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Shalnov [3]
3 years ago
9

Can I get help with this, please!!

Chemistry
1 answer:
kondor19780726 [428]3 years ago
8 0

Pressure in the submarine when the temperature is changed to 293 K is 108.9 K Pa

Explanation:

Pressure in the submarine = 108.9 kPa

Volume, V = 2.4 * 10^5 L

Pressure, P = 116k Pa

Temperature, T = 312 K

Ideal gas law: PV = nRT  or  n = PV / RT

So, moles of gas, n =116 KPa * 2.4 * 10 ^5L / 8.314 LK Pa K^-1 *312 K

= 1.073 *10^4 mol

when temperature is changed to 293K,

PV = nRT  or P = nRT / V

=1.073 *10^4 mol *8.314 LK Pa mol^-1 K^-1 *293 K / 2.4*10^5L

=108.9 K Pa

Pressure in the submarine when the temperature is changed to 293 K is 108.9 K Pa

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Answer:

b. Bakelite

Explanation:

Bakelite -

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6 0
3 years ago
The following question appears on a quiz: ""You fill a tank with gas at 60°C to 100 kPa and seal it. You decrease the temperatu
dusya [7]

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Explanation:

To calculate the final pressure of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

\frac{P_1}{T_1}=\frac{P_2}{T_2}

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P_2\text{ and }T_2 are the final pressure and temperature of the gas.

We are given:

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Hence, the final pressure will decrease ad the value is 85 kPa

8 0
3 years ago
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Zielflug [23.3K]

Answer: 7.693 L

Explanation:

To calculate the new volume, we use the equation given by Boyle's law. This law states that pressure is directly proportional to the volume of the gas at constant temperature.

The equation given by this law is:

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P_2\text{ and }V_2 are final pressure and volume.

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P_1=0.63atm\\V_1=12.70L\\P_2=105kPa=1.04atm(1kPa=0.009atm)\\V_2=?

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0.63\times 12.70mL=1.04\times V_2\\\\V_2=7.693L

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6 0
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3 years ago
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