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3 years ago

Which shows an isomer of the molecule below?

Chemistry

1 answer:

blsea [12.9K]3 years ago

3 0

Answer:

Explanation:

Hello,

In this case, the isomer of an organic compound is another organic compound having the same molecular formula but different structural formula, thus, the given compound's molecular formula is C₅H₈ since it is an alkyne due to the triple bond. Next, we analyze each option:

A. C₅H₁₂

B. C₅H₁₀

C. C₅H₁₀

D. C₅H₈

For that reason answer is D. based on the molecular formula as well as due to the presence of the triple bond unsaturation (alkyne as well).

Best regards.

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When Z-4,5-dimethyloct-4-ene is treated with hydrogen chloride, HCl, the result is:________.

Vika [28.1K]

Answer:

The correct option is c

Explanation:

The chemical equation for the reaction of Z-4,5-dimethyloct-4-ene and HCl is shown on the first uploaded image

Now looking at the product we see that there are two who has four different groups attached to them this carbon are known as chiral carbons hence the product formed is a pair of diastereomers

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3 years ago

Describe the day of June 21 at the North Pole in terms of daylight and darkness <br> ( science )

frosja888 [35]

On June 21, as seen from the North pole ...

-- the sun has been up, and it's been light outside,
for the past three months ... ever since March 21 .

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3 years ago

A student reacts 5.0 g of sodium with 10.0 g of chlorine and collect 5.24 g of sodium chloride. What is the percent yield of thi

Ede4ka [16]

Answer: The percent yield of this combination reaction is 41.3 %

Explanation : Given,

Mass of $Na$ = 5.0 g

Mass of $Cl_2$ = 10.0 g

Molar mass of $Na$ = 23 g/mol

Molar mass of $Cl_2$ = 71 g/mol

First we have to calculate the moles of $Na$ and $Cl_2$ .

$\text{Moles of }Na=\frac{\text{Given mass }Na}{\text{Molar mass }Na}$

$\text{Moles of }Na=\frac{5.0g}{23g/mol}=0.217mol$

and,

$\text{Moles of }Cl_2=\frac{\text{Given mass }Cl_2}{\text{Molar mass }Cl_2}$

$\text{Moles of }Cl_2=\frac{10.0g}{71g/mol}=0.141mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation will be:

$2Na+Cl_2\rightarrow 2NaCl$

From the balanced reaction we conclude that

As, 2 mole of $Na$ react with 1 mole of $Cl_2$

So, 0.217 moles of $Na$ react with $\frac{0.217}{2}=0.108$ moles of $Cl_2$

From this we conclude that, $Cl_2$ is an excess reagent because the given moles are greater than the required moles and $Na$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $NaCl$