How much heat energy is required to melt 337.1 g of HBr? The molar heat of fusion of HBr is 2.41 kJ/mol.

[tex]m=337,1g\\ M_{HBr}=81\frac{g}{mol} \ \ \Rightarrow \ \ n=\frac{m}{M}=\frac{337,1g}{81\frac{g}{mol}}=4,2mol[/tex]

If 2,41kJ --- require ----- 1mol
so
Q ---- require ----- 4,2mol

[tex]Q=\frac{4,2mol*2,41kJ}{1mol}=10,122kJ[/tex]

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ibiscollingwood ibiscollingwood · Answer 2 · 2022-02-24T13:40:59+01:00

The heat energy required is [tex]10.12kJ[/tex].

Heat of fusion :

The molar heat capacity of a chemical substance is the amount of energy that must be added, in the form of heat, to one mole of the substance

Given that, the molar mass of [tex]HBr[/tex] is [tex]337.1g[/tex]

Atomic mass of [tex]HBr[/tex] [tex]=81g/mol[/tex]

Number of mole [tex]=\frac{337.1}{81}=4.2mol[/tex]

The molar heat of fusion of [tex]HBr[/tex] is [tex]2.41 kJ/mol.[/tex]

It means that,

[tex]1[/tex] mol of[tex]HBr[/tex] require [tex]=2.41kJ[/tex] heat energy

[tex]4.2[/tex] mol of [tex]HBr[/tex] require [tex]=2.41*4.2=10.12kJ[/tex] heat energy

Learn more about the heat energy here:

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