write an equation to represent the oxidation of an alcohol.
identify the reagents that may be used to oxidize a given alcohol.
identify the specific reagent that is used to oxidize primary alcohols to aldehydes rather than to carboxylic acids.
identify the product formed from the oxidation of a given alcohol with a specified oxidizing agent.
identify the alcohol needed to prepare a given aldehyde, ketone or carboxylic acid by simple oxidation.
write a mechanism for the oxidation of an alcohol using a chromium(VI) reagent.
The reading mentions that pyridinium chlorochromate (PCC) is a milder version of chromic acid that is suitable for converting a primary alcohol into an aldehyde without oxidizing it all the way to a carboxylic acid. This reagent is being replaced in laboratories by Dess‑Martin periodinane (DMP), which has several practical advantages over PCC, such as producing higher yields and requiring less rigorous reaction conditions. DMP is named after Daniel Dess and James Martin, who developed it in 1983.
This page looks at the oxidation of alcohols using acidified sodium or potassium dichromate(VI) solution. This reaction is used to make aldehydes, ketones and carboxylic acids, and as a way of distinguishing between primary, secondary and tertiary alcohols.
Oxidizing the different types of alcohols
The oxidizing agent used in these reactions is normally a solution of sodium or potassium dichromate(VI) acidified with dilute sulfuric acid. If oxidation occurs, the orange solution containing the dichromate(VI) ions is reduced to a green solution containing chromium(III) ions. The electron-half-equation for this reaction is
Cr2O2−7+14H++6e−→2Cr3++7H2O
Answer:
Option F. 3moles
Explanation:
The equation for the reaction is given below:
2H2 + O2 —> 2H2O
from the equation,
2moles of H2 required 1mole of O2.
Therefore, 6moles of H2 will require = 6/2 = 3moles of O2
The relationship between pressure and solubility of the gas is given by Henry's law as:
where,
is the solubility of the gas.
is proportionality constant i.e. Henry's constant.
is pressure of the gas.
(given)
(given)
Substituting the values,
To convert 
to 
:
Molar mass of benzene, 
= 
Now for converting into 
:
Since, 
So, 
.
Hence, the solubility of benzene in water at 
in is 
.
Delta E = Ef - Ei
E = energy , h = plank constant , v = frequency
h= 6.626 * 10 ^-34 j*s , T = 10 ^ 12 , v = 74 * 10 ^12 Hz , Hz = s^-1
E = ( 6.626 * 10^ -34 j*s) ( 74 * 10 ^ 12 s^ -1 ) = 4.90 * 10 ^ -20 J
Delta E = Ef - Ei
-4.90 * 10 ^ -20 J = -2.18 * 10 ^ -18J ( 1/4 ^2 - 1/x ^2)
0.0225 = 0.0625 - ( 1/x ^ 2)
0.225 - 0.0625 = - 1/ x ^ 2
- 0.0400 = - 1/x ^2 = -1 / - 0.0400 = x^2
25 = x^2
x = 5
Answer is: 0,0095 mol of hydrogen gas will be produced in reaction.
Chemical reaction: Ca + 2HCl → CaCl₂ + H₂.
m(Ca) = 0,38 g.
n(H₂) = ?
n(Ca) = m(Ca) ÷ M(Ca).
n(Ca) = 0,38 g ÷ 40 g/mol
n(Ca) = 0,0095 mol.
from reaction: n(Ca) : n(H₂) = 1 : 1.
n(H₂) = n(Ca) = 0,0095 mol.
n - amount of substance.
