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2 years ago

A 1.20-L container contains 1.10 g of an unknown gas at STP. What is the molecular weight of the unknown gas?

Chemistry

1 answer:

SOVA2 [1]2 years ago

8 0

Answer:

M = 20.5 g/mol

Explanation:

Given data:

Volume of gas = 1.20 L

Mass of gas = 1.10 g

Temperature and pressure = standard

Solution:

First of all we will calculate the density.

Formula:

d = mass/ volume

d = 1.10 g/ 1.20 L

d = 0.92 g/L

Now we will calculate the molar mass.

d = PM/RT

0.92 g/L = 1 atm × M / 0.0821 atm.L/mol.K ×273.15 K

M = 0.92 g/L × 0.0821 atm.L/mol.K ×273.15 K / 1 atm

M = 20.5 g/mol

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2 years ago

In methane combustion, the following reaction pair is important: At 1500 K, the equilibrium constant Kp has a value of 0.003691

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Explanation:

Let us assume that the value of $K_{r}$ = $2.82 \times 10^{5} \times e^({\frac{-9835}{T}}) m^{6}/mol^{2}s$

Also at 1500 K, $K_{r}$ = $2.82 \times 10^{5} \times e^({\frac{-9835}{1500}}) m^{6}/mol^{2}s$

$K_{r} = 400.613 m^{6}/mol^{2}s$

Relation between $K_{p}$ and $K_{c}$ is as follows.

$K_{p} = K_{c}RT$

Putting the given values into the above formula as follows.

$K_{p} = K_{c}RT$

$0.003691 = K_{c} \times 8.314 \times 1500$

$K_{c} = 2.9 \times 10^{-7}$

Also, $K_{c} = \frac{K_{f}}{K_{r}}$

or, $K_{f} = K_{c} \times K_{r}$

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Thus, we can conclude that the value of $K_{f}$ is $1.1617 \times 10^{-4} m^{6}/mol^{2}s$ .

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3 years ago