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Answer:
the volume occupied by 3.0 g of the gas is 16.8 L.
Explanation:
Given;
initial reacting mass of the helium gas, m₁ = 4.0 g
volume occupied by the helium gas, V = 22.4 L
pressure of the gas, P = 1 .0 atm
temperature of the gas, T = 0⁰C = 273 K
atomic mass of helium gas, M = 4.0 g/mol
initial number of moles of the gas is calculated as follows;
[tex]n_1 = \frac{m_1}{M} \\\\n_1 = \frac{4}{4} = 1[/tex]
The number of moles of the gas when the reacting mass is 3.0 g;
m₂ = 3.0 g
[tex]n_2 = \frac{m_2}{M} \\\\n_2 = \frac{3}{4} \\\\n_2 = 0.75 \ mol[/tex]
The volume of the gas at 0.75 mol is determined using ideal gas law;
PV = nRT
[tex]PV = nRT\\\\\frac{V}{n} = \frac{RT}{P} \\\\since, \ \frac{RT}{P} \ is \ constant,\ then;\\\frac{V_1}{n_1} = \frac{V_2}{n_2} \\\\V_2 = \frac{V_1n_2}{n_1} \\\\V_2 = \frac{22.4 \times 0.75}{1} \\\\V_2 = 16.8 \ L[/tex]
Therefore, the volume occupied by 3.0 g of the gas is 16.8 L.
The volume occupied by 3.0 g of He is mathematically given as
V2=16.8L
Question Parameters:
If 4.0 g of helium gas occupies a volume of 22.4 L at 0 o C and a pressure of 1.0 atm, what volume does 3.0 g of H
Generally, the equation for the gas equation is mathematically given as
PV=nRT
When the reacting mass is 3g
n2=3/4
n2=0.75
Therefore
V1/n1=V2/n2
Where
n1=1
[tex]V2=\frac{22.4*0.75}{1}[/tex]
V2=16.8L
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