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
When a covalent bond<span> is present between two </span>atoms<span>, the </span>covalent radius<span> can be determined. When two </span>atoms<span> of the same </span>element<span> are covalently </span>bonded<span>, the </span>radius<span> of each </span>atom<span> will be half the distance between the two nuclei because they equally attract the electrons.</span>
Answer:
2445 L
Explanation:
Given:
Pressure = 1.60 atm
Temperature = 298 K
Volume = ?
n = 160 mol
Using ideal gas equation as:
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 08206 L.atm/K.mol
Applying the equation as:
1.60 atm × V = 160 mol × 0.08206 L.atm/K.mol × 298 K
<u>⇒V = 2445.39 L</u>
Answer to four significant digits, Volume = 2445 L
