W mol x 6 ?
Chemistry
1 answer:
Lesechka [4]3 years ago
4 0

Answer:

The answer is 130.953 g of hydrogen gas.

Explanation:

Hydrogen gas is formed by two atoms of hydrogen (H), so its molecular formula is H₂. We can calculate is molecular weight as the product of the molar mass of H (1.008 g/mol):

Molecular weight H₂= molar mass of H x 2= 1.008 g/mol x 2= 2.01568 g

Finally, we obtain the number of mol of H₂ there is in the produced gas mass (264 g) by using the molecular weight as follows:

mass= 264 g x 1 mol H₂/2.01568 g= 130.9731703 g

The final mass rounded to 3 significant digits is 130.973 g

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Answer:

M = 3.0 mol/L.

Explanation:

  • We can calculate the molarity of a solution using the relation:

<em>M = (mass x 1000) / (molar mass x V)</em>

  • M is the molarity "number of moles of solute per 1.0 L of the solution.
  • mass is the mass of the solute (g) (m = 87.75 g of NaCl).
  • molar mass of NaCl = 58.44 g/mol.
  • V is the volume of the solution (ml) (V = 500.0 ml).

∴ M = (mass x 1000) / (molar mass x V) = (87.75 g x 1000) / (58.44 g/mol x 500.0 ml) = 3.0 mol/L.

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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

Concentration of oxygen gas = \frac{0.0120}{1.00}=0.0120M

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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