Chemistry
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New answer posted
a year agoContributor-Level 10
(b) It is a nucleophilic substitution reaction. KOH (aq) provides OH- ion for the nucleophile attack.
New answer posted
a year agoContributor-Level 10
11.27
ortho-nitro phenol is more acidic than ortho-methoxy phenol.
Explanation: Due to strong –R and –I effect of NO2 group, electron density in the O-H bond decreases and hence the loss of a proton becomes easy.
Now after the loss of a proton, the o-nitrophenoxide ion left behind is stabilized by resonance and thus making o-nitro phenol a stronger acid.
In contrast, due to the +R effect of methoxy group increases the electron density in the O-H bond. Thereby making the loss of proton difficult.
Now, the o-methoxyphenoxide ion left after the loss of a proton is destabilized by resonance. The two negative charges repel each other, thereb
New answer posted
a year agoContributor-Level 10
(b) Iron (III) hexacyanidoferrate (II) (or ferriferrocyanide) Fe4 [Fe (CN)6]3 is the correct answer.
New answer posted
a year agoContributor-Level 10
11.26
The acidic nature of phenol can be represented by the following two reactions:-
(a) Phenols react with sodium to give sodium phenoxide, liberating H2.
(b) Phenols react with sodium hydroxide to give sodium phenoxide and water as a by-product.
The acidity of phenol is more than that of ethanol. This is because phenol after losing a proton becomes phenoxide ion which undergoes resonance and is stabilized whereas ethoxide ion does not.
The resonating structures of phenoxide ion are shown as below:
The lone pair of electrons on oxygen delocalizes into the benzene (mesomeric effect) which reduces the electron density in the O-H bon
New question posted
a year agoNew answer posted
a year agoContributor-Level 10
Mass of the compound = 0.468 g
Mass of barium sulphate= 0.668 g
% of sulphur = ![]()

New answer posted
a year agoContributor-Level 10
Mass of the compound = 0.3780 g
Mass of silver chloride = 0.5740 g
% of chlorine =![]()

New answer posted
a year agoContributor-Level 10
Step I: Calculation of volume of unused acid i.e. V2 =?
V1 = Volume of NAOH solution required = 60 cm3
N1 = Normality of NaOH solution = ½ N
N2 = Normality of H2SO4 = 1N
Applying N1V1 = N2V2
½ N x 60 cm3 = 1N x V2
Or V2 = 30 cm3
Step II: calculation of volume of acid used
Volume of acid added = 50 cm3
Volume of unused acid = 30 cm3
Volume of acid used = 50 – 30 = 20 cm3
Step III: Calculation of % of nitrogen
Mass of compound = 0.50 g
Volume of acid used = 20 cm3
Normality of acid used = 1 N
% of nitrogen = (1.4 x 20 x 1) / 0.50 = 56%
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