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Murrr4er [49]
3 years ago
9

What is the volume of 3.40 mol of gas at 33.3C and 22.2atm of pressure

Chemistry
1 answer:
Luda [366]3 years ago
5 0
The volume of 3.40 mol of gas at 33.3 C and 22.2 atm of pressure is 3.85 liter of gas. This problem can be solved by using the PV=nRT or V= nRT/V equation which is the relation between the molar volume, the temperature, and the pressure of gas. In this formula, P is the pressure, R is the universal gas constant ( 0.0821 atm L/mol K), n is the molecule amount, V is the molecular volume, and T is the temperature. The temperature used in this formula must be in Kelvin, therefore we have to convert the Celcius temperature into Kelvin temperature (33.3 C = 306.45 K).
<span>The calculation for the problem above: 3.4*0.0821*306.45/22.2 = 3.85 liter of gas.</span>
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masya89 [10]

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3 years ago
Assuming a car (with a 70-L) gas tank can hold approximately 50,000 (5.00 * 10^4) g of octane(C8H18) or 50,000 (5.00 * 10^4) g o
konstantin123 [22]

Answer:

- From octane: m_{CO_2}=1.54x10^5gCO_2

- From ethanol: m_{CO_2}=9.57x10^4gCO_2

Explanation:

Hello,

At first, for the combustion of octane, the following chemical reaction is carried out:

C_8H_{18}+\frac{25}{2} O_2\rightarrow 8CO_2+9H_2O

Thus, the produced mass of carbon dioxide is:

m_{CO_2}=5.00x10^4gC_8H_{18}*\frac{1molC_8H_{18}}{114gC_8H_{18}}*\frac{8molCO_2}{1molC_8H_{18}}*\frac{44gCO_2}{1molCO_2} \\\\m_{CO_2}=1.54x10^5gCO_2

Now, for ethanol:

C_2H_6O+3O_2\rightarrow 2CO_2+3H_2O

m_{CO_2}=5.00x10^4gC_2H_6O*\frac{1molC_2H_6O}{46gC_2H_6O}*\frac{2molCO_2}{1molC_2H_6O}*\frac{44gCO_2}{1molCO_2} \\\\m_{CO_2}=9.57x10^4gCO_2

Best regards.

3 0
3 years ago
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Explanation:

6 0
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sergejj [24]
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Answer: option C. Copper (II) chloride

Explanation:

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