Monday, August 23, 2021

Hydrogen: chemistry


  • Answers are in BOLD Option name.

 Q.1.  When COblank subscript 2  is bubbled through a solution of barium peroxide in water 

A) Oblank subscript 2 is released 

B) Carbonic acid is formed 

C) Hblank subscript 2Oblank subscript 2 is formed 

D) no reaction occurs 

Explanation-

BaO subscript 2 plus CO subscript 2 plus straight H subscript 2 straight O rightwards arrow BaCO subscript 3 downwards arrow plus straight H subscript 2 straight O subscript 2 


Q.2.  The hydrogen molecule is not very reactive under normal conditions because 

A) Hydrogen has a large number of isotopes 

B) The corresponding triangle S degree value is very low 

C) The bond dissociation energy of  the hydrogen molecule is high 

D) The hydrogen molecule has a low activation energy 

Explanation-

High bond dissociation energy of hydrogen molecule explains its less reactive nature.


Q.3.  Decomposition of H subscript 2 O subscript 2  can be prevented by addition of

A) Acetanilide

B) Benzene

C) Alkali metal oxide 

D) MnOblank subscript 2 

Explanation-

Acetanilide acts as stabilizer for straight H subscript 2 straight O subscript 2.


Q.4. Which is the correct order of boiling point of VA, VIA and VIIA hydride ?

A) NH subscript 3 space greater than space SbH subscript 3 space greater than space AsH subscript 3 space greater than space PH subscript 3

B) straight H subscript 2 straight O space greater than space straight H subscript 2 Te space greater than space straight H subscript 2 Se space greater than space straight H subscript 2 straight S

C) HF > HI > HBr > HCI

D) All of the above are correct 


Q.5.   HCI is added to following oxides . Which one would give H subscript 2 O subscript 2 ?

A) MnOblank subscript 2

B) PbOblank subscript 2

C) BaO

D) None of the above 

Explanation-

Only peroxides on action of dil. acid gives straight H subscript 2 straight O subscript 2 


Q.6.  Which of the following ions will cause hardness in water sample ?

A) Nablank to the power of plus

B) Kblank to the power of plus 

C) Brblank to the power of minus 

D) Cablank to the power of 2 plus end exponent 

Explanation-

Soluble Salts of calcium can cause hardness in water.


Q.7.  In context with the industrial preparation of hydrogen from water gas (CO+Hblank subscript 2) , which of the following is the correct statement ?

A) CO is oxidized to COblank subscript 2 with steam in the presence of a catalyst followed by absorption of COblank subscript 2 in alkali .

B) CO and Hblank subscript 2 are fractionally separated using differences in their densities.

C) CO is removed by absorption in aqueous Cublank subscript 2Clblank subscript 2 solution .

D) Hblank subscript 2 is removed through occlusion with Pd


Q.8.  A dilute solution of straight H subscript 2 straight O subscript 2  can be concentrated by 

A) Drying it over anhydrous CaClblank subscript 2

B) Drying it over concentrated Hblank subscript 2SOblank subscript 4

C) Drying it over anhydrous MgSOblank subscript 4 

D) Heating it under reduced pressure 

Explanation-

A dilute solution of straight H subscript 2 straight O subscript 2 can be concentrated by heating (distillation) under reduced pressure.


Q.9. Hydrides of metals are named like 

A) alkane 

B) alkene

C) alkyne

D) None is correct


Q.10. Consider the following statements : 

I : Rate of transfer  of Dblank to the power of plus  from Dblank subscript 2O  is  slower then that of Hblank to the power of minus b from Hblank subscript 2O

II :  Kblank subscript a for CHblank subscript 3COOH rightwards harpoon over leftwards harpoon CHblank subscript 3COOblank to the power of minus  + Hblank to the power of plus

III : Tritium is a  radioactive isotope . Select the correct statements .

A) I , II 

B) II , III 

C) I , III 

D) I, II , III 






Properties Of Bulk Matter: Physics

  • Answers are in BOLD Option name.


