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:
What mass (g) of barium iodide is contained in 188 mL of a barium iodide solution that has an iodide ion concentration of 0.532 M?
A) 19.6
B) 39.1
C) 19,600
D) 39,100
E) 276
The correct answer to the question is
B) 39.1 grams
Explanation:
To solve the question
The molarity ratio is given by
188 ml of 0.532 M solution of iodide.
Therefore we have number of moles = 0.188 × 0.532 M = 0.100016 Moles
To find the mass, we note that the Number of moles = 
from which we have
Mass = Number of moles × molar mass
Where the molar mass of Barium Iodide = 391.136 g/mol
= 0.100016 moles ×391.136 g/mol = 39.12 g
A negative formation enthalpy means that the reaction is exothermic, or that heat is released during the process.
(C,)
Answer:
The mass percentage of calcium carbonated reacted is 2.5%.
Explanation:
The reaction is:
Thus the Kp of the equilibrium will be:
Kp = partial pressure of carbon dioxide [as the other are solid]
Moles of calcium carbonate initially present = 
Let us apply ICE table to the equilibrium given:
Initial 0.2 0 0
Change -x +x +x
Equilibrium 0.2-x x x
Kp = partial pressure of carbon dioxide
Kp = Kc(RT)ⁿ
where n = difference in the number of moles of gaseous products and reactants
for given reaction n = 1
R = gas constant = 8.314 J /mol K
T = temperature = 800 ⁰C = 1073 K
Putting values
Kc =
Kc = ![\frac{[CO_{2}][CaO]}{[CaCO_{3}]}= \frac{x^{2} }{(0.2-x)}=1.3X10^{-4}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BCO_%7B2%7D%5D%5BCaO%5D%7D%7B%5BCaCO_%7B3%7D%5D%7D%3D%20%5Cfrac%7Bx%5E%7B2%7D%20%7D%7B%280.2-x%29%7D%3D1.3X10%5E%7B-4%7D)
On calculating
x = 0.005
where x = the moles of calcium carbonate dissociated or reacted.
Percentage of the moles or mass reacted = 
%
