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dolphi86 [110]
2 years ago
12

A sample of aluminum, which has a specific heat capacity of , is put into a calorimeter (see sketch at right) that contains of w

ater. The temperature of the water starts off at . When the temperature of the water stops changing it's . The pressure remains constant at .Calculate the initial temperature of the aluminum sample. Be sure your answer is rounded to significant digits.
Chemistry
1 answer:
Oksanka [162]2 years ago
5 0

Complete Question

A sample of aluminum, which has a specific heat capacity of 0.897 JB loc ! is put into a calorimeter (see sketch at right) that contains 200.0 g of water. The aluminum sample starts off at 85.6 °C and the temperature of the water starts off at 16.0 °C. When the temperature of the water stops changing it's 20.1 °C. The pressure remains constant at 1 atm. Calculate the mass of the aluminum sample.

Answer:

M=58g

Explanation:

From the question we are told that:

Heat Capacity H=0.897

Mass of water M=200g

Initial Temperature of Aluminium T_a=85.6

Initial Temperature of Water T_{w1}=16.0

Final Temperature of Water  T_{w2}=16.0

Generally

Heat loss=Heat Gain

Therefore

M*0.897*(85.6-20.1) =200*4.184*(20.1-16)

M=58g

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Explanation:

Step 1:

The balanced equation for the reaction.

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Step 2:

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Mass of PbCl2 = 0.2393 g

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concentration of Pb^2+, [Pb^2+] = 0.0159 M

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Step 3:

Determination of the number of mole PbCl2.

The number of mole of PbCl2 can be obtained as follow:

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Number of mole =Mass /Molar Mass

Number of mole of PbCl2 = 0.2393/278 = 8.61x10^-4 mole

Step 4:

Determination of Molarity of PbCl2.

At this stage we shall obtain the molarity of PbCl2. This is shown below:

Mole of PbCl2 = 8.61x10^-4 mole

Volume = 50mL = 50/1000 = 0.05L

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Molarity of PbCl2 = 0.01722 M

Step 5:

Determination of the equilibrium constant Kc.

PbCl2( s ) <=> Pb^2+(aq) + 2Cl^−(aq)

The equilibrium constant Kc for the equation above is given by:

Kc = [Pb^2+] [Cl^-]^2 / [PbCl2]

[Pb^2+] = 0.0159 M

[Cl^-] = 0.0318 M

[PbCl2] = 0.01722 M

Kc =?

Kc = [Pb^2+] [Cl^-]^2 / [PbCl2]

Kc = 0.0159 x (0.0318)^2/ 0.01722

Kc = 9.34x10^-4

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