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viktelen [127]
3 years ago
7

A 52.9g sample of brass, which has a specific heat capacity of 0.375·J·g^−1°C^−1, is put into a calorimeter (see sketch a

t right) that contains 100.0g of water. The temperature of the water starts off at 15.0°C. When the temperature of the water stops changing it's 18.4°C. The pressure remains constant at 1 atm. Calculate the initial temperature of the brass sample.
Chemistry
1 answer:
vaieri [72.5K]3 years ago
7 0

Answer: 90.04°C

Explanation: <u>Calorimeter</u> is a device measures the amount of heat of a chemical or physical process. An ideal calorimeter is one that is well-insulated, i.e., prevent the transfer  of heat between the calorimeter and its surroundings. So, the net heat change inside the calorimeter is zero:

q_{1}+q_{2}=0

Rearraging, it can be written as

q_{1}=-q_{2}

showing that the heat gained by Substance 1 is equal to the energy lost by Substance 2.

In our case, water is gaining heat, because its temperature has risen and so, brass is losing energy:

q_{water}=-q_{brass}

Calculating:

m_{w}.c_{w}.\Delta T=-[m_{b}.c_{b}.\Delta T]

100.4.18.(18.4-15)=-[52.9.0.375.(18.4-T)]

Note: final temperature is the same as the substances are in thermal equilibrium.

Solving:

418(3.4)= - 365.01 + 19.8375T

19.8375T = 1786.21

T = 90.04

The initial temperature for the sample of brass was 90.04°.

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Answer: A reversible reaction is a reaction that takes place in back and front directions. If the reaction were to reach equilibrium, the rate of forward direction would be equal to that of the reverse reaction.

Explanation:

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        A+B\leftrightharpoons C+D

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3 years ago
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STatiana [176]
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8 0
3 years ago
Consider the following reaction:
iren [92.7K]

Answer:

A. ΔG° = 132.5 kJ

B. ΔG° = 13.69 kJ

C. ΔG° = -58.59 kJ

Explanation:

Let's consider the following reaction.

CaCO₃(s) → CaO(s) + CO₂(g)

We can calculate the standard enthalpy of the reaction (ΔH°) using the following expression.

ΔH° = ∑np . ΔH°f(p) - ∑nr . ΔH°f(r)

where,

n: moles

ΔH°f: standard enthalpy of formation

ΔH° = 1 mol × ΔH°f(CaO(s)) + 1 mol × ΔH°f(CO₂(g)) - 1 mol × ΔH°f(CaCO₃(s))

ΔH° = 1 mol × (-635.1 kJ/mol) + 1 mol × (-393.5 kJ/mol) - 1 mol × (-1206.9 kJ/mol)

ΔH° = 178.3 kJ

We can calculate the standard entropy of the reaction (ΔS°) using the following expression.

ΔS° = ∑np . S°p - ∑nr . S°r

where,

S: standard entropy

ΔS° = 1 mol × S°(CaO(s)) + 1 mol × S°(CO₂(g)) - 1 mol × S°(CaCO₃(s))

ΔS° = 1 mol × (39.75 J/K.mol) + 1 mol × (213.74 J/K.mol) - 1 mol × (92.9 J/K.mol)

ΔS° = 160.6 J/K. = 0.1606 kJ/K.

We can calculate the standard Gibbs free energy of the reaction (ΔG°) using the following expression.

ΔG° = ΔH° - T.ΔS°

where,

T: absolute temperature

<h3>A. 285 K</h3>

ΔG° = ΔH° - T.ΔS°

ΔG° = 178.3 kJ - 285K × 0.1606 kJ/K = 132.5 kJ

<h3>B. 1025 K</h3>

ΔG° = ΔH° - T.ΔS°

ΔG° = 178.3 kJ - 1025K × 0.1606 kJ/K = 13.69 kJ

<h3>C. 1475 K</h3>

ΔG° = ΔH° - T.ΔS°

ΔG° = 178.3 kJ - 1475K × 0.1606 kJ/K = -58.59 kJ

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Answer:

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Explanation:

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5 0
3 years ago
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Answer:

Explanation:

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