Answer:
-24.76 kJ/g; -601.8 kJ/mol
Explanation:
There are two heat flows in this experiment.
Heat from reaction + heat absorbed by calorimeter = 0
q1 + q2 = 0
mΔH + CΔT = 0
Data:
m = 0.1375 g
C = 3024 J/°C
ΔT = 1.126 °C
Calculations:
0.1375ΔH + 3024 × 1.126 = 0
0.1375ΔH + 3405 = 0
0.1375ΔH = -3405
ΔH = -24 760 J/g = -24.76 kJ/g
ΔH = -24.76 kJ/g ×24.30 g/mol = -601.8 kJ/mol
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A process that involves rearrangement of the molecular or ionic structure of a substance, as opposed to a change in physical form or a nuclear reaction.
Answer:
ΔTb = 0.66 C
Explanation:
Given
Mass of KBr = 185 g
Mass of water = 1.2 kg
Kb = 0.51 C/m
Explanation:
The change in boiling point (ΔTb) is given by the product of molality (m) of the solution and the boiling point constant (Kb)
[tex]\Delta T_{b}= 0.51 C/m * 1.296 m = 0.66 C[\tex]
