Answer:
xy (-b+c+q) is the answer to this
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]
B positive and negative ion
First step is to balance the reaction equation. Hence we get
P4 + 5 O2 => 2 P2O5
Second, we calculate the amounts we start with
P4: 112 g = 112 g/ 124 g/mol – 0.903 mol
O2: 112 g = 112 g / 32 g/mol = 3.5 mol
Lastly, we calculate the amount of P2O5 produced.
2.5 mol of O2 will react with 0.7 mol of P2O5 to produce 1.4
mol of P2O5.
This is 1.4 * (31*2 + 16*5) = 198.8 g