94420 + Interview Questions in Chemistry in Class 12

1.	Fill in the blanks by choosing the appropriate word/words from those given in the brackets : (Natural, synthetic, step-growth, chain-growth, natural rubber, synthetic rubber, melamine, thermo, thermosetting, addition, condensation, hydrogen bonds, van der Waals', 373-573, 543, 100, 10, benzoyl peroxide, acetyl peroxide, amide, butadiene, natrium, dimethyl terephthalate) Bakelite is a common example of ..... plastics.
Answer» SOLUTION :THERMOSETTING

Discussion

2.	Fill in the blanks by choosing the appropriate word/words from those given in the brackets : (Natural, synthetic, step-growth, chain-growth, natural rubber, synthetic rubber, melamine, thermo, thermosetting, addition, condensation, hydrogen bonds, van der Waals', 373-573, 543, 100, 10, benzoyl peroxide, acetyl peroxide, amide, butadiene, natrium, dimethyl terephthalate) ............ polymer is used for making unbreakable plates and cups.
Answer» SOLUTION :MELAMINE

Discussion

3.	Fill in the blanks by choosing the appropriate word/words from those given in the brackets : (Natural, synthetic, step-growth, chain-growth, natural rubber, synthetic rubber, melamine, thermo, thermosetting, addition, condensation, hydrogen bonds, van der Waals', 373-573, 543, 100, 10, benzoyl peroxide, acetyl peroxide, amide, butadiene, natrium, dimethyl terephthalate) Depending upon the growth of chain, two types of polymerization are .......... and ............
Answer» SOLUTION :chain-growth, step-growth

Discussion

4.	Fill in the blanks by choosing the appropriate word/words from those given in the brackets : (Natural, synthetic, step-growth, chain-growth, natural rubber, synthetic rubber, melamine, thermo, thermosetting, addition, condensation, hydrogen bonds, van der Waals', 373-573, 543, 100, 10, benzoyl peroxide, acetyl peroxide, amide, butadiene, natrium, dimethyl terephthalate) Isoprene is the monomer of ............
Answer» SOLUTION :NATURAL RUBBER

Discussion

5.	Fill in the blanks by choosing the appropriate word/words from those given in the brackets : (Natural, synthetic, step-growth, chain-growth, natural rubber, synthetic rubber, melamine, thermo, thermosetting, addition, condensation, hydrogen bonds, van der Waals', 373-573, 543, 100, 10, benzoyl peroxide, acetyl peroxide, amide, butadiene, natrium, dimethyl terephthalate) Wool is an example of .............. polymers while nylon-6,6 is an example of ........ polymers.
Answer» SOLUTION :NATURAL, SYNTHETIC

Discussion

6.	Fill in the blanks by choosing the appropriate word/words from those given in the brackets : (Natural, synthetic, step-growth, chain-growth, natural rubber, synthetic rubber, melamine, thermo, thermosetting, addition, condensation, hydrogen bonds, van der Waals', 373-573, 543, 100, 10, benzoyl peroxide, acetyl peroxide, amide, butadiene, natrium, dimethyl terephthalate) Proteins are .......... polymers.
Answer» SOLUTION :NATURAL

Discussion

7.	Fill in the blanks by choosing the appropriate word from those given in the brackets. (isotropic, anisotropic, 0.255–0.414, 0.414–0.732, unit cell AB AB, .... ABC ABC ..., ...., 52.4, 60.4, F-centres, V-centres, 4, 2, hexagonal close packing, square close packing, 6, 8 glass, plastics, lead zirconate, barium titanate) A crystal lattice, a built of repititive units called......
Answer» SOLUTION :UNIT CELL

Discussion

8.	Fill in the blanks by choosing the appropriate word from those given in the brackets. (isotropic, anisotropic, 0.255–0.414, 0.414–0.732, unit cell AB AB, .... ABC ABC ..., ...., 52.4, 60.4, F-centres, V-centres, 4, 2, hexagonal close packing, square close packing, 6, 8 glass, plastics, lead zirconate, barium titanate) Out of barium titanate and lead zirconate, ......is antiferroelectric solid.
Answer» SOLUTION :LEAD ZIRCONATE

