94420 + Interview Questions in Chemistry in Class 12

1.	Decomposition of hydrogen peroxide in alkaline medium in presence of iodide (I^(-))catalyst occurs according to the equation:2H_(2)O_(2)underset("alkali")overset(I^(-))rarr2H_(2)O(l)+O_(2)(g) A proposed mechanism for this reaction involves the following steps : Step-I : H_(2)O_(2)(l)+I^(-)(aq)rarrH_(2)O(l)+IO^(-)(aq)("slow") Step-II : H_(2)O_(2)(l)+IO^(-)(aq)rarrH_(2)O(l)+IO^(-)(aq)+O_(2)(g)("fast")Write rate law for the reaction .
Answer» Solution :Rate EQUATION of a reaction=rate equation of the SLOWEST STEP of the reaction. Hence , the rate equation of the reaction `=k[H_(2)O_(2)][I^(-)]`

Discussion

2.	Decomposition of H_2O_2 was studied by titration against KMnO_4 solution. It was found that 0.4 mole of H_2O_2 was reduced to 0.2 mole in 20 minutes and to 0.1 mole in 40 minutes and to 0.05 mole after one hour. The order of reaction must be
Answer» 0 1 2 3 Solution :As `t_(1//2)` is INDEPENDENT of initial CONC, it is a reactionof 1ST ORDER.

Discussion

3.	Decomposition of H_2O_2 (Ist order) H_2O_2(aq)toH_2O(l)+1/2O_2(g) Can be monitored by titration method by pressure measurement.If {:("time(min.)",t=0,30,oo),("Pressure (mm of Hg)",200,375,400):} While when progress of same reaction was monitored by titration method or the volume of titrant consumed after 10 min was found to be 20 ml.Then complete the following table. {:("Time(min)",t=0,10,30),("Pressure(mm of Hg)",200,(b),(c )),("Volume of" KMnO_4,(a),20,(d)):} consumed (ml)
Answer» Solution :`k=1/(30"MIN")LN{(400-200)/(400-375)}=1/(30"min")ln{200/25}=(ln2)/T_(1//2)` `T_(1//2)=(30 "min")(ln2)/(ln8)=10` min So, we will have `{:("Time(min)",t=0,10,30),("Pressure (in MM of Hg)",200,300,375),("Volume of" KMnO_4"consumed (ML)",40,20,5):}`

Discussion

4.	Decomposition of H_2O_2 is retarded by :
Answer» Acetanilide Alcohol `H_3PO_4` All of these Answer :D

Discussion

5.	Decomposition of H_(2)O_(2) is slowed down by the addition of
Answer» ALCOHOL `MnO_(2)` ALKALI Pt Answer :A

Discussion

6.	Decomposition of H_(2)O_(2) is favouredby
Answer» acetanilide MNO TRACES of acids alcohol Solution :`MnO` CATALYSES the decomposition of `H_(2)O_(2)`.

Discussion

7.	Decomposition of H_2O_2 is accompanied by :
Answer» TRACES of acids Finely DIVIDED metals Acetanilide Alcohol Answer :B

Discussion

8.	Decomposition ofH_(2)O_(2) is a first order reaction. A solution of H_(2)O_(2) lebelled as "16.8 V" was left open. Due to this, some H_(2)O_(2) decomposed. To determine the new volume strength after 2.303 hours, 20 mL of this solution was diluted to 100 mL. 25 mL of this diluted solution was titrated against 37.5 mL of 0.02 M KMnO_(4) solution under acidic conditions [Given : STP is 1 atm and 273 K] Calculate the rate constant (in "hr"^(–1)) for decomposition of H_(2)O_(2).
Answer» 0.6 0.5 0.4 0.8 Answer :A

