Not sure if this is what you mean but l= A/w
(length equals area divided by width)
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
Yes I can help you in science
Answer:
Therefore it takes 8.0 mins for it to decrease to 0.085 M
Explanation:
First order reaction: The rate of reaction is proportional to the concentration of reactant of power one is called first order reaction.
A→ product
Let the concentration of A = [A]
![\textrm{rate of reaction}=-\frac{d[A]}{dt} =k[A]](https://tex.z-dn.net/?f=%5Ctextrm%7Brate%20of%20reaction%7D%3D-%5Cfrac%7Bd%5BA%5D%7D%7Bdt%7D%20%3Dk%5BA%5D)
![k=\frac{2.303}{t} log\frac{[A_0]}{[A]}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B2.303%7D%7Bt%7D%20log%5Cfrac%7B%5BA_0%5D%7D%7B%5BA%5D%7D)
[A₀] = initial concentration
[A]= final concentration
t= time
k= rate constant
Half life: Half life is time to reduce the concentration of reactant of its half.
Here 
To find the time takes for it to decrease to 0.085 we use the below equation
![\Rightarrow t=\frac{2.303}{k} log\frac{[A_0]}{[A]}](https://tex.z-dn.net/?f=%5CRightarrow%20t%3D%5Cfrac%7B2.303%7D%7Bk%7D%20log%5Cfrac%7B%5BA_0%5D%7D%7B%5BA%5D%7D)
Here , 
, [A₀] = 0.13 m and [ A] = 0.085 M
Therefore it takes 8.0 mins for it to decrease to 0.085 M
