7. Average Atomic Mass
8. Atomic Number
9. Isotopes
10.Ions
11. Mass number
When 0.424 moles of an unknown hydrocarbon (∆Hc = -8.21 kJ/mol) is burned in a bomb calorimeter (C = 1.12 kJ/°C), the change in the temperature is 3.10 °C.
The heat of combustion (∆Hc) for an unknown hydrocarbon is -8.21 kJ/mol. The heat released by the combustion of 0.424 moles of the hydrocarbon is:

According to the law of conservation of energy, the sum of the heat released by the combustion (Qc) and the heat absorbed by the bomb calorimeter (Qb) is zero.

Given the heat absorbed by the bomb calorimeter (Qb) and the heat capacity of the bomb calorimeter (C), we can calculate the temperature change (ΔT) using the following expression.

When 0.424 moles of an unknown hydrocarbon (∆Hc = -8.21 kJ/mol) is burned in a bomb calorimeter (C = 1.12 kJ/°C), the change in the temperature is 3.10 °C.
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Answer:
The answer to your question is V2 = 4.97 l
Explanation:
Data
Volume 1 = V1 = 4.40 L Volume 2 =
Temperature 1 = T1 = 19°C Temperature 2 = T2 = 37°C
Pressure 1 = P1 = 783 mmHg Pressure 2 = 735 mmHg
Process
1.- Convert temperature to °K
T1 = 19 + 273 = 292°K
T2 = 37 + 273 = 310°K
2.- Use the combined gas law to solve this problem
P1V1/T1 = P2V2/T2
-Solve for V2
V2 = P1V1T2 / T1P2
-Substitution
V2 = (783 x 4.40 x 310) / (292 x 735)
-Simplification
V2 = 1068012 / 214620
-Result
V2 = 4.97 l
potassium is just below the sodium in periodic table in s group !
so the valence electron of sodium and potassium is same and that is 1