Q.1. An iron bar of length l and cross-section A is heated from 0 degree C space t o space 100 degree C. If the rod is so held that it is not permitted to expand or bend, the force developed in it is:

A) directly proportional to length

B) inversely proportional to length

C) independent of length

D) inversely proportional to area of cross-section

Explanation-

Change in length increment l equals alpha divided by increment T


Stress = F over A space a n d space S t r a i n equals fraction numerator increment l over denominator l end fraction

N o w space Y equals F over A cross times fraction numerator l over denominator increment l end fraction equals F over A cross times fraction numerator l over denominator alpha l increment T end fraction
o r space F equals Y space alpha space A space increment space T.

Clearly F does not depend upon length.


Q.2. The radius of a spherical soap bubble is 0.5 mm. If the surface tension of the soap solution be 30 dyne/ cm, the excess pressure is :

A) 1500 space d y n e divided by c m squared

B) 1000 space d y n e divided by c m squared

C) 2400 space d y n e divided by c m squared

D) 1200 space d y n e divided by c m squared

Explanation-

For soap bubble, excess pressure is 

        P subscript i minus P subscript 0 equals fraction numerator 4 T over denominator r end fraction
space space space space space space space space space space space space space space space space space space space equals fraction numerator 4 cross times 30 over denominator 0.5 cross times 10 to the power of negative 1 end exponent end fraction equals 2400 space d y n e divided by c m squared 


Q.3. A thick rope of rubber of density 1.5 kg/m cubed and Y equals 5 cross times 10 to the power of 6 space N space m to the power of negative 2 end exponent metres in length when hung vertically will increase in length by :

(g=10 m s to the power of negative 2 end exponent )

A) 9.6 cross times 10 to the power of negative 5 end exponent space m

B) 9.6 space m

C) 19.2 cross times 10 to the power of negative 3 end exponent space m

D) 19.2 cross times 10 to the power of negative 5 end exponent space m 

Explanation-

As the weight of wire is at center of gravity

therefore space O n l y space h a l f space t h e space l e n g t h space o f space r o p e space g e t s space e x t e n d e d.
space space space space space space space space space space space Y equals F over A. fraction numerator open parentheses L divided by 2 close parentheses over denominator increment l end fraction.
therefore space space space increment l equals fraction numerator M g over denominator A Y end fraction. L over 2 equals fraction numerator A L p g. h over denominator 2 A y end fraction equals fraction numerator L squared p g over denominator 2 Y end fraction.
space space space space space space space space space space space space equals fraction numerator 64 cross times 1.5 cross times 10 over denominator 2 cross times 5 cross times 10 to the power of 6 end fraction equals 9.6 cross times 10 to the power of negative 5 end exponent space m 

Q.4. The density of ice is x g c m to the power of negative 3 end exponent and that of water is y g c m to the power of negative 3 end exponent. What is the change in volume when m gram of ice melts ?

A) m ( y - x ) c.c

B) fraction numerator y minus x over denominator m end fraction space c. c

C) m space x y space left parenthesis x minus y right parenthesis space c. c

D) m open parentheses 1 over y minus 1 over x close parentheses space c. c 

Explanation-

Vol. of m gram water formed equals m over y space c. c

Change in vol. on melting open parentheses m over y minus m over x close parentheses

                      equals m open parentheses 1 over y minus 1 over x close parentheses space c. c 


Q.5. Water rises to a height of 10 cm in capillary tube and mercuy falls to a depth of 3.42 cm in the same tube. If the density of mercury is 13.6 g  c m to the power of negative 3 end exponent and angle of contact is 135 degree, the ratio of the surface tensions for water and mercury is :

A) 1 : 5.57

B) 1 : 3.57

C) 1 : 6.57

D) 1 : 1.57

Explanation-

T subscript omega over T subscript m equals fraction numerator h subscript omega d subscript omega over denominator cos space theta subscript omega end fraction cross times fraction numerator cos space theta subscript m over denominator h subscript m cross times d subscript m end fraction equals fraction numerator 10 cross times 1 cross times cos space 135 degree over denominator 1 cross times 3.42 cross times 13.6 end fraction
space space space space space space space equals fraction numerator 10 cross times 1 over denominator 3.42 cross times 1.414 cross times 13.6 end fraction equals fraction numerator 1 over denominator 6.57 end fraction 


