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Fiesta28 [93]
3 years ago
15

A helium filled ballon had a volume of 8.50 L on the ground at 20.0 C and a pressure of 750.0 Torr. After the ballon was release

d, it rose to an altitude where the temperature was -40.0 C and the pressure was .550 atm. What is the new volume of the balloon in liters at the high altitude?
Chemistry
1 answer:
Marrrta [24]3 years ago
5 0

Answer:

V_2=12.1L

Explanation:

Hello!

In this case, according to the given data of volume, pressure and temperature, it is possible to infer this problem can be solved via the combined gas law:

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

Thus, regarding the question, we evidence we need V2, but first we make sure the temperatures are in Kelvins:

T_1=20+273=293K\\\\T_2=-40+273=233K

Then, we obtain:

V_2=\frac{P_1V_1T_2}{T_1P_2}\\\\V_2=\frac{0.987atm*8.50L*233K}{293K*0.550atm}\\\\V_2=12.1L

Best regards!

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Calculate ΔHrxn for the following reaction: Fe2O3(s)+3CO(g)→2Fe(s)+3CO2(g) Use the following reactions and given ΔH′s. 2Fe(s)+3/
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Answer:

The answer to your question is:  ΔHrxm = -23.9 kJ

Explanation:

Data:

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    CO(g)+1/2O2(g)→CO2(g)         ΔH = -282.7 kJ   (2)

Reaction:

                          Fe2O3(s)+3CO(g)→2Fe(s)+3CO2(g)  

We invert (1) and change the sign of  ΔH

        Fe2O3(s)   →  2Fe(s)+3/2O2(g)  ΔH = 824.2 kJ

We multiply (2) by 3

      3(  CO(g)+1/2O2(g)→CO2(g)         ΔH = -282.7 kJ)   (2)

      3CO(g)+3/2O2(g)→3CO2(g)         ΔH = -848.1 kJ

We add (1) and (2)

Fe2O3(s)   →  2Fe(s)+3/2O2(g)  ΔH = 824.2 kJ

3CO(g)+3/2O2(g)→3CO2(g)         ΔH = -848.1 kJ

Fe2O3(s) +  3CO(g)+3/2O2(g)  →  2Fe(s)+3/2O2 + 3CO2(g)    

Simplify

    Fe2O3(s)+3CO(g)→2Fe(s)+3CO2(g)    and ΔHrxm = -23.9 kJ

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