Chemistry ncert question

chemistry

#1
  1. list the uses of neon and argon gases ?
  2. in what way is a sol different from a gel ?
  3. what kind of defect are introduced by doping ?
  4. what are the expected products of hydrolysis of lactose ?
  5. how many atom are there in one unit cell of body centred cubic crystal ?
  6. how many unit cell are there in 1cm cube of aluminium ?

#2

1). List the uses of neon and argon gases ?

The noble gases are used in industry in arc welding, to dilute the oxygen in deep-sea divers’ gas tanks, and to fill light bulbs. Argon is used in arc welding and in common light bulbs, as it does not react with the metal at high temperatures.
Neon is also used to make high-voltage indicators and switching gear, lightning arresters, diving equipment and lasers. Liquid neon is an important cryogenic refrigerant. It has over 40 times more refrigerating capacity per unit volume than liquid helium, and more than 3 times that of liquid hydrogen.

Argon is a versatile industrial gas used in welding applications, such as the welding of specialty alloys, and in the production of light bulbs and lasers. As an inert gas, it can also be used to provide an oxygen- and nitrogen- free environment for heat treating processes.

2). In what way is a sol different from a gel ?

Sol is a colloidal suspension of very small solid particles in a continuous liquid medium. Solid dispersed in liquid.
Gels are defined as a substantially dilute cross-linked system, which exhibits no flow when in the steady-state.

3). What kind of defect are introduced by doping ?

4). What are the expected products of hydrolysis of lactose ?

Lactose + Water à Galactose + Glucose

5). How many atom are there in one unit cell of body centred cubic crystal ?

There is more than one atom per cubic unit cell in most cubic crystal systems.

6). How many unit cell are there in 1cm cube of aluminium ?

I think answer should be this.

Volume of one unit cell = side^3 = 354 pm^3 (1 pm = 10^–10 cm)
= 354 x 10^–10 cm^3 = 3.54 x 10^–8 cm^3
= 44.36 x 10^–24 cm^3
= 4.4 × 10^−23 cm^3

Total number of unit cells in 1.00 cm3
= total volume / size of each cell
= (1.00 cm^3)/( 4.4 × 10^−23 cm^3)
= 2.27 × 1022 unit cell


#3

ur right…
correct this whole this is correct…