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Vinil7 [7]
3 years ago
8

Which of the statements applies to oxidation?

Chemistry
1 answer:
Charra [1.4K]3 years ago
5 0

Answer:

The correct option is: (A) Ketones cannot be oxidized further

Explanation:

Oxidation refers to the gain of oxygen or the formation of carbon-oxygen bond (C-O bond).

Alcohols are organic compounds containing at least one hydroxyl group. These compounds can be further classified into <em>primary (R-CH₂-OH), secondary (R¹R²CH-OH) and tertiary alcohols (R¹R²R³C-OH).</em>

  • The <u>partial oxidation of </u><u>primary alcohol</u><u> gives </u><u>aldehyde</u> (R-CHO). Further <u>oxidation of aldehyds gives </u><u>carboxylic acid </u>(R-COOH). <u>Primary alcohol can also be oxidized directly to carboxylic acid.</u> <em>However, the carboxylic acid can not be further oxidized.</em>

R-CH_{2}-OH \overset{[O]}{\rightarrow} R-CHO \overset{[O]}{\rightarrow} R-COOH \\R-CH_{2}-OH \overset{[O]}{\rightarrow} R-COOH

  • <u>Oxidation of </u><u>secondary alcohols</u><u> gives </u><u>ketones</u> (R¹R²C=O), <em>which can not be oxidized further.</em>

R^{1}R^{2}CH-OH \overset{[O]}{\rightarrow} R^{1}R^{2}C=O

  • Whereas, tertiary alcohols are resistant to oxidation.

<u>Therefore, the correct statement regarding oxidation is </u><u>(A)</u><u> </u><em><u>Ketones cannot be oxidized further</u></em><u>.</u>

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A 2.0 mL sample of air in a syringe exerts a pressure of 1.02 atm at 22 C. If that syringe is placed into boiling water at 100 C
emmainna [20.7K]

Answer: The new volume of the air in the syringe is 7.3 mL.

Explanation:

Given: V_{1} = 2.0 mL,     P_{1} = 1.02 atm,          T_{1} = 22^{o}C

V_{2} = ?,               P_{2} = 1.27 atm,               T_{2} = 100^{o}C

Formula used to calculate the new volume in syringe is as follows.

\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}

Substitute the values into above formula as follows.

\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\\frac{1.02 atm \times 2.0 mL}{22^{o}C} = \frac{1.27 atm \times V_{1}}{100^{o}C}\\V_{1} = 7.3 mL

Thus, we can conclude that the new volume of the air in the syringe is 7.3 mL.

6 0
3 years ago
2C6H6 + 1502 →12CO2 + 6H20
solmaris [256]

Answer:

.926 moles

Explanation:

Rounding :

H2 0 = 18 gm/mole

   50 gm would then be   50 / 18 = 2.7777 moles of water

every two moles of 2 C6H6   produces   6 moles of water

   2.7777/6  * 2 = .926 moles

5 0
2 years ago
The chemical name for the compound formula NaCl is?
Mandarinka [93]

Answer:

The chemical name for the compound formula NaCl is sodium chloride.

7 0
4 years ago
Read 2 more answers
Be sure to answer all parts. What is the [H3O+] and the pH of a buffer that consists of 0.26 M HNO2 and 0.89 M KNO2? (K, of HNO2
Aleksandr-060686 [28]

Answer : The H_3O^+ ion concentration is, 1.12\times 10^{-3}M and the pH of a buffer is, 2.95

Explanation : Given,

K_a=7.1\times 10^{-4}

Concentration of HNO_2 (weak acid)= 0.26 M

Concentration of KNO_2 (conjugate base or salt)= 0.89 M

First we have to calculate the value of pK_a.

The expression used for the calculation of pK_a is,

pK_a=-\log (K_a)

Now put the value of K_a in this expression, we get:

pK_a=-\log (7.1\times 10^{-4})

pK_a=4-\log (7.1)

pK_a=3.15

Now we have to calculate the pH of the solution.

Using Henderson Hesselbach equation :

pH=pK_a+\log \frac{[Salt]}{[Acid]}

pH=pK_a+\log \frac{[KNO_2]}{[HNO_2]}

Now put all the given values in this expression, we get:

pH=3.15+\log (\frac{0.89}{0.26})

pH=2.95

The pH of a buffer is, 2.95

Now we have to calculate the H_3O^+ ion concentration.

pH=-\log [H_3O^+]

2.95=-\log [H_3O^+]

[H_3O^+]=1.12\times 10^{-3}M

The H_3O^+ ion concentration is, 1.12\times 10^{-3}M

4 0
3 years ago
What is the energy in joules of one photon of<br> microwaveradiation with a wavelength 0.122m?
pentagon [3]

<u>Answer:</u> The energy of photon is 162.93\times 10^{-26}J

<u>Explanation:</u>

The relation between energy and wavelength of light is given by Planck's equation, which is:

E=\frac{hc}{\lambda}

where,

E = energy of the light  = ?

h = Planck's constant  = 6.626\times 10^{-34}Js

c = speed of light = 3\times 10^8m/s

\lambda = wavelength of photon = 0.122 m

Putting values in above equation, we get:

E=\frac{(6.626\times 10^{-34}Js)\times (3\times 10^8m/s)}{0.122m}\\\\E=162.93\times 10^{-26}J

Hence, the energy of photon is 162.93\times 10^{-26}J

3 0
3 years ago
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