<span>3 elements
Nitrogen
Hydrogen
Oxygen
2 nitrogen 4 hydrogen and 3 oxygens
there are only 3 different elements</span>
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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<h3><u>Answer;</u></h3>
increase the partial pressure of CO2
<h3><u>Explanation;</u></h3>
2CO(g) + O2(g) -------> 2CO2(g) + heat
- Removel of Oxygen makes the equilibrium to move towards the left side (reactants) because the reactants concentration is decreasing. According to the LeChâtelier’s principle equilibrium moves towards the lower concentration.
- When the partial pressure of any of the gaseous reactants or of the products is increased, the position of equilibrium is shifted so as to decrease its partial pressure. In this case, the removal of oxygen causes a shift to to the left which causes an increase in the partial pressure of CO2.