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Vladimir79 [104]

4 years ago

7

A flexible container at an initial volume of 6.13 L contains 6.51 mol of gas. More gas is then added to the container until it r

eaches a final volume of 18.3 L. Assuming the pressure and temperature of the gas remain constant, calculate the number of moles of gas added to the container.

1 answer:

aivan3 [116]4 years ago

6 0

Answer:

the final mole of the flexible container = 12.92 moles

Explanation:

Given that :

initial volume of a flexible container = 6.13 L

initial mole of a flexible container = 6.51 mol

final volume of a flexible container = 18.3 L

final mole of a flexible container = ???

Assuming the pressure and temperature of the gas remain constant, calculate the number of moles of gas added to the container.

Therefore,

$n= \dfrac{V_2*n_1}{V_1}$

$n= \dfrac{18.3*6.51}{6.13}$

n = 19.43

$n=n_1+n_2$

19.43 = 6.51 + n₂

n₂ = 19.43 - 6.51

n₂ = 12.92 moles

Thus; the final mole of the flexible container = 12.92 moles

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4 years ago

Read 2 more answers

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3 years ago

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Answer :

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Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

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$2C(coal)+2H_2O(g)\rightarrow CH_4(g)+CO_2(g)$ $\Delta H_{rxn}=?$

The intermediate balanced chemical reaction will be,

(1) $C(coal)+H_2O(g)\rightarrow CO(g)+H_2(g)$ $\Delta H_1=29.7kJ$

(2) $CO(g)+H_2O(g)\rightarrow CO_2(g)+H_2(g)$ $\Delta H_2=-41kJ$

(3) $CO(g)+3H_2(g)\rightarrow CH_4(g)+H_2O(g)$ $\Delta H_3=-206kJ$

We are multiplying equation 1 by 2 and then adding all the equations, we get :

(b) The expression for enthalpy of reaction will be,

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$\Delta H_{rxn}=-187.6kJ/mol$

Therefore, the enthalpy of reaction is, -187.6 kJ/mol

(c) Now we have to calculate the total heat.

$\Delta H=\frac{q}{n}$

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$q=\Delta H\times n$

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$q=(-187.6kJ/mol)\times (83.33mol)=-2.251kJ$

Thus, the total heat will be, -2251 kJ

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