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Andrews [41]
3 years ago
8

In the presence of sulfuric acid, this alcohol is dehydrated to form an alkene through an E1 mechanism. In the box, draw the maj

or alkene product of this reaction.

Chemistry
1 answer:
Sophie [7]3 years ago
5 0

Answer:

figure is attached

Explanation:

When we treat alcohol with H₂SO₄ we get elimination as the major product.

As we can see in the given reaction that in step 1 the lone pair of electrons of oxygen attached to the alcohol make a bond with the hydrogen of H₂SO₄.

In the 2nd step H₂O gets detached from the parent ring which generated a positive charge on the ring.

In the 3rd step elimination of hydrogen from the carbon next to the carbonium carbon results into formation of an alkene.

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What do punnet squares do
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Answer:

The Punnett square is a square diagram that is used to predict the genotypes of a particular cross or breeding experiment. ... The diagram is used by biologists to determine the probability of an offspring having a particular genotype.

Explanation:

in biology it predicts the possible offspring provided the features of the mother or father

6 0
2 years ago
What measures are used to calculate the percent by volume of a solution?
Yuliya22 [10]
Titration is the method used.
4 0
3 years ago
A first-order reaction has a half-life of 16.7 s . How long does it take for the concentration of the reactant in the reaction t
andrew11 [14]

It takes 33.4 s for the concentration of A to fall to one-fourth of its original value.

A <em>half-life</em> is the time it takes for the concentration to fall to half its original value.

Assume the initial concentration is 1.00 mol/L. Then,

\text{1.00 mol/L }\stackrel{\text{1st half-life }}{\longrightarrow}\text{ 0.50 mol/L } \stackrel{\text{ 2nd half-life }}{\longrightarrow}\text{0.25 mol/L}\\

The concentration drops to one-fourth of its initial value in two half-lives.

∴ Time = 2 × 16.7 s = 33.4 s

5 0
3 years ago
The rate of effusion of an unknown gas was measured and found to be 11.9 mL/min. Under identical conditions, the rate of effusio
iren2701 [21]

Answer : The correct option is, (B) CO_2

Solution :

According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.

R\propto \sqrt{\frac{1}{M}}

or,

(\frac{R_1}{R_2})=\sqrt{\frac{M_2}{M_1}}       ..........(1)

where,

R_1 = rate of effusion of unknown gas = 11.9\text{ mL }min^{-1}

R_2 = rate of effusion of oxygen gas = 14.0\text{ mL }min^{-1}

M_1 = molar mass of unknown gas  = ?

M_2 = molar mass of oxygen gas = 32 g/mole

Now put all the given values in the above formula 1, we get:

(\frac{11.9\text{ mL }min^{-1}}{14.0\text{ mL }min^{-1}})=\sqrt{\frac{32g/mole}{M_1}}

M_1=44.2g/mole

The unknown gas could be carbon dioxide (CO_2) that has approximately 44 g/mole of molar mass.

Thus, the unknown gas could be carbon dioxide (CO_2)

5 0
3 years ago
1. The water cycle is nature's way of _______________water.
stellarik [79]
Regenerating is the answer
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3 years ago
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