The covalent compounds are soluble in:

In compound 1 the Tert butyl group occupies the equatorial position and the Bromine occupies the axial position and in compound 2 the Tert butyl occupies the axial and the bromine occupies equatorial positions. Compound 1 reacts faster than compound 2.

Explanation:

In cyclic organic compounds, substituents may occupy the axial or equatorial positions. The axial positions are aligned parallel to the symmetry axis of the ring while the equatorial positions are around the plane of the ring.

Bulky substituents have more room in the equatorial than in the axial position. This means that compound 1 is more stable than compound 2.

This is clear on the basis of stability of the molecules because compound 1 will react faster than compound 2 since the bulky tertiary butyl group in compound 1 occupy equatorial and not axial positions.

4 0

3 years ago

Write balanced chemical equation for the reactions used to prepare each of the following compounds from the given starting mater

creativ13 [48]

Explanation:

A balanced chemical equation is defined as the one that contains same number of atoms on both reactant and product side.

Whereas a chemical reaction that brings change in oxidation state of the reacting species is known as a redox reaction.

When there occurs decrease in oxidation number of an element then it shows the element has been reduced. And, if there occurs an increase in oxidation state of an element then it means the element has been oxidized.

(a) The reaction between nitrogen and hydrogen to form ammonia will be as follows.

$N_{2} + 3H_{2} \rightarrow 2NH_{3}$

So here, oxidation state of nitrogen is changing from 0 to -3. Whereas oxidation state of hydrogen is changing from 0 to +3.

Reaction between ammonia and nitric acid will be as follows.

$NH_{3} + HNO_{3} \rightarrow NH_{4}NO_{3}$

Since, it is a combination reaction. Therefore, no change in oxidation state of reacting species is taking place.

(b) Reaction for conversion of liquid bromine to hydrogen bromide is as follows.

$H_{2}(g) + Br_{2}(l) \rightarrow 2HBr(g)$

Here, oxidation state of hydrogen is changing from 0 to +1. Therefore, oxidation of hydrogen is taking place. On the other hand, oxidation state of bromine is changing from zero to -1. Therefore, reduction of bromine is taking place.

(c) Reaction between Zn and S will be as follows.

$Zn(s) + S(s) \rightarrow ZnS(s)$

Here, oxidation state of zinc is changing from 0 to +2. Hence, oxidation of zinc is taking place.

Oxidation state of sulfur is changing from 0 to -2. Therefore, reduction of sulfur is taking place.

7 0

3 years ago

If 50.0 g of formic acid (hcho2, ka = 1.8 x 10^-4) and 30.0 g of sodium formate (NaCHO2) are dissolved to make 500 ml. of soluti

gizmo_the_mogwai [7]

If 50.0 g of formic acid (HCHO2, ka = 1.8 x 10^-4) and 30.0 g of sodium formate (NaCHO2) are dissolved to make 500 ml. of solution, the ph of this solution is 3.35.

Therefore, option C is the correct option.

Given,

Given mass of sodium formate = 30 g

Given mass of formic acid = 50 g

Volume of sodium formate = 500 ml

Volume of formic acid = 500ml

Molar mass of sodium formate = 68 g

Molar mass of formic acid = 46 g

<h3>To calculate concentration of sodium formate and formic acid</h3>

The ratio of number of moles and the volume of solution is Molar concentration of substance.

Concentration of sodium formate

Cb = 30/(68×500)

= 0.00088m

Concentration of formic acid

Ca = 50/(46×500)

= 0.00217m

Now,

by using Henderson hesselbalch equation,

pH = pKa + log(Cb/Ca)

pKa = -log(1.8 × 10^(-4))

= 3.75

pH = 3.75 + log(0.00088/0.00217)

pH = 3.75 - 0.392

pH = 3.35

Thus, we calculated that the value of pH of solution of formic acid and sodium formate is 3.35.

learn more about pH:

brainly.com/question/13423434

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3 0

2 years ago