Discussion

9.	Decomposition of H_(2)O_(2) is a first-order reaction. A solution of H_(2)O_(2) labelled as 20 volumes was left open. Due to this some H_(2)O_(2) decomposed. To determine the new volume strength after 6 hours, 10 mL of this solution was diluted to 100 mL. 10 mL of this diluted solution was titrated against 25 mL of 0.025 m KMnO_(4) acidified solution. Calculate the rate constant for decomposition of H_(2)O_(2).
Answer» SOLUTION :N//A

Discussion

10.	Decomposition of H_(2)O_(2) follows a first order reaction . In fifty minutes the concentration of H_(2)O_(2) decreases from 0.5 to 0.125 M in one such decomposition . When the concentration of H_(2)O_(2) reaches from 0.05 M , the rate of formation of O_(2) will be
Answer» `6.93 xx 10^(-4) mol "min"^(-1)` `2.66 L "min"^(-1)` at STP `1.34 xx 10^(-2) mol "min"^(-1)` `6.93 xx 10^(-2) mol "min"^(-1)` Solution :In 50 minutes , concentration of `H_(2)O_(2)` becomes `(1)/(4)` of initial `implies 2 xx t_(1//2) = 50 ` minutes `"" t_(1//2) = 25` minutes `implies K = (0.693)/(25) ` PER minute `r_(H_(2) O_(2)) = (0.693)/(25) xx 0.05 = 1.386 xx 10^(-3)` `2H_(2)O_(2) to 2H_(2)O + O_(2) "" r_(O_(2)) = (1)/(2) xx r_(H_(2)O_(2))` `r_(O_(2)) = 0.693 xx 10^(-3) "" r_(O_(2)) = 6.93 xx 10^(-4)` mol/minute `xx` LITRE

Discussion

11.	Decomposition of H_(2)O_(2) follows a first order reaction. In fifty minutes, the concentration of H_(2)O_(2) decreases from 0.5 to 0.125 M in one such decomposition. When the concentration of H_(2)O_(2) reaches 0.05 M, the rate of formation of O_(2) will be
Answer» `6.93xx10^(-2)" mol min"^(-1)` `6.93xx10^(-4)"mol L"^(-1)"min"^(-1)` `2.66" L min"^(-1)" at STP"` `1.34xx10^(-2)" mol min"^(-1)` Solution :Decompostion of `H_(2)0_(2)` takes place as : `H_(2)O_(2) to H_(2)O + (1)/(2)IO_(2)` Decrease of concentration of `H_(2)O_(2)` from 0.5 to 0.125 in 50 MINUTES means two half lives `(05 to 025 to 0125)` i.e. `2xxt_(1//2) = 50 min or t_(1//2) = 25 min` For a 1st order reaction, `k = (0693)/(t_(1//2)) = (0693)/(25) min^(-1)` Rate of reaction = Rate of decomposition of `H_(2)O_(2),` i.e., `(d[H_(2)O_(2)])/(dt) = k [H_(2)O_(2)]` `= (0693)/(25) XX 005 = 1386 xx 10^(-3) mol L^(-1) min^(-1)` Rate of FORMATION of `O_(2) = (d[O_(2)])/(dt) = -(1)/(2)(d[H_(2)O_(2)])/(dt)` ` = (1)/(2) xx 1386 xx 10^(-3)` `= 693 xx 10 ^(-4) mol L^(-1) min^(-1)`

Discussion

12.	Decomposition of H_(2)O_(2) follows a first oeder reaction.In fifty minutes the concentration of H_(2)O_(2) decreases from 0.5 to 0.125 M in one such decomposition .When the concentration of H_(2)O_(2) reaches 0.05 M ,the rate of formation of O_(2) will be:
Answer» `1.34xx10^(-2)` mol `min^(1)` `6.93xx10^(-2)`mol `min^(-1)` `6.93xx10^(-4)` mol `min^(-1)` 2.66 L `min^(-1)` at STP Solution :Here solution is given assuming UNIT of rate is the option to be `mol^(-1)L^(-1) min^(-1)` For first order reaction K=`(1)/(50)` in `(0.5)/(0.125)` K=`(In4)/(50) min^(-1)` Now R=`K[H_(2)O_(2)]^(1)` `=2xx0.693xx10^(-3) mol L^(-1)min^(-1)` and `H_(2)O_(2)toH_(2)O+(1)/(2)O_(2)` So rate of reaction. `R=(R_(O_(2))("Rate formation of"O_(2)))/(((1)/(2)))` or `R_(O_(2))=6.93xx10^(-4)mol L^(-1)min^(-1)`

