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Naddika [18.5K]
2 years ago
11

A 50.00 g sample of an unknown metal is heated to 45.00°C. It is then placed in a coffee-cup calorimeter filled with water. The

calorimeter and the water have a combined mass of 250.0 g and an overall specific heat of 1.035 cal/g•°C. The initial temperature of the calorimeter is 10.00°C. The system reaches a final temperature of 11.08°C when the metal is added.

Chemistry
2 answers:
alex41 [277]2 years ago
8 0

Answer:

The three answers are in the attached picture.

Explanation:

True [87]2 years ago
6 0
First, in order to calculate the specific heat capacity of the metal in help in identifying it, we must find the heat absorbed by the calorimeter using:
Energy = mass * specific heat capacity * change in temperature
Q = 250 * 1.035 * (11.08 - 10)
Q = 279.45 cal/g

Next, we use the same formula for the metal as the heat absorbed by the calorimeter is equal to the heal released by the metal.

-279.45 = 50 * c * (11.08 - 45) [minus sign added as energy released]
c = 0.165

The specific heat capacity of the metal is 0.165 cal/gC
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Converting torr into atm as follows.

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PV = nRT\\0.95 atm \times 7.62 L = n \times 0.0821 L atm/mol K \times 360.5 K\\n = \frac{0.95 atm \times 7.62 L}{0.0821 L atm/mol K \times 360.5 K}\\= \frac{7.239}{29.59705}\\= 0.244 mol

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