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AnnyKZ [126]
3 years ago
11

What major product is formed when heptyne is treated with Br2 in CCl4?

Chemistry
2 answers:
melisa1 [442]3 years ago
8 0

<u>Answer: </u>

<em>Option A is the Answer. </em>

<u>Explanation:</u>

Treatment of alkyne with Bromine in the presence of CCl_4 produces a trans product that is, a alkenyl bromide.

CH_3 \ CH_2 \ CH_2 \ CH_2 \ CH_2 - C = CH \frac {(CCl_4)}{(Br_2 )} >  CH_3 \ CH_2 \ CH_2\ CH_2 \CH_2\ -CBr = CHBr

E-1, 2-dibromoheptene

It is a trans intermediate on further halogenations produces a tetrabromopentane.  

It is an electrophilic expansion response in which the triple bond breaks to turn into a twofold bond and creates a dibromoalkene (E-1,2 dibromoheptene), and on further expansion, the twofold bond turns into a solitary bond and delivers a tetrabromoalkane.

ANTONII [103]3 years ago
4 0

Answer:A 1,2-dibromoheptene

Explanation:

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2Al + 6HCl --&gt; 2AlCl3 + 3H2 Aluminium reacts with hydrochloric acid. How many grams of aluminum are necessary to produce 11 L
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8.8g of Al are necessaries

Explanation:

Based on the reaction, 2 moles of Al are required to produce 3 moles of hydrogen gas.

To solve this question we must find the moles of H2 in 11L at STP using PV = nRT. With these moles we can find the moles of Al required and its mass as follows:

<em>Moles H2:</em>

PV = nRT; PV/RT = n

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3 years ago
A student placed 10.5 g of glucose (C6H12O6) in a volumetric fla. heggsk, added enough water to dissolve the glucose by swirling
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<u>Answer:</u> The mass of glucose in final solution is 0.420 grams

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\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}        .........(1)

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Volume of solution = 100 mL

Putting values in equation 1, we get:

\text{Initial molarity of glucose}=\frac{10.5\times 1000}{180.16\times 100}\\\\\text{Initial molarity of glucose}=0.583M

To calculate the molarity of the diluted solution, we use the equation:

M_1V_1=M_2V_2

where,

M_1\text{ and }V_1 are the molarity and volume of the concentrated glucose solution

M_2\text{ and }V_2 are the molarity and volume of diluted glucose solution

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M_1=0.583M\\V_1=20.0mL\\M_2=?M\\V_2=0.5L=500mL

Putting values in above equation, we get:

0.583\times 20=M_2\times 500\\\\M_2=\frac{0.583\times 20}{500}=0.0233M

Now, calculating the mass of final glucose solution by using equation 1:

Final molarity of glucose solution = 0.0233 M

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Putting values in equation 1, we get:

0.0233=\frac{\text{Mass of glucose in final solution}\times 1000}{180.16\times 100}\\\\\text{Mass of glucose in final solution}=\frac{0.0233\times 180.16\times 100}{1000}=0.420g

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