Discussion

13.	Decomposition of H_2O_2 followsa first order reaction. In fifty minutes the concentration of H_2O_2 decreases from 0.5 to 0.125 M in one such decomposition . When the concentration of H_2O_2 reaches 0.05M , the rate of formation of O_2 will be-
Answer» `1.34 xx 10^(-2) mol.MIN^(-1)` `6.93 xx 10^(-2) mol.min^(-1)` `6.93 xx 10^(-4) mol.min^(-1)` `2.66L . "min"^(-1)` at STP Answer :C

Discussion

14.	Decomposition of H_(2)O_(2) can be prevented in presence of
Answer» GLYCEROL acetanilide phosphoric ACID all of these SOLUTION :All the given substances act as NEGATIVE CATALYST for decomposition of `H_(2)O_(2).`

Discussion

15.	Decomposition of ammonia on platinum surface follows the change 2NH_(3)(g) to N_(2)(g) + 3H_(2)(g) If the rate obeys (-d[NH_(3)])/(dt)=(k_(1)[NH_(3)])/(1+k_(2)[NH_(3)]), what will be the order for decomposition of NH_(3) if [NH_(3)] is very very less, K_(1)"and" K_(2) are constant?
Answer» ANSWER :1

Discussion

16.	Decomposition of benzene diazonium chloride by using (Cu_(2)CI_(2))/(HCI) to form chlorobenzene is
Answer» Rasching's reaction Sandmeyer's reactio Kolbe's reaction Cannizzaro's reaction Solution :

Discussion

17.	Decomposition of 3A(g)to2B(g)+2C(g) follows first order kinetics.Initially only A is present in the container.Pressure developed after 20 min. and infinite time are 3.5 and 4 atm respectively.Which of the following is true .
Answer» `t_(50%)=20` min `t_(75%)=40` min `t_(99%)=64//3` min `t_(87.5%)=60` min Solution :`{:(,3A(g)" "to,2B(g)" "+,2C(g)),(t=0,P_0," ", " "),(t=20,P_0-x,(2x)/3,(2x)/3),(t=oo, " ",(2P_0)/3,(2P_0)/3):}` `IMPLIES (4P_0)/3=4` `P_0=3` atm. `P_0+x/3=3.5implies t_(50%)=20` is the half life. `x=1.5 " "t_(75%)=2xx20=40` min. `t_(87.5%)=3xxt_(50%)=3xx20=60` min `t_(99%)=2/3xxt_(99.9%)=2/3xx10xxt_(50%)=400/3`

Discussion

18.	Decolourisation of acidified potassium permanganate occurs when H_2O_2 is added to it.This is due to:
Answer» OXIDATION of `KMnO_3` REDUCTION of `KMnO_4` Both oxidation and reduction of `KMnO_4` NONE of these Answer :B

Discussion

19.	Decolourisationof acidified permaganate in fast by Mohr'ssalt , butis solw by oxalate . Expalin .
Answer» Solution :Oxalate, `""^(-)OOC-COO^(-)`, on OXIDATION GIVES `CO_(2)`. It is involves breaking of C-C bond, which requires energy. Hence decolourisation of `MnO_(4)^(-)` by `C_(2)O_(4)^(2-)` is slow. On the other hard Mohr.s SALT is a COMPLEX salt containing `FE^(2+)`ions .

