Answer:
The new molar concentration of CO at equilibrium will be :[CO]=1.16 M.
Explanation:
Equilibrium concentration of all reactant and product:
![[CO_2] = 0.24 M, [H_2] = 0.24 M, [H_2O] = 0.48 M, [CO] = 0.48 M](https://tex.z-dn.net/?f=%5BCO_2%5D%20%3D%200.24%20M%2C%20%5BH_2%5D%20%3D%200.24%20M%2C%20%5BH_2O%5D%20%3D%200.48%20M%2C%20%5BCO%5D%20%3D%200.48%20M)
Equilibrium constant of the reaction :
![K=\frac{[H_2O][CO]}{[CO_2][H_2]}=\frac{0.48 M\times 0.48 M}{0.24 M\times 0.24 M}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BH_2O%5D%5BCO%5D%7D%7B%5BCO_2%5D%5BH_2%5D%7D%3D%5Cfrac%7B0.48%20M%5Ctimes%200.48%20M%7D%7B0.24%20M%5Ctimes%200.24%20M%7D)
K = 4

Concentration at eq'm:
0.24 M 0.24 M 0.48 M 0.48 M
After addition of 0.34 moles per liter of
and
are added.
(0.24+0.34) M (0.24+0.34) M (0.48+x)M (0.48+x)M
Equilibrium constant of the reaction after addition of more carbon dioxide and water:


Solving for x: x = 0.68
The new molar concentration of CO at equilibrium will be:
[CO]= (0.48+x)M = (0.48+0.68 )M = 1.16 M
Answer:
The answer to be filled in the respective blanks in question is
3 and 1
Explanation:
So, we know that the formation of cabon-dioxide mole and that of Adenosin-Tri-Phosphate (ATP) moles will be in the ratio of 3:1 i.e., three carbon-di-oxide moles and 1 ATP mole.
Therefore, we can say that one pyruvate mole when passed through citric acid cycle and pyruvate dehydrogenase yields carbon-di-oxide and ATP moles in the ratio 3:1
I’m pretty sure it is A at least that’s what we did at our school to test this
In this kind of exercises, you should use the "ideal gas" rules: PV = nRT
P should be in Pascal:
445mmHg = 59328Pa
1225mmHg = 163319Pa
V should be in cubic meter:
16L = 0.016 m3
R =

= constant

=

==> P1 * V1 = P2 * V2
V2 =

=
V2 = 0.00581 m3 = 5.81 L
3.84 - 1.43 = 2.41
2.41g of table sugar
% mass = ( (mass of element) / (total mass) ) * 100
% mass = (2.41 / 3.84) * 100
% mass = (0.6276) * 100
% mass = 62.76
62.76%