[sallyhansen]

Can you show us what you did for 5? Did you manage to set up the general rate law equation? Do you know what n and m represent?
Regarding question 7, you need to put in the equilibrium concentrations when solving for K, i.e. K = x/([0.225-x][0.150-x])
I do believe there is a typo in question 11 as the answer has Pb²⁺ while the question has Pb₂⁺ so either the question is wrong or the answer is wrong, I reckon it is the question. There is nothing much you can do about that, perhaps try the balance again with Pb²⁺ and see if you get the right answer

For number 5 I actually don't know what n and m represent

[iamback]

^^Do you remember the general rate law ?

Rate = k[X]^m[y]ⁿ

where is k is rate constant.

Here n is the coefficient of [OH], and m is coefficient of [ClO₂] in rate law.

Can you now solve it on your own ?

[sallyhansen]

^^Do you remember the general rate law ?

Rate = k[X]^m[y]ⁿ

where is k is rate constant.

Here n is the coefficient of [OH], and m is coefficient of [ClO₂] in rate law.

Can you now solve it on your own ?

For a question like this how would I calculate for X
1.6 × 10-48 = 4× 3

The answer is × = 7.4 × 10-17 mol/L but how do I get rid of an exponent with a 3É

[iamback]

You don't have calculate of for X. You are given the values of X and Y here in the question.

X and Y are the concentrations of [ClO₂] and [OH] respectively.

Form equations with the data given and solve.

I am forming one for you.

Rate = k[X]^m[Y]ⁿ

0.18 = k[0.10]^m[0.10]ⁿ

Now make equations like this and solve for m, n and k.

[sallyhansen]

You don't have calculate of for X. You are given the values of X and Y here in the question.

X and Y are the concentrations of [ClO₂] and [OH] respectively.

Form equations with the data given and solve.

I am forming one for you.

Rate = k[X]^m[Y]ⁿ

0.18 = k[0.10]^m[0.10]ⁿ

Now make equations like this and solve for m, n and k.

No the above question was a different question entirely

[iamback]

I solved it, and got the right answer, which you have mentioned in red letters.