Discussion

20.	Decolorises iodine solution. Identify (A) to (D) and give the reaction involved in each case.
Answer» SOLUTION :Theseobservations suggestthat thesubstance(A) issulphur (S), the gas (B) is nitrogen dioxide`(NO_(2))` , COMPOUND (C) is sodiumthiosulphute`(Na_(2)S_(2)O_(3))`and the gase(D) is sulphur dioxide `(SO_(2))`. The REACTION are as follows : `2Na_(2)underset((C))(S_(2)O_(3)) + underset("(violet)")(I_(2))to Na_(2)underset("colourless solution")(S_(4)O_(6)) + 2NaI`

Discussion

21.	Decoloriization of coloured solution by animal charcoal is an example of:
Answer» sorption desorption absorption ADSORPTION SOLUTION :adsorption

Discussion

22.	Decinormal solution of NaCl developed an osmotic pressure of 4.6 atmosphere at 300 K. Calcualte its degree of dissociatoin (R = 0.082" L atm K"^(-1)"mol"^(-1))
Answer» ANSWER :0.87

Discussion

23.	Decinormal solution of CH_3COOH ionised to an extent of 1.3%. pH of the solution is, (log 1.3 = 0.11)
Answer» 2.89 1.945 3.4 4.98 Answer :A

Discussion

24.	Decinirmal solution of NaCI develop an osmotic pressur of 4.6 atosphers at 300 K. Calculate the degreedessociation,
Answer» Solution :STEP I. Calculation os Van't Hoff factor (i) `PI=("inRT")/Vori=(piV)/(nRT),pi=4.6 atm, n=0.1 N(or 0.1 M), T=300 K` `V=1L, R=0.0821" L atm K"^(-1)mol^(-1)` ` i=((4.6" atm")xx(1L))/((0.1"mol"^(-1))xx(0.0821"L atm K"^(-1)mol^(-1))xx(300 K))=1.87` `Step II. Calculation of degree of dissociation `(alpha)` `Naoverset(("aq"))toNa^(+)(aq)THEREFORE" VALUE of "n=2` `alpha=(i-1)/(n-1)or alpha=(1.87-1)/(2-1)=0.87xx100=87%.`

Discussion

25.	Decide X, Y and Z correctly from the given options in the following sentences. (i) Thermal refining process is used for the refining of X metal. (ii) Zone refining process is used for purification of Y. (iii) Dyes can be separated by Z.
Answer» X = IRON, Y = Ge, Z = MOND Carbonyl process X = Iron, Y = Ge, Z = Van Arkel method X = Copper, Y = Si, Z = CHROMATOGRAPHIC method X = Copper, Y = Si, Z = LIQUATION Answer :C

Discussion

26.	Decide giving reason which one of the following pairsexhibits the property indicated : (i) Sc^(3+) or Cr^(3+) exhibits paramagnetism(ii) Vor Mn exhibits more number of oxidation states ( Atomic numbers : Sc = 21, Cr= 24, V = 23 , Mn = 25)
Answer» Solution :(i)`. _(21) Sc = [ Ar] 3d^(1) 4S^(2) , Sc^(3+) = [Ar] -` No unparied electron `. _(24)Cr)=[Ar] 3d^(5)4s^(1) , Cr^(3+)= [Ar] 3d^(3) - `Three unpaired electron Hence, `Cr^(3+)` exhibits PARAMAGNETISM. (ii) Mn exhibits more number of oxidation states. Reason. `._(23) V = 3d^(3) 4s^(2)`. Its ox. states can be` + 2, + 3, +4, +5` `. _(25) Mn = 3d^(5) 4s^(2)` . Its ox. statescan be ` +2, +3,+4,+5,+6,+7`

Discussion

27.	Decay of "_(92)U^(235) is ….., order reaction
Answer» ZERO First Second Third SOLUTION :The radioactive disintegration reactions are of first order because in this rate of disintegration depends on the CONCENTRATION term of radioactive MATERIAL only .

