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Darina [25.2K]
3 years ago
5

A metal sample weighing 129.00 grams and at a temperature of 97.8 degrees Celsius was placed in 45.00 grams of water in a calori

meter at 20.4 degrees Celsius. At equilibrium the temperature of the water and metal was 39.6 degrees Celsius. Calculate the specific heat of the metal. The specific heat of the water is 4.184 J/g/C
Chemistry
1 answer:
Nitella [24]3 years ago
4 0

Answer : The specific heat of metal is 0.481J/g^oC.

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of metal = ?

c_2 = specific heat of water = 4.184J/g^oC

m_1 = mass of metal = 129.00 g

m_2 = mass of water = 45.00 g

T_f = final temperature = 39.6^oC

T_1 = initial temperature of metal = 97.8^oC

T_2 = initial temperature of water = 20.4^oC

Now put all the given values in the above formula, we get

129.00g\times c_1\times (39.6-97.8)^oC=-45.00g\times 4.184J/g^oC\times (39.6-20.4)^oC

c_1=0.481J/g^oC

Therefore, the specific heat of metal is 0.481J/g^oC.

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1-propanol (P1° = 20.9 Torr at 25 °C) and 2-propanol (P2° = 45.2 Torr at 25 °C) form ideal solutions in all proportions. Let x1
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Answer:

y1 = 0.3162

y2 = 0.6838

Explanation:

ok let us begin,

first we would be defining the parameters;

at 25°C;

1-propanol P1° = 20.90 Torr

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From Raoults law:

P(1-propanol) = P⁰ × X(1-propanol)

P(1-propanol) = 20.9 torr × 0.45 = 9.405

P(1-propanol) = 9.405 torr

Also P(2-propanol) = P⁰ × X(2-propanol)

P(2-propanol) = 45.2 torr × 0.45

P(2-propanol) = 20.34 torr

but the total pressure = sum of individual pressures

total pressure = 9.405 + 20.34

total pressure = 29.745 torr

given that y1 and y2 represent the mole fraction of each in the vapor phase

y1 = P1 / total pressure

y1 = 9.405/29.745

y1 = 0.3162

Since y1 + y2 = 1

y2 = 1 - y1

∴ y2 = 1 -  0.3162

y2 = 0.6838

cheers, i hope this helps.

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g Consider a uniport system where a carrier protein transports an uncharged substance A across a cell membrane. Suppose that at
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Answer :  The ratio of the concentration of substance A inside the cell to the concentration outside is, 296.2

Explanation :

The relation between the equilibrium constant and standard Gibbs free energy is:

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-14.1\times 10^3J/mol =-(8.314J/K.mol)\times (298K)\times \ln (\frac{[A]_{inside}}{[A]_{outside}})

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