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New answer posted

a year ago

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P
Payal Gupta

Contributor-Level 10

This is a assertion and reason answer type question as classified in NCERT Exemplar

Correct option is (iii)

HNO3 forms an oxide layer on the surface of iron. This is also known as corrosion or rusting.

New answer posted

a year ago

0 Follower 5 Views

A
alok kumar singh

Contributor-Level 10

This is a Short Answer Type Question as classified in NCERT Exemplar

Sol:

f(x)    12− cosx,         ∀ x ∈RRange  ofcosx  is  [−1,1]Let        f(x)    =  y  =12− cosx⇒    2y−ycosx=1⇒  ycosx  =  2y−1⇒     cosx=2y−1y  =  2−1yNow−1≤cosx≤1⇒ −1≤2−1y≤1  ⇒−1−2≤−1y≤1−2⇒−3≤−1y≤−1⇒3≥1y≥1⇒13≤y≤1∴the  range  of  f=[13,1].

New answer posted

a year ago

0 Follower 7 Views

P
Payal Gupta

Contributor-Level 10

This is a assertion and reason answer type question as classified in NCERT Exemplar

Correct option is (iii)

Nitrogen gas has a complete octet structure for both atoms and is unreactive due to the presence of a strong triple bond. On the other hand,  phosphor has bonds with unstable angles strains compared to nitrogen,  therefore it burns quickly,  thus readily reacts. Nitrogen has a lower electron gain enthalpy than phosphorus.

New answer posted

a year ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

This is a Short Answer Type Question as classified in NCERT Exemplar

Sol:

Let x1 ,x2  ∈gof                                  

 gof{f(x1)}=gof{f(x2)}                             

⇒g(x1)=g(x2)                                      

 ∴
x1 =x2

Hence,f is one−one and g is onto.

New answer posted

a year ago

0 Follower 7 Views

P
Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

Correct option is (i) and (ii)

(i) The modest dispersion force causes attraction in noble gases.

(ii) In nature, xenon fluorides are reactive.

New answer posted

a year ago

0 Follower 9 Views

P
Pallavi Pathak

Contributor-Level 10

Explanation- given d= 0.1nm,  ?  =300 and n=1

According to bragg's law

2dsin ?  = n ?

2 *0.1*sin30=?

? =0.1nm=10-10

? =hmv=hp

P=h/ ?   = 6.62*10-3410-10 = 6.62 *10-24 kgm/s

k.E= 1/2mv2= 12m2v2m = p22m = 0.21eV

New answer posted

a year ago

0 Follower 7 Views

A
alok kumar singh

Contributor-Level 10

This is a Short Answer Type Question as classified in NCERT Exemplar

Sol:

Given  that, 

X={1,2,3}  and  Y={4,5}

∴X *?  Y={(1,4),(1,5),(2,4),(2,5),(3,4),(3,5)}

(i)f={(1,4),(1,5),(2,4),(3,5)}

f is not a function because f has no unique image.

(ii)g={(1,4),(2,4),(3,4)}

Since,g is a function as each element of the domain has a unique image.

(iii)h={(1,4),(2,5),(3,5)}

It is a function as each element of the domain has a unique image.

(iv)k={(1,4),(2,5)}

k is not a function as 3 does not have any image under the mapping.

New answer posted

a year ago

0 Follower 20 Views

P
Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

Correct option is (ii) and (iii)

Here fig (b) contains one S-S bond.

The oxidising behaviour of H2SO4 is represented by (ii) and (iii) among the aforementioned four. It oxidises HI in reaction (ii) and then reduces to SO2.

The oxidation state of sulphur's core atom drops from +6 to +4. It oxidises copper in (iii) and is reduced to SO2.

 

New answer posted

a year ago

0 Follower 6 Views

A
alok kumar singh

Contributor-Level 10

This is a Short Answer Type Question as classified in NCERT Exemplar

Sol:

Given function,f(x)=cos x    ∀ x∈ RLet          [−π2,π2]  ∈  f(x)f(−π2)  =  cos (−π2)  =  cos  π2  = 0cos (π2)  =  cos  π2  =0But         −π2  ≠  π2∴   ,f(x) is not one−one.Now,f(x)=cos x,∀ x∈ R is  not  onto  as  there  is  no  pre−image  for  any  real  number.Which  does  not  belong  to  the  intervals[−1,1],the  range  of  cosx.

New answer posted

a year ago

0 Follower 9 Views

P
Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

Correct option is (i)

Oxidation state of S=+6

(iii) Iron oxide with K2O and Al2O3 is used to increase the rate of attainment of equilibrium in Haber's process.

(iv) Change in enthalpy is negative for the preparation of SO3 by catalytic oxidation of SO2.

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