The solubility of nitrogen in water at 25 °C= 4.88 x 10⁻⁴ mol/L
<h3>Further explanation</h3>
Given
78% Nitrogen by volume
Required
The solubility of nitrogen in water
Solution
Henry's Law states that the solubility of a gas is proportional to its partial pressure
Can be formulated
S = kH. P.
S = gas solubility, mol / L
kH = Henry constant, mol / L.atm
P = partial gas pressure
In the standard 25 C state, the air pressure is considered to be 1 atm, so the partial pressure of N₂ -nitrogen becomes:
Vn / Vtot = Pn / Ptot
78/100 = Pn / 1
Pn = 0.78 atm
Henry constant for N₂ at 25 °c = 1600 atm/mol.L=6.25.10⁻⁴ mol/L.atm
The solubility :

<u>Answer:</u>
<em>0.264 g of
can be formed from 288 mg of
</em>
<u>Explanation:</u>
The balanced chemical equation is

The conversions are
Mass in mg
is converted to mass in g
Mass in g
is converted to moles
by dividing with molar mass
Moles
is converted to moles
by using the mole ratio of
is 9 : 6
Moles
is converted to mass
by multiplying with molar mass 
mass in mg
> mass in g
>moles
> moles
> mass 

=0.264g (Answer)
Answer:
4.43 g
Explanation:
The reaction between sodium chloride and flourine gas is given as;
NaCl + F2 --> NaF + Cl2
From the stochiometry of the equation;
1 mol of NaCl reacts eith 1 mol of F2 to form 1 mol of NaF and Cl2
Mass of 1 mol of F2 = 38g
Mass of 1 mol of sodium flouride, NaF = 42g
This means 38g of flourine reacted with NaCl to form 42g of NaF
xg of F2 would form 4.9g of NaF
38 = 42
x = 4.9
x = 4.9 * 38 / 42
x = 4.43 g
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