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2 years ago

if 0.00327 g of a gas dissolves in 0.376 L of water at 867 torr, what quantity of this gas (in grams) will dissolve at 759 torr?

Chemistry

1 answer:

Sedbober [7]2 years ago

7 0

Answer:

0.00286

Explanation:

Hope this helps!

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a glass container was initially charged with 1.50 mol of a gas sample at 3.75 atm and 21.7C. some of the gas was release as the

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Answer:

0.39 mol

Explanation:

Considering the ideal gas equation as:

$PV=nRT$

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L.atm/K.mol

At same volume, for two situations, the above equation can be written as:-

$\frac {{n_1}\times {T_1}}{P_1}=\frac {{n_2}\times {T_2}}{P_2}$

Given ,

n₁ = 1.50 mol

n₂ = ?

P₁ = 3.75 atm

P₂ = 0.998 atm

T₁ = 21.7 ºC

T₂ = 28.1 ºC

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T₁ = (21.7 + 273.15) K = 294.85 K

T₂ = (28.1 + 273.15) K = 301.25 K

Using above equation as:

$\frac{{n_1}\times {T_1}}{P_1}=\frac{{n_2}\times {T_2}}{P_2}$

$\frac{{1.50\ mol}\times {294.85\ K }}{3.75\ atm}=\frac{{n_2}\times {301.25\ K }}{0.998\ atm}$

$n_2=\frac{{1.50}\times {294.85}\times 0.998}{3.75\times 301.25}\ mol$

Solving for n₂ , we get:

n₂ = 0.39 mol

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Answer:

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