Discussion

9.	Fill in the blanks by choosing the appropriate word from those given in the brackets. (isotropic, anisotropic, 0.255–0.414, 0.414–0.732, unit cell AB AB, .... ABC ABC ..., ...., 52.4, 60.4, F-centres, V-centres, 4, 2, hexagonal close packing, square close packing, 6, 8 glass, plastics, lead zirconate, barium titanate) The electrons trapped in anion vacancies in metal excess defects are called........
Answer» SOLUTION :F-centres

Discussion

10.	Fill in the blanks by choosing the appropriate word from those given in the brackets. (isotropic, anisotropic, 0.255–0.414, 0.414–0.732, unit cell AB AB, .... ABC ABC ..., ...., 52.4, 60.4, F-centres, V-centres, 4, 2, hexagonal close packing, square close packing, 6, 8 glass, plastics, lead zirconate, barium titanate) For tetrahedral coordination, the radius ratio should be.......
Answer» SOLUTION :0.255-0.414

Discussion

11.	Fill in the blanks by choosing the appropriate word from those given in the brackets. (isotropic, anisotropic, 0.255–0.414, 0.414–0.732, unit cell AB AB, .... ABC ABC ..., ...., 52.4, 60.4, F-centres, V-centres, 4, 2, hexagonal close packing, square close packing, 6, 8 glass, plastics, lead zirconate, barium titanate) Amorphous solids are ........
Answer» SOLUTION :ISOTROPIC

Discussion

12.	Fill in the blanks by choosing the appropriate option. Cone. H_(2)SO_(4) chars paper, wood and sugar by removing (i) from them. It is also known as (ii) It is manufactured by (iii) process. It is a strong (iv) and (v) acid.
Answer» `{:("(i)","(II)","(III)","(iv)","(v)"),(H_(2)O,"oil of vitriol ","Contact ","oxidising","dibasic"):}` `{:("(i)","(ii)","(iii)","(iv)","(v)"),(O_(2),"oil of vitriol ","Oleum ","dehydrating","monobasic"):}` `{:("(i)","(ii)","(iii)","(iv)","(v)"),(H_(2)O,"oil of olay ","Solvay ","dehydrating","dibasic"):}` `{:("(i)","(ii)","(iii)","(iv)","(v)"),(SO_(2),"oil of winter green ","Contact ","oxidising","monobasic"):}` ANSWER :A

Discussion

13.	Fill in the blanks : (A) CH_(3)CH_(2)I overset(NaCN)to ? overset(OH^(-), "Partial hydrolysis")to? (B) C_(6)H_(5)overset(+)(N_(2))overset(-)(Cl) overset(CuCN)to ? underset(H^(+))overset(H_(2)O)to ?
Answer» Solution :(A) `CH_(3)CH_(2CN, CH_(3)CH_(2)CONH_(2)` (B) `C_(6)H_(5)CN, C_(6)H_(5)COOH`

Discussion

14.	Fill in the blanks. ""_92^235U + ""_0^1n to ""_52^137 A + ""_40^97B + ….
Answer» SOLUTION :`2 ""_0^1n`

Discussion

15.	Fill in the blanks: (a) ._(92)^(235)U + ._(0)^(1)n rarr ._(55)^(142)A + ._(37)^(92) B+.... (b) ._(34)^(82)Se rarr ...+2 ._(-1)e^(0)
Answer» Solution :`._(92)^(235)U+_(0)^(1)nrarr ._(52)^(137)A+_(40)^(97)B+X ._(0)n^(1)` (as the reaction is balancedwith respect to NUCLEAR charge, the missing particle must be neutral i.e. NEUTRON) Applying mass nuclear charge balancing `34=-2+z` `z=36" "` (This implies ELEMENT is kr) Applying mass NUMBER balancing `82=0+A` `A=82` Final balanced reaction is `._(34)^(82)Se rarr 2 ._(-1)^(0)e + _(36)^(82) Kr`

