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Firdavs [7]
3 years ago
14

2 attempts left Click the "draw structure" button to launch the drawing utility. Draw one product formed when the following dien

e is treated with one equivalent of HCl.

Chemistry
1 answer:
Yuliya22 [10]3 years ago
7 0

Answer:

2-chloro-1-methyl-cyclohex-1,4-diene.

Explanation:

Hello,

In this case, the addition of hydrochloric acid acts as an electrophilic atack in which the hydrogen bonded to the double-bonded carbon connected to the carbon with the methyl substitution is substituted by the chlorine from the hydrochloric acid, in such a way, 2-chloro-1-methyl-cyclohex-1,4-diene is produced as one equivalent of HCl is used therefore one substitution will be attained for chlorine, and hydrogen as a side product as shown on the attached picture.

Best regards.

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If you added 45,000 calories to water that was at 25 degrees C, and the ending temperature was 35 degrees C, how much water did
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<u>Answer:</u>

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<em><u></u></em>

<u>Explanation:</u>

Q=m\times c \times \Delta T

where

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c is the specific heat capacity (for water 1.0  cal/(g℃))  

m is the mass of water

Plugging in the values  

\\$45000 \mathrm{cal}=m \times 1.0 \frac{\mathrm{cal}}{\mathrm{g}^{\circ} \mathrm{C}} \times\left(35^{\circ} \mathrm{C}-25^{\circ} \mathrm{C}\right)$\\\\$45000 \mathrm{cal}=m \times 1.0 \frac{\mathrm{cal}}{\mathrm{g}^{\circ} \mathrm{C}} \times 10^{\circ} \mathrm{C}$\\\\$m=\frac{45000 \mathrm{cal}}{1.0 \frac{\mathrm{cal}}{\mathrm{g}^{\circ} \mathrm{C}} \times 10^{\circ} \mathrm{C}}$\\\\$m=4500 \mathrm{g}$\\\\Density of water $=\frac{\text { mass }}{\text { volume }}$

So,

Volume of water = mass/density

\\\\=\frac{4500 \mathrm{g}}{\frac{1.09}{\mathrm{mL}}}=4500 \mathrm{mL}$$

=4.5 L (Answer)

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