67 - General Chemistry Questions Answers

for second oxide ratio of metal and oxygen = 70:30

let molcular mass of metal = M so 70/30 = 2M/3*16 

or 7/3 = 2M/48 so M = 24*7/3 = 56

now for first oxide ratio of metal and oxygen = 72.4/27.6     (By weight)

so ratio by no. of atoms = (72.4/56)/(27.6/16) = 72.4*16/27.6*56 = 1158.4/1545.6 = 3/4

Third reaction is the sum of 1st and 2nd so energy of 3rd will be the sum of energies of 1st and 2nd.

delta G = n₁FE₁ + n₂FE₂ = 1F(0.771) + 2F(0.447) = F(0.771+0.894) = 1.665F = 1.665*96368 = 160452.7 J = 38203.03 cal = 38.20303 cal

so total energy of the reaction 

Fe³⁺(aq) + 3e^- -------> Fe(s) is 38.203 cal

and value of n = 3 so per mole energy = 38.203/3 

its simple because question is based on volume and gas covers the complete volume of the vessel in which it is placed.

v = {3RT/M}^1/2

so according to the given condition 

{3RT/32}^1/2 = {3R300/16}^1/2

or T/32 = 300/16

or T = 600

so t = 600-273 = 327⁰C

1)   no of miligm eq. = 10*0.1 = 1

so no. of gm. eq. = 1/1000

since acid is monobasic so no. of moles = 1/1000

so molecules = 1/1000 * 6.023 * 10²³

2)     0.1 litre soln. contains              4 gm. NaOH

       so 1 litre contains                      40 gm. NaOH

       so 1 litre soln. contains              40/40 moles of NaOH  = 1 mole so molarity = 1

 now 1 litre soln. contains                       40 gm NaOH

so 1*1038 gm soln. contains               40 gm. NaOH                   here 1038 gm/litre is density

so 1038 gm. soln. contains                  40 gm. NaOH

so 1038-40 gm. solvent contains        40 gm. NaOH

so 998 gm. solvent contains                 40 gm. NaOH = 1 mole 

so 0.998 Kg solvent contains                 1 mole NaOH

so molality = 1/0.998 

998 gm water contains                   40 gm. NaOH

so 998/18 mole water contains    40/40 mole NaOH

so mole fraction of NaOH = [40/40]/{[40/40]+[998/18]}

Third part is irrelevent without using water

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