14. A gas in a 400 ml flask exerts pressure of 81.0 kPa at 25.0°C. If the gas
was in a 250 ml container at a pressure of 101 kPa, the temperature would be
a) - 124 °C
b) - 41.2°C
c) 20.0°C
d) 232 °C

Answers

Answer 1

Answer: B would be the correct answer

Explanation: You will need to use P1V1/T1=P2V2/T2

So

P1= 81.0kPa /101 =0.799atm

V1=0.4L

T1=25.0C +273.15K= 298.15K

P2= 101kPa /101 = 1atm

V2=0.25L

T2= ?

T2=P2V2T1/P1V1

(1.0atm)(0.25L)(298.15K)/(0.799atm)(0.4L) =233.22K-273.15=-39.92C


Related Questions

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Pls make brainliest :}

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a standard solution is made by diluting a small volume of stock solution. if 2.42 ml of stock solution is mixed with 7.55 ml of water, what is the concentration of the diluted solution? the stock solution concentration is 0.197 m.

Answers

Taking into account the definition of dilution, the concentration of the diluted solution is 0.063 M.

Dilution

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Answer:

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Answer:

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Explanation:

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Answer: 105 mL

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We can use the formula M₁V₁ = M₂V₂ the total volume of water that will yield 0.800 M HCl. We can then subtract that from the initial volume to get how much water needs to be added.

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