Ncert Solutions Chemistry Class 11th

Get insights from 2k questions on Ncert Solutions Chemistry Class 11th, answered by students, alumni, and experts. You may also ask and answer any question you like about Ncert Solutions Chemistry Class 11th

Follow Ask Question
2k

Questions

0

Discussions

12

Active Users

29

Followers

New answer posted

6 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

Diamond has a crystalline lattice where each carbon atom undergoes sp3 hybridisation and linked to four other carbon atoms by using hybridised orbitals in tetrahedral fashion.  The C–C bond length is 154 pm. The structure extends in space and produces a rigid three- dimensional network of carbon atoms. It is very difficult to break extended covalent bonding and, therefore, diamond is a the hardest substance on the earth.

Graphite has layered structure in which the layers are held by van there Waals forces and distance between two layers is 340 pm. Each layer is composed of planar hexagonal rings of carbon atoms. C—C bond length

...more

New answer posted

6 months ago

0 Follower 32 Views

P
Payal Gupta

Contributor-Level 10

1.4 In order to answer the question, we need to know the balanced equation for the combustion of carbon in dioxygen/air, which can be written as:

C (s) + O2 (g) à CO2 (g)

(i) We can see from the above equation,

1 mole of carbon reacts with 1 mole of oxygen to produce 1 mol of carbon dioxide.

In air, combustion is complete.

Therefore, CO2 produced from combustion of 1 mole of carbon= Molar mass of CO2= 44 g

(ii) As only 16 g of dioxygen is available, it can combine only with 0.5 mole of carbon, i.e., dioxygen is the limiting reactant.

Hence, CO2 produced = 22 g

Here, dioxygen acts as the limiting reagent.

(iii) Here again, dioxygen is

...more

New answer posted

6 months ago

0 Follower

V
Vishal Baghel

Contributor-Level 10

(a) CO32-  (sp2) (b) Diamond (sp3) (c) Graphite (sp2)

New answer posted

6 months ago

0 Follower 16 Views

P
Payal Gupta

Contributor-Level 10

1.3 We are given that the percentage of iron by mass is 69.9% and the percentage of oxygen by mass is 30.1%.

Since we have relative moles of both elements, we can calculate the simpler molar ratio of iron to oxygen

= 1.25: 1.88

Divide by the smaller value to both

= 1.25/1.25: 1.88/1.25

= 1: 1.5

= 2: 3

So, now we can write the empirical formula of iron oxide as Fe2O3.

New question posted

6 months ago

0 Follower 2 Views

New answer posted

6 months ago

0 Follower 8 Views

V
Vishal Baghel

Contributor-Level 10

Electron deficient compounds are those in which the central atom in their molecule has the tendency to accept one or more electron pairs. They are also known as Lewis acid. BCl3 and SiCl4 both are electron deficient species.
Since, in BCl3, B atom has only six electrons. Therefore, it is an electron deficient compound.
In SiCl4 the central atom Si has 8 electrons but it can expand its covalency beyond 4 due to the presence of d-orbitals.

New answer posted

6 months ago

0 Follower

V
Vishal Baghel

Contributor-Level 10

Aluminium reacts with acid as well as base. This shows amphoteric nature of aluminium.
2Al (s) + 6HCl (dil.) 2AlCl3 (aq) + 3H2 (g)
2Al (s) + 2NaOH (aq) + 6H2O (l) 2Na+ [Al (OH)4] (aq) + 3H2 (g)

New answer posted

6 months ago

0 Follower 20 Views

P
Payal Gupta

Contributor-Level 10

1.2 Molar mass of Na2SO4= (2 x Atomic mass of Sodium) + Atomic mass of Sulphur + (4 x Atomic mass of Oxygen)

= (2x 23) + 32 + (4x 16)

= 46 + 32 + 64

= 142 g/mol

New answer posted

6 months ago

0 Follower 5 Views

V
Vishal Baghel

Contributor-Level 10

In BF3, boron is sp2 hybridized.
? shape of BF3 = planar.
In [BH4], boron is sp3 hybridized, thus the shape is tetrahedral.

New answer posted

6 months ago

0 Follower 6 Views

V
Vishal Baghel

Contributor-Level 10

On heating boric acid above 370 K, it forms metaboric acid, HBO2 which on further heating yields boric oxide B2O3.

H3B2O3 → HBO2 → B2O3

Get authentic answers from experts, students and alumni that you won't find anywhere else

Sign Up on Shiksha

On Shiksha, get access to

  • 65k Colleges
  • 1.2k Exams
  • 679k Reviews
  • 1800k Answers

Share Your College Life Experience

×
×

This website uses Cookies and related technologies for the site to function correctly and securely, improve & personalise your browsing experience, analyse traffic, and support our marketing efforts and serve the Core Purpose. By continuing to browse the site, you agree to Privacy Policy and Cookie Policy.