Discussion

16.	Fill in the blanks. ""_34^82Se to 2 ""_(-1)^0e + ….
Answer» SOLUTION :`""_36^82Kr`

Discussion

17.	Fill in teh blanks. i.H_(2)SO_(4)+HI rarr …..+SO_(2)+2H_(2)O ii.CaOCI_(2)+NaI+HCI rarr ….. +CaCI_(2)+H_(2)O+NaCI iii. NH_(3)+CI_(2)(excesse) rarr …. + HCI iv. KMnO_(4)+KCI+H_(2)SO_(4) rarr K_(2)SO_(4)+MnSO_(4)+…..+….. v. K_(2)Cr_(2)O_(7)+HCI rarr KCI+Cr4CI_(3)+.....+.... vi.CuSO_(4)+KI rarr Cu_(2)I_(2)+.....+.... vii. NH_(3)+NaOCI rarr N_(2)+NaCI+..... viii. CI_(2)+H_(2)O+HgO rarr HgCI_(2).HgO+..... ix.P_(4)+I_(2)+H_(2)O rarr H_(3)PO_(3)+....... x. NaBr+MnO_(2)+H_(2)S_(4)rarr NaHSO_(4)+......+H_(2)O+.....
Answer» Solution :(i) `I_(2)`, (II) `I_(2)` (iii) `NCl_(3)`, (iv) `Cl_(2),H_(2)O` (V)`H_(2)O,Cl_(2)`, (VI) `K_(2)SO_(4),I_(2)` (vii) `H_(2)O` , (viii) `HOCl` (ix)`HI`, (x)`MnSO_(4),Br_(2)`.

Discussion

18.	Fill in the blank space with a suitable answer selected from the list below. Write only the letter (A,B,C…..etc.) of the correct answer in the blanks {:((i) ._(6)^(12)C + ._(1)^(1)H to ._(7)^(13)N,A : "Projectile" "capture"),((ii) ._(13)^(27)Al + ._(1)^(1)H to ._(12)^(24)Mg + ._(2)^(4)He,B : "Spallation"),((iii) ._(92)^(235)U + ._(0)^(1)n to ._(56)^(140)Ba + ._(36)^(93)Kr + 3 ._(0)^(1)n, C : "Fusion"),((iv) ._(33)^(75)As + ._(1)^(2)H to ._(25)^(56)Mn + 9 ._(1)^(1)H + 12 ._(0)^(1)n, D : "Projectile" "capture" "and" "particle" "emission"),((v) ._(1)^(2)H + ._(1)^(3)H to ._(2)^(4)He + ._(0)^(1)n, E :"fission"):} Select the correct answers according to the given codes:
Answer» `{:(,(i),(ii),(III),(IV),(V)),((a),A,D,E,B,C):}` `{:(,(i),(ii),(iii),(iv),(v)),((b),D,C,A,E,B):}` `{:(,(i),(ii),(iii),(iv),(v)),((C ),A,B,C,D,E):}` `{:(,(i),(ii),(iii),(iv),(v)),((d ),E,D,C,B,A):}` Answer :a

Discussion

19.	………….filament is used to decompose titanium tetraoxide
Answer» SOLUTION :TUNGSTEN

Discussion

20.	Figure displays the plot of the compression factor Z versus P for a few gases which of the following statements is/are correct for a van der Waals' gas?
Answer» The PLOT I is applicable provided the van DER Waals' CONSTANT a is negligible The plot II is applicable provided the van der waals' constant b is negligible The plot III is applicable provided the van der waals' constant a and b are negligible the plot IV is applicable provided the temperature of theh gas is must higher than its critical temperature. Answer :A::B::C

Discussion

21.	Figure explains elevation in boiling point when a non-volatile solute is added to a solvent. Provide a thermodynamic explanation of the elevation in boiling point.
Answer» SOLUTION :N//A

Discussion

22.	Fifth group elements form hydrides of type AH_3.The hydrides have a lone pair of electrons.The hydrides are reducing in nature and the reducing power is related to the stability of A-H bonds.The hydrides are covalent and low boiling.Their boiling point depends on their ability to form hydrogen bond and their molecular size which which decide the intermolecular forces in the hydrides. Which of the following statements is true ?
Answer» Gas (G) REACTS with conc NaOH to form `OF_2` Pale Blue gas (E) OXIDISES ALKALINE KI to `KIO_3` Pale blue gas (E) in acidic medium reacts with `K_2Cr_2O_7` to form a bright blue coloured compound `(CrO_5)` (B) and (C ) both Solution :(A)`F_2` gives`O_2` gas (C )`H_2O_2` in acidic medium form `CrO_5 , "not" O_3` (B)`KI+3O_3 to KIO_3 + 3O_2` `underset((A))(XeF_2)overset(H_2)tounderset((B))(XE)+underset((C ))(2HF)` `XeF_2+H_2overset(H_2)to+underset((B))(Xe)+underset((C ))(HF)+underset((D))(O_2)` `underset(( D))(3O_2)hArrunderset((E))(2O_3),5O_3+2KOH to 2KO_3`(orange solid)+`5O_2+H_2O` `underset(( C))(HF)+KFtoKHF_2("molten")overset("ELECTROLYSIS")tounderset((G))(F_2)`

