216 - Chemistry Questions Answers

differential form of boyle's law

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Joshi sir comment

 dV/dP = -V/P

In [Ag(CN)2]- and [Ag(NH3)2]- , the number of π-bond respectively are

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Joshi sir comment

number of π-bond in [Ag(CN)2]is 4 and in [Ag(NH3)2], it is 0

the de- Broglie wavelength of a tennis ball of mass 200g and moving with a speed of 5m/h is of the order

10-20  , 10-30 ,   10-40

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Joshi sir comment

λ = h/mv

put values and solve but first convert all values in MKS

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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Joshi sir comment

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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Joshi sir comment

                        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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Joshi sir comment

no

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