In this item, I supposed, that we are determine the molar fraction of oxygen and carbon dioxide in the sample. This can be done by dividing their respective partial pressures by the total pressure of the sample.

O2 : mole fraction = (100.7 mmHg) / (763.00 mmHg) = 0.13

CO2 : mole fraction = (33.57 mmHg) / (763.00 mmHg) = 0.044

Answers: O2 = 0.13
CO2 = 0.044

3 0

2 years ago

Changes in ________ move matter from one state to another.

kipiarov [429]

Temperature. Water is an example. When water is at room temp. its liquid. When water is at boiling temp. It is a gas. And when water is at freezing temp. Its a solid.

7 0

3 years ago

Read 2 more answers

Help Please..... Points Available! I will give brainliest!!

patriot [66]

Answer:

I think Its B

Explanation:

4 0

3 years ago

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Gold forms a substitutional solid solution with silver. Compute the number of gold atoms per cubic centimeter for a silver-gold

Mama L [17]

Answer: The number of gold atoms per cubic centimeters in the given alloy is $1.83\times 10^{22}$

Explanation:

To calculate the number of gold atoms per cubic centimeters for te given silver-gold alloy, we use the equation:

$N_{Au}=\frac{N_AC_{Au}}{(\frac{C_{Au}M_{Au}}{\rho_{Au}})+(\frac{M_{Au}(100-C_{Au})}{\rho_{Ag}})}$

where,

$N_{Au}$ = number of gold atoms per cubic centimeters

$N_A$ = Avogadro's number = $6.022\times 10^{23}atoms/mol$

$C_{Au}$ = Mass percent of gold in the alloy = 42 %

$\rho_{Au}$ = Density of pure gold = $19.32g/cm^3$

$\rho_{Ag}$ = Density of pure silver = $10.49g/cm^3$

$M_{Au$ = molar mass of gold = 196.97 g/mol

Putting values in above equation, we get:

$N_{Au}=\frac{(6.022\times 10^{23}atoms/mol)\times 48\%}{(\frac{48\%\times 196.97g/mol}{19.32g/cm^3})+(\frac{196.97g/mol\times 58\%}{10.49g/cm^3})}\\\\N_{Au}=1.83\times 10^{22}atoms/cm^3$

Hence, the number of gold atoms per cubic centimeters in the given alloy is $1.83\times 10^{22}$

5 0

3 years ago