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3 years ago

A mothball, composed of naphthalene (c10h8), has a mass of 1.64 g . part a how many naphthalene molecules does it contain?

1 answer:

6 0

Answer is: 7,5·10²¹ <span>naphthalene molecules.
</span>m(C₁₀H₈) = 1,64 g.
n(C₁₀H₈) = m(C₁₀H₈) ÷ M(C₁₀H₈).
n(C₁₀H₈) = 1,64 g ÷ 128,17 g/mol.
n(C₁₀H₈) = 0,0125 mol.
N(C₁₀H₈) = n(C₁₀H₈) · Na.
N(C₁₀H₈) = 0,0125 mol · 6,022·10²³ 1/mol.
N(C₁₀H₈) = 7,5·10²¹.
Na - Avogadro number.

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Answer:

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Explanation:

Henry's law states that the solubility of a gas is directly proportional to its partial pressure. The equation may be written as:

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Where $k_H$ is Henry's law constant.

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Given initially:

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Also, at sea level, we have an atmospheric pressure of:

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According to Dalton's law of partial pressures, the partial pressure of oxygen is equal to the product of its mole fraction and the total pressure:

$p^o = \chi_{O_2} p$

Then the equation becomes:

$S_1 = k_H \chi_{O_2} p$

Solve for $k_H$ :

$k_H = \frac{S_1}{\chi_{O_2} p} = \frac{2.67\cdot 10^{-4} M}{0.209\cdot 1.00 atm} = 0.001278 M/atm$

Now we're given that at an altitude of 12,000 ft, the atmospheric pressure is now:

$p = 0.657 atm$

Apply Henry's law using the constant we found:

$S_2 = k_H \chi_{O_2} p = 0.001278 M/atm\cdot 0.209\cdot 0.657 atm = 1.75\cdot 10^{-4} M$

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1.26 * 10^4 + 2.50 * 10^4 in sceintific notation

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