All categories

3 years ago

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!

You might be interested in

If 16.00 g of O₂ reacts with 80.00 g NO, how many the excess reactant are left over? (enter only the value, round to whole numbe

pishuonlain [190]

Answer:

Explanation:

We will need a balanced equation with masses, moles, and molar masses of the compounds involved.

1. Gather all the information in one place with molar masses above the formulas and masses below them.

Mᵣ: 30.01 32.00 46.01

2NO + O₂ ⟶ 2NO₂

Mass/g: 80.00 16.00

2. Calculate the moles of each reactant

$\text{moles of NO} = \text{80.00 g NO} \times \dfrac{\text{1 mol NO}}{\text{30.01 g NO}} = \text{2.666 mol NO}\\\\\text{moles of O}_{2} = \text{16.00 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.5000 mol O}_{2}$

3. Calculate the moles of NO₂ we can obtain from each reactant

From NO:

The molar ratio is 2 mol NO₂:2 mol NO

$\text{Moles of NO}_{2} = \text{2.333 mol NO} \times \dfrac{\text{2 mol NO}_{2}}{\text{2 mol NO}} = \text{2.333 mol NO}_{2}$

From O₂:

The molar ratio is 2 mol NO₂:1 mol O₂

$\text{Moles of NO}_{2} = \text{0.5000 mol O}_{2}\times \dfrac{\text{2 mol NO}_{2}}{\text{1 mol Cl}_{2}} = \text{1.000 mol NO}_{2}$

4. Identify the limiting and excess reactants

The limiting reactant is O₂ because it gives the smaller amount of NO₂.

The excess reactant is NO.

5. Mass of excess reactant

(a) Moles of NO reacted

The molar ratio is 2 mol NO:1 mol O₂

$\text{Moles reacted} = \text{0.500 mol O}_{2} \times \dfrac{\text{2 mol NO}}{\text{1 mol O}_{2}} = \text{1.000 mol NO}$

(b) Mass of NO reacted

$\text{Mass reacted} = \text{1.000 mol NO} \times \dfrac{\text{30.01 g NO}}{\text{1 mol NO}} = \text{30.01 g NO}$

(c) Mass of NO remaining

Mass remaining = original mass – mass reacted = (80.00 - 30.01) g = 50 g NO

5 0

3 years ago

Select the correct answer.

alexandr402 [8]

Option D
Step by step explanation:

8 0

4 years ago

Read 2 more answers

Melamine, C3N3(NH2)3, is used in adhesives and resins. It is manufactured in a two-step process: CO(NH2)2(l) → HNCO(l) + NH3(g)

svetlana [45]

Answer:

43.13Kg of melamine

Explanation:

The problem gives you the mass of urea and two balanced equations: $CO(NH_{2})_{2}_{(l)}=HNCO_{(l)}+NH_{3}_{(g)}$

$6HNO_{l}=C_{3}N_{3}(NH_{2})_{3}_{(l)}+3CO_{2}_{(g)}$

First we need to calculate the number of moles of urea that are used in the reaction, so:

molar mass of urea = $60.06\frac{g}{mol}*\frac{1kg}{1000g}=0.06006\frac{Kg}{mol}$

The problem says that you have 161.2Kg of urea, so you take that mass of urea and find the moles of urea:

161.2Kg of urea $*\frac{1molofurea}{0.06006Kgofurea}=$ 2684 moles of urea

Now from the stoichiometry you have:

2684 moles of urea $*\frac{1molofHNCO}{1molurea}*\frac{1molofmelamine}{6molesofHNCO}$ = 447 moles of melamine

The molar mass of the melamine is $126.12\frac{g}{mol}$ so we have:

$447molesofmelamine*\frac{126.12g}{1molofmelamine}$ = 5637.64 g of melamine

Converting that mass of melamine to Kg:

5637.64 g of melamine * $\frac{1Kg}{1000g}$ = 56.38 Kg of melamine, that is the theoretical yield of melamine.

Finally we need to calculate the mass of melamine with a yield of 76.5%, so we have:

%yield = 100*(Actual yield of melamine / Theoretical yield of melamine)

Actual yield of melamine = $\frac{76.5}{100}*56.38Kg$ = 43.13Kg of melamine

8 0

3 years ago

In two or more complete sentences describe how atomic radius changes as you move down a group in the periodic table and why.

FrozenT [24]

in two of they got harder

8 0

3 years ago

What is the pH of a solution with a [H+] of 0.0001M?

Sauron [17]

Answer:

Explanation:

The pH of a solution can be found by using the formula

$pH = - log [ {H}^{+} ]$

From the question we have

$ph = - log(0.0001) \\ = 4$

We have the final answer as

Hope this helps you

6 0

3 years ago