Discussion

28.	Decarboxylation of which yield 1,1,2,2-tetrabromoethane :
Answer» `CH_3COOH` `CH_2Br_2CBr_2COOH` `HCBr_2CBr_2COOH` `CH_3BrCHBrCOOH` ANSWER :C

Discussion

29.	Decarboxylation of monocarboxylic acids can be carried out by :
Answer» KOH `CA(OH)_(2)` Sodalime Red P. Answer :C

Discussion

30.	Decarboxylation of isobutyric acid leads to:
Answer» Isobutane Propane Butane NONE of these Answer :B

Discussion

31.	Decarboxylationof isobutyricacidgivesn - alkanebutreducationof isobutyricacidwithphsphorusandhydrogeniodinegives :
Answer» n- PROPANE isombutane n- butane NONE of these ANSWER :B

Discussion

32.	Decarboxylation of malonic acid gives:
Answer» HCHO COOH-COOH `CH_3COOH` `CH_4` ANSWER :D

Discussion

33.	Decarboxylation of glycine yields
Answer» `CH_4` `CH_3COOH` `CH_3NH_2` `ETHANAMIDE` ANSWER :C

Discussion

34.	Decarboxylation of carboxylic acid takes place:
Answer» in soda lime TREATMENT in quick lime treatment in KOLBE's ELECTROLYSIS in all of the above Answer :D

Discussion

35.	Debye, Huckel and Onsager equation for strong electrolytes is lambda_(c)=lambda_(infty)(A+Blambda_(infty))sqrt(C). The slope value is
Answer» `lambda_(INFTY)` `(A+Blambda_(infty))` A `SQRT(C)` ANSWER :B

Discussion

36.	Debye Huckel Onsager theory predicts ?
Answer» <P>VARIATION of velocity constant with T. Variation of EQUIVALENT CONDUCTANCE with concentration Variation of equivalent constant with P Variation of equivalent conductance with T Answer :B

Discussion

37.	Debye an unit of dipole moment is of the order of :
Answer» `10^(-10)` ESU CM `10^(-18)` esu cm `10^(-6 )` esu cm `10^(-12)` esu cm` Answer :B

Discussion

38.	Debye constants A and B depend on ________.
Answer» NATURE of the solvent temperature concentration of the solvent both (a) and (B) ANSWER :D

Discussion

39.	Debrominaton of d-2, 3-dibromobutane gives :
Answer» TRANS -2- butene Cis - 2-butene 1-butene 2-butyne Answer :B

Discussion

40.	Debrominationof mesodibromobutane will give the product as :
Answer» n-butane 2-butyne cis-2-butene trans-2-butene Solution :N//A

Discussion

41.	Deaths due to liquor poisoning are due to the presence of
Answer» ETHYL ALCOHOL kcn methyl alcohol carbonic ACID Answer :C

Discussion

42.	Deamination of arylamines via diazonium salts is best achieved by_____
Answer» SOLUTION :Hypophosphorus ACID `(H_(3)PO_(2))`.

Discussion

43.	Dead burnt plaster is obtained by heating Plaster of Paris. It is :
Answer» `CaSO_(4)1/2H_(2)O` `CaSO_(4)2H_(2)O` `CaSO_(4)` `CaHCO_(3)` ANSWER :C

Discussion

44.	Dead burntplaster is
Answer» `CaSO_(4)` `CaSO_(4)1/2H_(2)O` `CaSO_(4)H_(2)O` `CaSO_(4)*2H_(2)O` ANSWER :C

Discussion

45.	Dead burnt is:
Answer» `CaSO_4` `Na_2CO_3` ANHYDROUS `Na_2So_4` Anhydrous `CuSo_4` ANSWER :A

Discussion

46.	Deacon's process is used to manufacture …………. .
Answer» `Cl_(2)`<BR>`F_(2)` `Br_(2)` `I_(2)` SOLUTION :`Cl_(2)`

Discussion

47.	Deacon.sprocess is used for the manufacture of:
Answer» BLEACHING powder Sulphuric acid Nitric acid Chlorine Answer :D

Discussion

48.	De-electronation in electrochemical cell is called as
Answer» Anodic OXIDATION Cathodic oxidation Anodic reduction Cathodic reduction Solution :LOSS of ELECTRONS (oxidation ) takes place at ANODE.

