Ncert Solutions Chemistry Class 11th
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New answer posted
5 months agoContributor-Level 10
Time duration, t = 2 ns = 2 x 10-9 s
Frequency, ν = 1 / t = 1 / 2 x 10-9 s = 109 / 2 s-1
Energy of one photon, E = hν = 6.626 x 10-34Js) x (109 / 2 s-1) = 3.25 x 10-25 J
No. of photons = 2.5 x 105
Energy of source = 3.3125 x 10-25 J x 2.5 x 1015 = 8.28 x 10-10 J
New answer posted
5 months agoContributor-Level 10
Energy of one photon, E = hν = 6.626 x 10-34Js) x (3 x 108 / 2 ms-1) / 600 x 10-9
= 3.31 x 10-19 J
No. of photons = (3.15 x 10-18) / 3.31 x 10-19 = 9.52 ≈ 10.
New question posted
5 months agoNew answer posted
5 months agoContributor-Level 10
(a) Frequency of emission, ν = c/λ = (3.0 x 108 ms-1) / (616 x 10-9 m) = 4.87 x 1014 s-1
(b) Speed of radiation, c = 3 x 108 ms-1
Distance travelled by this radiation in 30s = 3 x 108 ms-1 x 30 s = 9.0 x 109 m
(c) Energy of quantum, E = hν =hc/λ = [ (6.626 x 10-34Js) x (3 x 108 ms-1)] / (616 x 10-9 m) = 32.27 x 10-20 J
(d) Number of quanta present if it produces 2 J of energy
&n
New answer posted
5 months agoContributor-Level 10
This is a short answer type question as classified in NCERT Exemplar
If arenes would undergo electrophilic addition reaction then they would lose their aromaticity which would lead to comparatively less stability. However, alkenes undergo electrophilic addition reaction via carbocation intermediate.
New answer posted
5 months agoContributor-Level 10
Power of the laser E = Nhv = Nh c/λ, where N is the number of photos emitted
= [ (5.6 x 1024) x (6.626 x 10-34Js) x (3 x 108 ms-1)] / (337.1 x 10-9 m)
= 3.3 x 106 J
New answer posted
5 months agoContributor-Level 10
The given radiations in increasing order of wavelength are:
Cosmic rays < X-rays < radiation from microwave oven < amber light from traffic signal < radiation from FM radio
New answer posted
5 months agoContributor-Level 10
This is a long answer type question as classified in NCERT Exemplar
The bond energy of HCl is higher than that of HBr thus it is not cleaved by free radical mechanism to exhibit peroxide effect. However in case of HI the bond energy is so low that the iodine radical forms readily and after formation it combines to form an iodine molecule.
New answer posted
5 months agoContributor-Level 10
Let the no. of electrons in the ion= x
∴ the no. of the protons= x + 3 (as the ion has three units positive charge)
and the no. of neutrons= x + (x*30.4 / 100) = x+ 0.304 x
Now, mass no. of ion = No. of protons + No. of neutrons = (x + 3) + (x + 0.304x)
∴ 56 = (x + 3) + (x + 0.304x)
=>2.304x = 56 – 3 = 53
=>x = 53 / 2.304 = 23
Atomic no. of the ion (or element) = 23 + 3 = 26
The element with atomic number 26 is iron (Fe) and the corresponding ion is Fe3+.
New answer posted
5 months agoContributor-Level 10
Let the no. of electron in the ion = x
∴ The no. of protons = x – 1 (as the ion has one unit negative charge)
and the no. of neutrons = x + (x*11.1 / 100) = 1.111 x
Mass of the ion = No. of protons + No. of neutrons
(x – 1) + (1.111 x)
Given mass of the ion = 37
∴ (x – 1) + (1.111 x) = 37
=> 2.111 x = 37 + 1 = 38
x = 38 / 2.111 = 18
No. of electrons = 18; No. of protons = 18 – 1 = 17
Atomic no. of the ion = 17; atom corresponding to ion = Cl
Symbol of the ion = 3717Cl–
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