Answer:
13.85 kJ/°C
-14.89 kJ/g
Explanation:
<em>At constant volume, the heat of combustion of a particular compound, compound A, is − 3039.0 kJ/mol. When 1.697 g of compound A (molar mass = 101.67 g/mol) is burned in a bomb calorimeter, the temperature of the calorimeter (including its contents) rose by 3.661 °C. What is the heat capacity (calorimeter constant) of the calorimeter? </em>
<em />
The heat of combustion of A is − 3039.0 kJ/mol and its molar mass is 101.67 g/mol. The heat released by the combustion of 1.697g of A is:

According to the law of conservation of energy, the sum of the heat released by the combustion and the heat absorbed by the bomb calorimeter is zero.
Qcomb + Qcal = 0
Qcal = -Qcomb = -(-50.72 kJ) = 50.72 kJ
The heat capacity (C) of the calorimeter can be calculated using the following expression.
Qcal = C . ΔT
where,
ΔT is the change in the temperature
Qcal = C . ΔT
50.72 kJ = C . 3.661 °C
C = 13.85 kJ/°C
<em>Suppose a 3.767 g sample of a second compound, compound B, is combusted in the same calorimeter, and the temperature rises from 23.23°C to 27.28 ∘ C. What is the heat of combustion per gram of compound B?</em>
Qcomb = -Qcal = -C . ΔT = - (13.85 kJ/°C) . (27.28°C - 23.23°C) = -56.09 kJ
The heat of combustion per gram of B is:

Thus we can balance the oxygen atoms by putting a prefix of 25/2 on the left side. To obtain a equation containing whole numbers, we multiply the entire equation by 2. This gives the final equation. 2 C8H18 + 25 O2 ---> 16 CO2 +18 H2O.
The number of atoms of each element :
C : 1 atom
H : 3 atoms
Br = 1 atom
<h3>Further explanation</h3>
Given
Bromomethane-CH₃Br
Required
The number of atoms
Solution
The empirical formula is the smallest comparison of atoms of compound forming elements.
A molecular formula is a formula that shows the number of atomic elements that make up a compound.
The number of atoms in a compound is generally indicated as a subscript after the atom
C : 1 atom
H : 3 atoms
Br = 1 atom
Total 5 atoms
The electron configuration that belongs to the atom with the lowest first ionization energy is francium.
<h3>What is ionization energy? </h3>
Ionization energy is defined as the minimum amount of energy required to remove the most loosely electron present in outermost shell.
<h3>Ionization energy across period</h3>
Ionization energy increase as we move from left to right in the period. This can be explained as when we move from left to right along period new electron is added to the same shell which increase the nuclear charge. Hence results int he decrease in size. Due to this decrease in size more energy is required to remove electron from outermost shell.
<h3>Ionization energy along group</h3>
Ionization energy decrease as we move from top to bottom along group. This can be explained as we move from top to bottom new electron is added to new shell. Due to addition of new shell the size of atom increases which results in the decrease in the nuclear charge. Due to this less amount of energy is needed to remove an electron.
Thus, we concluded that the electron configuration that belongs to the atom with the lowest first ionization energy is francium.
learn more about ionization energy:
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