94420 + Interview Questions in Chemistry in Class 12

1.	Compound (A) C_(5)H_(10)O forms a phenyl hydrogen and gives negative Tollen's and iodoform tests. Compound (A) on reduction gives n-pentane. The compound (A) is
Answer» `CH_(3)CH_(2)CH_(2)CH_(2)OVERSET(O)overset(\|\|)(C)-H` `CH_(3)CH_(2)CH_(2)overset(O)overset(\|\|)(C)CH_(3)` `CH_(3)CH_(2)overset(O)overset(\|\|)(C)-CH_(2)CH_(3)` `CH_(3)CH_(2)CH_(2)CH_(2)CH_(2)OH`\ Answer :C

Discussion

2.	Compound A, C_5H_(10)O, forms a phenyl hyrazone, given negative Tollen's and iodoform tests and is reduced to n - pentane. What is the compound , A ?
Answer» Solution :Compound A FORMS hydrazone, so it is a CARBONYL compound. It is a ketone as it does not REDUCE Tollens reagent. The ketone does not have methyl group as it does not respond to iodoform test. Thus, the compound is pentanone-3. The STRUCTURE of the compound is `CH_(3)CH_(2)COCH_(2)CH_(3)`.

Discussion

3.	Compound (A) C_(5)H_(10)O forms a phenyl hydrazone and gives negative Tollen's and iodoform tests. Compound (A) on reduction gives n-pentane. Compound (A) is :
Answer» `CH_(3)CH_(2)CH_(2)CH_(2)-overset(O)overset(\|\|)(C ) - H` `CH_(3) CH_(2) CH_(2) - overset(O)overset(\|\|)(C ) - CH_(3)` `CH_(3) CH_(2) - overset(O)overset(\|\|)(C ) - CH_(2)CH_(3)` `CH_(3) CH_(2) CH_(2) CH_(2) CH_(2) - OH` SOLUTION :Choice (1)can GIVE tollen.s test as it is an aldehyde and choice (2) can respond to iodoform test as it is a methly ketone. Choice (3) is an alcohol so it cannot form phenyl HYDRAZONE. The only possible OPTION here is choice (3) which can give phenyl hydrazone and also not respond to either tollen.s or Iodoform test

Discussion

4.	Compound (A) C_5H_10O Forms a phenyl hydrazone and gives nagitevTollen's and iodoform test.COMPOUND (A) on reduction gives n- pentane.Compound (A) is :
Answer» A PRIMARY alcohol An aldehyde Acetone Secondary alcohol Answer :C

Discussion

5.	Compound (A) C_5H_10O Forms a phenyl hydrazone and gives nagitevTollen.s and iodoform test.COMPOUND (A) on reduction gives n- pentane.Compound (A) is :
Answer» A PRIMARY alcohol An aldehyde Acetone Secondary alcohol Answer :C

Discussion

6.	Compound (A) C_5 H_8 O_2 liberated CO_2 on reaction with sodium bicarbonate. It exists in two forms neither of which is optically active. It yielded compound (B), C_5 H_(10) O_2 on hydrogenation. Compound (B) can be separated into enantimorphs. Write structures of (A) and (B).
Answer» SOLUTION :

Discussion

7.	Compound A, C_(3)H_(10)O_(s), gives a tetraacetate with (CH_(3),CO)_(2),O, and oxidation of A with Br_(2)-H_(2)O gives an acid, C_(5)H_(10)O_(6). Reduction of 'A' with HI gives iso-pentane. There are two possible structures for compound 'A' which can be distinguished by using
Answer» `HIO_(4)` excess of P,HI phenylhdrazine TOLLENS' reagent Solution : I FROMS DIPHENYL (osazonederivative). II does not .