Discussion

23.	Fifth group elements form hydrides of type AH_3.The hydrides have a lone pair of electrons.The hydrides are reducing in nature and the reducing power is related to the stability of A-H bonds.The hydrides are covalent and low boiling.Their boiling point depends on their ability to form hydrogen bond and their molecular size which which decide the intermolecular forces in the hydrides. The boiling poins of the hydrides of V-group elements are in the order :
Answer» `NH_3 gt PH_3 gt AsH_3 gt SbH_3` `NH_3 gt AsH_3 gt SbH_3 gt PH_3` `SbH_3 gt NH_3 gt AsH_3 gt PH_3 ` `AsH_3 gt SbH_3 gt NH_3 gt PH_3` Solution :DUE to INCREASE in size from `PH_3` to `SbH_3` the VAN der Waals FORCES increase.So their boiling points increase from `PH_3` to `SbH_3`.But does to intermolecular hydrogen bonding, `NH_3` has very high boiling point but LESS than that of `SbH_3`.

Discussion

24.	Figure displays the plot of the compression factor Z veres p for a few gases Which of the following statements is//are correct for a van-der waals gas:
Answer» The plot I is applicable provided the vander waals constant a is NEGLIGIBLE. The plot II is a applicable provided the vander waals constant `b` is negligible. The plot III is applicable provided the vander waals constants `a` and `b` are negligible. The plot IV is applicable provided the TEMPERATURE of the gas is much higher than its critical temperature Solution :The vander waals equation of state is: (for 1 MOLE of gas) `(P +(a)/(V_(m)^(2))) (V_(m) -b) = RT` When a is negligible, then `Z = (pV_(m))/(RT) = 1+(b)/(RT) P` that is `Z` increases with increaser in `p`. when `b` is negligible, then `Z = (pV_(m))/(RT) = 1 - (a)/(VRT)` increasing `p` implies decreases in `V`, which is turn, implies increase in the value of `a//VRT` and hence decrease in the value of `Z`. The curve IV is applicable provided temperature of the gas is near bit LARGER than it's critical temperature Hence, teh choice (a),(b) and (c ) are correct.

Discussion

25.	Microbes are present in
Answer» Wool Silk Nails Skin

Discussion

26.	Which is not a fibrous protein?
Answer» Wool Insulin Nails Skin

Discussion

27.	Fibrous protein are present in:
Answer» myosin albumin collagen FIBROIN ANSWER :A::C::D

Discussion

28.	FexO contains one Fe(III) for every three Fe(II). What is x?
Answer» `2/3` `8/9` `3/4` `5/3` Solution :Let Number of `FE^(3+)` IONS = y Number of `Fe^(3+)` ions = 3y `thereforey + 3y = x thereforex = 4Y` `FEXO` us electrically NEUTRAL, hence we get, `3(y) = 3(3y) = 2 xx 1 ` `3y + 6y = 2` `9y = 2` `y = 2/9` `y = 2/9` `x = 4 cdot (2/9) = 8/9`.

Discussion

29.	Fibrous protein are present in
Answer» wool haemoglobin albumin hyroglobulin Answer :A

Discussion

30.	Few simple chemical tests are given below to differentiate between the pairs of compounds.Which of the following tests is not correct for differentiation ?
Answer» Propanal and PROPANONE - SILVER MIRROR test Acetophenone and benzophenone - Iodoform test Ethanal and propanal - Fehling's test Benzoic acid and ethyl benzoate - Sodium bicarbonate test SOLUTION :Both ethanal and propanal will give silver mirror test and red PRECIPITATE with Fehling's solution. They can be differentiated by iodoform test.

Discussion

31.	Few polymers are matched with their uses. Point out the wrong match .
Answer» a.Polyesters- FABRIC , tyre cords , SAFETY belts b.Nylon 6- ROPES , tyre cords , fabrics c.Bakelite-Packaging industry , lubricant d.Teflon utensils - Oil seals , gaskets , non-stick SOLUTION :Bakelite is used for making combs, phongraph records, electrical switches and handles of cooking utensils.

