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madreJ [45]
3 years ago
12

Name the following alkyne. Please help me <3

Chemistry
1 answer:
Alexandra [31]3 years ago
8 0

Answer:

D. 7-methyl-3-octyne

Explanation:

1) Identify the parent chain and name it like an alkane.

• The longest chain of carbons, which consists of the functional group (which is the alkyne group in this case: C≡C).

• There are 8 carbons in the longest chain, so it is called an octane.

2) Now, identify the location of the functional group.

• The location number of the functional group should be as low as possible. Thus although we could count from the right, we start counting from the left since the functional group is closer to the left. From the left, the functional group would be at carbon 8 while from the left, it is on carbon 3.

• Replace 'ane' with 'yne' in octane for the alkyne group.

• This would give us 3-octyne.

3) Lastly, add in the name of the branch.

• Here we have one branch, -CH₃. This is read as methyl.

• Identify the location number of the branch by counting the number of carbons in the same direction as when we counted the location number of the functional group. The methyl branch has a location number of 7.

• Adding the name of the branch before the parent chain, we would arrive at 7-methyl-3-octyne as the IUPAC name of the alkyne.

Further Explanation:

A) This option is incorrect as there are only 8 carbons in the parent chain. Although there are 9 carbons in total, the 9th carbon is taken care of in '7-methyl'.

B) Location number of the functional group should be as low as possible, so start counting the number of carbons from the left!

C) Since the functional group is an alkyne, the word 'octane' should be 'octyne' instead.

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Goshia [24]

Answer:

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4 0
3 years ago
The freezing point of benzene is 5.5°C. What is the freezing point of a solution of 2.60 g of naphthalene (C10H8) in 675 g of be
Mrac [35]

<u>Answer:</u> The freezing point of solution is 5.35°C

<u>Explanation:</u>

The equation used to calculate depression in freezing point follows:

\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}

To calculate the depression in freezing point, we use the equation:

\Delta T_f=iK_fm

Or,

\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

where,

Freezing point of pure solution = 5.5°C

i = Vant hoff factor = 1 (For non-electrolytes)

K_f = molal freezing point elevation constant = 4.90°C/m

m_{solute} = Given mass of solute (naphthalene) = 2.60 g

M_{solute} = Molar mass of solute (naphthalene) = 128.2 g/mol

W_{solvent} = Mass of solvent (benzene) = 675 g

Putting values in above equation, we get:

5.5-\text{Freezing point of solution}=1\times 4.90^oC/m\times \frac{2.60\times 1000}{128.2g/mol\times 675}\\\\\text{Freezing point of solution}=5.35^oC

Hence, the freezing point of solution is 5.35°C

3 0
3 years ago
A 8.96-L sample of gas has a pressure of 1.86 atm and a temperature of 94 °C. The sample is allowed to expand to a volume of 11.
Varvara68 [4.7K]

Answer:

Explanation:

Explanation:

All you have to do here is use the ideal gas law equation, which looks like this

P

V

=

n

R

T

−−−−−−−−−−

Here

P

is the pressure of the gas

V

is the volume it occupies

n

is the number of moles of gas present in the sample

R

is the universal gas constant, equal to

0.0821

atm L

mol K

T

is the absolute temperature of the gas

Rearrange the equation to solve for

T

P

V

=

n

R

T

⇒

T

=

P

V

n

R

Before plugging in your values, make sure that the units given to you match those used in the expression of the universal gas constant.

In this case, the volume is given in liters and the pressure in atmospheres, so you're good to go.

Plug in your values to find

T

=

3.10

atm

⋅

64.51

L

9.69

moles

⋅

0.0821

atm

⋅

L

mol

⋅

K

T

=

251 K

−−−−−−−−−

The answer is rounded to three

8 0
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