Lets organise the data given in the question
[ClO₂] (m) [OH⁻] (m) initial rate (m/s)
<span>0.060 0.030 0.0248
</span><span> 0.020 0.030 0.00276
</span><span> 0.020 0.090 0.00828
rate equation as follows
rate = k [</span>ClO₂]ᵃ [OH⁻]ᵇ
where k - rate constant
we need to find order with respect to ClO₂ therefore lets take the 2 equations where OH⁻ is constant.
1) 0.00276 = k [0.020]ᵃ[0.030]ᵇ
2) 0.0248 = k [0.060]ᵃ[0.030]ᵇ
divide first equation from the second
0.0248/0.00276 = [0.060/0.020]ᵇ
8.99 = 3ᵇ
8.99 rounded off to 9
9 = 3ᵇ
b = 2
order with respect to ClO₂ is 2
Answer : The number of grams of solute in 500.0 mL of 0.189 M KOH is, 5.292 grams
Solution : Given,
Volume of solution = 500 ml
Molarity of KOH solution = 0.189 M
Molar mass of KOH = 56 g/mole
Formula used :

Now put all the given values in this formula, we get the mass of solute KOH.


Therefore, the number of grams of solute in 500.0 mL of 0.189 M KOH is, 5.292 grams
Answer: 
Explanation:- Volume percentage is the ratio of volume of solute to the volume of solution defined in terms of percentage.

Given: volume of solute = 29.4 ml
Volume of solution= 359 ml

Assuming we have 100 g of sample
30.45/MW of N 14g = 2.175
69.55/MW of O 16g = 4.34
4.34/2.185 = 2
for every 1 mole of N we have 2 moles of O
so the empirical formula would be NO2
without having the molecular weight of the entire molecule the molecular formula can not be determined with the information in your question