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3 years ago

A mixture of He

Chemistry

1 answer:

aksik [14]3 years ago

3 0

Answer : The partial pressure of nitrogen gas in the mixture is, 9.88 atm

Explanation :

According to the Dalton's Law, the total pressure of the gas is equal to the sum of the partial pressure of individual gases.

Formula used :

$p_T=p_{He}+p_{Ar}+p_{N_2}$

where,

$p_T$ = total pressure of gas = 13.6 atm

$p_{He}$ = partial pressure of helium gas = 1831 torr = 2.41 atm

$p_{Ar}$ = partial pressure of argon gas = 997 torr = 1.31 atm

Conversion used: (1 atm = 760 torr)

$p_{N_2}$ = partial pressure of nitrogen gas = ?

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

$13.6=2.41+1.31+p_{N_2}$

$p_{N_2}=9.88atm$

Thus, the partial pressure of nitrogen gas in the mixture is, 9.88 atm

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Sliva [168]

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$\text{Molecules of } \text{P}_{2}\text{O}_{5} = \text{0.002 043 mol } \text{P}_{2}\text{O}_{5} \times \frac{6.022 \times 10^{23}\text{ molecules }\text{P}_{2}\text{O}_{5}}{\text{1 mol } \text{P}_{2}\text{O}_{5}}\\$

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3 years ago

Calculate ∆G ◦ r for the decomposition of mercury(II) oxide 2 HgO(s) → 2 Hg(ℓ) + O2(g) ∆H◦ f −90.83 − − (kJ · mol−1 ) ∆S ◦ m 70.

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Explanation:

ΔG of a reaction is:

ΔGr = ΔHr - TΔSr (1)

For the reaction:

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ΔHr: 2ΔHf Hg(l) + ΔHf O₂(g) - 2ΔHf HgO(s)

As ΔHf of Hg(l) and ΔHf O₂(g) are 0:

ΔHr: - 2ΔHf HgO(s) = 181,66 kJ/mol

In the same way ΔSr is:

ΔSr= 2ΔS° Hg(l) + ΔS° O₂(g) - 2ΔS° HgO(s)

ΔSr= 2* 76,02J/Kmol + 205,14 J/Kmol - 2*70,19 J/Kmol

ΔSr= 216,8 J/Kmol = 0,216 kJ/Kmol

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ΔGr = 181,66 kJ/mol - 298K*0,216kJ/Kmol

ΔGr = +117,3 kJ/mol ≈ 4. +117,1 kJ/mol

I hope it helps!

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