Discussion

49.	de Broglie (1924) predicted that small particles such as electrons should show wave -like properties along with paticle character. The wave length(lamda) associated with particle of mass m and moving with velocity v is given as lamda=h/(mv) where 'h' is plank's constant. The wave nature was confirmed by Davisson andGermer's experiment and modified equation for calculation of lamda can be given as: lamda=-h/(sqrt(2Em)) where E= kinetic energy of particle. lamda=h/(sqrt(2dVm)), where d= change of particle accelerated potnetial of V volt. If the kinetic energy of free electron is doubled,. Its de Broglie wavelengthh changes bythe
Answer» `sqrt(2)` `1/(sqrt(2))` `2` `1/2` Solution :`K.E.=1/2 mv^(2)=E` `:.v=sqrt((2xxE)/m)` `:.(v_(2))/(v_(1))=sqrt(2)` Also `lamda_(1)=1/(mv_(1)),lamda_(2)=1/(mv_(2))` `(lamda_(2))/(lamda_(1))=(v_(1))/(v_(2))implieslamda_(2)=((v_(1))/(v_(2)))xxlamda_(1)=1/(sqrt(2))xxlamda_(1)` Hence B is the correct ANSWER.

Discussion

50.	de Broglie (1924) predicted that small particles such as electrons should show wave -like properties along with paticle character. The wave length(lamda) associated with particle of mass m and moving with velocity v is given as lamda=h/(mv) where 'h' is plank's constant. The wave nature was confirmed by Davisson andGermer's experiment and modified equation for calculation of lamda can be given as: lamda=-h/(sqrt(2Em)) where E= kinetic energy of particle. lamda=h/(sqrt(2dVm)), where d= change of particle accelerated potnetial of V volt. The ratio of de-Broglie chi wavelength of molecules of H_(2) and He at 27^(@)C adn 127^(@)C respectively is
Answer» 1.633 0.612 1.265 0.79 Solution :`LAMDA=h/(MV)` `lamda_(H_(1))=h/(m_(H_(2))sqrt((3RT_(1))/(mH_(2))))` `lamda_(H_(e))=h/(m_(H_(e))sqrt((3RT_(2))/(mH_(e))))` `(lamda_(H_(2)))/(lamda_(H_(2)))=sqrt((T_(2)xxm_(H_(e)))/(T_(xxm_(H_(2))))` `=sqrt((400xx4xx10^(-3))/(300xx2xx10^(-3)))` `=sqrt(8/3)` `=1.633` Hence a is the correct OPTION.

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

Decomposition of H_2O_2 was studied by titration against KMnO_4 solution. It was found that 0.4 mole of H_2O_2 was reduced to 0.2 mole in 20 minutes and to 0.1 mole in 40 minutes and to 0.05 mole after one hour. The order of reaction must be

Decomposition of H_2O_2 is retarded by :

Decomposition of H_(2)O_(2) is slowed down by the addition of

Decomposition of H_(2)O_(2) is favouredby

Decomposition of H_2O_2 is accompanied by :

Decomposition of H_(2)O_(2) follows a first order reaction . In fifty minutes the concentration of H_(2)O_(2) decreases from 0.5 to 0.125 M in one such decomposition . When the concentration of H_(2)O_(2) reaches from 0.05 M , the rate of formation of O_(2) will be

Decomposition of H_(2)O_(2) follows a first order reaction. In fifty minutes, the concentration of H_(2)O_(2) decreases from 0.5 to 0.125 M in one such decomposition. When the concentration of H_(2)O_(2) reaches 0.05 M, the rate of formation of O_(2) will be

