Chemistry

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New answer posted

a year ago

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V
Vishal Baghel

Contributor-Level 10

Kindly consider the following figure

New answer posted

a year ago

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V
Vishal Baghel

Contributor-Level 10

 ln (K2K1)=EaR (1T1−1T2)

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

a year ago

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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=0.0423 L

VH2=0.007462*22.7=0.0846 L Vtotal≅127mL

New answer posted

a year ago

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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

a year ago

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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

a year ago

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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=151.93≅152g/mol

New answer posted

a year ago

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A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

a year ago

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V
Vishal Baghel

Contributor-Level 10

Kindly consider the following figure

New answer posted

a year ago

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V
Vishal Baghel

Contributor-Level 10

  [Co (H2O)6]Cl2⇒Co2+⇒3d7⇒t2g5eg2

Number of unpaired electrons = 3

[Cr (H2O)6]Cl3⇒Cr3+⇒3d3⇒t2g3eg0

Number of unpaired electrons = 3

New question posted

a year ago

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