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scoray [572]
3 years ago
10

What is the total number of years that must pass before only 25.00 grams of an original 100.0-gram sample of C-14 remains unchan

ged?
(1) 2,865 y (3) 11,460 y
(2) 5,730 y (4) 17,190 y
Chemistry
2 answers:
Karolina [17]3 years ago
8 0
The answer is (3) 11,460 y. For this problem, you must know that the half life of C-14 is 5740 y and that nuclear decay processes are first order reactions (which means that the half life remains constant). For 25.00g to be left of a 100.0g sample, two half life must have elapsed (100*0.5*0.5=25). Each half life is 5730 y, so the total time elapsed is 5730 * 2 = 11,460 y.
REY [17]3 years ago
6 0

Answer:

The answer is 3; 11,460

Explanation:

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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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