Discussion

32.	Few drops of HNO_(3) are added to II group before proceeding to III group in order to
Answer» CONVERT `Fe^(+2)` to `Fe^(+3)` convert `Fe^(+3)` to `Fe^(+2)` PPT III group convert `Fe^(2+)` to Fe Solution :`3FE^(2+)+HNO_(3)+3H^(+) to NO uarr +3Fe^(3+)+2H_(2)O`

Discussion

33.	Few drop of HNO_(3) are added to group if before precoodinh to group III in order to :
Answer» Covert `Fe^(2+)` to `Fe^(3+)` CONVERT` Fe^(3+)` to `Fe^(2+)` PPT GROUP III None of these Solution :`Fe^(2+)` is oxidised in `Fe^(3+)` in order toprecipitate `Fe(OH)_(3)`

Discussion

34.	FeSO_(4) solution mixed with (NH_(4))_(2)SO_(4) solution (in molar ratio 1 : 1) gives the test of Fe^(2+) ion but CuSO_(4) solution mixed with liquid NH_(3) (in the molar ratio 1 : 4) does not give the test of Cu^(2+).Explain why.
Answer» Solution :`FeSO_(4)` solution does not FORM any complex with `(NH_(4))_(2)SO_(4)`. INSTEAD, they form a double salt, `FeSO_(4).(NH_(4))_(2)SO_(4).6H_(2)O` which dissociates COMPLETELY into ions in the solution. But `CuSO_(4)` combines with `NH_(3)` to form the complex `[Cu(NH_(3))_(4)]SO_(4)` in which the complex ion `[Cu(NH_(3))_(4)]^(2+)` does not dissociate to give `Cu^(2+)` ions.

Discussion

35.	FeSO_(4) solution mixed with (NH_(4))_(2)SO_(4) solution in 1 : 1 molar ratio gives the test of Fe^(2+) ion but CuSO_(4) solution mixed with aqueous ammonia in 1 : 4 molar ratio does not give the test of Cu^(2+) ion. Explain, why ?
Answer» Solution :`FeSO_(4)` solution mixed with `(NH_(4))_(2)SO_(4)` solution forms a double salt, having the formula `FeSO_(4).(NH_(4))_(2)SO_(4).6H_(2)O` (Mohr salt) which ionises in the solution to give `Fe^(2+)` ions. HENCE, it gives the tests of `Fe^(2+)` ions. `CuSO_(4)` solution mixed with aqueous ammonia forms a complex salt, with the formula `[Cu(NH_(3))_(4)]SO_(4)`. The complex ion, `[Cu(NH_(3))_(4)]^(2+)` does not ionise to give `Cu^(2+)` ions. Hence, it does not give the tests of `Cu^(2+)` ion.

Discussion

36.	FeSO_(4) solution mixed with (NH_(4))_(2)SO_(4) solution in 1:1 molar ratio gives the test of Fe^(2+)ion but CuSO_(4) solution mixed with aqueous ammonia in 1:4molar ratio does not give the test of Cu^(2+) ion. Explain why?
Answer» SOLUTION :When `FeSO_(4)` solution is mixed with `(NH_(4))_(2)SO_(4)` solution in `1:1` molar ratio, a double salt, `FeSO_(4).(NH_(4))_(2)SO_(4).6H_(2)O` (Mohr salt), is formed. It dissociates completely in the solution to form the COMPONENT IONS `(Fe^(2+), NH_(4)^(+), SO_(4)^(2-))`. Thus, the solution gives the tests of `Fe^(2+)` ion. When `CuSO_(4)` solution is mixed with aqueous ammonia in `1:4` molar ratio, a coordination compound with the formula, `[Cu(NH_(3))_(4)]SO_(4)`, is formed. It ionises in the solution to form the complex cation `[Cu(NH_(3))_(4)]^(2+)`, which does not DISSOCIATED further to produce `Cu^(2+)` ions. Hence it does not give the tests of `Cu^(2+)` ion.

Discussion

37.	FeSO_(4) solution mixed with (NH_(4))_(2)SO_(4) solution in 1 : 1 molar ratio gives the test of Fe^(2+) but CuSO_(4)solution mixed with aqueous ammonia in 1 : 4 molar ratio does not give the test of Cu^(2+) ion. Explain why ?
Answer» Solution :`FeSO_(4)` solution MIXED with `(NH_(4))_(2)SO_(4)` solution in 1 : 1 molar ratio forms a double salt, `FeSO_(4). (NH_(4))_(2)SO_(4).6 H_(2)O` (Mohr salt) which ionizes in the solution to GIVE `Fe^(2+)` ions. Hence, it gives the tests of `Fe^(2+)` ions. `CuSO_(4)` solution mixed with `(NH_(4))_(2)SO_(4)` solution in 1 : 1 molar ratio forms a complex salt, with the formula `[Cu(NH_(3))_(4)]SO_(4)`. The complex ion, `[Cu(NH_(3))_(4)]^(2+)` does not ionize to give `Cu^(2+)` ions. Hence, it does not give the tests of `Cu^(2+)` ion.

