52671 + Interview Questions in Chemistry in Class 11

1.	The structural formula of 2-methyl-2-butene is
Answer» `CH_(3)- CH= C(CH_(3))-CH_(3)` `CH_(3) - CH = CH- CH_(3)` `CH_(3) - CH_(2) - C(CH_(3)) = CH_(2)` `CH_(3) - CH (CH_(3)) - CH= CH_(2)` ANSWER :A

Discussion

2.	Structrue of ICl_(2)^(-) is
Answer» TRIGONAL octahedral square planar distorted trigonal pyramidal Answer :D

Discussion

3.	Strontia is
Answer» SrO MgO `SrCO_(3)` `SrCO_(4)` Solution :Sro is CALLED Stontia

Discussion

4.	Strongest reducing agent is"_______________".
Answer» `F^(-)`<BR>`CL^(-)` `Br^(-) ` `I^(-)` ANSWER :D

Discussion

5.	Strongest reducing agent and strongest oxidising agent are respectively
Answer» `CS and Cl_2` `LI and Cl_2` `Cs and F_2` `Cs and O_2` ANSWER :C

Discussion

6.	Strongest nucleophile in polar-protic solvent is
Answer» `CH_(3)O^(-)` `CH_(3)SH` `CH_(3)-S^(-) ` `CH_(3)OH` ANSWER :C

Discussion

7.	Strongest oxidant among the following is
Answer» `B^(+3)` `AL^(3+)` `Ga^(+3)` `TL^(+3)` ANSWER :D

Discussion

8.	Strongest conjugate base is
Answer» `Cl^(-)`<BR>`Br^(-)` `F^(-)` `I^(-)` SOLUTION :Their conjugate acids are HCI, HBr, HF and HI Out of these HF is the weakest due to maximum bond strength (and hydrogen bonding). Therefore, its conjugate BASE,` F^(-)` is the strongest

Discussion

9.	Strongest hydrogen bonds can be formed by
Answer» `HF` `H_2O` `NH_3` `HCL` ANSWER :A

Discussion

10.	Strongest bond is between
Answer» CsF NACL Both (A) and (B) None of the above. Solution :Ionic bond is strongest when the DIFFERENCE in electronegativities is maximum. Cs is more electropositive than Na and F is more electronegative than CL. Therefore, bond in CsF will have more ionic CHARACTER than NaCl and the bond in CsF will be STRONGER than in NaCl.

Discussion

11.	Stronger deactivator group is ……… .
Answer» SOLUTION :`-NO_(2)` GROUP

Discussion

12.	Strong reducing behaviour of H_(3)PO_(2) is due to
Answer» LOW OXIDATION STATE of phosphorus Presence of two - OH groups and ONE P-H bond Presence OFONE - OH group and two P-H bonds High electron gain enthalpy of phosphorus Answer :C

Discussion

13.	Strong reducing behaviour of H_(3)PO_(2)is due to
Answer» low OXIDATION state of phosphorus PRESENCE of two -OH groups and one P-H bond presence of one -OH groups and two P-H bonds high ELECTRON GAIN enthalpy of phosphorus Solution :`H-underset(H)underset(\|)OVERSET(OH)overset(\|)P=O`

Discussion

14.	Strength or concentration of H_(2)O_(2) solution is expressed in the following ways: In terms of volume of O_(2) gas: The volume (in mL) of O_(2) gas produced at STP on heating one volume (or 1 mL) of H_(2)O_(2) solution is known as the concentration of H_(2)O_(2) solution. The commercial samples are marked as "10 volume, 20 volume 30 volume.etc.. 34g of H_(2)O_(2) is present in 1120 m. of solution is called
Answer» 10 VOL SOLUTION 20 vol solution 34 vol solution 32 vol solution Solution :`M = 100/112 rArr 11.2 vol to 1M`

Discussion

15.	Strength of the sample of H_(2)O_(2) is generally expressed in terms of volume strength . It means the volume of oxygen liberated at NTP by heating one volume of H_(2)O_(2) . The concentration of H_(2)O_(2) ina solution can also expressed as percentage of H_(2)O_(2) in solution . Normality of this solution can be calculated if the equivalent mass of H_(2) O_(2) is known . Percentage strength of the above H_(2)O_(2) solution is
Answer» `0.126` `0.272` `0.544` `0.136` ANSWER :B

