102 - Physical Chemistry Questions Answers

M of H₂SO₄ = 1/200

so [H⁺] = 2*1/200 = 1/100 = 10-2

now calculate pH

no of hydroxide ion in pure water = 10^-7

so according to the given condition no. of hydroxide ions in solution = 2*10^-7

so pOH = -log2*10^-7

so pH = 14 - pOH 

HCN ------------------->  H⁺ + CN^-

0.1                                 0          0

0.1-x                              x           x

pH = 5.2

so -log[H⁺] = 5.2 

solve for [H⁺], it will give x

then calculate K_a

2 heat of atomization of C + 4 heat of atomization of H - 4 bond energy of C-H - 1 bond energy of C=C = heat of formation of C₂H₄

2*171 + 4*52.1 - 4*99.3 - x = 12.5

solve

n is the no. of electrons in anode or cathode reaction.

If no. of electrons are different then make these no. common

for ex. if reactions are 

             Al ---------------> Al⁺⁺⁺ + 3e^-

and     Cu⁺⁺ + 2e^- ------------> Cu

then multiply (1) to 2 and (2) to 3

after that electrons will be 6 in both reactions so n = 6

Integral enthalpy 

please check this word if wrong submit the correct question again

question is incorrect 

because (du/dv)s represents pressure and T is temperature

I think question is incomplete, correct it first

CH₃COOH  = CH₃COO^- + H+

     1                     0                 0

    0.8                 0.2               0.2

CH₃COO^- + H⁺ = CH₃COOH

0.2             0.2               0.8

0.06            0.06           0.8+0.14

0.06             0.06              0.94                      sum = 1.06

so i = 1.06/1 = 1.06 

By formula πV = nRT

π = hdg, n=w/M

given 

0.2%wt/vol solution in acetone means w= 0.2 gm, V = 100 ml= 100/1000 litre = 0.1*10^-3 m³ ,   h = 23.1/1000 m, d = 800 kg/m³ ,  

now solve