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iragen [17]
2 years ago
6

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 He occupy under

the same conditions?
Chemistry
1 answer:
WINSTONCH [101]2 years ago
4 0

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;

n_1 = \frac{m_1}{M} \\\\n_1 = \frac{4}{4} = 1

The number of moles of the gas when the reacting mass is 3.0 g;

m₂ = 3.0 g

n_2 = \frac{m_2}{M} \\\\n_2 = \frac{3}{4} \\\\n_2 = 0.75 \ mol

The volume of the gas at 0.75 mol is determined using ideal gas law;

PV = nRT

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

Therefore, the volume occupied by 3.0 g of the gas is 16.8 L.

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What mass of oxygen reacts with 3.6 g of magnesium to form magnesium oxide?
levacccp [35]

Answer is: the mass of the oxygen is 2.37 grams.

Balanced chemical reaction: 2Mg + O₂ → 2MgO.

m(Mg) = 3.6 g; mass of magnesium.

n(Mg) = m(Mg) ÷ M(Mg).

n(Mg) = 3.6 g ÷ 24.3 g/mol

n(Mg) = 0.149 mol.; amount of the magnesium.

n(O₂) = ?

From chemical reaction: n(Mg) : n(O₂) = 2 : 1.

n(O₂) = n(Mg) ÷ 2.

n(O₂) = 0.149mol ÷ 2.

n(O₂) = 0.075 mol; amount of the oxygen.

m(O₂) = m(O₂) · M(O₂).

m(O₂) = 0.075 mol · 32 g/mol.

m(O₂) = 2.37 g; mass of the oxygen.

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3 years ago
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6 0
2 years ago
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