Discussion

8.	Compound (A) C_(10)H_(22)0_(2) is insoluble in aq. NaOH bu not is NaHCO_(3). Treatment of (A) with DMSO (CH_(3)-overset(S)overset(\|\|)(S)-CH_(3)) in alkali give (B) C_(11)H_(14)O_(2). Treatment of (A) with strong alkali alone give an isomeric compound (C). When (A) is reflux with HI, CH_(3)I is obtained, compound (B) is insoluble in alkali and decolurises Br_(2)//C Cl_(4). (B) on treating with strong base gives (D), an isomer of (B). Ozonolysis (C) of gives (E), C_(8)H_(8)O and isomer of vanilline. Ozolysis of (D) gives (F) C_(9)H_(10)O_(3), which is identical with product of methylation of vanilline (4-hydroxy-3-methoxy benzaldehyde). Compound (E) and (F) are respectively:
Answer» NONE of these Answer :A

Discussion

9.	Compound (A) C_(10)H_(22)0_(2) is insoluble in aq. NaOH bu not is NaHCO_(3). Treatment of (A) with DMSO (CH_(3)-overset(S)overset(\|\|)(S)-CH_(3)) in alkali give (B) C_(11)H_(14)O_(2). Treatment of (A) with strong alkali alone give an isomeric compound (C). When (A) is reflux with HI, CH_(3)I is obtained, compound (B) is insoluble in alkali and decolurises Br_(2)//C Cl_(4). (B) on treating with strong base gives (D), an isomer of (B). Ozonolysis (C) of gives (E), C_(8)H_(8)O and isomer of vanilline. Ozolysis of (D) gives (F) C_(9)H_(10)O_(3), which is identical with product of methylation of vanilline (4-hydroxy-3-methoxy benzaldehyde). Structure of compound (A) is:
Answer» ANSWER :C

Discussion

10.	Compound (A) C_(10)H_(22)0_(2) is insoluble in aq. NaOH bu not is NaHCO_(3). Treatment of (A) with DMSO (CH_(3)-overset(S)overset(\|\|)(S)-CH_(3)) in alkali give (B) C_(11)H_(14)O_(2). Treatment of (A) with strong alkali alone give an isomeric compound (C). When (A) is reflux with HI, CH_(3)I is obtained, compound (B) is insoluble in alkali and decolurises Br_(2)//C Cl_(4). (B) on treating with strong base gives (D), an isomer of (B). Ozonolysis (C) of gives (E), C_(8)H_(8)O and isomer of vanilline. Ozolysis of (D) gives (F) C_(9)H_(10)O_(3), which is identical with product of methylation of vanilline (4-hydroxy-3-methoxy benzaldehyde). Compound (B) is:
Answer» ANSWER :B

Discussion

11.	Compound (A), C_(10)H_(12)O gives off hydrogen on treatment with sodium metal and decolourises Br_(2) in C Cl_(4) to give (B), C_(10)H_(12)OBr. (A) on treatment with I_(2) in NaOH gives iodoform and an acid (C) after acidification. Give the structure of (A) to (C) and also of all geometrical and optical isomers of (A). Answer the following based on the above. Which of the following statements regarding A are incorrect ?
Answer» In the presence of acidic MEDIUM AFORMS a 5 membred ring. A has 2 GEOMETRICAL isomers. A has 4 pairs of diasteereomers A has 2 chiral centres Answer :D

Discussion

12.	Compound (A), C_(10)H_(12)O gives off hydrogen on treatment with sodium metal and decolourises Br_(2) in C Cl_(4) to give (B), C_(10)H_(12)OBr. (A) on treatment with I_(2) in NaOH gives iodoform and an acid (C) after acidification. Give the structure of (A) to (C) and also of all geometrical and optical isomers of (A). Answer the following based on the above. Compound B is
Answer» ANSWER :A

Discussion

13.	Compound (A), C_(10)H_(12)O gives off hydrogen on treatment with sodium metal and decolourises Br_(2) in C Cl_(4) to give (B), C_(10)H_(12)OBr. (A) on treatment with I_(2) in NaOH gives iodoform and an acid (C) after acidification. Give the structure of (A) to (C) and also of all geometrical and optical isomers of (A). Answer the following based on the above. Compound A is
Answer» ANSWER :A

