Answer :
- Carbonyl group : It is a functional group composed of a carbon atom that double bonded to oxygen atom. It is represented as

Carboxylic group : It is the class of organic compound in which the carboxylic (-COOH) group is attached to a hydrocarbon is known as carboxylic.
The general formula of carboxylic is,
. According to the IUPAC naming, the carboxylic are named as alkanoic acids.
Aldehyde group : It is the class of organic compound in which the (-CHO) group is attached to a hydrocarbon is known as aldehyde.
The general representation of aldehyde is,
. According to the IUPAC naming, the aldehyde are named as alkanals.
Ketone group : It is the class of organic compound in which the (-CO) group is directly attached to the two alkyl group of carbon is known as ketone.
The general representation of ketone is,
. According to the IUPAC naming, the ketone are named as alkanone.
Ester group : It is the class of organic compound in which the (-COO) group is directly attached to the two alkyl group of carbon is known as ester.
The general representation of ester is,
. According to the IUPAC naming, the ester are named as alkyl alkanoate.
Answer:
92.6
Explanation:
6 mol x 18.02 g of H2o --> 3 mol x 58.33 g Mg(OH)2
108.12 g of h2o --> 174.99 of Mg(OH)2
g of H2O is 150 g of Mg(OH)2
150g x 108.12g / 174.99 =
92.67
220 grams of sugar would be in 2 liters of orange juice
The gravitational force between the objects depends on the mass of the objects and the distance between them.
Answer:
pH=8.676
Explanation:
Given:
0.75 M 
0.20 M 
The objective is to calculate the pH of the buffer using the kb for 
Formula used:
![pOH=pka+log\frac{[salt]}{[base]}\\](https://tex.z-dn.net/?f=pOH%3Dpka%2Blog%5Cfrac%7B%5Bsalt%5D%7D%7B%5Bbase%5D%7D%5C%5C)
pH=14-pOH
Solution:
On substituting salt=0.75 and base=0.20 in the formula

pH=14-pOH
On substituting the pOH value in the above expression,
pH=14-5.324
Therefore,
pH=8.676