Mol sulfuric acid = 19 g * (1 mol) / (98.1 g) = 0.19367 mol
mol H2O = 0.19367 mol H2SO4 * (2 H2O) / (1 H2SO4)
= 0.387359 mol H2O
grams H2O = 0.387359 mol H2O * (18 g)/(1 mol)
= 6.97 g
The answer is 7.0 grams of water
Answer:
Rate = k [OCl] [I]
Explanation:
OCI+r → or +CI
Experiment [OCI] M I(-M) Rate (M/s)2
1 3.48 x 10-3 5.05 x 10-3 1.34 x 10-3
2 3.48 x 10-3 1.01 x 10-2 2.68 x 10-3
3 6.97 x 10-3 5.05 x 10-3 2.68 x 10-3
4 6.97 x 10-3 1.01 x 10-2 5.36 x 10-3
The table above able shows how the rate of the reaction is affected by changes in concentrations of the reactants.
In experiments 1 and 3, the conc of iodine is constant, however the rate is doubled and so is the conc of OCl. This means that the reaction is in first order with OCl.
In experiments 3 and 4, the conc of OCl is constant, however the rate is doubled and so is the conc of lodine. This means that the reaction is in first order with I.
The rate law is given as;
Rate = k [OCl] [I]
C) Nucleus...... Hope it helps, Have a nice day :)
