The question is basically asking what is happening to the energy (that is in the form of heat) when it is being absorbed by an object. The energy being absorbed from the heat source is being turned into kinetic energy. This can be explained by temperature change. As you add more heat to an object, the temperature rises. Since temperature is the average kinetic energy of all of the molecules in an object, we can say that as temperature rises so does the kinetic energy of the molecules in the object. Due to the fact that heat is causing the temperature to increase, we can say that the energy from the heat is being turned into kinetic energy.
I hope this helps. Let me know in the comments if anything is unclear.
The question is incomplete, here is the complete question:
The rate of certain reaction is given by the following rate law:
![rate=k[H_2]^2[NH_3]](https://tex.z-dn.net/?f=rate%3Dk%5BH_2%5D%5E2%5BNH_3%5D)
At a certain concentration of ![H_2 and [tex]I_2, the initial rate of reaction is 0.120 M/s. What would the initial rate of the reaction be if the concentration of [tex]H_2 were halved.Answer : The initial rate of the reaction will be, 0.03 M/sExplanation :Rate law expression for the reaction:[tex]rate=k[H_2]^2[NH_3]](https://tex.z-dn.net/?f=H_2%20and%20%5Btex%5DI_2%2C%20the%20initial%20rate%20of%20reaction%20is%200.120%20M%2Fs.%20What%20would%20the%20initial%20rate%20of%20the%20reaction%20be%20if%20the%20concentration%20of%20%5Btex%5DH_2%20were%20halved.%3C%2Fp%3E%3Cp%3E%3Cstrong%3EAnswer%20%3A%20The%20initial%20rate%20of%20the%20reaction%20will%20be%2C%200.03%20M%2Fs%3C%2Fstrong%3E%3C%2Fp%3E%3Cp%3E%3Cstrong%3EExplanation%20%3A%3C%2Fstrong%3E%3C%2Fp%3E%3Cp%3E%3Cstrong%3ERate%20law%20expression%20for%20the%20reaction%3A%3C%2Fstrong%3E%3C%2Fp%3E%3Cp%3E%5Btex%5Drate%3Dk%5BH_2%5D%5E2%5BNH_3%5D)
As we are given that:
Initial rate = 0.120 M/s
Expression for rate law for first observation:
![0.120=k[H_2]^2[NH_3]](https://tex.z-dn.net/?f=0.120%3Dk%5BH_2%5D%5E2%5BNH_3%5D)
....(1)
Expression for rate law for second observation:
![R=k(\frac{[H_2]}{2})^2[NH_3]](https://tex.z-dn.net/?f=R%3Dk%28%5Cfrac%7B%5BH_2%5D%7D%7B2%7D%29%5E2%5BNH_3%5D)
....(2)
Dividing 2 by 1, we get:
![\frac{R}{0.120}=\frac{k(\frac{[H_2]}{2})^2[NH_3]}{k[H_2]^2[NH_3]}](https://tex.z-dn.net/?f=%5Cfrac%7BR%7D%7B0.120%7D%3D%5Cfrac%7Bk%28%5Cfrac%7B%5BH_2%5D%7D%7B2%7D%29%5E2%5BNH_3%5D%7D%7Bk%5BH_2%5D%5E2%5BNH_3%5D%7D)
Therefore, the initial rate of the reaction will be, 0.03 M/s
Answer:
With Br2 - Bromobenzene
With Cl2 - Chlorobenzene
With HNO3- Nitrobenzene
With H2SO4 - Benzenesulphonic acid
With HCOCl - Benzoyl chloride
With 1-chloro-2,2-dimethylpropane - 2,2dimethyl-1-phenyl propane
Explanation:
The common thread joining all these reactions is that they are all electrophillic reactions. They are so called because the attacking agents in each reagent is an electrophile. Electrophiles are species that have electron deficient centers and are known to attack molecules that are high in electron density at regions of high electron density.
The benzene molecule has rich electron density. Any substituents that donates electrons to the ring improves the likelihood that benzene will undergo electrophillic substitution reactions while electron withdrawing substituents decrease the likelihood that benzene will undergo electrophillic substitution reactions.
The names of the compounds formed when benzene undergoes electrophillic reaction with the attacking agents listed in the question are displayed in the answer section.
Answer:
2 CH2 + 3 O2 = 2 CO2 + 2 H2O
Explanation:
This is what I think that you meant by the question listed. When balancing a chemical equation, you want to make sure that there are equal amounts of each element on each side.
Originally, the equation's elements looked like this: 1 C on left & 1 C on right; 2 H on left & 2 H on right; 2 O on left and 3 O on right. Because these are not balanced, you need to add coefficients.
When adding coefficients, you need to make sure that all of the elements stay balanced, not just one that you are trying to fix. I know that some equations are really difficult to balance, and when that is the case, there are equation balancing websites that can help out.
However, what always helps me is making a chart and continuing to keep up with the changes I am making. It is a trial and error process.
