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stepan [7]
3 years ago
15

Which ketone in each pair is more reactive?

Chemistry
1 answer:
KiRa [710]3 years ago
4 0

Answer:

a. 2-heptanone is more reactive than 4-heptanone

b. chloromethyl phenyl ketone is more reactive than bromomethyl phenyl ketone

Explanation:

The reactivity of the carbonyl compound (ketone ) is affected by the steric effect. The steric effect is a hindrance that occurs in the structure or reactivity of a molecule, which is affected by the physical size and the proximity of the adjacent parts of the molecule.

Between 2-heptanone or 4-heptanone, 2-heptanone is more reactive than 4-heptanone. This is because 2-heptanone is less affected by the steric hindrance, unlike the 4-heptanone.

Similarly, the reactivity of the carbonyl compound (ketone) is also affected by the polarity on the carbon compound, which is associated with how electronegative the substituent attached is to the carbonyl compound. From the periodic table, the electronegativity of the Halogen family decreases down the group. Therefore chlorine is more electronegative than bromine.

As such, chloromethyl phenyl ketone is more reactive than bromomethyl phenyl ketone.

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2. Which of the following accurately pairs
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Answer:

C. Electron-negative charge

Explanation:

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Symbol= e⁻

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It was discovered by j. j. Thomson in 1897 during the study of cathode ray properties.

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8 0
3 years ago
You have a 2.0 mL sample of acetic acid (molar mass 60.05 g/mol) of unknown concentration. You titrate it to its endpoint with 2
Katyanochek1 [597]

<u>Answer:</u> The mass of acetic acid used is 0.12 grams

<u>Explanation:</u>

To calculate the concentration of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is CH_3COOH

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=1\\M_1=?M\\V_1=2.0mL\\n_2=1\\M_2=0.1M\\V_2=20.0mL

Putting values in above equation, we get:

1\times M_1\times 2.0=1\times 0.1\times 20.0\\\\M_1=\frac{1\times 0.1\times 20.0}{1\times 2.0}=1M

To calculate the mass of solute, we use the equation used to calculate the molarity of solution:

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

Given mass of acetic acid = ? g

Molar mass of acetic acid = 60.05 g/mol

Molarity of solution = 1 M

Volume of the solution = 2.0 mL

Putting values in above equation, we get:

1mol/L=\frac{\text{Mass of acetic acid}\times 1000}{60.05g/mol\times 2.0}\\\\\text{Mass of acetic acid}=\frac{1\times 60.05\times 2}{1000}=0.12g

Hence, the mass of acetic acid used is 0.12 grams

5 0
3 years ago
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