Answer:
Q = 1.68 Kcal
Explanation:
∴ m H2O = 12.0 g
∴ T1 = 35°C ≅ 308 K
∴ T2 = 175°C ≅ 448 K
∴ ΔT = T2 - T1 = 175 - 35 = 140 °C
∴ Cp H2O = 1 cal /g.°C
⇒ Q = (12.0 g)(1 cal/g°C)(140 °C)
⇒ Q = (1680 cal)*(Kcal/1000 cal)
⇒ Q = 1.68 Kcal
Answer:
47.47 g of CO₂ is the amount formed.
Explanation:
The reaction is:
2C₆H₅COOH(aq) + 15O₂(g) → 14CO₂(g) + 6H₂O(l)
Let's apply the formula for the percent yield
Percent yield of reaction = (Produced yield/Theoretical yield) . 100
First of all we convert the moles of CO₂ to mass: 1.30 mol . 44 g /1 mol = 57.2 g. So now, we replace:
(Produced yield / 57.2 g ). 100 = 83
Produced yield / 57.2 g = 83 / 100
Produced yield / 57.2 g = 0.83
Produced yield = 0.83 . 57.2g → 47.47 g of CO₂
Answer:
a) Keq = 4.5x10^-6
b) [oxaloacetate] = 9x10^-9 M
c) 23 oxaloacetate molecules
Explanation:
a) In the standard state we have to:
ΔGo = -R*T*ln(Keq) (eq.1)
ΔGo = 30.5 kJ/moles = 30500 J/moles
R = 8.314 J*K^-1*moles^-1
Clearing Keq:
Keq = e^(ΔGo/-R*T) = e^(30500/(-8.314*298)) = 4.5x10^-6
b) Keq = ([oxaloacetate]*[NADH])/([L-malate]*[NAD+])
4.5x10^-6 = ([oxaloacetate]/(0.20*10)
Clearing [oxaloacetate]:
[oxaloacetate] = 9x10^-9 M
c) the radius of the mitochondria is equal to:
r = 10^-5 dm
The volume of the mitochondria is:
V = (4/3)*pi*r^3 = (4/3)*pi*(10^-15)^3 = 4.18x10^-42 L
1 L of mitochondria contains 9x10^-9 M of oxaloacetate
Thus, 4.18x10^-42 L of mitochondria contains:
molecules of oxaloacetate = 4.18x10^-42 * 9x10^-9 * 6.023x10^23 = 2.27x10^-26 = 23 oxaloacetate molecules
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