Answer:
Approximately 
.
Explanation:
Start by finding the concentration of 
at equilibrium. The solubility equilibrium for
.
The ratio between the coefficient of and that of 
is 
. For
Let the increase in 
concentration be 
. The increase in concentration would be 
. Note, that because of the 
of 
, the concentration of
- The concentration of would be
. - The concentration of would be
.
Apply the solubility product expression (again, note that in the equilibrium, the coefficient of is two) to obtain:
![\begin{aligned}&\rm \left[Ag^{+}\right]^2 \cdot \left[{CO_3}^{2-}\right] = K_{\text{sp}} \\ & \implies (0.057 + x)^2\cdot x = 8.1 \times 10^{-12} \end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%26%5Crm%20%5Cleft%5BAg%5E%7B%2B%7D%5Cright%5D%5E2%20%5Ccdot%20%5Cleft%5B%7BCO_3%7D%5E%7B2-%7D%5Cright%5D%20%3D%20K_%7B%5Ctext%7Bsp%7D%7D%20%5C%5C%20%26%20%5Cimplies%20%280.057%20%2B%20x%29%5E2%5Ccdot%20x%20%3D%208.1%20%5Ctimes%2010%5E%7B-12%7D%20%5Cend%7Baligned%7D)
.
Note, that the solubility product of , 
is considerably small. Therefore, at equilibrium, the concentration of
Apply this approximation to simplify 
:
.
.
Calculate solubility (in grams per liter solution) from the concentration. The concentration of is approximately 
, meaning that there are approximately 
of
.
As a result, the maximum solubility of in this solution would be approximately .
Equations represents the law of conservation of mass is 2H₂O → 2H₂ + O₂
Law of conservation of mass stated that energy neither be created nor destroyed called as law of conservation of mass and as the same no of each atom present in the both side of the reaction this reaction represents the law of conservation of mass so in the equation 2H₂O → 2H₂ + O₂ the energy cannot be created nor destroyed water can form 2 mole of hydrogen and 1 mole of oxygen which form water
Know more about law of conservation of mass
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D.
A more accurate answer would be -50 m/s. 50 m/s is the change in speed.
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Answer: d. Remove one-half of the initial CaCO3.
Explanation: Le Chatelier's principle states that changes on the temperature, pressure, concentration and volume of a system will affect the reaction in an observable way. So in the reaction above:
A decrease in temperature will shift the equilibrium to the left because the reaction is exothermic, which means heat is released during the reaction. In other words, when you decrease temperature of a system, the equilibrium is towards the exothermic reaction;
A change in volume or pressure, will result in a production of more or less moles of gas. A increase in volume or in the partial pressure of CO2, the side which produces more moles of gas will be favored. In the equilibrium above, the shift will be to the left.
A change in concentration will tip the equilibrium towards the change: in this system, removing the product will shift the equilibrium towards the production of more CaCO3 to return to the equilibrium.
So, the correct answer is D. Remove one-half of the initial CaCO3.
