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
Answer:
0.45 M
Explanation:
Now we need to make use of the dilution formula
C1V1= C2V2
C1= initial molarity of the solution which is the unknown
V1 = initial volume of the solution which is 175 ml
C2 =final molarity of the solution= 0.315 M
V2= final volume of solution= 250 ml
From C1V1= C2V2 we have;
C1= C2V2/V1
C1= 0.315 × 250/ 175
C1= 0.45 M
B. 1, 1, 1, 2
Explanation:
You only need to balance the NaNO3 on the right. Since there is 2 NO3 on the left, you need to put a 2 in front of the NaNO3 on the right. Everything else is already balanced so the only coefficient needed is 2 in front of the NaNO3.
