A. The longest carbon chain is eight, and it has two methyl groups attached to carbon three, and a special group attached to carbon five. Its two names could be:
3-dimethyl-5-(1-methylethyl)octane
3-dimethyl-5-isopropyloctane
Both of these are correct. This is an alkane, because it has all single bonds.
B. This has a triple bond contained between carbons 2 and 3, and has a methyl group off carbon 4. The longest chain is 5. It’s name is:
4-methyl-2-pentyne
This is an alkene, because of the double bond.
C. This has a double bond contained between carbons 2 and 3, and has a methyl off of four and an methyl off of six. The longest chain is eight (follow the longest chain of carbons).
4,6-dimethyl-2-octene
This is an alkene, because of the double bond.
D. This has an ethyl group at 1 and a methyl group at 2 (rotate the compound to make it as clean as possible, in this case, the ring is flipped and rotated to make it alphabetical with the smallest numbers possible). The two names are:
1-ethyl-2-methylbenzene
ortho-ethylmethylbenzene
Both are correct, the ortho prefix telling the location of the ethyl and methyl groups. This is an aromatic structure because of its double bonded ring.
E. The longest chain is nine, and has methyls at three, five, and seven, along with a propyl at five. The name is:
3,5,7-trimethyl-5-propylnonane
This is an alkane, due to the single bonds.
Hope this helps!
<span>The best reason I can think of for why we believe that air is a mixture is that freezing air slowly yields different liquids at different temperatures. Liquid nitrogen has a different boiling point than liquid oxygen. They also freeze at different temperatures. If air were only 1 compound, then air in its entirety would have a single boiling point and a single freezing point. </span>
Answer:
c
Explanation:
yxyyyxyxxyxxy yy y gg ff ff d d f t r rr rr rr rr r r t tt t
Answer:
the value of equilibrium constant for the reaction is 8.5 * 10⁷
Explanation:
Ti(s) + 2 Cl₂(g) ⇄ TiCl₄(l)
equilibrium constant Kc = ![\frac{1}{[Cl_2]^2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BCl_2%5D%5E2%7D)
Given that,
We are given:
Equilibrium amount of titanium = 2.93 g
Equilibrium amount of titanium tetrachloride = 2.02 g
Equilibrium amount of chlorine gas = 1.67 g
We calculate the No of mole = mass / molar mass
mass of chlorine gas = 1.67 g
Molar mass of chlorine gas = 71 g/mol
mole of chlorine = 1.67 / 71
= 7.0L
Concentration of chlorine is = no of mole / volume
= 0.024 / 7
= 3.43 * 10⁻³M
equilibrium constant Kc = ![\frac{1}{[Cl_2]^2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BCl_2%5D%5E2%7D)
= ![\frac{1}{[3.43 * 10^-^3]^2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5B3.43%20%2A%2010%5E-%5E3%5D%5E2%7D)
= 8.5 * 10⁷