Answer :
= -1622.8 J/K
= -94.6 J/K
= 0 J/K
Explanation : Given,
= -483.6 kJ
The given chemical reaction is:
First we have to calculate the value of .
Entropy of system = -1622.8 J/K
As we know that:
Entropy of system = -Entropy of surrounding = 1622.8 J/K
and,
Entropy of universe = Entropy of system + Entropy of surrounding
Entropy of universe = -1622.8 J/K + (1622.8 J/K)
Entropy of universe = 0
The immediate product of neutron absorption by Ag-107 is silver atom with a mass of 108, Ag-108.
<h3>What is radioactivity?</h3>
Radioactivity is the spontaneous emission of radiation as wall as particles and energy l by the nucleus of elements due to their disintegration.
The radioactive particles that are usually emitted include:
- alpha particles
- beta particles and
- neutrons
The neutron has a mass of 1.
When silver isotope having a mass of 107 absorbs a neutron, the silver isotope produces will have a mass of 108, Ag-108.
Learn more about radioactivity at: brainly.com/question/25750315
Answer:
12426torr
Explanation:
The following data were obtained from the question:
n = 0.63 mole
V = 750mL = 750/1000 = 0.75L
T = -35.6°C = -35.6 + 273 = 237.4K
R =0.082atm.L/Kmol
P =?
Using the ideal gas equation PV = nRT, the pressure can be obtained as follows:
PV = nRT
P = nRT/V
P = (0.63 x 0.082 x 237.4)/0.75
P = 16.35atm
Now let us convert this pressure (i.e 16.35atm) to a pressure in torr. This is illustrated below:
1atm = 760torr
16.35atm = 16.35 x 760 = 12426torr
Therefore, the pressure of the gas is 12426torr
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
