67 - General Chemistry Questions Answers

6.3g of hydrated oxalic acid is treated with M/20 Mg(OH)2. The volume of Mg(OH)2 used for complete neutralisation is

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Joshi sir comment

gm eq of oxalic acid = 6.3/63 = 0.1

gm eq of Mg(OH) = (1/20)*2*V

now compare the two

 

 

In a hydrogen atom electron moves around the circular orbits of radii R and 4R respectively. The ratio of the time taken by them to complete one revolution is

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Joshi sir comment

mv2/r = kqe/r2

so v2α 1/r

or (r/t)2α 1/r

or t2 α r3

now solve

 

The equivalent weight of N2 and NH3 for the given reaction respectively will be (if M1 and M2 are molar masses of NH3 and N2)

N2 => NH3

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e w of N= 28/6

e w of NH3 = 17/3

The density of a gas is 3.80 g/l at STP. Calculate its density at 27°C and 700 torr pressure.

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Joshi sir comment

P = dRT/M

for first condition

760 = 3.8R273/M

and 700 = dR300/M

divide and get the answer

 

Two oxides of a metal M contain respectively 22.53% and 30.38% of O2. If the formula of first oxide is MO, the formula of second oxide is

(a) M2O3         (b) MO3         (c) M2O         (d) M2O4

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                        M                          O

                     77.47                   22.53

m w               M                           16

formula for first oxide is MO

so 77.47/M = 22.53/16 

so M = 77.47*16/22.53 = 55

 

                    M                         O

               69.62                   30.38

m w         55                         16

so M:O = 69.62/55:30.38/16 = 1.27:1.90

1.9/1.27 = 1.5

so formula will be M2O3

if radius of first bohr orbit of hydrogen atom is X, then de Broglie wavelength of electron in 4 th orbit is

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Joshi sir comment

radius of fourth orbit = 16X

now put 2πr = 4λ

 

Is there is any  relation between bond order and bond angle

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no

the correct increasing order of lattice energy of the above compounds is - AlF3 , Al2O3 and AlN

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AlF3              5924 kJ/mol

Al2O3           15916 kJ/mol  

AlN               9506.8 kJ/mol

24.5g KClO3 heated to give O2, which is allowed to react completely with H2 to form H2O. This H2 comes from Zn and H2SO4 reaction. The amount of Zn required for the purpose is

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Joshi sir comment

2KClO3 ----------------------> 2KCl + 3O2

3O2 + 6H2 --------------------> 6H2O

6Zn + 6H2SO4 -------------------> 6ZnSO4 + 6H2

so 2 mole  KClO3  -----------------------------  6 mole Zn

now solve

If the potential energy of hydrogen electron is -3.02eV in which of the following energy levels is electron present =

1st , 2nd , 3rd  or  4th 

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Joshi sir comment

Energy = -13.6/n2 ,  Potential Energy = -27.2/n2

for n = 3 we get -3.02  

in this answer infinite is zero potential energy level

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