Oxygen and glucose and energy. <span />
Answer:
See Explanation
Explanation:
Let us consider the first two reactions, the initial concentration of CO was held constant and the concentration of Hbn was doubled.
2.68 * 10^-3/1.34 * 10^-3 = 6.24 * 10^-4/3.12 * 10^-4
2^1 = 2^1
The rate of reaction is first order with respect to Hbn
Let us consider the third and fourth reactions. The concentration of Hbn is held constant and that of CO was tripled.
1.5 * 10^-3/5 * 10^-4 = 1.872 * 10^-3/6.24 * 10^-4
3^1 = 3^1
The reaction is also first order with respect to CO
b) The overall order of reaction is 1 + 1=2
c) The rate equation is;
Rate = k [CO] [Hbn]
d) 3.12 * 10^-4 = k [5 * 10^-4] [1.34 * 10^-3]
k = 3.12 * 10^-4 /[5 * 10^-4] [1.34 * 10^-3]
k = 3.12 * 10^-4/6.7 * 10^-7
k = 4.7 * 10^2 mmol-1 L s-1
e) The reaction occurs in one step because;
1) The rate law agrees with the experimental data.
2) The sum of the order of reaction of each specie in the rate law gives the overall order of reaction.
Since all of those percents add up to 100, you can just directly convert that to grams. So now you can use 2 grams H, 32.7 grams S, and 65.3 grams O. Use that info and convert that to moles for an answer of 2mol H, 1mol S, and 4mol O. In every empirical question you need to divide each quantity of moles by the lowest number. In this case, that number is one, so they stay the same, but it's important to remember that step. You're final chemical formula would be H2SO4 and the answer to your question would be that the subscript for oxygen is 4. Hope this helped!
Brainless answer:
I think it is ionic.. I am not sure
