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3 years ago

How many moles of C8H18 contain 9.25 × 1024 molecules of C8H18 ?

1 answer:

3 0

<span>One mole of any material contains 6.02 * 10e23 atoms or molecules. This is the Avogadro number. So 9.25 * 10e24 atoms are contained in (9.25*10e24 / 6.02 * 10e23) = 15.36 moles. So the answer is that moles of C8H18 is 15.36 moles</span>

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3 years ago

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The rate of effusion of an unknown gas was measured and found to be 11.9 mL/min. Under identical conditions, the rate of effusio

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Answer : The correct option is, (B) $CO_2$

Solution :

According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.

$R\propto \sqrt{\frac{1}{M}}$

or,

$(\frac{R_1}{R_2})=\sqrt{\frac{M_2}{M_1}}$ ..........(1)

where,

$R_1$ = rate of effusion of unknown gas = $11.9\text{ mL }min^{-1}$

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Now put all the given values in the above formula 1, we get:

$(\frac{11.9\text{ mL }min^{-1}}{14.0\text{ mL }min^{-1}})=\sqrt{\frac{32g/mole}{M_1}}$

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The unknown gas could be carbon dioxide $(CO_2)$ that has approximately 44 g/mole of molar mass.

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3 years ago

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3 years ago

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I NEED HELP PLEASE, THANKS! :)

krok68 [10]

Answer:

$\large \boxed{\text{2.20 g Pb}}$

Explanation:

They gave us the masses of two reactants and asked us to determine the mass of the product.

This looks like a limiting reactant problem.

1. Assemble the information

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ: 239.27 32.00 207.2

2PbS + 3O₂ ⟶ 2Pb + 2SO₃

m/g: 2.54 1.88

2. Calculate the moles of each reactant

$\text{Moles of PbS} = \text{2.54 g PbS } \times \dfrac{\text{1 mol PbS}}{\text{239.27 g PbS}} = \text{0.010 62 mol PbS}\\\\\text{Moles of O}_{2} = \text{1.88 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.058 75 mol O}_{2}$

3. Calculate the moles of Pb from each reactant

$\textbf{From PbS:}\\\text{Moles of Pb} = \text{0.010 62 mol PbS} \times \dfrac{\text{2 mol Pb}}{\text{2 mol PbS}} = \text{0.010 62 mol Pb}\\\\\textbf{From O}_{2}:\\\text{Moles of Pb} =\text{0.058 75 mol O}_{2} \times \dfrac{\text{2 mol Pb}}{\text{3 mol O}_{2}}= \text{0.039 17 mol Pb}\\\\\text{PbS is the $\textbf{limiting reactant}$ because it gives fewer moles of Pb}$

4. Calculate the mass of Pb

$\text{ Mass of Pb} = \text{0.010 62 mol Pb} \times \dfrac{\text{207.2 g Pb}}{\text{1 mol Pb}} = \textbf{2.20 g Pb}\\\\\text{The reaction produces $\large \boxed{\textbf{2.20 g Pb}}$}$

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3 years ago

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2 years ago