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marissa [1.9K]
3 years ago
11

A mixture of krypton and argon gas is expanded from a volume of 33.0L to a volume of 61.0L , while the pressure is held constant

at 58.0atm . Calculate the work done on the gas mixture. Be sure your answer has the correct sign (positive or negative) and the correct number of significant digits.
Chemistry
1 answer:
Andrej [43]3 years ago
4 0

Answer:

W = +1624J

Explanation:

V1 = 33.0L

V2 = 61.0L

P = 58 atm

Work = force * distance (∇x)

Work = F.∇x

But force = pressure * area

F = P.A

Work = P.∇x.A

Work = P∇v

Work = P (V₂ - V₁)

Work = 58 * (61 - 33)

Work = 58 * 28

Work = 1624J

The work done is +1624J

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See explanation.

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In this case, according to the described chemical reaction, we first write the corresponding equation to obtain:

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m_{Cu(NO_3)_2}=1.29gAg*\frac{1molAg}{107.87gAg}*\frac{1molCu(NO_3)_2}{2molAg}*\frac{187.55gCu(NO_3)_2}{1molCu(NO_3)_2}   \\\\m_{Cu(NO_3)_2}=1.12gCu(NO_3)_2

Part 2 of 3: here, the molar mass of copper is 63.55 g/mol and the mole ratio of silver to copper is 2:1, the mass of the former that was used to start the reaction was:

m_{Cu}=1.29gAg*\frac{1molAg}{107.87gAg}*\frac{1molCu}{2molAg}*\frac{63.55gCu)_2}{1molCu}   \\\\m_{Cu}=0.380gCu

Part 3 of 3: here, the molar mass of silver nitrate is 169.87 g/mol and their mole ratio 2:2, thus, the mass of initial silver nitrate is:

m_{AgNO_3}=1.29gAg*\frac{1molAg}{107.87gAg}*\frac{2molAgNO_3}{2molAg}*\frac{169.87gAgNO_3}{1molAgNO_3}   \\\\m_{AgNO_3}=2.03gAgNO_3

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