Class 12th

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New answer posted

5 months ago

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P
Payal Gupta

Contributor-Level 10

Kindly consider the below image

New answer posted

5 months ago

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A
alok kumar singh

Contributor-Level 10

Uses of catalyst is very specific for a particular reaction.

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5 months ago

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A
alok kumar singh

Contributor-Level 10

Let we take  1 l  of solution

Mass of solute = Volume * Density

= 0.5  m l * 1 . 0 5 g m / m l  

= 0.525 gram

Mass of solution = 1 kg. [considering very dilute solution]

Mass of solvent = 1000 – 0.525 = 999.475 gram

i = 1 . 9  

New answer posted

5 months ago

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P
Payal Gupta

Contributor-Level 10

Ce3+ [Xe]4f15d0

Ce4+ [Xe]4f05d0  (Noble gas configuration)

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5 months ago

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A
alok kumar singh

Contributor-Level 10

Let a moles of SO2Cl2 is taken

Then no. of moles of H2SO4 = a moles

No. of moles of HCl = 2a moles

No. of moles of NaOH required = 2a + 2a = 4a = 16

 

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P
Payal Gupta

Contributor-Level 10

In sulphide ore, depressants selectively prevent impurity from coming to the fourth.

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5 months ago

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V
Vishal Baghel

Contributor-Level 10

Answer (1)

e E = e d V d r = m ω 2 r

? V = d V = R 2 R m e ω 2 r d r

= ? V = 3 m ω 2 R 2 2 e

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A
alok kumar singh

Contributor-Level 10

[ C u ( e n ) 2 ( S C N ) 2 ]

More stable isomers = 3 (trans isomers)

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P
Payal Gupta

Contributor-Level 10

Invertase → Cane sugar to glucose and fructose

Zymase → Glucose to ethanol

Diastase → Starch to Maltose

Maltase → Maltose to glucose

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5 months ago

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P
Payal Gupta

Contributor-Level 10

l=120604000=0.015A

Thus l2 = l - LL

= 0.015 – 0.006

= 0.009 = 9mA

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