Can someone tell me if I did right?? Or if you have any ideas on what I should put it would be helpful
1 answer:
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Answer:
Explanation:
NaOH + HNO₃ ⟶ NaNO₃ + H₂O
There are two energy flows in this reaction.
Data:
V(base) = 100.0 mL; c(base) = 0.300 mol·L⁻¹
V(acid) = 100.0 mL; c (acid) = 0.300 mol·L⁻¹
T₁ = 35.00 °C; T₂ = 37.00 °C
Calculations:
(a) q₁
We have equimolar amounts of NaOH and HNO₃
n = 0.0300 mol
q₁ = 0.0300ΔH
(b) q₂
V = 100.0 mL + 100.0 mL = 200.0 mL
m = 200.0 g
ΔT = T₂ - T₁ = 37.00 °C – 35.00 °C = 2.00 °C
q₂ = 200.0 × 4.184 × 2.00 = 1674 J
(c) ΔH
0.0300ΔH + 1674 = 0
0.0300ΔH = -1674
ΔH = -1674/0.0300
ΔH = -55 800 J/mol
ΔH = -55.8 kJ/mol