Answer:
15.0 mL
Explanation:
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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>
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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:

Hydrogen is grouped with the alkali metals because they are grouped based on their properties and behavior, and like the other elements in Group 1 (1A), it only has one valence.
When an electron in a quantum system drops from a higher energy level to a lower one, the system<u> emit a photon.</u>
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The energy of the electron drops when it transitions levels, as well as the atom releases photons. The emission of the photon occurs as the electron transitions from an energy state to a lower state. The photon energy represents precisely the energy that would be lost when an electron moves to a level with less energy.
When such an excited electron transitions from one energy level to another, this could emit a photon. The energy drop would be equivalent to the power of the photon that is released. In electron volts, the energy of an electron, as well as its associated photon (emitted or absorbed) has been stated.
Therefore, when an electron in a quantum system drops from a higher energy level to a lower one, the system<u> emit a photon.</u>
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Answer:
C₆H₈O₆
Explanation:
First off, the<u> percent of oxygen by mass</u> of vitamin C is:
- 100 - (40.9+4.58) = 54.52 %
<em>Assume we have one mol of vitamin C</em>. Then we would have <em>180 grams</em>, of which:
- 180 * 40.9/100 = 73.62 grams are of Carbon
- 180 * 4.58/100 = 8.224 grams are of Hydrogen
- 180 * 54.52/100 = 98.136 grams are of Oxygen
Now we <u>convert each of those masses to moles</u>, using the <em>elements' respective atomic mass</em>:
- C ⇒ 73.62 g ÷ 12 g/mol = 6.135 mol C ≅ 6 mol C
- H ⇒ 8.224 g ÷ 1 g/mol = 8.224 mol H ≅ 8 mol H
- O ⇒ 98.136 g ÷ 16 g/mol = 6.134 mol O ≅ 6 mol O
So the molecular formula for vitamin C is C₆H₈O₆