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
- 2k
Questions
- 0
Discussions
- 0
Active Users
- 29
Followers
Have a question about Ncert Solutions Chemistry Class 11th? Ask our experts
Keep it short & simple. Type complete word. Avoid abusive language.
An ion with mass number 37 possesses one unit of negative charge. If the ion contains 11.1% more neutrons than the electrons, find the symbol of the ion.
An ion with mass number 37 possesses one unit of negative charge. If the ion contains 11.1% more neutrons than the electrons, find the symbol of the ion.
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.
Upvote 
An unsaturated hydrocarbon ‘A’ adds two molecules of H2 and on reductive ozonolysis gives butane-1,4-dial, ethanal and propanone. Give the structure of ‘A’, write its IUPAC name and explain the reactions involved.
An unsaturated hydrocarbon ‘A’ adds two molecules of H2 and on reductive ozonolysis gives butane-1,4-dial, ethanal and propanone. Give the structure of ‘A’, write its IUPAC name and explain the reactions involved.
This is a long answer type question as classified in NCERT Exemplar
The structure of A is
The reaction involved are as follows:
Upvote An element with mass number 81 contains 31.7% more neutrons as compared to protons. Assign the atomic symbol.
An element with mass number 81 contains 31.7% more neutrons as compared to protons. Assign the atomic symbol.
An element can be identified by its atomic number only. Let us find the atomic number.
Let the number of protons = x
Number of neutrons = x + (x×31.7/100) = 1.371x
Now, Mass no. of element = no. of protons + no. neutrons
⇒ x + 1.317x = 81
⇒ x = 34.958
x ? 35
∴ No.
Upvote Symbols 3579Br and 79Br can be written, whereas symbols 7935Br and 35Br are not acceptable. Answer briefly.
Symbols 3579Br and 79Br can be written, whereas symbols 7935Br and 35Br are not acceptable. Answer briefly.
The composition of any atom can be represented by using the normal element symbol (X) with super-script on the left hand side as the atomic mass number (A) and subscript (Z) on the left hand side as the atomic number (i.e., AZX).
No two elements can have the same atomic number. However, the mass numb
Upvote In Rutherford experiment, generally the thin foil of heavy atoms like gold, platinum etc. have been used to be bombarded by the α-particles. If a thin foil of light atoms like aluminium etc. is used, what difference would be observed from the above results?
In Rutherford experiment, generally the thin foil of heavy atoms like gold, platinum etc. have been used to be bombarded by the α-particles. If a thin foil of light atoms like aluminium etc. is used, what difference would be observed from the above results?
Heavy metals have a heavy nucleus and contain a large amount of positive change in their nucleus. By using heavy metals like gold and platinum in Rutherford’s experiment, a large number of α-particles get deflected and experience a repulsion thus finding it hard for these α-particles to retrace thei
Upvote In Millikan’s experiment, the charge on the oil droplets was found to be -1.282 x 10-18C. Calculate the number of electrons present in it.
In Millikan’s experiment, the charge on the oil droplets was found to be -1.282 x 10-18C. Calculate the number of electrons present in it.
Charge on oil droplet = – 1.282 x 10-18C
Charge on an electron = – 1.602 x 10-19C
Number of electrons = q /e = (– 1.282 x 10-18C) / (– 1.602 x 10-19C) = 8
Upvote A certain particle carries 2.5 x 10-16 C of static electric charge. Calculate the number of electrons present in it.
A certain particle carries 2.5 x 10-16 C of static electric charge. Calculate the number of electrons present in it.
Static electric charge (q) = 2.5 x 10-16 C
Charge on one electron (e) = 1.602 x 10-19 C
No. of electrons present = (2.5 x 10-16 C) / (1.602 x 10-19 C) = 1560
Upvote The diameter of zinc atom is 2.6 Å. Calculate
(a) The radius of zinc atom in pm
(b) Number of atoms present in a length of 1.6 cm if the zinc atoms are arranged side by side length wise.
The diameter of zinc atom is 2.6 Å. Calculate
(a) The radius of zinc atom in pm
(b) Number of atoms present in a length of 1.6 cm if the zinc atoms are arranged side by side length wise.
(a) The diameter of zinc atom is 2.6 Å =2.6×10−10m.
The radius of Zn atom is (2.6×10−10) / 2=1.3×10−10m=130×10−12m=130 pm.
(b) The number of Zn atoms present on 1.6 cm of length are 1.6 / (2.6×10−8) =6.154×107.
Upvote 2 × 108 atoms of carbon are arranged side by side. Calculate the radius of carbon atom if the length of this arrangement is 2.4 cm.
2 × 108 atoms of carbon are arranged side by side. Calculate the radius of carbon atom if the length of this arrangement is 2.4 cm.
The length of the arrangement = 2.4 cm
Total number of carbon atoms present = 2 *108
Diameter of each C-atom = (2.4 cm) / (2 x 108) = 1.2 x 10-8 cm
Radius of each C-atom = ½ x 1.2 x 10-8 cm = 6.0 x 10-9 cm = 0.06 nm
Upvote If the diameter of carbon atom is 0.15 nm, calculate the number of carbon atoms which can be placed side by side in a straight line across a length of a scale of length 20 cm long.
If the diameter of carbon atom is 0.15 nm, calculate the number of carbon atoms which can be placed side by side in a straight line across a length of a scale of length 20 cm long.
Length of scale = 20 cm = 20 x 107 nm = 2 x 108 nm
Diameter of carbon atom = 0.15 nm
∴ Number of carbon atoms which can be placed side by side in a straight line across a length of a scale of length 20 cm = (2 x 108 nm) / (0.15 nm) = 1.33 x 109
Upvote