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 is: <span>requirement of valine is 0,0222 mol for a100 kilograms adult.
</span>m(C₅H₁₁NO₂) = 2,60 g.
M(C₅H₁₁NO₂) = 5 · 12 + 11 · 1 + 1 · 14 + 2 · 16 · g/mol.
M(C₅H₁₁NO₂) = 117 g/mol.
n(C₅H₁₁NO₂) = m(C₅H₁₁NO₂) ÷ M(C₅H₁₁NO₂).
n(C₅H₁₁NO₂) = 2,60 g ÷ 117 g/mol.
n(C₅H₁₁NO₂) = 0,0222 mol.
n - amount of substance.
m - mass of substance.
M - molar mass of substance.
Answer:
Explained
Explanation:
the various ways of input of CO_2 into the atmosphere.
1.Dissolved CO 2 in the ocean is released back into the atmosphere by heating ocean surface water
2.Plant and animal respiration, an exothermic reaction involving the breakdown into CO 2 and water of organic molecules.
3. Degradation of fungi and bacteria which are responsible for breaking down carbon compounds in dead animals and plants(fossils) and convert carbon into CO2 when oxygen or methane is present.
4.Combustion of organic matter (that includes deforestation and combustion of fossil fuels) oxidizing to produce CO2;
5. Cement production when calcium carbonate (limestone) is heated to produce calcium oxide(lime), cement component, and CO2 are released;
Iron, or Fe, is the only element here. The rest are alloys. I hope this helps!