A 4.0 g sample of a solid changes from 20.0 ºC to 36.0 ºC upon addition of 25.0 cal of heat. What is the specific heat of this s
1 answer:
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Explanation:
The reaction equation will be as follows.
So, according to this equation, 1 mole = 2 mol HBr = 1 mol
Therefore, calculate the number of moles of calcium hydroxide as follows.
No. of moles of =
=
= 30 mmol
Similarly, calculate the number of moles of HBr as follows.
No. of moles of HBr =
=
= 30 mmol
This means that the limiting reactant is HBr.
So, no. of moles of =
= 15 mmol
Hence, calculate the amount of heat released as follows.
Heat released in the reaction(q) =
as, m = mass of solution
and, Density =
or, mass = Density × Volume
= 1.08 g/ml \times (50 + 50) ml
= 108 g
where, s = specific heat of solution = 4.18 j/g.k
and, change in temperature =
=
Hence, the heat released will be as follows.
q =
q =
= 1354.32 joule
or, = 1.354 kJ (as 1 kJ = 1000 J)
Also, =
=
= -90.267 kJ/mol
Thus, we can conclude that the enthalpy change for the given reaction is -90.267 kJ/mol.