Q.6. The isothermal bulk modulus of an ideal gas at a pressure P is:

A) P

B) y P

C) P/2

D) P/y

Explanation-

 For isothermal process PV + constant 

Differentiating both sides,

p d V space plus space V d p space equals space 0 space space space space
space
space space space space space space space space space space space space space space space P d v space equals space minus space V space d p space space 
rightwards double arrow space space space space space space P equals negative fraction numerator V d p over denominator d V end fraction.
o r space space space space space space P equals negative fraction numerator d P over denominator open parentheses d V divided by V close parentheses end fraction
o r space sin c e space b u l k space m o d u l u s comma space K equals negative fraction numerator d P over denominator d V divided by V end fraction.
therefore space space space space space space space space space space space space space space space space space space K plus P 


Q.7. The reading of a pressure meter attached with a closed water pipe is 3.5 cross times 10 to the power of 5 space N divided by m squared. On opening the valve, the reading of the pressure meter is reduced to 3 cross times 10 to the power of 5 space N divided by m squared. Calculate the speed of water flowing in the pipe.

A) 10 m/s

B) 20 m/s

C) 30 m/s

D) 40 m/s.

Explanation-

According to Bernoulli's theorem,

  P subscript 1 over p plus 1 half space v subscript 1 squared equals P subscript 2 over p plus v subscript 2 squared
fraction numerator 3.5 cross times 10 to the power of 5 over denominator 10 cubed end fraction plus 0 equals fraction numerator 3 cross times 10 to the power of 5 over denominator 10 cubed end fraction plus 1 half v subscript 2 squared
350 minus 300 equals 1 half v subscript 2 squared
space space space space space space space space space v subscript 2 squared equals 50 cross times 2 equals 100
space space space space space space space space space v subscript 2 squared equals 10 space m divided by s. 


Q.8. A circular tub filled with liquid is uniformly tapering with base area 10 to the power of negative 2 end exponent space m squared and top area 2 cross times 10 to the power of negative 3 end exponent space m squared. Its depth is 0.4 m as shown in fig. What is the thrust on its base ? open parentheses g equals 10 space m s to the power of negative 2 end exponent comma space p equals 900 space k g space m to the power of negative 3 end exponent close parentheses space. 

A) 3.6 N

B) 7.2 N

C) 9 N

D) 14.4 N

Explanation-

Here thrust at bottom equals space h d g space cross times space A

                     F equals 0.4 cross times 900 cross times 10 cross times 10 to the power of negative 3 end exponent
space space space equals 3.6 space N 


Q.9. An incompressible liquid flows through a horizontal tube bifurcating at the end as shown

The areas are expressed in m squared and velocity in m space s to the power of negative 1 end exponent. What is the value of v subscript 0 ?

A) 3 m space s to the power of negative 1 end exponent

B) 1.5 m space s to the power of negative 1 end exponent

C) 1 m space s to the power of negative 1 end exponent

D) 1.25 m space s to the power of negative 1 end exponent

Explanation-

Here a v subscript 0 equals a subscript 1 v subscript 1 plus a subscript 2 v subscript 2 cross times 3 equals a cross times 1.5 plus 1.5 a cross times v subscript 2
o r space space space space space space space space v subscript 2 equals 1 space m s to the power of negative 1 end exponent 


Q.10.  A log of wood floats in water with 1/5th of its volume above the surface of water. The density of wood is :

A) 0.8 cross times 10 cubed space k g space m to the power of negative 3 end exponent

B) 8 cross times 10 cubed space k g space m to the power of negative 3 end exponent

C) 0.08 cross times 10 cubed space k g space m to the power of negative 3 end exponent

D) 10 cubed space k g space m to the power of negative 3 end exponent 

Explanation-

Here the wt. of the body = Wt. of water displaced by immersed portion of the body

        space space space space space space space space space space space space space space space space V space cross times p g equals 4 over 5 space V. space d. space g
therefore space space space space space space space space space space space space space space space space space space space p equals 4 over 5. space d equals 08 cross times 10 cubed space k g space m to the power of negative 3 end exponent.