Discussion

16.	Strength of the sample of H_(2)O_(2) is generally expressed in terms of volume strength . It means the volume of oxygen liberated at NTP by heating one volume of H_(2)O_(2) . The concentration of H_(2)O_(2) ina solution can also expressed as percentage of H_(2)O_(2) in solution . Normality of this solution can be calculated if the equivalent mass of H_(2) O_(2) is known . Volume strength of above H_(2)O_(2) solution is ?
Answer» `0.448` `0.632` `0.896` `0.556` ANSWER :C

Discussion

17.	Strength of the sample of H_(2)O_(2) is generally expressed in terms of volume strength . It means the volume of oxygen liberated at NTP by heating one volume of H_(2)O_(2) . The concentration of H_(2)O_(2) ina solution can also expressed as percentage of H_(2)O_(2) in solution . Normality of this solution can be calculated if the equivalent mass of H_(2) O_(2) is known . 25 mL of H_(2)O_(2) solution was added to the excess of acidified KI solution . The iodine so liberated required 40 mL of 0.1 N Na_(2)S_(2)O_(3) solution . what is normality of H_(2)O_(2) solution ?
Answer» `0.08` `0.02` `0.16` `0.20` ANSWER :C

Discussion

18.	Strength or concentration of H_(2)O_(2) solution is expressed in the following ways: In terms of volume of O_(2) gas: The volume (in mL) of O_(2) gas produced at STP on heating one volume (or 1 mL) of H_(2)O_(2) solution is known as the concentration ofH_(2)O_(2) solution. The commercial samples are marked as "10 volume, 20 volume 30 volume.etc.. A5.0 mL solution of Hydrogen peroxide liberates 1.27g of iodine from an acidified Kl solution. The percentage strength of H_(2)O_(2) is
Answer» `11.2` `5.6` `1.7` `3.4` Solution :`H_(2)O_(2)+2KI to 2KOH+I_(2)` `W_(H_(2)O_(2))=(1.27)/(254)xx34 rArr ((W)/(V))%=3.4%`

Discussion

19.	Strength of acidity is in order
Answer» `II gt I gt III gt IV` `III gt IV gt I gt II` `I gt IV gt III gt II` `IV gt III gt I gt II` ANSWER :B

Discussion

20.	Strength of a weak acid or a weak base depends upon its
Answer» Temperature Nature of SOLVENT DEGREE of DISSOCIATION All the above Solution :Conceptual

Discussion

21.	Strength of 10 volume hydrogen peroxide solution means
Answer» `30.35 GL^(-1)` `17 gL^(-1)` `34 gL^(-1)` `68 gL^(-1)` Solution :`underset(68 g)(2H_(2))O_(2)to2H_(2)O+underset(22.4" L at STP")(O_(2))` `22.4 L O_(2)` is given by 68 g of `H_(2)O_(2)` `10 L O_(2)` is given by `(68)/(22.4)xx10=30.35 g L^(-1)`

Discussion

22.	Stomach acid, a dilute solution of HCl can be neutralised by reaction with Aluminium hydroxideAl(OH)_(3) + 3HCl(aq)to + AICI_(3) + 3H_(2)O How many millilitres of 0.1 M Al(OH)_(3) solution are needed to neutralise 21 mL of 0.1 M HCI?
Answer» 14 mL 7 ml 21 mL none of these Solution :` Mxx V = M_(2) XXv_(2)` `:’0.1M AL(OH)_(3) gives 3 XX 0.1 = 0.3 M OH^(-) ions` `0.3 xx V_(1) = 0.1xx 21` `V_(1)=(0.1xx21)/(0.3)=7ml`

Discussion

23.	Stomach acid, a dilute solution of HCl can be neutralised by reaction with aluminig hydroxide Al (OH)_(3) + 3HCl (aq) to AlCl _(3) + 3H _(2)O How many milliliters of 0.1 M Al (OH) _(3) solution are needed to neutralise 21 mL of 0.1 M HCl ?
Answer» 1) 14 mL 2) 7 mL 3) 21 mL 4) none of these Solution :`M _(1) XX V _(1) = M_(2) xx V _(2)` `because 0.1 M AL (OH) _(3)` GIVES `3 xx 0.1 =0.3 M OH ^(-)` ions `0.3 xx V _(1) = 0.1 xx 21` `V _(1) = (0.1 xx 21)/( 0.3) = 7 ml`