Discussion

14.	Compound 'A' (C_(10)H_(12)O) evolves H_(2) gas with Na metal. It reacts with Br_(2) // C CI_(4) to give B( C_(10)H_(12)Br_(2)O) with I_(2)//NaOH it forms iodoform and acid C[C_(9)H_(8)O_(2)]. A has geometrical and optical isomers. The structure of A and C should be.
Answer» Ph. CH=CHCOOH Solution :COMPOUND A (`C_(10)H_(12)O`) -OH GROUPS reacts with Na to liberate `H_(2)` C = C and `C -= C` groups decolourise `Br_(2) // C Cl_(4)`, `A=C_(6)H_(5)-CH=CH-overset(OH)overset(\|)CH-CH_(3)` Compound `C(C_(9)H_(8)O_(2))`, `A overset(I_(2)+NaOH)to Ph-CH=CH-COOH+CHI_(3)`

Discussion

15.	Compound A an hydrohalogenation gives B, which is treated with alc. KCN and followed by acid hydrolysis gives 2- methyl butanoic acid. The compound A will be
Answer» ANSWER :C

Discussion

16.	Compound (A), an alkene with molecule formula (C_(5)H_(10)) exists in various strictures. On mononcholorination and dichlorination, it again shows variousstructures andstreoisomers. The total number of stereoisomerism shown by isomer in Q.No 26 is:
Answer» `2` `3` `4` `5` Solution :Since compound `(A)` has `1 D.U` and is an ALKANE, so it should be cyclic. The number of cyclic isomers of `(A)` is fice. TOTAL number of steoroisomers shown by `(III) = 3`.

Discussion

17.	Compound (A), an alkene with molecule formula (C_(5)H_(10)) exists in various strictures. On mononcholorination and dichlorination, it again shows variousstructures andstreoisomers. Whichisomer of (A) shows styereoisomerism ?
Answer» SOLUTION :Since compound `(A)` has `1 D.U` and is an alkane, so it should be cyclic. The NUMBER of cyclic isomers of `(A)` is fice. Isomer `(III)` shows both geometrical and OPTICAL isomners.

Discussion

18.	Compound (A), an alkene with molecule formula (C_(5)H_(10)) exists in various strictures. On mononcholorination and dichlorination, it again shows variousstructures andstreoisomers. Two isomers (IV) and (V) on monochlorination give four products. The isomers (IV) and (V) are:
Answer» SOLUTION :Since COMPOUND `(A)` has `1 D.U` and is an alkane, so it should be CYCLIC. The NUMBER of cyclic isomers of `(A)` is fice.

Discussion

19.	Compound (A), an alkene with molecule formula (C_(5)H_(10)) exists in various strictures. On mononcholorination and dichlorination, it again shows variousstructures andstreoisomers. Isomer (III) on monochlorination gives three products. The isomer (III) is:
Answer» Solution :SINCE COMPOUND `(A)` has `1 D.U` and is an ALKANE, so it should be CYCLIC. The number of cyclic isomers of `(A)` is FICE.

Discussion

20.	Compound (A), an alkene with molecule formula (C_(5)H_(10)) exists in various strictures. On mononcholorination and dichlorination, it again shows variousstructures andstreoisomers. Another isomer (II) on monochlorination gives two products. The isomer (II) is:
Answer» Solution :Since compound `(A)` has `1 D.U` and is an alkane, so it should be CYCLIC. The number of cyclic ISOMERS of `(A)` is FICE.

Discussion

21.	Compound (A), an alkene with molecule formula (C_(5)H_(10)) exists in various strictures. On mononcholorination and dichlorination, it again shows variousstructures andstreoisomers. One of the cyclic isomers of (A) on monochlorination gives one product. The isomer (I) is:
Answer» Solution :Since compound `(A)` has `1 D.U` and is an ALKANE, so it should be cyclic. The number of cyclic ISOMERS of `(A)` is FICE.

