

=

= 0.134 mol
mol of oxygen:
ratio of

= 2 : 15
= 1 : 7.5
: . mol of

= 0.134mol * 7.5
= 1.01 mol
Mass of Oxygen = mol * Mr
= 1.01 mol * (16*2) g/mol
= 32.22g
Note: Mr is molar mass
Answer:
11%
Explanation:
1) Calculate van 't Hoff factor:
Δt = i Kf m
0.31 = i (1.86) (0.15)
i = 1.111
2) Calculate value for [H+]:
CCl3COOH ⇌ H+ + CCl3COO¯
total concentration of all ions in solution equals:
(1.11) (0.15) = 0.1665 m
This is a molality, but we will act as if it a molarity since we will assume the density of the solution is 1.00 g/cm3, which makes the molarity equal to the molality.
0.1665 = (0.15 − x) + x + x
x = 0.0165 M
3) Calculate the percent dissociation:
0.0165/ 0.15 = 11 %
The answer is Br₂ since catalysts are regenerated at the end of a reaction.
Answer:
The molecular formula of the given compound is C6H12O6, which is also called Glucose.
Explanation:
The molar mass of glucose is 180.18.