Discussion

24.	Stomach acid a dilute solution of HCI can be neutralised by reaction with Aluminium hydroxide.Al(OH)_3 +3HCl(aq) to AlCl_3 +3H_2O how many millitres of 0.1 MAl(OH)_3solution are needed to neutralise 21 ml of0.1M HCl?
Answer» 14 mL 7 mL 21 mL none of these ANSWER :B

Discussion

25.	Stetement-1: Boat fromis the least stable conformation of cyclohexane. Statement-2: Boat from is eclipsed from.
Answer» Statement-1 is TRUE, Statement-2 is True, Statement-2 is a CORRECT EXPLANATION for Statement-1. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1 Statement-1 is True, Statement-2 is FALSE. Statement-1 is False, Statement-2 is True. Solution :N//A

Discussion

26.	Stereoisomers, which can be interconverted simply by rotation about sigma bonds, are conformational isomers while those, which can be converted only by breaking and remaking of bonds and not simply by rotation, are called configurational isomers. The angle between C-C and C – H bonds on adjacent carbon atoms in any conformation is called dihedral angle. The cyclic compounds most commonly found in nature containing six membered rings can exist in a conformation that is almost completely free of strain. The most stable conformation of cyclohexane is chair form. According to Bayer strain theory, the greater deviation from the normal tetrahedral angle, greater is the angle strain or torsional strain and hence lesser is the stability of the cycloalkane. Which among the following conformation of cyclohexane is the most stable form ?
Answer» CHAIR FORM Half chair forms Twist boat form Boat form SOLUTION :Chair form is staggered form

Discussion

27.	Stereoisomers, which can be interconverted simply by rotation about sigma bonds, are conformational isomers while those, which can be converted only by breaking and remaking of bonds and not simply by rotation, are called configurational isomers. The angle between C-C and C – H bonds on adjacent carbon atoms in any conformation is called dihedral angle. The cyclic compounds most commonly found in nature containing six membered rings can exist in a conformation that is almost completely free of strain. The most stable conformation of cyclohexane is chair form. According to Bayer strain theory, the greater deviation from the normal tetrahedral angle, greater is the angle strain or torsional strain and hence lesser is the stability of the cycloalkane. Dihedral angle between two methyl groups of n-butane in the gauche and anti forms are :
Answer» `60^@, 0^@` `60^@, 180^@` `0^@, 160^@` `180^@, 60^@` SOLUTION :`60^@, 180^@`

Discussion

28.	Stereoisomers, which can be interconverted simply by rotation about sigma bonds, are conformational isomers while those, which can be converted only by breaking and remaking of bonds and not simply by rotation, are called configurational isomers. The angle between C-C and C – H bonds on adjacent carbon atoms in any conformation is called dihedral angle. The cyclic compounds most commonly found in nature containing six membered rings can exist in a conformation that is almost completely free of strain. The most stable conformation of cyclohexane is chair form. According to Bayer strain theory, the greater deviation from the normal tetrahedral angle, greater is the angle strain or torsional strain and hence lesser is the stability of the cycloalkane. Dihedral angle in staggered (gauche) and eclipsed conformation are:
Answer» `60^@` and `0^@` `0^@` and `60^@` `60^@, 120^@` `120^@, 60^@` SOLUTION :`60^@` and `0^@`

Discussion

29.	Do structural isomers possess similar chemical properties?
Answer» ANSWER :T

Discussion

30.	Stephen reduction (SnCl_(2)//HCl) converts cyanides to
Answer» Aldehydes Ketones Amines Acids Answer :A

Discussion

31.	STEP-(I): Salt 'A' +Zn +dil. H_(2)SO_(4) rarr 'X" (g) (Rotten egg's smell) STEP-(II): 'X" gas + (("filter paper moistened"),("with" Pb(OAc)_(2)))rarr 'Y' (Black colour) STEP-(III): 'X' (gas) +H_(2)SO_(3) rarr 'Z' (white turbidity) Here 'Y' and 'Z' are:
Answer» `PbS` and `PbSO_(4)` `PbS` and `S` `ZNS` and `PbSO_(4)` `ZnS` and `ZnSO_(4)` Answer :B