Discussion

22.	Compound (A), an alkene with molecule formula (C_(5)H_(10)) exists in various strictures. On mononcholorination and dichlorination, it again shows variousstructures andstreoisomers.
Answer» `5` `6` `7` `8` Solution :Since COMPOUND `(A)` has `1 D.U` and is an alkane, so it should be cyclic. The number of cyclic isomers of `(A)` is fice. Total number of cyclic STRUCTURES including steroisomers for `(A) = 7`.

Discussion

23.	Composition of Ziegler-Natta catalyst is
Answer» `(Et_(3))_(3) AL* TiCl_(2)` `(Me)_(3) Al* TiCl_(2)` `(Et)_(3) Al* TiCl_(4)` `(Et)_(3) Al* PtCl_(4)` ANSWER :C

Discussion

24.	Composition of urotropine is _____.
Answer» SOLUTION :`(CH_2)_6N_4`

Discussion

25.	Composition of sample wurtzite is Fe_(0.93)O_(1.0). What percentage of iron is present in the form of Fe(III) ?
Answer» Solution : `Fe_(0.93)O_(1.0)` is a non-stoichiometric compound and contains MIXTURE of `Fe^(2+)` and `Fe^(3+)` ions. Let x `Fe^(3+)` ions are present in the compound. This MEANS that `xFe^(2+)` ions have been replaced by `Fe^(3+)` ions. No. of `Fe^(2+)` ions in the compound = 0.93 – x For electrical neutrality, total positive charge on cations = Total NEGATIVE charge on anions. `:.` 2 (0.93 - x) + 3X = 2 or 1.86 +x=2 or x=0.14 `:.` Percentage of `Fe^(3+)` ions = (0.14)/(0.93)XX 100 = 15.05%`

Discussion

26.	Composition of glass is
Answer» silica, LIME, NaCl Silica,lime, `Na_(2)CO_(3)` Silica,NaCl lime, `Na_(2)CO_(3)` Solution :Ordinary GLASS is OBTAINED from silica, lime and `Na_(2)CO_(3)`.

Discussion

27.	Composition of copper matte is20% Cu + 80%Cu_2S and FeS. Is it true or false?
Answer» SOLUTION :`50%.CU + 50% Cu_2S` and FES

Discussion

28.	The composition of 'Copper Matte' is :
Answer» SOLUTION :`50% CU + 50% Cu_2S` and FES

Discussion

29.	Composition of copper matel is _____.
Answer» SOLUTION :`50% CU + 50% Cu_2S` and FES

Discussion

30.	Composition of Brass is :
Answer» Cu(60%) - Zn(40%) Cu(80%) - Zn(20%) Cu(80%) - Zn(10%)-Sn(10%) None of the above. Answer :B

Discussion

31.	Composition of borax (Tincal) is
Answer» `Na_2B_(4)O_(7).4H_(2)O` `Na_(2)B_(4)O_(7).10H_(2)O` `NaBO_(2)` `Na_(2)BO_(3)` ANSWER :B

Discussion

32.	Composition ofAzurite mineral of
Answer» `CuCO_(3)CUO` `Cu(HCO_(3))_(2).Cu(OH)_(2)` `Cu(OH)_(2).2CuCO_(3)` `CuCO_(3).2CU(OH)_(2)` ANSWER :C

Discussion

33.	Composition of bleaching powder is
Answer» `Ca(OCl)_2. CaCl_2. Ca(OH)_2. H_2O` `Ca(OCl)_2. CaO. Ca(OH)_2` `Ca(OCl_2)_2. CaO. Ca(OH)_2. H_2O` `Ca(OCl)_2. CaCl_2. Ca(OH)_2. 2H_2O` SOLUTION :`Ca(OCl)_2. CaCl_2. Ca(OH)_2. 2H_2O`

