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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
The ratio which denotes the relation between the two different units is said to conversion factors.
According to conversion factors, the relation between calorie and kilojoule is given as, one Calorie is equal to 4.184 kJ.
1 Cal = 4.184 kJ
Now, according to above relation, convert 300 calories into kilojoules.
300 calories =
=
Thus, kilojoules present in soda is equal to