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NISA [10]
3 years ago
11

From the list of available reagents, show how you would bring about this synthesis. Enter the letters of the reagents necessary,

with no commas or spaces, in the order that they are to be used. If a step is to be performed more than once in succession, enter its letter only once. For example: acega, not acegga

Chemistry
1 answer:
HACTEHA [7]3 years ago
3 0

Answer:

The reagents -a)HNO_{3},H_{2}SO_{4}\,\,b)H_{2},Pd,C\,\,c)CH_{3}COCl

Explanation:

-COCH_{3} is an electron donating group which directs the electrophile to meta position reduction of nitro group with H_{2},Pd,C produces amine.

Acylation of amine with CH_{3}COCl produces required product.

The complete reaction is as follows.

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How many acetone molecules are in a bottle of acetone with a volume of 445 mL ? (density of acetone = 0.788 g/cm3).
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<h2>Hello!</h2>

The answer is: 3.63x10^{24}particles

<h2>Why?</h2>

First, we need to calculate the molecular weight of the acetone (CH3COCH3)

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C=12.011g/mol\\O=15.99g/mol\\H=1.008g/mol

Then,

CH3COCH3=12.011+(1.008)*3+12.011+15.999+12.011+(1.008)*3=58.08g/mol

Second, we need to calculate the mass of the acetone in the bottle

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445cm^{3}*\frac{0.788g}{cm^{3}}=350.66g

Third, we need to calculate the number of moles:

350.66g*\frac{1mol}{58.08g}=6.03moles

Fourth, we need to calculate the number of atoms:

Remember, 1 mol=6.022x10^{23}particles

Therefore,

6.03moles*\frac{6.022x10^{23}particles }{1mol}=3.63x10^{24}particles

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