B. nuclear to thermal to mechanical to electrical
That depends. there are 2 possible answers.
H
C - C = C - H gives a different answer on the right than on the left.
One the left side, the second Carbon is attached to a double bond and has but one hydrogen attached to it.
The Carbon on the right of the double bond has 2
H
C- C = C - H
H
I'm not sure what you should put. It's one of those things that I would repeat my argument and submit it.
Answer:
The correct answer is option D.
Explanation:
Rate of the reaction is a change in the concentration of any one of the reactant or product per unit time.

Rate of the reaction:
![R=-\frac{1}{1}\times \frac{d[NO_2]}{dt}=-\frac{1}{1}\times \frac{d[CO]}{dt}](https://tex.z-dn.net/?f=R%3D-%5Cfrac%7B1%7D%7B1%7D%5Ctimes%20%5Cfrac%7Bd%5BNO_2%5D%7D%7Bdt%7D%3D-%5Cfrac%7B1%7D%7B1%7D%5Ctimes%20%5Cfrac%7Bd%5BCO%5D%7D%7Bdt%7D)
Rate of decrease in nitrogen dioxide concentration is equal to the rate of decrease in carbon monoxide.
Given rate expression of the reaction:
![R = k[NO2]^2[CO]](https://tex.z-dn.net/?f=R%20%3D%20k%5BNO2%5D%5E2%5BCO%5D)
Rate of the reaction on doubling concentration of nitrogen dioxide and carbon monoxide : R'
![R'=k(2\times [NO_2])^2(2\times [CO])=8\times k[NO2]^2[CO]=8R](https://tex.z-dn.net/?f=R%27%3Dk%282%5Ctimes%20%5BNO_2%5D%29%5E2%282%5Ctimes%20%5BCO%5D%29%3D8%5Ctimes%20k%5BNO2%5D%5E2%5BCO%5D%3D8R)
Doubling the concentrations of nitrogen dioxide and carbon monoxide simultaneously will increase the rate of the reaction by a factor of eight.
Hence, none of the given statements are true.
Following the Law of Conservation of Mass, you simply add the mass of both substances. Thus, 160 grams + 40 grams = 200 grams. So, even if initially, they are in liquid and solid form, they would still have the same mass even if they change phases, owing to that they are in a closed space.