Class 11th

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New answer posted

10 months ago

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Payal Gupta

Contributor-Level 10

During removal of temporary hardness of water.

Mg (HCO3)2BoilMg (OH)2+2CO2Ca (HCO3)2BoilCaCO3+H2O+CO2

New answer posted

10 months ago

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A
alok kumar singh

Contributor-Level 10

B2H6 has 4 2 c -2e bonds and 2 3c-2e bonds.

Bridging (B-H) bonds have more value of bond- length then terminal (B – H) bonds

Bridging bonds are in one plane, but terminal bonds are in perpendicular plane.

Due to presence of (3c-2e) bonds, it behaves as electrons deficient and prone to get attached by  lewis      base.

3 N a B H 4 + 4 B F 3 ? ? 3 N a B F 4 + 2 B 2 H 6

New answer posted

10 months ago

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Payal Gupta

Contributor-Level 10

Except Te, all are metals.

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10 months ago

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alok kumar singh

Contributor-Level 10

Due to higher extent of polarization by Li+ and Mg2+, LiCl and Mgcl2 have covalent character. Therefore they are soluble in ethanol.

Due to very high value of lattice energy, LiF is having very less solubility in water.

New answer posted

10 months ago

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A
alok kumar singh

Contributor-Level 10

The highest industrial consumption of hydrogen gas is in the synthesis of ammonia gas (Having manufacturing of N-based fertizers)

New answer posted

10 months ago

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A
alok kumar singh

Contributor-Level 10

The highest industrial consumption of hydrogen gas is in the synthesis of ammonia gas (Having  manufacturing of N-based fertizers)

New answer posted

10 months ago

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A
alok kumar singh

Contributor-Level 10

% of C in organic compound

= 1 2 1 4 * W C O 2 W . O . C . * 1 0 0  

= 9 5 2 . 5 6 2 1 . 6 4 8 = 4 4 %  

% o f H = 2 1 8 * W H 2 O W . O . C . * 1 0 0  

= 2 0 1 8 * 0 . 4 4 2 8 0 . 4 9 2 * 1 0 0  

8 8 . 5 6 8 . 8 5 6 = 1 0 %  

= 100 – 54 = 46%

 

New answer posted

10 months ago

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Payal Gupta

Contributor-Level 10

In 4d orbital, n = 4 and l=2

Radial nodes = nl1

Radial nodes = 4 – 2 – 1 = 1

And angular nodes,  l=2

New answer posted

10 months ago

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Payal Gupta

Contributor-Level 10

L = 1 m

ΔL=0.4*103m

m=1kg

d = 0.4 * 103 m

FA=yΔLL

y=FLAΔL= (mg)*1 (πd24)*0.4*103

Δy=0.1*0.199*1012=1.99*1010

= 1.99

New answer posted

10 months ago

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P
Payal Gupta

Contributor-Level 10

Loss in P.E. = Gain in k.E

2 mg R = 12 (12mR2+mR2)ω2

ω=8g3R=4g2*3R

x=g2=5

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