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3 years ago

What is the mass of oxygen gas in a 12.2 L

Chemistry

1 answer:

baherus [9]3 years ago

4 0

Answer:

grams O₂ = 134 grams

Explanation:

PV = nRT => n = PV/RT

P = 8.15atm

V = 12.2 Liters

R = 0.08206L·atm/mol·K

T = 16.0°C + 273 = 289K

n = (8.15atm)(12.2L)/(0.08206L·atm/mol·K)(289K) = 4.2 moles O₂

grams O₂ = 4.2 moles O₂ x 32g/mol = 134 grams O₂

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WILL GIVE BRAINLIST

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expeople1 [14]

The question is incomplete, the complete question is:

The solubility of slaked lime, $Ca(OH)_2$ , in water is 0.185 g/100 ml. You will need to calculate the volume of $2.50\times 10^{-3}$ M HCl needed to neutralize 14.5 mL of a saturated

<u>Answer:</u> The volume of HCl required is 290mL, the mass of $Ca(OH)_2$ is 0.0268g, the moles of

<u>Explanation:</u>

Given values:

Solubility of $Ca(OH)_2$ = 0.185 g/100 mL

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Using unitary method:

In 100 mL, the mass of $Ca(OH)_2$ present is 0.185 g

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The equation used is:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ......(1)

Given mass of $Ca(OH)_2$ = 0.0268 g

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Plugging values in equation 1:

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$Ca(OH)_2+2HCl\rightarrow CaCl_2+2H_2O$

By the stoichiometry of the reaction:

Moles of $OH^-$ = Moles of $H^+$ = 0.000724 mol

The formula used to calculate molarity:

$\text{Molarity of solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (mL)}}$ .....(2)

Moles of HCl = 0.000724 mol

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Putting values in equation 2, we get:

$2.50\times 10^{-3}mol=\frac{0.000724\times 1000}{\text{Volume of solution}}\\\\\text{Volume of solution}=\frac{0.000725\times 1000}{2.50\times 10^{-3}}=290mL$

Hence, the volume of HCl required is 290mL, the mass of $Ca(OH)_2$ is 0.0268g, the moles of

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Answer:

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