Answer:
Heat transfer = 3564 Jolues
The same value
Explanation:
The heat of combustion is the heat released per 1 mole of substunce experimenting the combustion at standard conditions of pressure and temperature ( 1 atm, 298 K):
Qtransfer = - mol x ΔHºc Qtransfer
So look up in appropiate reference table ΔHºc and solve the problem:
ΔHºc = - 891 kJ/mol
Qtransfer = - (4 x 10³ mol x -891 kJ/mol ) = 3564 J
if the combustion were achieved with 100 % excess air, the result will still be the same. As long as the standard conditions are maintained, the heat of combustion remains constant. In fact in many cases the combustion is performed under excess oxygen to ensure complete combustion.
2 AL and 6 HCL AND 2 ALCL3 and 3 H2
Answer:
Lithium oxide, Li₂O.
Explanation:
Hello!
In this case, according to the given amounts, it is possible to write down the chemical reaction as shown below:
Which means that the metallic oxide has the following formula: M₂O. Next, we can set up the following proportional factors according to the chemical reaction:
Thus, we perform the operations in order to obtain:
So we solve for x as shown below:
Whose molar mass corresponds to lithium, and therefore, the metallic oxide is lithium oxide, Li₂O.
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