Discussion

32.	STEP-(I): Salt 'A' +Zn +dil. H_(2)SO_(4) rarr 'X" (g) (Rotten egg's smell) STEP-(II): 'X" gas + (("filter paper moistened"),("with" Pb(OAc)_(2)))rarr 'Y' (Black colour) STEP-(III): 'X' (gas) +H_(2)SO_(3) rarr 'Z' (white turbidity) Salt 'A' and gas 'X' may be:
Answer» `Na_(2)SO_(3)` and `SO_(2)` `NaHSO_(3)` and `H_(2)S` `Na_(2)SO_(3)` and `H_(2)S` `NaHSO_(3)` and `SO_(2)` Answer :B::C

Discussion

33.	STEP-(I): Salt 'A' +Zn +dil. H_(2)SO_(4) rarr 'X" (g) (Rotten egg's smell) STEP-(II): 'X" gas + (("filter paper moistened"),("with" Pb(OAc)_(2)))rarr 'Y' (Black colour) STEP-(III): 'X' (gas) +H_(2)SO_(3) rarr 'Z' (white turbidity) The correct statement about the above experimental obserbations is/are:
Answer» In step-1,salt 'A' gets reduced to 'X' gas which turms lime water milky. X' gas also TURNS filter paper moistened with CADMIUM acetate to yellow colouration Y' is insoluble in water but with `O_(3)` it gets completely oxidised and CHANGES to white colour precipitate In step-III white turbidity is formed due to redox reaction (comproportionation) of 'X' gasand `H_(2)SO_(3)`. Answer :B::C::D

Discussion

34.	STEP-(I): Na_(2)CO_(3)(aq)overset("excess" SO_(2))rarr 'A' overset(Na_(2)CO_(3))rarr 'B' STEP-(II): The compound 'D' when treatedwith dilute H_(2)SO_(4) gives get with pungent smell and turns lime water milky. the compound 'D' is:
Answer» `BaCO_(3)` `BaSO_(3)` `BaSO_(4)` `Na_(2)SO_(3)` Answer :B

Discussion

35.	STEP-(I): Na_(2)CO_(3)(aq)overset("excess" SO_(2))rarr 'A' overset(Na_(2)CO_(3))rarr 'B' STEP-(II): E may be:
Answer» `BaCO_(3)` `BaSO_(3)` `BaSO_(4)` `BaS_(2)O_(3)` Answer :C

Discussion

36.	STEP-(I): Na_(2)CO_(3)(aq)overset("excess" SO_(2))rarr 'A' overset(Na_(2)CO_(3))rarr 'B' STEP-(II): Compound 'A' and 'B' are:
Answer» `NaHCO_(3)` and `Na_(2)CO_(3)` `NaHCO_(3)` and `Na_(2)SO_(3)` `NaHSO_(3)` and `Na_(2)SO_(3)` `NaHSO_(3)` and `Na_(2)CO_(3)` Answer :C

Discussion

37.	STEM stand for
Answer» SCANNING TRANSMISSION electron MICROSCOPE scanning tunneling electron microscope both A and B None of these. Solution :STEM STAND both for scanning tunneling electron microscope and scanning transmission electron microscope.

Discussion

38.	Steel is denser than water but a blade made from it floats on the surface of water. Why ?
Answer» Solution :This is because surface of WATER ACTS as a THIN film which is capable of resisting small temsile force which is DUE to surface tension.

Discussion

39.	Steam distillation is used to purify a compound which is steam volatile and insoluble in water. The impurities should not be steam volatile. It is based on the principle that liquid will boil when partial sure. p=p_(1)+p_(2). It reduces the boiling point of liquid.("Weight of water distilled")/("Wt of substance distilled")=("M.wt of water"xx"V.P. of steam")/("M.Wt of substance"xx "V.P. of aniline")Calculate weight of aniline distilled if weight of water distilled is 100g when P_("organic compound")=100mm and P_(H_(2)O)=200mm
Answer» <P>250 G 258 g 100 g 25.8 g SOLUTION :`(omega_(1))/(omega_(2))=(P_(1).18)/(P_(2).93)` suffix 1 for `H_(2)O` , suffix 2 for ANILINE, MW of aniline = 93. `(100)/(omega_(2))=(200xx18)/(100xx93)rArr omega_(2)=258.33`

Discussion

40.	Steam distillation is used to purify a compound which is steam volatile and insoluble in water. The impurities should not be steam volatile. It is based on the principle that liquid will boil when partial sure. p=p_(1)+p_(2). It reduces the boiling point of liquid.("Weight of water distilled")/("Wt of substance distilled")=("M.wt of water"xx"V.P. of steam")/("M.Wt of substance"xx "V.P. of aniline")Which of the following is steam volatile
Answer» o-nitrophenol p-nitrophenol p-hydroxy BENZALDEHYDE ETHANOL SOLUTION :O-nitro phenol is STEAM volatitle SINCE it has intra molecular H - bonding.

