This is an incomplete question, here is a complete question.
The Henry's law constant for oxygen dissolved in water is 4.34 × 10⁹ g/L.Pa at 25⁰C.If the partial pressure of oxygen in air is 0.2 atm, under atmospheric conditions, calculate the molar concentration of oxygen in air-saturated and oxygen saturated water.
Answer : The molar concentration of oxygen is, 
Explanation :
As we know that,
where,
= molar solubility of 
= ?
= partial pressure of = 0.2 atm = 1.97×10⁻⁶ Pa
= Henry's law constant = 4.34 × 10⁹ g/L.Pa
Now put all the given values in the above formula, we get:
Now we have to molar concentration of oxygen.
Molar concentration of oxygen = 
Therefore, the molar concentration of oxygen is,
Answer : The value of 
for the given reaction is, 0.36
Explanation :
Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.
The equilibrium expression for the reaction is determined by multiplying the concentrations of products and divided by the concentrations of the reactants and each concentration is raised to the power that is equal to the coefficient in the balanced reaction.
As we know that the concentrations of pure solids and liquids are constant that is they do not change. Thus, they are not included in the equilibrium expression.
The given equilibrium reaction is,
The expression of will be,
![K_c=\frac{[BrCl]^2}{[Br_2][Cl_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BBrCl%5D%5E2%7D%7B%5BBr_2%5D%5BCl_2%5D%7D)
First we have to calculate the concentration of 
.
Now we have to calculate the value of for the given reaction.
Therefore, the value of for the given reaction is, 0.36
<span>rutherfordium element # 104</span>
Answer:
11.31 g.
Explanation:
Molarity is defined as the no. of moles of a solute per 1.0 L of the solution.
M = (no. of moles of solute)/(V of the solution (L)).
<em>∴ M = (mass/molar mass)of NaCl/(V of the solution (L)).</em>
<em></em>
<em>∴ mass of NaCl remained after evaporation of water = (M)(V of the solution (L))(molar mass)</em> = (0.45 M)(0.43 L)(58.44 g/mol) = <em>11.31 g.</em>
