Answer:
The approximate molar enthalpy of combustion of this substance is -66 kJ/mole.
Explanation:
First we have to calculate the heat gained by the calorimeter.
where,
q = Heat gained = ?
c = Specific heat = 
ΔT = The change in temperature = 3.08°C
Now put all the given values in the above formula, we get:
Now we have to calculate molar enthalpy of combustion of this substance :
where,
= enthalpy change = ?
q = heat gained = 8.2544kJ
n = number of moles methane = 
Therefore, the approximate molar enthalpy of combustion of this substance is -66 kJ/mole.
It would be false sulfur has 6
The percent composition of this compound :
Mg = 72.182%
N = 27.818%
<h3>Further explanation</h3>
Given
9.03 g Mg
3.48 g N
Required
The percent composition
Solution
Proust stated the Comparative Law that compounds are formed from elements with the same Mass Comparison so that the compound has a fixed composition of elements
Total mass of the compound :
= 9.03 g + 3.48 g
= 12.51 g
The percent composition :
Mg : 9.03/ 12.51 g x 100% = 72.182%
N : 3.48 / 12.51 g x 100% = 27.818%
Given teh equation adn the heat of reaction, reaction 2's heat of reaction can be obtained by simply multiplying teh heat of reaction of 1 by 3. The final answer is -6129 kJ.
