The given question is incomplete. The complete question is as follows.
Citrate synthase catalyzes the reaction
Oxaloacetate + acetyl-CoA 
citrate + HS-CoA
The standard free energy change for the reaction is -31.5 kJ*mol^-1
(
a) Calculate the equilibrium constant for this reaction a 37degrees C
Explanation:
(a). It is known that
, relation between change in free energy (
) of a reaction and equilibrium constant (K) is as follows.
where, T = temperature in Kelvin
The given data is as follows.
T = 310 K, 
(as 1 kJ = 1000 J)
Now, putting the given values into the above formula as follows.
ln K = 
= 
ln K = 12.22
K = antilog (12.22)
= 
Therefore, we can conclude that value of equilibrium constant for the given reaction is .
Statement 1 is false because heat is the movement of molecules.
Statement 2 is true because matter is composed of elementary particles which are VERY small.
Statement 3 is false because the kinetic energy of a particle determines its heat and it is impossible for a particle to be completely motionless.
Statement 4 is false because neutrons do not have a charge and make up a large portion of all matter.
Answer:
The molar solubility of lead bromide at 298K is 0.010 mol/L.
Explanation:
In order to solve this problem, we need to use the Nernst Equaiton:
![E = E^{o} - \frac{0.0591}{n} log\frac{[ox]}{[red]}](https://tex.z-dn.net/?f=E%20%3D%20E%5E%7Bo%7D%20-%20%5Cfrac%7B0.0591%7D%7Bn%7D%20log%5Cfrac%7B%5Box%5D%7D%7B%5Bred%5D%7D)
E is the cell potential at a certain instant, E⁰ is the cell potential, n is the number of electrons involved in the redox reaction, [ox] is the concentration of the oxidated specie and [red] is the concentration of the reduced specie.
At equilibrium, E = 0, therefore:
![E^{o} = \frac{0.0591}{n} log \frac{[ox]}{[red]} \\\\log \frac{[ox]}{[red]} = \frac{nE^{o} }{0.0591} \\\\log[red] = log[ox] - \frac{nE^{o} }{0.0591}\\\\[red] = 10^{ log[ox] - \frac{nE^{o} }{0.0591}} \\\\[red] = 10^{ log0.733 - \frac{2x5.45x10^{-2} }{0.0591}}\\\\](https://tex.z-dn.net/?f=E%5E%7Bo%7D%20%20%3D%20%5Cfrac%7B0.0591%7D%7Bn%7D%20log%20%5Cfrac%7B%5Box%5D%7D%7B%5Bred%5D%7D%20%5C%5C%5C%5Clog%20%5Cfrac%7B%5Box%5D%7D%7B%5Bred%5D%7D%20%3D%20%5Cfrac%7BnE%5E%7Bo%7D%20%7D%7B0.0591%7D%20%5C%5C%5C%5Clog%5Bred%5D%20%3D%20%20log%5Box%5D%20-%20%20%5Cfrac%7BnE%5E%7Bo%7D%20%7D%7B0.0591%7D%5C%5C%5C%5C%5Bred%5D%20%3D%2010%5E%7B%20log%5Box%5D%20-%20%20%5Cfrac%7BnE%5E%7Bo%7D%20%7D%7B0.0591%7D%7D%20%5C%5C%5C%5C%5Bred%5D%20%3D%2010%5E%7B%20log0.733%20-%20%20%5Cfrac%7B2x5.45x10%5E%7B-2%7D%20%20%7D%7B0.0591%7D%7D%5C%5C%5C%5C)
[red] = 0.010 M
The reduction will happen in the anode, therefore, the concentration of the reduced specie is equivalent to the molar solubility of lead bromide.
That specific kind is called gas.
When you put water in a freezing temprature it becomes solid.
When you boil water it becomes gas
Answer:
The molality of the glycerol solution is 2.960×10^-2 mol/kg
Explanation:
Number of moles of glycerol = Molarity × volume of solution = 2.950×10^-2 M × 1 L = 2.950×10^-2 moles
Mass of water = density × volume = 0.9982 g/mL × 998.7 mL = 996.90 g = 996.90/1000 = 0.9969 kg
Molality = number of moles of glycerol/mass of water in kg = 2.950×10^-2/0.9969 = 2.960×10^-2 mol/kg