Discussion

41.	Steam distillation is used to purify a compound which is steam volatile and insoluble in water. The impurities should not be steam volatile. It is based on the principle that liquid will boil when partial sure. p=p_(1)+p_(2). It reduces the boiling point of liquid.("Weight of water distilled")/("Wt of substance distilled")=("M.wt of water"xx"V.P. of steam")/("M.Wt of substance"xx "V.P. of aniline")Isolation of essential oils from flowers etc is done by
Answer» Steam DISTILLATION Distillation Fractional distillation Distillation under reduced pressure Solution :By steam distillation essential OILS are ISOLATED from FLOWERS.

Discussion

42.	The term anthesis is used for
Answer» SOLUTION :ESSENTIAL OILS

Discussion

43.	Steam distillation method is useful for the purification of
Answer» Are insolute in water Are VOLATILE in steam Have non-steam volatile impurities All of the above Answer :D

Discussion

44.	Steam distillation is used for separation and purification of those compounds which are.............with water but................ .
Answer» ANSWER :IMMISCIBLE, VOLATILE is STEAM

Discussion

45.	Steam distillation in principle is comparable to distillation under reduced pressure. Do you agree to it ?
Answer» SOLUTION :YES.

Discussion

46.	Status coated with white lead on long exposure to atmosphere turn black and the original colour can be restored on treatment with H_(2)O_(2) . Why ?
Answer» SOLUTION :On long exposure to atmosphere , white lead is converted into black PbS DUE to the action of `H_(2)S` PRESENT in the atmosphere . As a result , statues TURN black . `PbO_(2) + 2H_(2)S to PbS + 2H_(2)O` On treatment of these blackened statues with `H_(2)O_(2)` , black PbS gets oxidised to white `PbSO_(4)` and the colour is RESTORED. `PbS + 4H_(2)O_(2) to PbSO_(4) + 4H_(2)O`

Discussion

47.	Statuse and paintings coated with white lead turn black on long exposure to atmosphere. The original colour can be restored by treating them with H_(2)O_(2). The reason behind this is
Answer» blackened statues get coated with PBS which on reaction with `H_(2)O_(2)` is oxidised to WHITE `PbSO_(4)` `H_(2)O_(2)` DISSOLVES the COATING white lead and exposes the inner surface. White lead reacts with `H_(2)O_(2)` to form white `PbSO_(4)` blackened STATUSE get coated with lead sulphate which reacts with `H_(2)O_(2)` to give PbS. Solution :The statues and paintings coated with white lead get blackened due to action of `H_(2)S` present in traces in atmosphere. When these are treated with `H_(2)O_(2)`, PbS is oxidised to `PbSO_(4)` which is colourless or white. `underset("black")(PbS)+4H_(2)O_(2)to underset("white")(PbSO_(4))+4H_(2)O`

Discussion

48.	Statues coated with white lead turn black on exposure to air Its original colour is restored on treatment with H_2O_2. Explain.
Answer» SOLUTION :Statues turn black DUE to the formation of lead sulphide. Hydrogen PEROXIDE oxidises black COLOURED lead sulphide to white coloured lead sulphate, there by restoring the COLOUR. `PbS+4H_2O_2toPbSO_4+4H_2O`

Discussion

49.	Steam distillation at low pressure: (i) In soap industry (ii) Water pump, vacuum pump, air pump are used. (iii) High boiling point containing liquid are not decompose. (iv) The liquid is decomposed. (v) The liquid is boiled at low temperature
Answer» SOLUTION :(i-T), (ii-F), (iii-T), (iv-F), (v-T)

Discussion

50.	Statues and monuments in India such as Tajmahal are affected by acid rain. How?
Answer» Solution :Due to large NUMBER of chemical factories, power PLANTS and other industries, air around the monuments are highly polluted with oxides of SULPHUR and nitrogen. This causes ACID rain which corrodes metallic surfaces exposed to it. Marble of Taj Mahal is slowly eaten AWAY by acid rain.