Discussion

38.	FeSO_4 solution mixed with (NH_4)_2 SO_4 solution in 1:1 molar ratio gives the test of Fe^(2+) but CuSO_4solution mixed with aqueous ammonia in 1:4 molar ratio does not give the test Cu^(2+) ion. Explain why.
Answer» SOLUTION :`FeSO_4` Solution mixed with `(NH_4)_2 SO_4` solution in 1:1 molar ratio forms a double salt. `FeSO_4 . (NH_4)_2 SO_4 . 6 H_2 O` (MOHR salt) which ionizes in the solution to give `Fe^(2+)` ions. Hence, it gives the tests of `Fe^(2+)` ions. `CuSO_4` Solution mixed with aqueous ammonia in 1: 4 molar ratio forms a complex salt, with the FORMULA `[Cu(NH_3)_4 ] SO_4`.The complex ION, `[Cu(NH_3)_4]^(2+)` does not ionize to give `Cu^(2+)` ions. Hence, it does not give the tests of `Cu^(2+)` ion.

Discussion

39.	FeSO_(4)+KCN(excess) to "complex"" "X Which of the following option incorrect regarding complex X ?
Answer» It has `d^(2)sp^(3)` hybridisation. It is an inner orbital complex. It is an diamagnetic complex. It is an high SPIN complex. Solution :`K_(4)[Fe(CN)_(6)]implies d^(6) "(LOW spin)" to "Diamagnetic"`

Discussion

40.	FeSO_(4)solution gives brown colour ring in testing nitrates or nitrites. This is due to formation of
Answer» `[Fe(H_(2)O)_(5)NO]^(2+) `[Fe(H_(2)O)_(5)NO_(2)]^(2+) `[Fe(H_(2)O)_(4)NO_(2)]^(2+) `[Fe(H_(2)O)_(4)NO]^(2+) Solution :brown ring is formed due to `[Fe(H_(2)O)_(5)NO]^(2+)`

Discussion

41.	FeSO_4 is used in brown ring test for nitrates and nitrites. In this test a freshly preparedFeSO_4 solution is mixed with eolution containingNO_(2)^(-) or NO_(3)^(-) and the conc. H_2SO_4 is run down the side of the tube. If the mixture gets hot is shaken I) The brown colour disappear II) No is evolved III) A yellow solution of Fe_(2)(SO_4)_3 is formed
Answer» I,II,III correct I,III correct II,III correct Only I correct Solution :On heating brown RING decomposes and `Fe^(+2)` oxidizes to `Fe^(+3)`

Discussion

42.	FeSO_(4) on heatinggives "______"
Answer» `SO_(2)` and `O_(2)` `SO_(2)` and `SO_(3)` `SO_(2)` `SO_(3)` ANSWER :B

Discussion

43.	FeSO_(4) is a very good absorber for NO, the new compound formed by this process is found to contain number of unpaired electrons:
Answer» 4 5 3 6 Solution :`Fe^(3+)(aq)+NO+SO_(4)^(2-)(aq) to [Fe(H_(2)O)_(5)(NO)]^(2+)+SO_(4)^(2-)` `mu_(E f f)=3.89` BM HENCE, no of unpaired electron=3.

Discussion

44.	FeSO_4 + H_2O_2 is called …………………….. .
Answer» SOLUTION :Fenton.s REAGENT

Discussion

45.	FeSO_(4) , forms brown ring with
Answer» `NO_3` NO `N_2O` `N_2O_3` ANSWER :B

Discussion

46.	FeS_(2) is __________.
Answer» magnetite pyrite limestone haematite Answer :B

Discussion

47.	Fertilizer having the highest nitrogen percentage is:
Answer» CALCIUM cyanamide Urea Ammonium nitrate Ammonium sulphate Answer :B

Discussion

48.	The crystals of ferrous sulphate on heating give:
Answer» `SO_3` `SO_2` `Fe_2O_3` All Answer :D

Discussion

49.	Ferrous sulphate isomorphous with:
Answer» `CuSO_4` `ZnSO_4` `MnSO_4` `NiSO_4` ANSWER :B

Discussion

50.	Ferrous sulphate is called as:
Answer» GREEN vitriol White vitriol Jeweller.s rouge Glauber.s salt Answer :A

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

Fill in the blanks by choosing the appropriate option. Cone. H_(2)SO_(4) chars paper, wood and sugar by removing (i) from them. It is also known as (ii) It is manufactured by (iii) process. It is a strong (iv) and (v) acid.

