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A mixture of hydrogen (2.02 g) and chlorine (35.90 g) in a container at 300 K has a total gas pressure of 748 mm Hg. What is the

Llana [10]

The partial atmospheric pressure (atm) of hydrogen in the mixture is 0.59 atm.

<h3>How do we calculate the partial pressure of gas?</h3>

Partial pressure of particular gas will be calculated as:

p = nP, where

P = total pressure = 748 mmHg
n is the mole fraction which can be calculated as:
n = moles of gas / total moles of gas

Moles will be calculated as:

n = W/M, where
W = given mass
M = molar mass

Moles of Hydrogen gas = 2.02g / 2.014g/mol = 1 mole

Moles of Chlorine gas = 35.90g / 70.9g/mol = 0.5 mole

Mole fraction of hydrogen = 1 / (1+0.5) = 0.6

Partial pressure of hydrogen = (0.6)(748) = 448.8 mmHg = 0.59 atm

Hence, required partial atmospheric pressure of hydrogen is 0.59 atm.

To know more about partial pressure, visit the below link:
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3 0

1 year ago

The theory of evolution was proposed by

yulyashka [42]

The theory of evolution was proposed by Darwin.

4 0

3 years ago

Read 2 more answers

The isotopes K-37 and K-42 have the same

Masteriza [31]

Answer:

the same number of protons

Explanation:

5 0

2 years ago

Read 2 more answers

Determine the expected diffraction angle for the first-order diffraction from the (111) set of planes for FCC nickel (Ni) when m

faust18 [17]

Answer:

56°

Explanation:

First calculate $a:$

$a=2 R \sqrt{2}=2(0.1246) \sqrt{2}=0.352 \mathrm{nm}$

The interplanar spacing can be calculated from:

$d_{111}=\frac{a}{\sqrt{1^{2}+1^{2}+1^{2}}}=\frac{0.352}{\sqrt{3}}=0.203 \mathrm{nm}$

The diffraction angle is determined from:

$\sin \theta=\frac{n \lambda}{2 d_{111}}=\frac{1(0.1927)}{2(0.2035)}=0.476$

Solve for $\theta$

$\theta=\sin ^{-1}(0.476)=28^{\circ}$

The diffraction angle is:

$2 \theta=2\left(28^{\circ}\right)=56^{\circ}$

4 0

3 years ago

How many liters of water vapor are in 36.21 g?

Elden [556K]

Answer:

45.02 L.

Explanation:

Firstly, we need to calculate the no. of moles of water vapor.

n = mass / molar mass = (36.21 g) / (18.0 g/mol) = 2.01 mol.

We can calculate the volume of knowing that 1.0 mole of a gas at STP occupies 22.4 L.

Using cross multiplication:

1.0 mole of CO occupies → 22.4 L.

2.01 mole of CO occupies → ??? L.

∴ The volume of water vapor in 36.21 g = (22.4 L)(2.01 mole) / (1.0 mole) = 45.02 L.

4 0

2 years ago