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Anastaziya [24]
4 years ago
12

I need alot of help.

Chemistry
1 answer:
GaryK [48]4 years ago
6 0
Of course! I can help, what is it that you need help on and I shall try my best to answer!
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A student burns 1.50 mol C3H8 according to the following reaction:
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The balloon in the previous problem will burst if its volume reaches 400. L. Given the initial conditions specified in that prob
NARA [144]

This is an incomplete question, here is complete question.

A metrological balloon contains 250 L of He at 22 C and 740 mmHg.

The balloon in the previous problem will burst if its volume reaches 400 L. Given the initial conditions specified in that problem, at what temperature, in degrees Celsius, will the balloon burst if its pressure at that bursting point is 0.475 atm.

Answer : The final temperature will be, -44.4^oC

Explanation :

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas = 740 mmHg  = 0.974 atm

P_2 = final pressure of gas = 0.475 atm

V_1 = initial volume of gas = 250 L

V_2 = final volume of gas = 400 L

T_1 = initial temperature of gas = 22^oC=273+20=293K

T_2 = final temperature of gas = ?

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

\frac{0.974atm\times 250L}{293K}=\frac{0.475atm\times 400L}{T_2}

T_2=228.6K=228.6-273=-44.4^oC

Thus, the final temperature will be, -44.4^oC

7 0
4 years ago
Help with a and b if so thanks please asap!!
Anastaziya [24]

Answer:

A.  0.01  B. 650

Explanation:

I love helping people.

7 0
4 years ago
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