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solniwko [45]
4 years ago
11

A 50 L cylinder is filled with argon gas to a pressure of 10130.0 kPa at 300°C. How many moles of argon gas does the cylinder co

ntain? (Given: R = 8.314 L∙kPa/K∙mol) 2519.142 mol
Chemistry
1 answer:
Vlad1618 [11]4 years ago
7 0
To answer this question, we will use the general gas law which states that:
PV = nRT where:
P is the pressure of the gas = <span>10130.0 kPa
</span>V is the volume of the gas = 50 liters
n is the number of moles that we want to calculate
R is the gas constant = <span>8.314 L∙kPa/K∙mol
T is the temperature = 300+273 = 573 degree kelvin

Substitute with the givens in the equation to get the number of moles as follows:
</span><span>10130 * 50 = n * 8.314 * 573
506500 = 4763.922 n
n = </span>506500 / 4763.922
n = 106.3199 moles
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Answer:

5.75

Explanation:

Weak acid and strong base,

Make it a two part problem

The first will be Partial neutralization of the weak acid, while the second is the equilibrium and final pH

1.Parameters

HC2H3O2 =10mL x 0.75M = 7.5 mmol

NaOH =22mL x 0.30M = 6.6 mmol

R HC2H3O2 + OH- --> C2H3O2- + H2O

I 7.5 mmol 6.6 mmol 0 mmol ignore

C -6.6 mmol -6.6 mmol +6.6 mmol

E 0.9 mmol 0.0 mmol 6.6 mmol

2.) HC2H3O2 --> H+ + C2H3O2-

Initial concentrations

[HC2H3O2]: 0.9 mmol / 32mL = 0.0281M

[H+]:

[C2H3O2-]: 6.6 mmol x 32mL = 0.206M

Concentrations at equilibrium

[HC2H3O2]: 0.0281M - x

[H+]: [C2H3O2-]: 0.206M + x

If x is small and can be ignored except [H+]

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1.3 x 10-5 = (x)(0.206)

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3 years ago
A 0.907 mole sample of ideal gas at 12c occupies a volume of 4.25L. What is the pressure of gas?
mel-nik [20]

Pressure of the ideal gas=505.7kPa

Given:

No of moles=0.907mole

Temperature of the gas=12^{\circ} \mathrm{C}

Volume of the gas=4.25L

To find:

Pressure of the gas

<u>Step by Step Explanation: </u>

Solution:

According to the ideal gas equation

P V=n R T and from this pressure is derived as

P=\frac{n R T}{V}

Where P=Pressure of the gas

V=Volume of the gas=4.25L

n=No of the moles=0.907mole

R=Gas constant=8.314J / m o l^{-1} K^{-1}

T=Temperature of the gas=12^{\circ} \mathrm{C} =273+12=285K

Substitute these known values in the above equation we get

P=\frac{0.907 \times 8.314 \times 285}{4.25}

P=\frac{2149.13}{4.25}

P=505.7kPa

Result:

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Explanation:

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