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lara31 [8.8K]
2 years ago
11

6) A volume of 473 mL of oxygen was collected at 27°C. What volume would the oxygen occupy at

Chemistry
1 answer:
sertanlavr [38]2 years ago
6 0

Answer : The volume of oxygen occupy at 173° would be, 703.2 mL

Explanation :

Charles' Law : It states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

\frac{V_1}{T_1}=\frac{V_2}{T_2}

where,

V_1\text{ and }T_1 are the initial volume and temperature of the gas.

V_2\text{ and }T_2 are the final volume and temperature of the gas.

We are given:

V_1=473mL\\T_1=27^oC=(27+273)K=300K\\V_2=?\\T_2=173^oC=(173+273)K=446K

Now put all the given values in above equation, we get:

\frac{473mL}{300K}=\frac{V_2}{446K}\\\\V_2=703.2mL

Therefore, the volume of oxygen occupy at 173° would be, 703.2 mL

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A) First start out with the 0.300 mol of C6H12O6...
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*The significant figures (sig figs) at still three, the 2 is a conversion counting number and does not count*

B) First change 2.00 g of C2H5OH to moles of C2H5OH...
The molecular mass of C2H5OH is...
2(12.01 g/mol) + 5(1.008 g/mol) + 16.00 g/mol + 1.008 g/mol = 46.07 g/mol
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2.00 g C2H5OH * (1 mol C2H5OH / 46.07 g C2H5OH) = 0.0434 mol C2H5OH

Second, you can change the moles of C2H5OH to moles of C6H12O6..
0.0434 mol C2H5OH * (1 mol C6H12O6 / 2 mol C6H12O6) = 0.0217 mol C6H12O6

Third, change moles of C6H12O6 to grams...
MM = 6(12.01 g/mol) + 12(1.008 g/mol) + 6(16.00 g/mol) = 180.16 g/mol
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2.00 g C2H5OH * (1 mol C2H5OH / 46.07 g C2H5OH) * (2 mol CO2 / 2 mol C2H5OH) * (44.01 g CO2 / 1 mol CO2) = 1.91 g CO2

I hope this helped and I am sorry that I talked to much, I just didn't want to miss anything!
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what is the mole fraction of neon in a mixture that contains 0.628 g of helium, 11.491 g of neon, and 7.613 g of argon?​
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Answer:

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Explanation:

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Unknown:

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The mole fraction of an element is the number of moles of that element to the total number of moles in the gas mixture.

We need to calculate the number of moles of each element first;

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