Class 12th

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New answer posted

11 months ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

 ln (K2K1)=EaR (1T11T2)

ln (K2K1)=5326118.3* (10310*300)

Where, K2 is at 310 K and K1 at 300K

ln (K2K1)=6.9=3*ln10

ln (K2K1)=ln103

K2 = K1 * 103

K1 = K2 * 10-3

 x = 1

New answer posted

11 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

Case 1: 1 14x211

4x2 – 1 1or4x211

x224orx20

So x  (, 12] [12, ) {0}

New answer posted

11 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

At anode (oxidation)

2H2O O2 (g) + 4H+ + 4e-

At cathode (Reduction)

2H+ + 2e- H2 (g)

No. of gm- equivalents = i*t96500=0.1*2*60*6096500=0.00746

VO2=0.007464*22.7=0.0423L

VH2=0.007462*22.7=0.0846L Vtotal127mL

New answer posted

11 months ago

0 Follower 8 Views

V
Vishal Baghel

Contributor-Level 10

Mass = d * v = 1.02 * 1.2 = 1.224gm

Moles of acetic acid = 0.0204 moles in 2L

So molality = 0.0102 mol/kg

Δ T f = i * K f * m

i = 1 + a for acetic acid

0.0198  = (1 + a) * 1.85 * 0.0102

α = 0.04928 = 5%

New answer posted

11 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

85gm NH3 = 5 moles of NH3

Enthalpy change for 1 mol = 23.4 kJ

Then enthalpy change for 5 mol = 23.4 * 5 = 117 kJ

New answer posted

11 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

Let perimeter of Δ is x and that of square is 22 – x

 

now area =34 (x3)2+ (22x4)2

for maximum or minimum,  dAdx=0

=2233+4

now side of a Δ=x3

=2233 (3+4)

=669+4

New answer posted

11 months ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

Assuming ideal behaviour,  P=dRTM

P=100760atm, T=257+273=530K

d = 0.46gm/L

M=0.46*0.082*530100*760=151.93152g/mol

New answer posted

11 months ago

0 Follower 21 Views

A
alok kumar singh

Contributor-Level 10

Let A, A' be (, 2) AB and A'B subtends π4 angle at (0, 0) slope of OA = 2α

 

slope of OB = 32

tanπ4=|2α321+2α32|

1+3α=± (43α2α)

α+3α=± (43α2α)α=10, 25

now distance between A'A, (10, 2) &  (25, 2)is525

New answer posted

11 months ago

0 Follower 16 Views

A
alok kumar singh

Contributor-Level 10

an+2=2an+1an+1

a2=2a1a0+1

a2 = 1

a3 = 3

a4 = 6

So for n 2,an=n(n1)2

n=2n(n1)27n=172+373+674+......

Let S = 172+373+674+.....

S7=173+374+....SS7=172+273+374+....

67S17=173+273+....67S649S=172+173+....

3649S=172117

S=172*76*4936

S=7216

New answer posted

11 months ago

0 Follower 29 Views

A
alok kumar singh

Contributor-Level 10

Use aRb = a is related to b, belongs to A iff a belongs to A.

In simple terms, aRb is true if both a & b belongs to the same set.

For reflexive

aRa, a A, so it is true.

For symmetric

Let aRb be true

Þ a & b belongs to the same set.

Þ b & a also belongs to the same set

Þ bRa will be true

For transitive

Let aRb and bRc be true.

aRb Þ a, b belongs to the same set

bRc Þ b, c belongs to the same set

Þ (a, c) belongs to the same set

Þ so aRc will be true.

So R is an equivalence relation.

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