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fomenos
3 years ago
5

At 35.0°c and 3.00 atm pressure, a gas has a volume of 1.40 l. what pressure does the gas have at 0.00°c and a volume of 0.950 l

? which equation should you use?.

Chemistry
2 answers:
Alja [10]3 years ago
5 0

Answer:

The answer is 3.92 sign. digits.

Explanation:

Leona [35]3 years ago
4 0

Answer : The pressure of gas will be, 3.918 atm and the combined gas law is used for this problem.

Solution :

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 = 3 atm

P_2 = final pressure of gas = ?

V_1 = initial volume of gas = 1.40 L

V_2 = final volume of gas = 0.950 L

T_1 = initial temperature of gas = 35^oC=273+35=308K

T_2 = final temperature of gas = 0^oC=273+0=273K

Now put all the given values in the above equation, we get the final pressure of gas.

\frac{3atm\times 1.40L}{308K}=\frac{P_2\times 0.950L}{273K}

P_2=3.918atm

Therefore, the pressure of gas will be, 3.918 atm and the combined gas law is used for this problem.

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A sample of gaseous arsine (AsH3) in a 460 mL flask at 332 Torr and 223 K, is heated to 437 K, at which temperature arsine decom
Taya2010 [7]

Answer:

28/95 = 29.,5 % of Arsine decomposed

Explanation:A sample of gaseous arsine (AsH3) in a 460 mL flask at 332 Torr and 223 K, is heated to 437 K, at which temperature arsine decom- poses to solid arsenic and hydrogen gas. The flask is then cooled to 273 K, at which tem- perature the pressure in the flask is 488 Torr. What percentage of arsine molecules have de- composed?

Answer in units of %.

initial pressure 332 Torr initial volume 0.46 L initial temperature 223K

final pressure 488 Torr final volume 0.46 L final 273 K

Torr is 1/760 atm 332 torr = 0.437 atm 488 Torr =0.642 atm

PV = nRT so n=RT/PV

INITIAL n= 0.082 X 223/(0.437)(0.46) = 91 moles

final n= 0.082 X 273 / (.437)(488) = 105 moles

2AsH3----------> 2As + 3H2

x moles of Arsine decomposed to make 1.5 moles of H2

the final number of moles was

(91 -X)+ 1.5 X = 105 moles

91 + 0.5 X = 105

0.5 X = 14

X =28

CHECK

if 28 moles of Arsine , then the container would have

91 --28 + 1.5(28) = 91 +14 =105 check

so 28/95 = 29.,5 % of Arsine decomposed

Your answer

(quit)

polyalchemVirtuoso

Answer:

Explanation:

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