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konstantin123 [22]

2 years ago

9

Calculate the mass of oxygen (in mg) dissolved in a 5.00 L bucket of water exposed to a pressure of 1.13 atm of air. Assume the

mole fraction of oxygen in air to be 0.21 and the Henry's law constant for oxygen in water at this temperature to be 1.3 x 10-3 M/atm. Group of answer choices

1 answer:

AleksAgata [21]2 years ago

6 0

Answer:

50 mg

Explanation:

First, we have to calculate the partial pressure of O₂ (pO₂) using the following expression.

pO₂ = P × X(O₂) = 1.13 atm × 0.21 = 0.24 atm

where,

P: total pressure

X(O₂): mole fraction of oxygen

Then, we can calculate the concentration of O₂ in water (C) using Henry's law.

C = k × pO₂ = (1.3 × 10⁻³ M/atm) × 0.24 atm = 3.1 × 10⁻⁴ M

where,

k: Henry's constant for O₂

The mass of oxygen in a 5.00 L bucket with a concentration of 3.1 × 10⁻⁴ M is: (MM 32.0 g/mol)

$5.00L.\frac{3.1 \times 10^{-4}mol}{L} .\frac{32.0 \times 10^{3}mg}{mol} =50mg$

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The correct answer is ClO, ClO3-, ClO- and ClO4-

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In ClO2-, oxygen contains 8 electrons in its valence shell and oxygen will share one electron with chlorine to complete the octet of Cl. In other four, we can clearly see that there are more or less than 8 electrons in the outer shell of oxygen so we can clearly say that ClO, ClO3-, ClO- and ClO4- are disobeying the octet rule.

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To calculate the percentage of lead in sample, we use the equation:

$\%\text{ of lead}=\frac{\text{Mass of lead}}{\text{Mass of sample}}\times 100$

<u>For $PbS$ :</u>

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Putting values in above equation, we get:

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<u>For $PbCO_3$ :</u>

Mass of $PbCO_3$ = 267.2 g

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Putting values in above equation, we get:

$\%\text{ of lead}=\frac{\text{Mass of lead}}{\text{Mass of sample}}\times 100=\frac{207.2g}{267.2g}\times 100=77.55\%$

The percentage of lead in the $PbCO_3$ is 77.55 %.

<u>For $PbCl(OH)$ :</u>

Mass of $PbCl(OH)$ = 259.7 g

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Putting values in above equation, we get:

$\%\text{ of lead}=\frac{\text{Mass of lead}}{\text{Mass of sample}}\times 100=\frac{207.2g}{259.7g}\times 100=79.78\%$

The percentage of lead in the $PbCl(OH)$ is 79.78 %.

<u>For $Pb_2Cl_2CO_3$ :</u>

Mass of $Pb_2Cl_2CO3_$ = 545.3 g

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Putting values in above equation, we get:

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Hence, from this we conclude that, the compound PbS contains the highest percentage of lead (by mass).

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