155 gas a5 points) A 5.00 g sample of water Vapor, initially at 155°C is cooled...

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Answer 1

Answer #1

0°C AH 155°C H20 H20 fusion AH heat AH heat ΔΗ condensation 1000C H20 H20

ΔH of vapor cooler to 155 ^oC to 100 ^oC = 1.84 J/g ^oC x 5g x(100 -155) = -506 J

ΔH of Condensation = -2260 J/g x 5g = -11300 J

ΔH of water cooled 100 ^oC to ^oC = 4.18 J/g ^oC x 5g x (0-100) = -2090 J

ΔH of fusion = -336 J/g x 5g = -1680J

ΔH of ice cooled 0 ^oC to -55 ^oC = 2.09 J/g ^oC x 5g x (-55 -0) = -574.75J

Sum of ΔH = (-506 J) + (-11300 J) + (-2090 J) + (-1680 J) + (-574.75 J) = -16150.75 J = -16.1 kJ

Answer 2

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