Decomposition of H_(2)O_(2) follows a first oeder reaction.In fifty minutes the concentration of H_(2)O_(2) decreases from 0.5 to 0.125 M in one such decomposition .When the concentration of H_(2)O_(2) reaches 0.05 M ,the rate of formation of O_(2) will be:

Decomposition of H_2O_2 followsa first order reaction. In fifty minutes the concentration of H_2O_2 decreases from 0.5 to 0.125 M in one such decomposition . When the concentration of H_2O_2 reaches 0.05M , the rate of formation of O_2 will be-

Decomposition of H_(2)O_(2) can be prevented in presence of

Decomposition of ammonia on platinum surface follows the change 2NH_(3)(g) to N_(2)(g) + 3H_(2)(g) If the rate obeys (-d[NH_(3)])/(dt)=(k_(1)[NH_(3)])/(1+k_(2)[NH_(3)]), what will be the order for decomposition of NH_(3) if [NH_(3)] is very very less, K_(1)"and" K_(2) are constant?

Decomposition of benzene diazonium chloride by using (Cu_(2)CI_(2))/(HCI) to form chlorobenzene is

Decomposition of 3A(g)to2B(g)+2C(g) follows first order kinetics.Initially only A is present in the container.Pressure developed after 20 min. and infinite time are 3.5 and 4 atm respectively.Which of the following is true .

Decolourisation of acidified potassium permanganate occurs when H_2O_2 is added to it.This is due to:

Decolourisationof acidified permaganate in fast by Mohr'ssalt , butis solw by oxalate . Expalin .

Decolorises iodine solution. Identify (A) to (D) and give the reaction involved in each case.

Decoloriization of coloured solution by animal charcoal is an example of:

Decinormal solution of NaCl developed an osmotic pressure of 4.6 atmosphere at 300 K. Calcualte its degree of dissociatoin (R = 0.082" L atm K"^(-1)"mol"^(-1))

Decinormal solution of CH_3COOH ionised to an extent of 1.3%. pH of the solution is, (log 1.3 = 0.11)

Decinirmal solution of NaCI develop an osmotic pressur of 4.6 atosphers at 300 K. Calculate the degreedessociation,

Decide X, Y and Z correctly from the given options in the following sentences. (i) Thermal refining process is used for the refining of X metal. (ii) Zone refining process is used for purification of Y. (iii) Dyes can be separated by Z.

Decide giving reason which one of the following pairsexhibits the property indicated : (i) Sc^(3+) or Cr^(3+) exhibits paramagnetism(ii) Vor Mn exhibits more number of oxidation states ( Atomic numbers : Sc = 21, Cr= 24, V = 23 , Mn = 25)

Decay of "_(92)U^(235) is ….., order reaction

Decarboxylation of which yield 1,1,2,2-tetrabromoethane :

Decarboxylation of monocarboxylic acids can be carried out by :

Decarboxylation of isobutyric acid leads to:

Decarboxylationof isobutyricacidgivesn - alkanebutreducationof isobutyricacidwithphsphorusandhydrogeniodinegives :

Decarboxylation of malonic acid gives:

Decarboxylation of glycine yields

Decarboxylation of carboxylic acid takes place:

Debye, Huckel and Onsager equation for strong electrolytes is lambda_(c)=lambda_(infty)(A+Blambda_(infty))sqrt(C). The slope value is

Debye Huckel Onsager theory predicts ?

Debye an unit of dipole moment is of the order of :

Debye constants A and B depend on ________.

Debrominaton of d-2, 3-dibromobutane gives :

Debrominationof mesodibromobutane will give the product as :

Deaths due to liquor poisoning are due to the presence of

Deamination of arylamines via diazonium salts is best achieved by_____

Dead burnt plaster is obtained by heating Plaster of Paris. It is :

Dead burntplaster is

Dead burnt is:

Deacon's process is used to manufacture …………. .

Deacon.sprocess is used for the manufacture of:

De-electronation in electrochemical cell is called as