Class 12th

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New answer posted

a year ago

0 Follower 5 Views

P
Payal Gupta

Contributor-Level 10

In acidic solution

3Mn+6O42−+4H+→2Mn+7O4−+Mn+4O2+2H2O

Difference in oxidation state of Mn is product

= 7 – 4 = 3

New answer posted

a year ago

0 Follower 1 View

P
Payal Gupta

Contributor-Level 10

Baryte – BaSO4 [Sulphate based ore]

Galena – PbS

Zinc blende – ZnS

Copper pyrite – CuFeS2

New answer posted

a year ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

S = { n : 1 ≤ n ≤ 5 0 & n = o d d }  

= { 1 , 3 , 5 , . . . . . . , 4 9 ( 2 5     t e r m s ) }  

| A | = | 1 0 4 − 1 1 0 − a 0 1 | = 1 + a2

∴ ∑ a ∈ S d e t ( a d j       A ) = 1 0 0 λ ⇒ ∑ ( 1 + a 2 ) 2 = 1 0 0 λ ⇒ λ = 2 2 1

                

New answer posted

a year ago

0 Follower 7 Views

P
Payal Gupta

Contributor-Level 10

Number of moles of CH3COOH Adsorbed on 0.6 gm wood charcoal.

= ( 0 . 2 * 2 0 0 ) * 1 0 − 3 − ( 0 . 1 * 2 0 0 ) * 1 0 − 3                

= 20 * 10-3 moles

Mass of CH3COOH absorbed = 20 * 10-3 * 60 = 1.2 gm

Mass of CH3COOH adsorbed per gram = 1 . 2 0 . 6 = 2 g of charcoal.

New answer posted

a year ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

K = 1 R * c e l l       c o n s t a n t

= 0.152 * 10-3 * 1750

 = 266 * 10-3 cm-1

New answer posted

a year ago

0 Follower 7 Views

P
Payal Gupta

Contributor-Level 10

According to Henry's law

PGas=KH*xGas

=0.835=1.67*103* [nCO2nCO2+55.5]

As nCO2<<55.5

⇒nCO2=27.78*10−3moles

Mass of CO2 gas = nCO2* (MM)CO2

= 1222 * 10-3 gm

New answer posted

a year ago

0 Follower 9 Views

A
alok kumar singh

Contributor-Level 10

For inconsistent, Δ = 0 , | 1 1 1 α 2 a 3 1 3 α 5 | = 0

a = 1

∴ Δ 1 = | 1 1 1 − 1 2 3 4 3 5 | = 1 ( 1 ) + 1 ( 1 2 + 5 ) + 1 ( − 3 − 8 ) = 7 ≠ 0              

New answer posted

a year ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

  6 1 1 = Required probability         

 

After solving, we get n = 4

New answer posted

a year ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

Surface area, S = 4pr2

? d s d t = 4 ? . 2 r d r d t = 8 ? d r d t = c o n s t a n t = k ( s a y )                

? d s d t = k ? s = k t + c

? 4 ? r 2 = k t + c        

Initially t = 0, r = 3

c = 36 p

When t = 5, r = 7, k = 32p

When t = 9, r = r, r = 9

 

New answer posted

a year ago

0 Follower 3 Views

P
Payal Gupta

Contributor-Level 10

Since student guesses only two wrong. So there are three possibilities

(i) both wrong in section A

(ii) both wrong in section B

(iii) one wrong in each section A and B.

∴ Required possibilities =

=4C4*6C4(34)4*(14)4(34)2+4C3*6C5(34)3(14)5*14*34 +4C2*6C6*(34)2(14)2*(14)6

=27410[15*27+24*3+2]=27*479410

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