Dissolving sugar in water
based on the law of conservation of energy its it atoms hold by strong chemical bonds
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]
<u>Answer:</u> The standard potential of the cell is 0.77 V
<u>Explanation:</u>
We know that:
The substance having highest positive 
reduction potential will always get reduced and will undergo reduction reaction.
The half reaction follows:
<u>Oxidation half reaction:</u> 
<u>Reduction half reaction:</u> 
( × 2)
To calculate the 
of the reaction, we use the equation:
Substance getting oxidized always act as anode and the one getting reduced always act as cathode.
Putting values in above equation follows:
Hence, the standard potential of the cell is 0.77 V
