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Natali [406]
2 years ago
12

How many moles of helium are in a Superman ballon of volume 8.5 L at 92.3 kPa and 287 K?

Chemistry
1 answer:
kirza4 [7]2 years ago
7 0

Answer : The number of moles of helium are, 0.329 mol.

Explanation :

To calculate the moles of helium gas we are using ideal gas equation.

PV=nRT

where,

P = Pressure of He gas = 92.3 kPa = 0.911 atm

V = Volume of He gas = 8.5 L

n = number of moles He = ?

R = Gas constant = 0.0821L.atm/mol.K

T = Temperature of He gas = 287 K

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

0.911atm\times 8.5L=n\times (0.0821L.atm/mol.K)\times 287K

n=0.329mol

Thereofore, the number of moles of helium are, 0.329 mol.

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Convert 338 L at 63.0 atm to its new volume at standard pressure.
taurus [48]

The new volume at standard pressure of 1 atm is 21294 liters.

Explanation:

Data given:

Initial volume of the gas V1 = 338 liters

initial pressure on the gas P1 = 63 atm

standard pressure as P2 = 1 atm

Final volume at standard pressure V2 =?

The data given shows that Boyle's law equation is to used:

P1V1 = P2V2

rearranging the equation to calculate V2,

V2 = \frac{P1V1}{P2}

Putting the values in the equation:

V2 = \frac{338X63}{1}

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as the pressure on the gas is reduced to 1 atm the volume of the gas increased incredibly to 21294 litres.

7 0
2 years ago
A 0.72-mol sample of PCl5 is put into a 1.00-L vessel and heated. At equilibrium, the vessel contains 0.40 mol of PCl3(g) and 0.
Sonja [21]

Answer:

Equilibrium constant for PCl_5 is 0.5

Equilibrium constant for decomposition of NO_2 is 1.79 \times 10^{-14}

Explanation:

PCl_5 dissociates as follows:

                    PCl_5 \rightleftharpoons PCl_3+Cl_2

initial          0.72 mol     0         0

at eq.     0.72 - 0.40   0.40      0.40

Expression for the equilibrium constant is as follows:

k=\frac{[PCl_3][Cl_2]}{[PCl_5]}

Substitute the values in the above formula to calculate equilibrium constant as follows:

k=\frac{[0.40/1][0.40/1]}{0.32/1} \\=\frac{0.40 \times 0.40}{0.32} \\=0.5

Therefore, equilibrium constant for PCl_5 is 0.5

Now calculate the equilibrium constant for decomposition of  NO_2

It is given that 3.3 \times 10^{-3} \% is decomposed.

NO_2 decomposes as follows:

                                  2NO_2 \rightleftharpoons 2NO + O_2

initial                            1.0 M       0           0

at eq. concentration of  NO_2   is:

[NO_2]_{eq}=1-(0.000066) = 0.999934\ M

[NO]_{eq}=6.6 \times 10^{-5}\ M

[O_2]_{eq}=3.3\times 10^{-5} = 3.3\times 10^{-5}\ M      

Expression for equilibrium constant is as follows:

K=\frac{[NO]^2[O_2]}{[NO_2]^2}

Substitute the values in the above expression

K=\frac{[6.6\times 10^{-5}]^2[3.3 \times 10^{-5}]}{[0.999934]^2} \\=1.79\times 10^{-14}

Equilibrium constant for decomposition of NO_2 is 1.79 \times 10^{-14}

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