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Elenna [48]
3 years ago
7

Which shows an isomer of the molecule below?

Chemistry
1 answer:
blsea [12.9K]3 years ago
3 0

Answer:

D.

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

6 0
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 . 

-- The sun won't set, and it won't be dark outside,
for another three months ... until September 21.

-- Here at the North pole, it stays daylight for six months straight.
Today, on June 21, we're exactly halfway through the period of
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7 0
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

From the reaction, we conclude that

As, 2 mole of Na react to give 2 mole of NaCl

So, 0.217 mole of HCl react to give 0.217 mole of NaCl

Now we have to calculate the mass of NaCl

\text{ Mass of }NaCl=\text{ Moles of }NaCl\times \text{ Molar mass of }NaCl

Molar mass of NaCl = 58.5 g/mole

\text{ Mass of }NaCl=(0.217moles)\times (58.5g/mole)=12.7g

Now we have to calculate the percent yield of this reaction.

Percent yield = \frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100

Actual yield = 5.24 g

Theoretical yield = 12.7 g

Percent yield = \frac{5.24g}{12.7g}\times 100

Percent yield = 41.3 %

Therefore, the percent yield of this combination reaction is 41.3 %

4 0
3 years ago
Make an observation about the average height of the beanbag for each mass dropped. How does it compare with your calculated kine
Gekata [30.6K]

Answer:

the other person is correct

Explanation:

3 0
3 years ago
What is the boiling point of 0.464 m lactose in water? (Kb of water = 0.512 oC/m). Enter your answer to 3 decimal places.
kirill [66]

Answer:

Boiling point for the solution is 100.237°C

Explanation:

We must apply colligative property of boiling point elevation

T° boiling solution - T° boiling pure solvent = Kb . m

m = molalilty (a given data)

Kb = Ebulloscopic constant (a given data)

We know that water boils at 100°C so let's replace the information in the formula.

T° boiling solution - 100°C = 0.512 °C/m . 0.464 m

T° boiliing solution = 0.512 °C/m . 0.464 m + 100°C → 100.237 °C

3 0
3 years ago
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