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2 years ago

Measurements show that unknown x compound has the following composition:

Chemistry

1 answer:

Fudgin [204]2 years ago

7 0

The empirical formula of the compound with the percent composition C 18.1%, H 2.27%, Cl 79.8% is C₂H₃Cl₃.

<h3>What is an empirical formula?</h3>

It is the minimum ratio between the elements that form a compound.

Step 1: Divide each percentage by the molar mass of the element.

C: 18.1/12.01 = 1.51

H: 2.27/1.01 = 2.25

Cl: 79.8/35.45 = 2.25

Step 2: Divide all the numbers by the smallest one.

C; 1.51/1.51 = 1

H: 2.25/1.51 ≈ 1.5

Cl: 2.25/1.51 ≈ 1.5

Step 3: Multiply all the numbers by 2 so all of them are whole.

C: 1 × 2 = 2

H: 1.5 × 2 = 3

Cl: 1.5 × 2 = 3

The empirical formula is C₂H₃Cl₃.

The empirical formula of the compound with the percent composition C 18.1%, H 2.27%, Cl 79.8% is C₂H₃Cl₃.

Learn more about empirical formula here: brainly.com/question/1603500

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in a second experiment, the total pressure is observed to be 58 atm. what is the partial pressure of the water vapor produced?

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Value for ideal gas constant, R: 0.0821 L.atm/mol.K

In a second experiment, the total pressure is observed to be 58 atm. what is the partial pressure of the water vapor produced?

Answer : The partial pressure of the water vapor is, 20.01 atm

Explanation :

First we have to calculate the moles of $C_3H_5N_3O_9$

$\text{Moles of }C_3H_5N_3O_9=\frac{\text{Given mass }C_3H_5N_3O_9}{\text{Molar mass }C_3H_5N_3O_9}=\frac{227g}{227g/mol}=1mol$

Now we have to calculate the moles of $CO_2,O_2,N_2\text{ and }H_2O$

The balanced chemical reaction is:

$4C_3H_5N_3O_9(l)\rightarrow 12CO_2(g)+O_2(g)+6N_2(g)+10H_2O(g)$

From the balanced chemical reaction we conclude that,

As, 4 moles of $C_3H_5N_3O_9$ react to give 12 moles of $CO_2$

So, 1 moles of $C_3H_5N_3O_9$ react to give $\frac{12}{4}=3$ moles of $CO_2$

and,

As, 4 moles of $C_3H_5N_3O_9$ react to give 1 moles of $O_2$

So, 1 moles of $C_3H_5N_3O_9$ react to give $\frac{1}{4}=0.25$ moles of $O_2$

and,

As, 4 moles of $C_3H_5N_3O_9$ react to give 6 moles of $N_2$

So, 1 moles of $C_3H_5N_3O_9$ react to give $\frac{6}{4}=1.5$ moles of $N_2$

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As, 4 moles of $C_3H_5N_3O_9$ react to give 10 moles of $H_2O$

So, 1 moles of $C_3H_5N_3O_9$ react to give $\frac{10}{4}=2.5$ moles of $H_2O$

Now we have to calculate the mole fraction of water.

$\text{Mole fraction of }H_2O=\frac{\text{Moles of }H_2O}{\text{Moles of }H_2O+\text{Moles of }CO_2+\text{Moles of }O_2+\text{Moles of }N_2}$

$\text{Mole fraction of }H_2O=\frac{2.5}{2.5+3+0.25+1.5}=0.345$

Now we have to calculate the partial pressure of the water vapor.

According to the Raoult's law,

$p_{H_2O}=X_{H_2O}\times p_T$

where,

$p_{H_2O}$ = partial pressure of water vapor gas = ?

$p_T$ = total pressure of gas = 58 atm

$X_{H_2O}$ = mole fraction of water vapor gas = 0.345

Now put all the given values in the above formula, we get:

$p_{H_2O}=X_{H_2O}\times p_T$

$p_{H_2O}=0.345\times 58atm=20.01atm$

Therefore, the partial pressure of the water vapor is, 20.01 atm

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<u>Answer:</u> The value of $K_c$ for the given reaction is 1.435

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

Given mass of $N_2O_4$ = 9.2 g

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Volume of solution = 0.50 L

Putting values in above equation, we get:

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For the given chemical equation:

$N_2O_4(g)\rightleftharpoons 2NO_2(g)$

<u>Initial:</u> 0.20

<u>At eqllm:</u> 0.20-x 2x

We are given:

Equilibrium concentration of $N_2O_4$ = 0.057

Evaluating the value of 'x'

$\Rightarrow (0.20-x)=0.057\\\\\Rightarrow x=0.143$

The expression of $K_c$ for above equation follows:

$K_c=\frac{[NO_2]^2}{[N_2O_4]}$

$[NO_2]_{eq}=2x=(2\times 0.143)=0.286M$

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