<u>Answer:</u> The rate law of the reaction is ![\text{Rate}=k[HgCl_2][C_2O_4^{2-}]^2](https://tex.z-dn.net/?f=%5Ctext%7BRate%7D%3Dk%5BHgCl_2%5D%5BC_2O_4%5E%7B2-%7D%5D%5E2)
<u>Explanation:</u>
Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.
For the given chemical equation:

Rate law expression for the reaction:
![\text{Rate}=k[HgCl_2]^a[C_2O_4^{2-}]^b](https://tex.z-dn.net/?f=%5Ctext%7BRate%7D%3Dk%5BHgCl_2%5D%5Ea%5BC_2O_4%5E%7B2-%7D%5D%5Eb)
where,
a = order with respect to 
b = order with respect to 
Expression for rate law for first observation:
....(1)
Expression for rate law for second observation:
....(2)
Expression for rate law for third observation:
....(3)
Expression for rate law for fourth observation:
....(4)
Dividing 2 from 1, we get:

Dividing 2 from 3, we get:

Thus, the rate law becomes:
![\text{Rate}=k[HgCl_2]^1[C_2O_4^{2-}]^2](https://tex.z-dn.net/?f=%5Ctext%7BRate%7D%3Dk%5BHgCl_2%5D%5E1%5BC_2O_4%5E%7B2-%7D%5D%5E2)
For example we are going to use this unbalanced chemical reaction:
H₂ + O₂ → H₂O.
First, calculate number of atoms (hydrogens and oxygens) on left and right. There is two oxygen and two hydrogen on left and two hydrogen and one oxygen on right.
You can not change molecular formula of compound, only you can put coefficient in fron of compound to balance reaction.
Put 2 in front water to balance oxygen (now you have two oxygens on left and right). But now you have four hydrogens on right, so you must put 2 in fron hydrogen on the left.
2H₂ + O₂ → 2H₂O.
I think it’s either solid or water (if that’s a choice)
A because Energy is absorbed
The answer is b .............................................