Discussion

34.	Composition of a sample of Wustite is Fe_(0-93)O_(1-0). What percentage of iron is present in the form of Fe (III)?
Answer» Solution :`Fe_(0.93) O_(1-0)` is non-stoichiometric is a mixture of `Fe^(2+)` and `Fe^(3+)` ions. LET atoms of`Fe^(3+)`ions are present in the compound. This means that x`Fe^(3+)` ions have been replaced by`Fe^(2+)` ions. No. of `Fe^(2+)` ions = 0.93 - x For electrical neutrality, Positive charge on the compound = NEGATIVE charge on the compound 2 (0.93 - x) + 3x = 2 1.86 + x = 2 x = 0.14 age of `Fe^(3+)` ions = `(0.14)/(0.93) XX 100` = 15.05

Discussion

35.	Components of gaseous mixture useful for sea divers
Answer» `O_2` and He `O_2` and `H_2` `O_2` and `N_2` `O_2` and `CO_2` ANSWER :A

Discussion

36.	Components of buffer solution are 0.1 M HCN and 0.2 M NaCN. What is the pH of the solution
Answer» 9.61 6.15 2 4.2 Solution :`PH = pK_(a) + LOG[("SALT")/("ACID")]` `= 9.30 + log[(0.2)/(0.1)] = 9.30 + 0.3010 = 9.6`.

Discussion

37.	Components of alloy 'Invar' are
Answer» STEEL and chromium vanadium and manganese tungsten and chromium steel and nickel Answer :B

Discussion

38.	Components of a binary mixture of two liquids A and B were being separated by distillation. After some time separation of components stopped and composition of vapour phase became same as that of liquid phase. Both the components started coming in the distillate.Explain why this happened.
Answer» Solution :SINCE both the components are appearing in the DISTILLATE and COMPOSITION of liquid and vapour is same, this SHOWS that liquids have formed azeotropic mixture and hence cannot be SEPARATED at this stage by fractional distillation.

Discussion

39.	Components of a binary mixture of two liquids A and B were being separated by distillation. After some time separation of components stopped and composition of vapour phase became same as that of liquid phase. Both the components started coming in the distillate. Explain why this happened.
Answer» Solution :SINCE both the components are appearing in the distillate and composition of liquid and vapour is same, this SHOWS that liquids have formed azeotropic mixture and hence cannot be separated under these CONDITIONS.

Discussion

40.	Components of a non-idea) binary solution cannot be completely separated by fracti.o.nal distillation. Why?
Answer» SOLUTION :It FORMS AZEOTROPIC MAXTURE.

Discussion

41.	Components, Aand B, respectively of an ideal binary solution. If X_A represents the mole fraction of component A, the total pressure of the solution will be :
Answer» `p_A + x_A (p_B - p_A)` `p_A + x_A (p_A - p_B)` `p_B + x_A (p_B-p_A)` `p_B + x_A (p_A-p_B)` ANSWER :D

Discussion

42.	Complte the following reaction equation (ii)
Answer» SOLUTION :

Discussion

43.	{:("Complex","Type"),((A)CoCl_(3).3NH_(3),(i)"Anionic complex"),((B)Na_(2)ZnCl_(4),(ii)"Cationic complex"),((C)PtCl_(4).5NH_(3),(iii)"Neutral complex"):} The correct match is
Answer» A - II, B-iii, C-i A-iii, B-, C-ii A-ii, B-i, C-iii A-iii, B-ii, C-i Answer :B

Discussion

44.	Complexes with halide ligands are generally:
Answer» HIGH SPIN COMPLEXES Low spin complexes, Both (a) and (B) None Answer :A

Discussion

45.	Complexes with CN^- ligands are usually:
Answer» HIGH spin COMPLEXES LOW spin complexes Both (a) and (b) None Answer :B

