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Alex787 [66]
2 years ago
6

What is the pressure exerted by 1.0 mol of a gas with temperature of 20 C and a volume of 7.5 L?

Chemistry
1 answer:
Fofino [41]2 years ago
5 0

We have to assume the gas behaves similar to an ideal gas and we can use the ideal gas law equation

PV = nRT

Where

P - pressure

V - volume - 7.5 L

n - number of moles - 1.0 mol

R - universal gas constant - 0.0821 L.atm/mol.K

T - temperature in kelvin

20 degrees celcius + 273 = 293 K

Substituting the values in the equation

P x 7.5 L = 1.0 mol x 0.0821 L.atm/mol.K x 293 K

P = 3.2 atm

The pressure exerted is 3.2 atm

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Answer:

The precipitate is CuS.

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Explanation:

<u>Step 1: </u>Data given

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Ksp (CuS) = 1.3 × 10-36

Ksp (FeS) = 6.3 × 10-18

Step 2:  Calculate precipitate

CuS → Cu^2+ + S^2-         Ksp= 1.3*10^-36

FeS → Fe^2+ + S^2-      Ksp= 6.3*10^-18

Calculate the minimum of amount needed to form precipitates:

Q=Ksp

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Ksp (CuS) = 1.3*10^-36 = 0.036M *[S2-]

[S2-]= 3.61*10^-35 M

<u>For Iron</u>  we have: Ksp=[Fe2+]*[S2-]

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[S2-]= 1.43*10^-16 M

CuS will form precipitates before FeS., because only 3.61*10^-35 M Sulfur Ions are needed for CuS. For FeS we need 1.43*10^-16 M Sulfur Ions which is much larger.

The precipitate is CuS.

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<h3>Answer:</h3>

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