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:
.
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
.
Answer : The products of the chemical reaction are,
and 
Explanation :
Balanced chemical reaction : It is defined as the reaction in which the number of atoms of individual elements present on reactant side must be equal to the product side.
When sodium bromide react with silver nitrate then it react to give sodium nitrate and silver bromide as a product.
The balanced chemical reaction will be:

The given reaction is a double-displacement reaction in which the cation of two reactants molecule exchange their places to give two different products.
The species present on the left side of the right arrow is the reactant and the species present on the right side of the right arrow is the product.
In the balanced chemical reaction,
and
are reactants.
and
are products.
Hence, the products of the chemical reaction are,
and 
Answer:
The response can be defined as follows:
Explanation:
ATP is a hydrolysis energy-saving money of the cell. It is used to support the cell's endothermic processes.
![ATP\ +\ H_{2}O\ \rightleftharpoons \ ADP\ +\ P_{i}\\\\Q\ = \ \frac{[ADP][P_{i}]}{[ATP][H_{2}O]}\\\\ADP= 0.250 \ M\\\\P_i = 0.010 \ M\\\\ATP = 0.150 \ M\\\\H_2O = 55.55 \ M\\\\\Delta G \ =\ \Delta G^{o}\ +\ 2.303\ RT\ \log\ Q\\\\R = 8.314 \frac{J}{mol\ K}\\\\T = 298\ K\\\\](https://tex.z-dn.net/?f=ATP%5C%20%2B%5C%20H_%7B2%7DO%5C%20%5Crightleftharpoons%20%5C%20ADP%5C%20%2B%5C%20P_%7Bi%7D%5C%5C%5C%5CQ%5C%20%3D%20%5C%20%5Cfrac%7B%5BADP%5D%5BP_%7Bi%7D%5D%7D%7B%5BATP%5D%5BH_%7B2%7DO%5D%7D%5C%5C%5C%5CADP%3D%200.250%20%5C%20M%5C%5C%5C%5CP_i%20%3D%200.010%20%5C%20M%5C%5C%5C%5CATP%20%3D%200.150%20%5C%20M%5C%5C%5C%5CH_2O%20%3D%2055.55%20%5C%20M%5C%5C%5C%5C%5CDelta%20G%20%5C%20%3D%5C%20%5CDelta%20G%5E%7Bo%7D%5C%20%2B%5C%202.303%5C%20RT%5C%20%5Clog%5C%20Q%5C%5C%5C%5CR%20%3D%208.314%20%5Cfrac%7BJ%7D%7Bmol%5C%20K%7D%5C%5C%5C%5CT%20%3D%20298%5C%20%20K%5C%5C%5C%5C)
Use the relatio P1V1=P2V2 ( p= pressure, v= volume)- assuming number of moles of gas and temperature are kept constant.
1.) A. Larger- pressure increases from 500mmHg to 750mmHg.
2.) B. Smaller
3.) B. 6.0mL