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:
magnification makes everything smaller so you can see smaller things up close and study them more
Explanation:
Answer:
d = 0.98 g/L
Explanation:
Given data:
Density of acetylene = ?
Pressure = 0.910 atm
Temperature = 20°C (20+273 = 293 K)
Solution:
Formula:
PM = dRT
R = general gas constant = 0.0821 atm.L/mol.K
M = molecular mass = 26.04 g/mol
0.910 atm × 26.04 g/mol = d × 0.0821 atm.L/mol.K×293 K
23.7 atm.g/mol = d × 24.1 atm.L/mol
d = 23.7 atm.g/mol / 24.1 atm.L/mol
d = 0.98 g/L
Answer:
a. Pb 208
b. About 21.7 minutes
c. only a trace amount
Explanation:
It under goes beta decay.
There should be virtually nothing after an hour
Answer : The value of 
is, 
Explanation :
The formula used for is:
............(1)
where,
= Gibbs free energy for the reaction
= standard Gibbs free energy
R = gas constant = 8.314 J/mole.K
T = temperature = 
Q = reaction quotient
First we have to calculate the .
Formula used :
Now put all the given values in this formula, we get:
Now we have to calculate the value of 'Q'.
The given balanced chemical reaction is,
The expression for reaction quotient will be :
In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.
Now put all the given values in this expression, we get
Now we have to calculate the value of by using relation (1).
Now put all the given values in this formula, we get:
Therefore, the value of is,