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

The structural formula of 2-methyl-2-butene is

Structrue of ICl_(2)^(-) is

Strontia is

Strongest reducing agent is"_______________".

Strongest reducing agent and strongest oxidising agent are respectively

Strongest nucleophile in polar-protic solvent is

Strongest oxidant among the following is

Strongest conjugate base is

Strongest hydrogen bonds can be formed by

Strongest bond is between

Stronger deactivator group is ……… .

Strong reducing behaviour of H_(3)PO_(2) is due to

Strong reducing behaviour of H_(3)PO_(2)is due to

Strength of acidity is in order

Strength of a weak acid or a weak base depends upon its

Strength of 10 volume hydrogen peroxide solution means

Stomach acid, a dilute solution of HCl can be neutralised by reaction with Aluminium hydroxideAl(OH)_(3) + 3HCl(aq)to + AICI_(3) + 3H_(2)O How many millilitres of 0.1 M Al(OH)_(3) solution are needed to neutralise 21 mL of 0.1 M HCI?

Stomach acid, a dilute solution of HCl can be neutralised by reaction with aluminig hydroxide Al (OH)_(3) + 3HCl (aq) to AlCl _(3) + 3H _(2)O How many milliliters of 0.1 M Al (OH) _(3) solution are needed to neutralise 21 mL of 0.1 M HCl ?

Stomach acid a dilute solution of HCI can be neutralised by reaction with Aluminium hydroxide.Al(OH)_3 +3HCl(aq) to AlCl_3 +3H_2O how many millitres of 0.1 MAl(OH)_3solution are needed to neutralise 21 ml of0.1M HCl?

Stetement-1: Boat fromis the least stable conformation of cyclohexane. Statement-2: Boat from is eclipsed from.

Do structural isomers possess similar chemical properties?

Stephen reduction (SnCl_(2)//HCl) converts cyanides to

STEP-(I): Salt 'A' +Zn +dil. H_(2)SO_(4) rarr 'X" (g) (Rotten egg's smell) STEP-(II): 'X" gas + (("filter paper moistened"),("with" Pb(OAc)_(2)))rarr 'Y' (Black colour) STEP-(III): 'X' (gas) +H_(2)SO_(3) rarr 'Z' (white turbidity) Here 'Y' and 'Z' are:

STEP-(I): Salt 'A' +Zn +dil. H_(2)SO_(4) rarr 'X" (g) (Rotten egg's smell) STEP-(II): 'X" gas + (("filter paper moistened"),("with" Pb(OAc)_(2)))rarr 'Y' (Black colour) STEP-(III): 'X' (gas) +H_(2)SO_(3) rarr 'Z' (white turbidity) Salt 'A' and gas 'X' may be:

STEP-(I): Na_(2)CO_(3)(aq)overset("excess" SO_(2))rarr 'A' overset(Na_(2)CO_(3))rarr 'B' STEP-(II): The compound 'D' when treatedwith dilute H_(2)SO_(4) gives get with pungent smell and turns lime water milky. the compound 'D' is:

STEP-(I): Na_(2)CO_(3)(aq)overset("excess" SO_(2))rarr 'A' overset(Na_(2)CO_(3))rarr 'B' STEP-(II): E may be:

STEP-(I): Na_(2)CO_(3)(aq)overset("excess" SO_(2))rarr 'A' overset(Na_(2)CO_(3))rarr 'B' STEP-(II): Compound 'A' and 'B' are:

STEM stand for

Steel is denser than water but a blade made from it floats on the surface of water. Why ?

The term anthesis is used for

Steam distillation method is useful for the purification of

Steam distillation is used for separation and purification of those compounds which are.............with water but................ .

Steam distillation in principle is comparable to distillation under reduced pressure. Do you agree to it ?

Status coated with white lead on long exposure to atmosphere turn black and the original colour can be restored on treatment with H_(2)O_(2) . Why ?

Statuse and paintings coated with white lead turn black on long exposure to atmosphere. The original colour can be restored by treating them with H_(2)O_(2). The reason behind this is

Statues coated with white lead turn black on exposure to air Its original colour is restored on treatment with H_2O_2. Explain.

Steam distillation at low pressure: (i) In soap industry (ii) Water pump, vacuum pump, air pump are used. (iii) High boiling point containing liquid are not decompose. (iv) The liquid is decomposed. (v) The liquid is boiled at low temperature

Statues and monuments in India such as Tajmahal are affected by acid rain. How?