The final temperature, t₂ = 30.9 °C
<h3>Further explanation</h3>
Given
24.0 kJ of heat = 24,000 J
Mass of calorimeter = 1.3 kg = 1300 g
Cs = 3.41 J/g°C
t₁= 25.5 °C
Required
The final temperature, t₂
Solution
Q = m.Cs.Δt
Q out (combustion of compound) = Q in (calorimeter)
24,000 = 1300 x 3.41 x (t₂-25.5)
t₂ = 30.9 °C
<span>Answer:
Nothing is balanced in your final equation: not H, not O, not Cr, not I and your charges aren't either.
Start with your 2 half reactions:
I- --> IO3-
Cr2O72- --> 2 Cr3+
Balance O by adding H2O:
I- + 3 H2O --> IO3-
Cr2O72- --> 2 Cr3+ + 7H2O
Balance H by adding H+:
I- + 3 H2O --> IO3- + 6 H+
Cr2O72- + 14 H+ --> 2 Cr3+ + 7H2O
Balance charge by adding e-:
I- + 3 H2O --> IO3- + 6 H+ + 6 e-
Cr2O72- + 14 H+ + 6 e- --> 2 Cr3+ + 7H2O
Since the numbers of electrons in your two half reactions are the same, just add them and simplify to give:
Cr2O72- + I- + 8 H+ --> IO3- + 2 Cr3+ + 4 H2O</span>
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