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 .
Elements<span>simplest form of matter that can exist under normal laaboratory conditions</span>
Q=m(c∆t +heat of fusion + heat of evaporation)
m= 44g
c= 4.186 J/g.C
∆t= 107-(-8) =115 C
heat of fusion= 333.55 J/g
heat of evaporation=2260 J/g
Q=44(4.186*115 + 333.55 + 2260)
Q= 135297.36 J
Explanation:
Hydrogen does not obey the octet rule. Boron does not always
obey the octet rule and in fact forms Lewis acids such as BF3 which
only has 6 electrons.
Answer:
∇T = 51.68°C
Explanation:
Mass = 150g
Heat Energy (Q) = 1.0*10³J
Change in temperature ∇T = ?
Q = mc∇T
Q = heat energy
M = mass
C = specific heat capacity of the gold = 0.129j/g°C
∇T = change in temperature
Q = Mc∇T
1.0*10³ = 150 * 0.129 * ∇T
1000 = 19.35∇T
Solve for ∇T
∇T = 1000 / 19.35
∇T = 51.679°C = 51.68°C
The change in temperature of gold was 51.68°C