Discussion

46.	Complexes in which the metal atom or ion is linked to only one type of ligands are called …… whereas those in which metal atom is linked to more than one type of ligands are called …… .
Answer» SOLUTION :homoleptic, HETEROLEPTIC

Discussion

47.	Complexes with bidentate ligands are called:
Answer» Ligands Chelates Complexes None Answer :B

Discussion

48.	Complexes given below show:
Answer» OPTICAL isomerism coordinate isomerism Geometrical isomerism Bridge isomerism Answer :C

Discussion

49.	Complexes formed in the following methods are: (I) Mond's process for purification of nickel (II) Removal of unreacted AgBr from photographic plate (III) Removal of lead poisoning from the body
Answer» `{:(I,II,III),(Ni(CO)_(4),[Ag(CN)_(2)]^(-),[Pb(EDTA)]^(2-)):}` `{:(I,II,III),(Ni(CO)_(4),[Ag(S_(2)O_(3))_(2)]^(3-),[Pb(EDTA)]^(2-)):}` `{:(I,II,III),(Ni(CO)_(6),Ag(S_(2)O_(3))_(2)]^(3-),([Pb(EDTA)]^(4-)):}` `{:(I,II,III),(Ni(CO)_(6),[Ag(S_(2)O_(3))]^(-),[Pb(EDTA)]^(2-)):}` SOLUTION :(I) Mond.s process for PURIFICATION of Ni-`[Ni(CO)_(4)]` (II) REMOVAL of unreacted AgBr from photographic PLATE-`[Ag(S_(2)O_(3))_(2)]^(3-)` (III) Removal of lead poisoning from body-`[Pb(EDTA)]^(2-)`.

Discussion

50.	Complexes expected to be coloured in solution is/are
Answer» `K_(5) [CrF_(6)]` `[Co (NH_(3))_(6)] Cl_(3)` `K_(3) [FeCl_(6)]` `K_(3)[CoCl_(6)]` ANSWER :B::D

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

Compound (A) C_(5)H_(10)O forms a phenyl hydrogen and gives negative Tollen's and iodoform tests. Compound (A) on reduction gives n-pentane. The compound (A) is

Compound A, C_5H_(10)O, forms a phenyl hyrazone, given negative Tollen's and iodoform tests and is reduced to n - pentane. What is the compound , A ?

Compound (A) C_(5)H_(10)O forms a phenyl hydrazone and gives negative Tollen's and iodoform tests. Compound (A) on reduction gives n-pentane. Compound (A) is :

Compound (A) C_5H_10O Forms a phenyl hydrazone and gives nagitevTollen's and iodoform test.COMPOUND (A) on reduction gives n- pentane.Compound (A) is :

Compound (A) C_5H_10O Forms a phenyl hydrazone and gives nagitevTollen.s and iodoform test.COMPOUND (A) on reduction gives n- pentane.Compound (A) is :

Compound (A) C_5 H_8 O_2 liberated CO_2 on reaction with sodium bicarbonate. It exists in two forms neither of which is optically active. It yielded compound (B), C_5 H_(10) O_2 on hydrogenation. Compound (B) can be separated into enantimorphs. Write structures of (A) and (B).

Compound A, C_(3)H_(10)O_(s), gives a tetraacetate with (CH_(3),CO)_(2),O, and oxidation of A with Br_(2)-H_(2)O gives an acid, C_(5)H_(10)O_(6). Reduction of 'A' with HI gives iso-pentane. There are two possible structures for compound 'A' which can be distinguished by using

Compound 'A' (C_(10)H_(12)O) evolves H_(2) gas with Na metal. It reacts with Br_(2) // C CI_(4) to give B( C_(10)H_(12)Br_(2)O) with I_(2)//NaOH it forms iodoform and acid C[C_(9)H_(8)O_(2)]. A has geometrical and optical isomers. The structure of A and C should be.

