Answer:solid: still boi liquid: bouncy boi has: crazy boi
Explanation:
<span>1,3-cylohexadiene i synthesized starting from cyclohexane in following 4 steps.
1) Free Radical Substitution Rxn: Halogenation of cyclohexane in the presence of UV yield chlorocyclohexane.
2) Elimination Rxn: Dehydrohalogenation of chlorocyclohexane yields cyclohexene.
3) Halogenation of Cyclohexene (
Electrophillic Addition Rxn) gives 1,2-dihalocyclohexane.
4) Elemination Rxn: When dibromocyclohexane is treated with KOH and heated it gives 1,3-cyclohexadiene as shown below,</span>
Answer:
that's because....
group 1 (e.g Na, K) those tend to lose one electron to gain noble gas electron configuration.
they can achieve that by just losing one electron from their outer shell.
as you go down the group 1, element gets bigger in size, which means there is more space between nucleus (which is in center of atom) and electron of outer shell. the more far away they are the less attraction force between them.
so its easier for potassuim to lose one electron than for lithuim.
so that means potassium will easily give up 1 electron to react with non metal or other element therefore it is more reactive than lithuim
but in case of non metal, the opposite happens but simple to understand.
as you go down the group 7 (halogen- Cl, Br, I) element will get bigger therefore force between nucleus and outer electron is getting smaller. they have to gain 1 electron in order to fill the outer shell (to gain noble gas electron configuration.)
as florine is more smaller in size than clorine it is more reactive because florine has more tendency to pull extra electron from metal or other element towards its side. so it easily gain 1 electron to react.
Answer:

Explanation:
Hello there!
Unfortunately, the question is not given in the question; however, it is possible for us to compute the equilibrium constant as the problem is providing the concentrations at equilibrium. Thus, we first set up the equilibrium expression as products/reactants:
![K=\frac{[NO_2]^2}{[NO]^2[O_2]}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BNO_2%5D%5E2%7D%7B%5BNO%5D%5E2%5BO_2%5D%7D)
Then, we plug in the concentrations at equilibrium to obtain the equilibrium constant as follows:

In addition, we can infer this is a reaction that predominantly tends to the product (NO2) as K>>>>1.
Best regards!
Answer:

Explanation:
Hello,
In this case, we can consider that the given heat of combustion is indeed the heat of reaction since it corresponds to the combustion of propane, which is computed by using the heat formation of all the involved species as shown below:

Thus, since the heat of formation of gaseous carbon dioxide is -393.5 kJ/mol, water -241.8 kJ/mol and oxygen 0 kJ/mol, the heat of formation of propane is:

Best regards.