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Gemiola [76]
2 years ago
6

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

partial atmospheric pressure (atm) of hydrogen in the mixture?
Chemistry
1 answer:
Llana [10]2 years ago
3 0

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:
brainly.com/question/15302032

#SPJ1

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Consider two different ions. The anion has a valence of -2. The cation has a valence of +2. The two ions are separated by a dist
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Answer:

Force of attraction = 35.96 \times 10^{27}N

Explanation:

Given: charge on anion = -2

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Distance = 1 nm = 10^{-9} m

To calculate: Force of attraction.

Solution: The force of attraction is calculated by using equation,

F = \dfrac{k \times q_1 q_2}{ \r^2} ---(1)

where, q represents the charge and the subscripts 1 and 2 represents cation and anion.

k = 8.99 \times 10^9 \ Nm^{2}C^{-2}

F = force of attraction

r = distance between ions.

Substituting all the values in the equation (1) the equation becomes

F = \dfrac{8.99 \times 10^9 \times 2 \times 2}{ \left ( 10^-9 \right )^2 }

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3 years ago
What is the concentration (in M) of a 225ml potassium sulfate solution that contains 4.15g of potassium?
mars1129 [50]

The concentration of solution in M or mol/L can be calculated using the following formula:

C=\frac{n}{V} .... (1)

Here, n is number of moles and V is volume of solution in L.

The molecular formula of potassium sulfate is K_{2}SO_{4} thus, there are 2 moles of potassium in 1 mol of potassium sulfate.

1 mol of potassium will be there in 0.5 mol of potassium sulfate.

Mass of potassium is 4.15 g, molar mass is 39.1 g/mol.

Number of moles can be calculated as follows:

n=\frac{m}{M}

Here, m is mass and M is molar mass

Putting the values,

n=\frac{(4.15 g}{(39.1 g/mol}=0.1061 mol

Thus, number of moles of  K_{2}SO_{4} will be 0.1061\times 0.5=0.053 mol.

The volume of solution is 225 mL, converting this into L,

1 mL=10^{-3}L

Thus,

225 mL=0.225 L

Putting the values in equation (1),

C=\frac{(0.053 mol}{0.225 L}=0.236 M

Therefore, concentration of potassium sulfate solution is 0.236 M.


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