Compound A an hydrohalogenation gives B, which is treated with alc. KCN and followed by acid hydrolysis gives 2- methyl butanoic acid. The compound A will be

Compound (A), an alkene with molecule formula (C_(5)H_(10)) exists in various strictures. On mononcholorination and dichlorination, it again shows variousstructures andstreoisomers. The total number of stereoisomerism shown by isomer in Q.No 26 is:

Compound (A), an alkene with molecule formula (C_(5)H_(10)) exists in various strictures. On mononcholorination and dichlorination, it again shows variousstructures andstreoisomers. Whichisomer of (A) shows styereoisomerism ?

Compound (A), an alkene with molecule formula (C_(5)H_(10)) exists in various strictures. On mononcholorination and dichlorination, it again shows variousstructures andstreoisomers. Two isomers (IV) and (V) on monochlorination give four products. The isomers (IV) and (V) are:

Compound (A), an alkene with molecule formula (C_(5)H_(10)) exists in various strictures. On mononcholorination and dichlorination, it again shows variousstructures andstreoisomers. Isomer (III) on monochlorination gives three products. The isomer (III) is:

Compound (A), an alkene with molecule formula (C_(5)H_(10)) exists in various strictures. On mononcholorination and dichlorination, it again shows variousstructures andstreoisomers. Another isomer (II) on monochlorination gives two products. The isomer (II) is:

Compound (A), an alkene with molecule formula (C_(5)H_(10)) exists in various strictures. On mononcholorination and dichlorination, it again shows variousstructures andstreoisomers. One of the cyclic isomers of (A) on monochlorination gives one product. The isomer (I) is:

Compound (A), an alkene with molecule formula (C_(5)H_(10)) exists in various strictures. On mononcholorination and dichlorination, it again shows variousstructures andstreoisomers.

Composition of Ziegler-Natta catalyst is

Composition of urotropine is _____.

Composition of sample wurtzite is Fe_(0.93)O_(1.0). What percentage of iron is present in the form of Fe(III) ?

Composition of glass is

Composition of copper matte is20% Cu + 80%Cu_2S and FeS. Is it true or false?

The composition of 'Copper Matte' is :

Composition of copper matel is _____.

Composition of Brass is :

Composition of borax (Tincal) is

Composition ofAzurite mineral of

Composition of bleaching powder is

Composition of a sample of Wustite is Fe_(0-93)O_(1-0). What percentage of iron is present in the form of Fe (III)?

Components of gaseous mixture useful for sea divers

Components of buffer solution are 0.1 M HCN and 0.2 M NaCN. What is the pH of the solution

Components of alloy 'Invar' are

Components of a binary mixture of two liquids A and B were being separated by distillation. After some time separation of components stopped and composition of vapour phase became same as that of liquid phase. Both the components started coming in the distillate.Explain why this happened.

Components of a binary mixture of two liquids A and B were being separated by distillation. After some time separation of components stopped and composition of vapour phase became same as that of liquid phase. Both the components started coming in the distillate. Explain why this happened.

Components of a non-idea) binary solution cannot be completely separated by fracti.o.nal distillation. Why?

Components, Aand B, respectively of an ideal binary solution. If X_A represents the mole fraction of component A, the total pressure of the solution will be :

Complte the following reaction equation (ii)

{:("Complex","Type"),((A)CoCl_(3).3NH_(3),(i)"Anionic complex"),((B)Na_(2)ZnCl_(4),(ii)"Cationic complex"),((C)PtCl_(4).5NH_(3),(iii)"Neutral complex"):} The correct match is

Complexes with halide ligands are generally:

Complexes with CN^- ligands are usually:

Complexes in which the metal atom or ion is linked to only one type of ligands are called …… whereas those in which metal atom is linked to more than one type of ligands are called …… .

Complexes with bidentate ligands are called:

Complexes given below show:

Complexes formed in the following methods are: (I) Mond's process for purification of nickel (II) Removal of unreacted AgBr from photographic plate (III) Removal of lead poisoning from the body

Complexes expected to be coloured in solution is/are