Answer: 0.002 M
Explanation:
The balanced chemical equation for ionization of 
in water is:
According to stoichiometry :
1 mole of ionizes to give =2 mole of 
ions
0.001 mole of ionizes to give = 
mole of ions
Thus of a solution of 0.001 M aqueous sulfuric acid is 0.002 M
Answer:
mass of X extracted from the aqueous solution by 50 cm³ of ethoxy ethane = 3.33 g
Explanation:
The partition coefficient of X between ethoxy ethane (ether) and water, K is given by the formula
K = concentration of X in ether/concentration of X in water
Partition coefficient, K(X) between ethoxy ethane and water = 40
Concentration of X in ether = mass(g)/volume(dm³)
Mass of X in ether = m g
Volume of ether = 50/1000 dm³ = 0.05 dm³
Concentration of X in ether = (m/0.05) g/dm³
Concentration of X in water = mass(g)/volume(dm³)
Mass of X in water left after extraction with ether = (5 - m) g
Volume of water = 1 dm³
Concentration of X in water = (5 - m/1) g/dm³
Using K = concentration of X in ether/concentration of X in water;
40 = (m/0.05)/(5 - m)
(m/0.05) = 40 × (5 - m)
(m/0.05) = 200 - 40m
m = 0.05 × (200 - 40m)
m = 10 - 2m
3m = 10
m = 10/3
m = 3.33 g of X
Therefore, mass of X extracted from the aqueous solution by 50 cm³ of ethoxy ethane = 3.33 g
Answer:
D
Explanation:
31 / 2.8 = 11.0714286 L per mole of helium
3.5 / 4 = 0.875 moles
2.8 + 0.875 = 3.675 moles
11.0714286 x 3.675 = 40.6875 L
