Answer:
See explanation below
Explanation:
The question is incomplete, however, I found a question very similar to this, and I'm assuming this is the question you are asking to answer. If it's not, please tell me which one it is. Here's a tip for you to get an idea of how to solve.
Picture 1, would be the original question. Picture 2 is the answer of it.
Now, This is a E1 reaction where this type of reactions are taking place in two steps. The first step is the formation of the carbon cation, this step is always slow. The secon step is the addition of a nucleophyle, or, in this case, formation of a pi bond, and we get a alkene.
Hope this can help you
2KClO3 --> 2KCl + 3O2
3 moles of oxygen are produced when 2 mol of potassium chlorate (KClO3) decompose.
Answer:
Your coefficients (the numbers in front of the molecule) will be the following from left to right.
1. <u>1 - 2 - 1 - 2</u>
2. <u>2 - 1 - 2 - 2 - 1</u>
3. <u>2 - 4 - 1</u>
4. <u>2 - 4 - 3</u>
5. <u>2 - 2 - 2 - 1</u>
6. <u>1 - 1 - 1</u>
7. <u>2 - 1 - 2</u>
8. <u>3 - 1 - 2 - 3</u>
9. <u>3 - 1 - 2 - 3</u>
10. <u>2 - 1 - 1 - 1</u>
Explanation:
To balance this equations first count how many times an element is on each side and then see what needs to be changed in order to balance them.
Answer: E=∆H*n= -40.6kj
Explanation:
V(CO) =15L=0.015M³
P=11200Pa
T=85C=358.15K
PV=nRT
n=(112000×0.015)/(8.314×358.15)
n(Co)= 0.564mol
V(Co)= 18.5L = 0.0185m³
P=744torr=98191.84Pa
T= 75C = 388.15k
PV=nRT
n= (99191.84×0.0185)/(8.314×348.15)
n(H2) = 0.634mol
n(CH30H) =1/2n(H2)=1/2×0.634mol
=0.317mol
∆H =∆Hf{CH3OH}-∆Hf(Co)
∆H=-238.6-(-110.5)
∆H = 128.1kj
E=∆H×n=-40.6kj.
