Answer:
The half-reaction for the oxidation of the manganese in
to
.:

Explanation:

Let us say the medium in which reaction is taking place is an acidic medium. And balancing in an acidic mediums done as;
Step 1: Balance all the atom beside oxygen and hydrogen atom;

Manganese and carbon are balanced.
Step 2: Balance oxygen atom adding water on the required side:


Step 3: Now balance hydrogen atom by adding hydrogen ion on the required side:

A) mass of solute = 17 g
mass of solvent = 183 g
% = mass of solute / total x 100
% = 17 / ( 17 + 183 ) x 100
% = ( 17 / 200 ) x 100
% = 0.085 x 100
= 8.5 %
________________________________
b) mass of solute = 30.0 g
mass of solvent =300.0 g
% = mass of solute / total x 100
% = 30.0 / ( 30.0 + 300.0 ) x 100
% = (30.0 / 330.0 ) x 100
% = 0.090 x 100
<span>= 9.0 %
</span>__________________________________________
hope this helps!
Explanation:
For the given reaction 
Now, expression for half-life of a second order reaction is as follows.
....... (1)
Second half life of this reaction will be
. So, expression for this will be as follows.
=
...(2)
where
is the final concentration that is,
here and
is the initial concentration.
Hence, putting these values into equation (2) formula as follows.
=
=
...... (3)
Now, dividing equation (3) by equation (1) as follows.
=
= 3
or,
= 3
Thus, we can conclude that one would expect the second half-life of this reaction to be three times the first half-life of this reaction.
CU is the element symbol for Copper.