Answer:
The heat of combustion is -25 kJ/g = -2700 kJ/mol.
Explanation:
According to the Law of conservation of energy, the sum of the heat released by the combustion reaction and the heat absorbed by the bomb calorimeter is equal to zero.
Qcomb + Qcal = 0
Qcomb = - Qcal
The heat absorbed by the calorimeter can be calculated with the following expression.
Qcal = C × ΔT
where,
C is the heat capacity of the calorimeter
ΔT is the change in temperature
Then,
Qcomb = - Qcal
Qcomb = - C × ΔT
Qcomb = - 1.56 kJ/°C × 3.2°C = -5.0 kJ
Since this is the heat released when 0.1964 g o quinone burns, the energy of combustion per gram is:
The molar mass of quinone (C₆H₄O₂) is 108 g/mol. Then, the energy of combustion per mole is:
D. is the answer
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Answer:
26,167.47m
Explanation:
Problem:
We are to solve this problem using dimensional analysis:
from the given conversion factors-
0.621mi = 1.00km
1km = 1000m
From the problem, we are groing from miles to metres:
16.25mi x 
x 
The mi will cancel out and also the km. This leaves us with the desired unit we want which is meters, m.
= 26,167.47m
Answer:
A. 28 years
Explanation:
Applying,
R = R'(2ᵃ/ⁿ).............. Equation 1
Where R = Original sample, R' = Sample left after decay, a = Total time taken to decay, n = half life.
From the question,
Given: R = 12 g, R' = 6 g, a = 28 years.
Substitute into equation 1 and solve for n
12 = 6(2²⁸/ⁿ)
12/6 = 2²⁸/ⁿ
2²⁸/ⁿ = 2
Equation the base,
28/n = 1
n = 28 years.
Hence the half-life is 28 years
