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 .
Answer:-
2328.454 grams
Explanation:-
Volume V = 18.4 litres
Temperature T = 15 C + 273 = 288 K
Pressure P = 1.5 x 10^ 3 KPa
We know universal Gas constant R = 8.314 L KPa K-1 mol-1
Using the relation PV = nRT
Number of moles of oxygen gas n = PV / RT
Plugging in the values
n = (1.5 x 10^3 KPa ) x ( 18.4 litres ) / ( 8.314 L KPa K-1 mol-1 x 288 K)
n = 11.527 mol
Now the balanced chemical equation for this reaction is
2KNO3 --> 2KNO2 + O2
From the equation we can see that
1 mol of O2 is produced from 2 mol of KNO3.
∴ 11.527 mol of O2 is produced from 2 x 11.527 mol of KNO3.
= 23.054 mol of KNO3
Molar mass of KNO3 = 39 x 1 + 14 x 1 + 16 x 3 = 101 grams / mol
Mass of KNO3 = 23.054 mol x 101 gram / mol
= 2328.454 grams
Answer:
The critical temperature of a substance is the temperature at and above which vapour of the substance cannot be liquefied, no matter how much pressure is applied.
Average atomic mass is the weighted average atomic masses with regard to the relative abundance of the isotopes
average atomic mass of Li = relative abundance of Li-6 x mass of Li-6 + relative abundance of Li-7 x mass of Li-7
average atomic mass of Li = (7.42% x 6.0151 a.m.u) + (92.58% x 7.0160 a.m.u)
= 0.446 + 6.495
= 6.941 amu
average atomic mass of Li is 6.941 amu
