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3 years ago

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The law of conservation of momentum states that if objects exert forces only on each other, their total momentum is ____________

. Group of answer choices

1 answer:

iren2701 [21]3 years ago

8 0

Answer:

Conserved.

Explanation:

Momentum can be defined as the multiplication (product) of the mass possessed by an object and its velocity. Momentum is considered to be a vector quantity because it has both magnitude and direction.

Mathematically, momentum is given by the formula;

$Momentum = Mass * Velocity$

The law of conservation of momentum states that the total linear momentum of any closed system would always remain constant with respect to time.

This ultimately implies that, the law of conservation of momentum states that if objects exert forces only on each other, their total momentum is conserved.

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As a scuba diver descends under water, the pressure increases. At a total air pressure of 2.71 atm and a temperature of 25.0 C,

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Answer:

$1.32\times 10^{-3} mol/L$ is the solubility of nitrogen gas in a diver's blood.

Explanation:

Henry's law states that the amount of gas dissolved or molar solubility of gas is directly proportional to the partial pressure of the liquid.

To calculate the molar solubility, we use the equation given by Henry's law, which is:

$C_{N_2}=K_H\times p_{liquid}$

where,

$K_H$ = Henry's constant = $6.26 \times 10^{-4}mol/L.atm$

$p_{N_2}$ = partial pressure of nitrogen

$p_{N_2}= P\times \chi_{N_2}$ (Raoult's law)

$=2.71 atm\times 0.78=2.1138 atm$

$C_{N_2}=6.26 \times 10^{-4}mol/L.atm\times 2.1138 atm$

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$1.32\times 10^{-3} mol/L$ is the solubility of nitrogen gas in a diver's blood.

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3 years ago

Determine the mole fractions and partial pressures of CO2, CH4, and He in a sample of gas that contains 1.20 moles of CO2, 1.79

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Answer : The mole fraction and partial pressure of $CH_4,CO_2$ and $He$ gases are, 0.267, 0.179, 0.554 and 1.54, 1.03 and 3.20 atm respectively.

Explanation : Given,

Moles of $CH_4$ = 1.79 mole

Moles of $CO_2$ = 1.20 mole

Moles of $He$ = 3.71 mole

Now we have to calculate the mole fraction of $CH_4,CO_2$ and $He$ gases.

$\text{Mole fraction of }CH_4=\frac{\text{Moles of }CH_4}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}$

$\text{Mole fraction of }CH_4=\frac{1.79}{1.79+1.20+3.71}=0.267$

and,

$\text{Mole fraction of }CO_2=\frac{\text{Moles of }CO_2}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}$

$\text{Mole fraction of }CO_2=\frac{1.20}{1.79+1.20+3.71}=0.179$

and,

$\text{Mole fraction of }He=\frac{\text{Moles of }He}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}$

$\text{Mole fraction of }He=\frac{3.71}{1.79+1.20+3.71}=0.554$

Thus, the mole fraction of $CH_4,CO_2$ and $He$ gases are, 0.267, 0.179 and 0.554 respectively.

Now we have to calculate the partial pressure of $CH_4,CO_2$ and $He$ gases.

According to the Raoult's law,

$p_i=X_i\times p_T$

where,

$p_i$ = partial pressure of gas

$p_T$ = total pressure of gas = 5.78 atm

$X_i$ = mole fraction of gas

$p_{CH_4}=X_{CH_4}\times p_T$

$p_{CH_4}=0.267\times 5.78atm=1.54atm$

and,

$p_{CO_2}=X_{CO_2}\times p_T$

$p_{CO_2}=0.179\times 5.78atm=1.03atm$

and,

$p_{He}=X_{He}\times p_T$

$p_{He}=0.554\times 5.78atm=3.20atm$

Thus, the partial pressure of $CH_4,CO_2$ and $He$ gases are, 1.54, 1.03 and 3.20 atm respectively.

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3 years ago