217 - Chemistry Questions Answers

corner atoms = 8*1/8 = 1

diagonal atoms = 4*2 = 8

total = 9

in second case calculate similarly

E_cell = E⁰_cell - (0.0591/n)  log {[Ni⁺⁺]³ / [Au⁺⁺⁺]²}

Reactions are 3Ni ------> 3Ni⁺⁺   +   6e^-

and      2Au⁺⁺⁺ +   6e^- ---------->   2Au  

now put E⁰_cell = E⁰_c - E⁰_a = 1.5 - (-0.25)      here you should remember that you have to take only reduction potentials not oxidation potentials

                                               = 1.75

n = 6  and concentrations if given otherwise take these two equal to zero.

HCl, HNO₃ , HCOOH are mono basic acids

COOH-COOH, H₂SO₄ are di basic acids

similarly others 

E = hc/λ

so E is inversely proportional to λ 

for balmer series of hydrogen atom 

 1/λ = R ( 1/2²  -   1/3²)     

solve for λ

for n^th level values of l will be 0, 1,  2,  3, 4, 5, .....................(n-1) 

for l = 0   means s subshell no. of electrons = 2      = 2(2*0+1) 

for l = 1 means p subshell no. of electrons = 6        = 2(2*1+1)

for l = 2 means d subshell no. of electrons = 10      = 2(2*2+1)

similarly for others 

No. of gm. equivalent of KMnO₄ = No. of gm. equivalent of H₂O₂

so 1*10/1000   =     x  /  17             (17 is eq. wt. of H₂O₂) 

so  x = 0.17

so % purity = 0.17*100/0.2 = 85 

N₁V₁ = N₂V₂

so 10*10 = 0.1*v 

or v = 1000 ml

so extra water added will be 990 ml

10 % w/v acetic acid means 10 gm. in 100 ml. soln.

so 1000 ml soln. contains 100 gm acetic acid

so 1 litre contains 100/60 moles of acetic acid

so M = 100/60 = 1.67

so N = M * basicity = 1.67*1 = 1.67 

for max transitional energy we will use 

E = hcR_H {1/1²  - 1/∞²}

so E will be proportional to R_H