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Tomtit [17]
3 years ago
11

Please help! Help me solve problems about naming structures with IUPAC rules

Chemistry
1 answer:
lianna [129]3 years ago
7 0
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!
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elena-s [515]

Answer:

B. 67.6/Hg(200.5)= .337. 10.8/S(32.1)= .336. 21.6/O(16)= 1.35--> .337/.336= 1 .336/.336= 1 1.35/.336= 4. Formula= HgSO4

Explanation:

3 0
3 years ago
Mg(OH)2 + 2HNO3 → Mg(NO3)2 + 2H20
LenaWriter [7]

Answer:

A. 1350

You multiply 18.21HNO3* 1mol MgN2O6 * 148.30MgN2O6

Then divide it by the 2mol HNO3 to get 1350

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3 years ago
A single 250 gram
Gala2k [10]

2.2 x 10^-2

0.055 / 250 = 0.00022 - This would be 2.2 x 10^-4, but the question is asking for percent, not proportion, so multiply by 100% to get the percentage.

0.00022 * 100% = 0.022% = 2.2 * 10^-2

6 0
2 years ago
Write the Henderson-Hasselbalch equation for a propanoic acid solution ( CH3CH2CO2H , pKa=4.874 ) using the symbols HA and A− ,
zysi [14]

Answer:

a) [A⁻]/[HA] = 0.227

b) [A⁻]/[HA] = 0.991

c) [A⁻]/[HA] = 2.667

Explanation:

In the Henderson-Hasselbalch equation, HA stands from an acid an A⁻ stands from its conjugate base, as follows:

  • CH₃CH₂CO₂H = HA
  • CH₃CH₂CO₂⁻ = A⁻

pH = pka + Log [A⁻]/[HA]

pH = 4.874 + Log[CH₃CH₂CO₂⁻]/[CH₃CH₂CO₂H]

  • (a)

4.23 = 4.874 + Log [A⁻]/[HA]

-0.644 = Log [A⁻]/[HA]

10^{-0.644} = [A⁻]/[HA]

0.227 = [A⁻]/[HA]

  • (b)

4.87 = 4.874 + Log [A⁻]/[HA]

-0.004 = Log [A⁻]/[HA]

10^{-0.004} = [A⁻]/[HA]

0.991 = [A⁻]/[HA]

  • (c)

5.30 = 4.874 + Log [A⁻]/[HA]

0.426 = Log [A⁻]/[HA]

10^{0.426} = [A⁻]/[HA]

2.667 = [A⁻]/[HA]

6 0
3 years ago
Solve for b, when c= a/b and a = 5.8 and c= 7.4
vova2212 [387]

Answer:

b= 42.92

Explanation:

5.8 x 7.4 = 42.92

4 0
3 years ago
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