Fill in the blanks : (A) CH_(3)CH_(2)I overset(NaCN)to ? overset(OH^(-), "Partial hydrolysis")to? (B) C_(6)H_(5)overset(+)(N_(2))overset(-)(Cl) overset(CuCN)to ? underset(H^(+))overset(H_(2)O)to ?

Fill in the blanks. ""_92^235U + ""_0^1n to ""_52^137 A + ""_40^97B + ….

Fill in the blanks: (a) ._(92)^(235)U + ._(0)^(1)n rarr ._(55)^(142)A + ._(37)^(92) B+.... (b) ._(34)^(82)Se rarr ...+2 ._(-1)e^(0)

Fill in the blanks. ""_34^82Se to 2 ""_(-1)^0e + ….

………….filament is used to decompose titanium tetraoxide

Figure displays the plot of the compression factor Z versus P for a few gases which of the following statements is/are correct for a van der Waals' gas?

Figure explains elevation in boiling point when a non-volatile solute is added to a solvent. Provide a thermodynamic explanation of the elevation in boiling point.

Figure displays the plot of the compression factor Z veres p for a few gases Which of the following statements is//are correct for a van-der waals gas:

Microbes are present in

Which is not a fibrous protein?

Fibrous protein are present in:

FexO contains one Fe(III) for every three Fe(II). What is x?

Fibrous protein are present in

Few simple chemical tests are given below to differentiate between the pairs of compounds.Which of the following tests is not correct for differentiation ?

Few polymers are matched with their uses. Point out the wrong match .

Few drops of HNO_(3) are added to II group before proceeding to III group in order to

Few drop of HNO_(3) are added to group if before precoodinh to group III in order to :

FeSO_(4) solution mixed with (NH_(4))_(2)SO_(4) solution (in molar ratio 1 : 1) gives the test of Fe^(2+) ion but CuSO_(4) solution mixed with liquid NH_(3) (in the molar ratio 1 : 4) does not give the test of Cu^(2+).Explain why.

FeSO_(4) solution mixed with (NH_(4))_(2)SO_(4) solution in 1 : 1 molar ratio gives the test of Fe^(2+) ion but CuSO_(4) solution mixed with aqueous ammonia in 1 : 4 molar ratio does not give the test of Cu^(2+) ion. Explain, why ?

FeSO_(4) solution mixed with (NH_(4))_(2)SO_(4) solution in 1:1 molar ratio gives the test of Fe^(2+)ion but CuSO_(4) solution mixed with aqueous ammonia in 1:4molar ratio does not give the test of Cu^(2+) ion. Explain why?

FeSO_(4) solution mixed with (NH_(4))_(2)SO_(4) solution in 1 : 1 molar ratio gives the test of Fe^(2+) but CuSO_(4)solution mixed with aqueous ammonia in 1 : 4 molar ratio does not give the test of Cu^(2+) ion. Explain why ?

FeSO_4 solution mixed with (NH_4)_2 SO_4 solution in 1:1 molar ratio gives the test of Fe^(2+) but CuSO_4solution mixed with aqueous ammonia in 1:4 molar ratio does not give the test Cu^(2+) ion. Explain why.

FeSO_(4)+KCN(excess) to "complex"" "X Which of the following option incorrect regarding complex X ?

FeSO_(4)solution gives brown colour ring in testing nitrates or nitrites. This is due to formation of

FeSO_(4) on heatinggives "______"

FeSO_(4) is a very good absorber for NO, the new compound formed by this process is found to contain number of unpaired electrons:

FeSO_4 + H_2O_2 is called …………………….. .

FeSO_(4) , forms brown ring with

FeS_(2) is __________.

Fertilizer having the highest nitrogen percentage is:

The crystals of ferrous sulphate on heating give:

Ferrous sulphate isomorphous with:

Ferrous sulphate is called as: