94420 + Interview Questions in Chemistry in Class 12

1.	The co-ordination number and oxidation number of X in the compound[X(SO_(4))(NH_(3))_(5)] will be :
Answer» 10 and `+3` 1 and `+6` 6 and `+4` 6 and `+2` Solution :`[overset(+2)X(Soverset(-2)O_(4))(Noverset(0)H_(3))5]`

Discussion

2.	The co-ordination number and oxidation number of X in the following compound [X(SO_4)(NH_3)_5]C1 will be:
Answer» 10 and +3 2 and +6 6 and +3 6 and +4 Answer :C

Discussion

3.	The co-ordination compounds with bidentate ligands are called as
Answer» complexes ligands chelates double salts Answer :C

Discussion

4.	The CO in a 20L sample of gas was converted to CO_2 by passing the gas over iodine pentoxide heated to 150^@C: I_2O_5(s)+5CO(g)to5CO_2(g)+I_2(g) The iodine distilled at this temperature and was collected in an absorber containing 10 mL of 0.011 M Na_2S_2O_3.The excess hypo was back titrated with 5 mL of 0.001 M I_2 solution. What must be the milligrams of Co in 1 L of the original gas sample ?
Answer» 0.35 MG 0.70 mg 0.07 mg 3.5 mg Solution :MEQ of `l_2` LIBERATED =10 X 0.011 x 1-5 x 0.001 x 2=0.1 Millimoles of CO=`0.1/2 XX5 =0.25` Weight of CO in 1L =`0.25/20xx28 =0.35 mg`

Discussion

5.	The CO in a 20L sample of gas was converted to CO_2 by passing the gas over iodine pentoxide heated to 150^@C: I_2O_5(s)+5CO(g)to5CO_2(g)+I_2(g) The iodine distilled at this temperature and was collected in an absorber containing 10 mL of 0.011 M Na_2S_2O_3.The excess hypo was back titrated with 5 mL of 0.001 M I_2 solution. What is the weight (in mg) of iodine produced due to the reaction with I_2O_5(s) when it reacted with CO(g) in 20L sample of gas ?
Answer» 6.35 MG 127 mg 12.7 mg 0.635 mg Solution :WEIGHT of `I_2=0.1/2xx(2xx127)=12.7 mg`

Discussion

6.	The co-ordination compound of which one of the following compositions will produce two equivalents of AgCl on reaction with aqueous silver nitrate solution
Answer» `CoCl_(3).3NH_(3)` `CoCl_(3).6NH_(3)` `CoCl_(3).4NH_(3)` `CoCl_(3).5NH_(3)` Solution :The chloride ions which are present in the ionisation sphere GET precipitated as `AgCl` on reaction with `AgNO_(3).` THUS depending on the number of chloride present in the ionisation sphere will determine the no. of equivalents of AgCl produced. `[Co(NH_(3))_(5)Cl]Cl_(2)+2AgNO_(3)RARR[Co(NH_(3))_(5)Cl](CO_(3))_(2)+2AgCl.`

Discussion

7.	The CN of Ni^(+2)=4 NiCl_(2)+KCN_("excess")toA (cyano complex) NiCl_(2)+conc.KCl_("(excess)")toB (Chloro complex) The IUPAC name of A and B are
Answer» POTASSIUM tetracyano nickelate (II), Potassium TETRACHLORO nickelate(II) Tetracyano potassium nickelate(II), Tetrachloro potassium nickelate (II) Tetracyano nickel (II), Tetrachloro nickel (II) Potassium tetracyano nickel (II), Potassium tetracholoro nickel (II) Solution :`NiCl_(2)+underset(("excess"))(4KCN)toK_(2)[underset((A))(Ni)(CN)_(4)]+2KCl` `K_(2)[Ni(CN)_(4)]to` Potassium tetrachloronickelate II `NiCl_(2)+underset(("excess"))(2KCl)toK_(2)[underset((B))Ni(Cl)_(4)]` `K_(2)[Ni(Cl)_(4)]to` Potassium tetrachloronickelate (II)

Discussion

8.	The CN of Ni^(+2)=4 NiCl_(2)+KCN_("excess")toA (cyano complex) NiCl_(2)+conc.KCl_("(excess)")toB (Chloro complex) The hybridisation of A and B are
Answer» `dsp^(2),sp^(3L)` `sp^(3),sp^(3)` `dsp^(2), dsp^(2)` `sp^(3)d^(2)` both Solution :(A) Hybridisation of NI in `K_(2)[Ni(CN)_(4)]` is `dsp^(2)`. (B) Hybridisation of Ni in `K_(2)[Ni(CN)_(4)]` is `sp^(3)`.

Discussion

9.	The C.N. of a M^(2+) in MX_(2) is 8. Hence, C.N. of X^(– ) is -
Answer» 8 6 4 2 Answer :C

Discussion

10.	The CN of Ni^(+2)=4 NiCl_(2)+KCN_("excess")toA (cyano complex) NiCl_(2)+conc.KCl_("(excess)")toB (Chloro complex) Predict the magnetic nature of A and B
Answer» Both the DIAMAGNETIC Both are paramagnetic A is diamagnetic and B is PARA magnetic with 1 unpaired `e^(-)` A is diamagnetic and B in para magnetic with 2 unpaired `e^(-)` Solution :`K_(2)[underset((A))Ni(CN)_(4)]to` Diamagnetic is SFL, all electron gent paired. `K_(2)[underset((B))Ni(Cl)_(4)]to` Paramagnetic, 2 unpaired electrons is WFL.

Discussion

11.	The CN of copper in complex formed by adding excess of NH_(3) to CuSO_(4) is
Answer» SOLUTION :The COMPOUND formed is `[CU(NH_(3))_(4)]SO_(4)`

Discussion

12.	The clouds consist of charged particles of water dispersed in air. Some of them are + vely charged, others are negatively charged. When + vely charged clouds come closer they have cause lightening and thundering whereas when + ve and –ve charged colloids closer they cuase heavy rain by aggregation of minute particles. It is possible to cause artificial rain by throwing electrified sand or silver iodide from an aeroplane and thus coagulation the mist hanging in air. Electrical chimneys are made on the principle of
Answer» Electroosmosis Electrophoresis Coagulation All of these Answer :B

Discussion

13.	The clouds consist of charged particles of water dispersed in air. Some of them are + vely charged, others are negatively charged. When + vely charged clouds come closer they have cause lightening and thundering whereas when + ve and –ve charged colloids closer they cuase heavy rain by aggregation of minute particles. It is possible to cause artificial rain by throwing electrified sand or silver iodide from an aeroplane and thus coagulation the mist hanging in air. Clouds are colloids solution of
Answer» LIQUID in gas gas in liquid liquid in liquid solid in liquid Answer :A

Discussion

14.	The ClO radicals decay by second-order reaction. If the initial concentration is2.5 xx 10^(-5) "mole" dm^(-3), calculate its (a) first half-life, (b) second half-life, and (c) concentration of ClO radical after 4 min. The rate constant of the process is 2.25 xx 10^7 dm^3 "mol"^(-1) s^(-1)
Answer» Solution :(a) 1.778 MILLISECONDS (b) 3.556 milliseconds(c)`7.69 XX 10^(-6) "mol" DM^(-3)`

Discussion

15.	The calomel present in an electrode is :
Answer» `Hg_2Cl_2` + HG `Hg_2Cl` Hg + HGCL `HgCl_2` ANSWER :B

Discussion

16.	The cleavage of carbon-oxygen bond in ether takes place in ….
Answer» ALKALINE MEDIUM ACIDIC medium Neutral medium ALCOHOL medium Solution :Acidic medium

Discussion

17.	The cleavage of an ethyl methyl ether with cold hydrogen iodide will give
Answer» a MOLECULE each of an methyl IODIDE and water a molecule each of an ethyl iodide and water a molecule each of ETHANOL and an methyl iodide a molecule each of an ethyl iodide, methyl iodide and water Answer :C

Discussion

18.	The cleaning action of soap and detergent in water is due to the formation of
Answer» micelle salt base acid SOLUTION :A spericla aggregate of soap MOLECULES in the soap solution in water is CALLED micelle when soap is dissolved in water it forms a colloidal suspencsion in which the soap moleculs CLUSTER together ot form sperical micelles

Discussion

19.	The cleaning action of soap or detergent is due to
Answer» LYOPHILIC SOLUTION FORMATION EMULSIFICATION precipitation of dirt lyophobic solution formation Answer :B

Discussion

20.	The cleavage of an aryl-alkyl ether with hydrogen halide will give :
Answer» A molecule each of an alkyl HALIDE and water A molecule each of an ARYL halide and water A molecule each of an alkyl halide, aryl halide and water A molecule each of PHENOL and an alkyl halide Answer :D

Discussion

21.	The clean water should have BOD value
Answer» LESS than 5 PPM 10 ppm 20 ppm 100ppm. Answer :A

Discussion

22.	The Clausius-Clapeyron equation may be given as:
Answer» `log_(10)((P_(2))/(P_(1)))=(DeltaH_(vap))/(2.303R)[(1)/(T_(1))-(1)/(T_(2))]` `(dP)/(dT)=(Q)/(TDeltaV)` `(dP)/(dT)=(q)/(TDeltaV)` `(dP)/(dT)=(DELTAV)/(DeltaS)` Answer :A::B::C

Discussion

23.	The classification of drugs in the following manner is most conveniant for doctors
Answer» BASED on pharmacological effect Based on drug action Based on CHEMICAL STRUCTURE Based on MOLECULAR targets Answer :A

Discussion

24.	The class of drugs used for the treatment of stress is
Answer» analgesics antiseptic ANTIHISTAMINE tranquilizers Answer :D

Discussion

25.	The class of drugs used for the treatment of stree is
Answer» Analgesics Antiseptic Antihistamine Tranquillizers ANSWER :D

Discussion

26.	The class of drugs used for the treatment of cut or wound is :
Answer» Tranquillizers Antiseptics Antihistamines Antipyretics Answer :B

Discussion

27.	The class of chemical compounds used for the treatment of stress are called
Answer» Analgesics Tramquilizers Antihistamines Antibiotics Answer :B

Discussion

28.	The class of compounds that are reduced to primary alcohols and also respond to Fehling.s solution are known as:
Answer» ALIPHATIC aldehydes Aliphatic ketones Aromatic amines Aromatic ketones Answer :A

Discussion

29.	The claisen schmidt condensation product of benzaldehyde and acetone is
Answer» `C_(6)H_(5)CH=C(CH_(3))_(2)` `C_(6)H_(5)CH_(2)-overset(O)overset("\|\|")C-CH=CH_(2)` `C_(6)H_(5)-overset(O)overset("\|\|")C-CH=CH-CH_(3)` `C_(6)H_(5)-CH=CH-overset(O)overset("\|\|")C-CH_(3)` Solution :`{:(C_(6)H_(5)CHO+CH_(3)COCH_(3) to C_(6)H_(5)-overset(OH)overset(\|)(CH)-CH_(2)-overset(O)overset("\|\|")C-CH_(3)),(""overset(" ")underset(" ")DELTA DARR -H_(2)O),(""C_(6)H_(5)-CH=CH-overset(O)overset("\|\|")C-CH_(3)):}`

Discussion

30.	The cis form of the complex [RhCl_(3)(py)_(3)] is called …… isomer whereas its trants form is called …… isomer.
Answer» SOLUTION :FACIAL (FAC) and meridianal (MER)

Discussion

31.	The circulation of blood in human body supplies O_(2) and releases CO_(2) the concentration of O_(2) and CO_(2) is variable but on an average, 100 ml blood contains 0.02 g of O_(2) and 0.08 g of CO_(2). The volume of O_(2) and CO_(2) at 1 atm and at body temperature 37^(@)C, assuming 10 It blood in human body, is
Answer» 2 It, 4It 1.5 It, 4.5 It 1.59 It, 4.62 It 3.82 It, 4.62 It Solution :`because 100 ML` BLOOD has `0.02 g P_(2) and 0.08 g CO_(2)` `therefore 10,000 ml` blood has `2g O_(2) and 8 g CO_(2)` Using PV=nRT For `O_(2), 1xxV_(o_(2))=(2)/(32) xx 0.0821xx310 rArr V_(o_(2))=1.59` litre For `CO_(2), 1xx V_(co_(2))=(8)/(44) xx0.0821xx310 rArr V_(Co_(2))=4.62` litre Hence, (C) is the CORRECT answer.

Discussion

32.	The circulation of blood in human body supplies O_2 and releases CO_2 The concentration of O_2 and CO_2 is variable but on the average, 100 mL blood contains 0.02 g of O_2 and 0.08 g CO_2 . The volume of O_2 and CO_2 at 1 atm and body temperature 37^@C , assuming 10 litre blood in human body is :
Answer» 2 LITRE, 4 litre 1.5 litre, 4.5 litre 1.59 litre, 4.62 litre 3.82 litre, 4.62 litre Answer :C

Discussion

33.	The Cl-C-Cl angle in 1,1,2,2-tetra-chloroethene and tetrachloromethane respectively will be about:
Answer» `120^@` and `190^@5` `90^@` and `109.5^@ `109.5^@` and `90^@` `109.5^@` and `120^@` ANSWER :D

Discussion

34.	The chromyl chloride testresponds pourly with the chlorides of Pb,Ag so Sn but failwith the chlorides of
Answer» `HG` `As` `BI` `CU` ANSWER :a

Discussion

35.	Chromatographic technique was first introduced by ___________.
Answer» Leibig Tswett Kekule Wohler Answer :B

Discussion

36.	The choice of reducing agent for metal oxide is decided by thermodynamic principles of metallurgy. Delta G must be negative. Ellingham diagram is a plot of Delta_(f)G^(0) vs T for the formation of oxides of various elements. All the plots slope upwards when temperature increases. Each plot is a straight line except when some change in phase takes place. A metal will reduce the oxide of other metals which lie above it in this diagram. Delta G^(0) vs T plot in the Ellingham's diagram slopes downward for the reaction.
Answer» `MG + (1)/(2)O_(2) rarr MgO` `2Ag + (1)/(2) O_(2) rarr Ag_(2)O` `C + (1)/(2) O_(2) rarr CO` `CO + (1)/(2) O_(2) rarr CO_(2)` Answer :C

Discussion

37.	The choice of reducing agent for metal oxide is decided by thermodynamic principles of metallurgy. Delta G must be negative. Ellingham diagram is a plot of Delta_(f)G^(0) vs T for the formation of oxides of various elements. All the plots slope upwards when temperature increases. Each plot is a straight line except when some change in phase takes place. A metal will reduce the oxide of other metals which lie above it in this diagram. At a temperature below 1623 K. Al cannot reduce
Answer» FeO CrO MGO NiO Solution :AI can.t REDUCE MgO as MG is more ELECTROPOSITIVE.

Discussion

38.	The choice of a reducing agent in a particular case depends the thermodynamic factors. How far do you agree with this statement? Support your opinion with examples.
Answer» Solution :If the free energy of oxidation of the reducing agent is more negative than that of the metal to be reduced, reduction will take place. Most reactive METALS like ALKALI metals are reduced only by electrolytic reduction. Less reactive metals like Cu, SN, Pb, etc., can be reduced by carbon or carbon monoxide. Moderrately active metals like Mn, CR, FE, etc., are reduced with the help of Al.

Discussion

39.	The choice of areducing agent in the extraction of a particular case depends on thermodynamic factor. Explain.
Answer» SOLUTION :THERMODYNAMIC CONSIDERATIONS are very importantin decidingthe temperature and reductant in the extraction of a metal. `Fe_(2) O_(3)` is REDUCED to metal using CO asreagentat 823 K. `SnO_(2)` is reducedusingcoke at 1473 K.

Discussion

40.	The choice of a reducing agent in a particular case depends on thermodynamic factor. How far do you agree with this statement? Support your opinion with example.
Answer» Solution :The STATEMENT is correct. We CONSIDER the relation : `DeltaG=DeltaH-TDeltaS` `DeltaG^(@)=-RTlnK` For a PARTICULAR reducing AGENT to work with a metallic oxide. Example : The VALUE of `DeltaG` should be negative.

Discussion

41.	Thechoiceof areducingagentin aparticularcasedepends on thermodynamicfactor. How far do youagreewiththisstatement ?Supportyouropinionwithtwoexamples.
Answer» SOLUTION :Thermodynamic factorhelpsus in choosingasuitablereducingagentfor thereductionofaparticularmetaloxidetothemetallic stateasdiscussedbelow. It isevidentthatmetalsfor whichthestandardfreeenergyofformation`(DELTA _f G ^(@)) `oftheiroxidesislessnegative.Inotherwords,anymetalwillreducetheoxidesof other metalswhichlieaboveit in theEllinghamdiagrambecausethe STANDARD freeenergychange` (Delta _r G ^(@)) `ofthe combinedredox reactionwill be- vebyan amountequaltothedifferencein`Delta _f G^(@) `oftwometaloxides. Thus,both AlandZn can reduce FeOtoFebutFe cannotreduce`Al_ 2O_ 3 `toAl orZnOtoZn.Similarly, C canreduceZnOtoZn but notCO.

Discussion

42.	The choice of a reducing agent in a particular case depends on thermodynamic factor. How far do you agree with this statement ? Support your opinion with two examples.
Answer» Solution :The thermodynamic factor helps us in choosing a suitable reducing agent for the reduction of a particular metal oxide to metal. The feasibility of thermal reduction can be predicted on the basis of `Delta_fG^@` vs T plots for the formation of oxides, known as Ellingham diagram. From the diagram, it can be predicted that metals for which the standard FREE energy of formation of their oxides is more negative can reduce those metal oxides for which the standard free energy of formation of their respective oxides is less negative. In other words, a metal will reduce the oxides of other metals which lie above it in Ellingham diagram because the standard free energy change, `(Delta_rG^@)` of the combined redox reaction will be - ve by an amount EQUAL to the difference in `Delta_fG^@` of TWO two metal oxides. For EXAMPLE, both Al and Zn can reduce FeO to Fe but Fe cannot reduce `Al_2O_3` to Al or ZnO to Zn. Similarly, C can reduce ZnO to Zn but not CO. Thus, the choice of a particular reducing agent depends on thermodynamic factor.

Discussion

43.	The choice of a reducing agent in a particular case depends on thermodynamic factor. How far do you agree with this statement? Support your opinion with two examples.
Answer» Solution :Thermodynamic factor helps us in choosing a suitable reducing agent for the reduction of a particular metal oxide to the metallic state as discussed below. Refer to the Ellingham diagram Fig. 6.1. From the diagram, it is evident that metals for which tha standard free ENERGY of formation`(triangle_(f)G^(0))` of their oxides is more negative can reduce those metal oxides for which the standard free energy of formation of `(triangle_(f)G^(0))` of their oxides is less negative. In other words, any metal will reduce the oxides of other metals which lie above it in the Ellingham diagram. This is so because the standard free energy change `(triangle_(f)G^(0))` of the combined redox REACTION will be -ve. Thus, both AL and Zn can reduce FeO to Fe but Fe cannot reduce `Al_(2)O_(3)` to Al or ZnO to Zn. SIMILARLY, C can reduce ZnO to Zn but CO cannot.

Discussion

44.	The choice of a reducing agent in a particular case depends on thermo -dynamic factor. Explain with two examples.
Answer» Solution :The choice of the a reducing agent in a particular CASE depends on thermodynamic factor. This fact can be explained by considering the following EXAPLES. `rarr` out of Cand CO , carbon nonoxide (CO) is a better reducting agent at 673K. `rarr` At 983K (or) CO , carbon nonoxide (CO) is a better reducting agent at 673K . `rarr`The above observations are form Ellingham diagram. Zinc is not extracted from Zinc oxide through reduction using CO. Explanation : `2Zn+O_(2)rarr2ZnO,DeltaG^(@)=-650KJ` `2CO+O_(2)rarr2CO_(2),DeltaG^(@)=-450KJ` `2ZnO+2COrarr2Zn+2CO_(2')DeltaG^(@)=200KJ` `DeltaG^(@)` = Positive indicatis that the reaction is not feasible. The extraction of copper from pyrites is more difficult than that from its oxide ore through reduction. Explanation : Pyrites is `(Cu_(2)S)` cannot be reduced by carbir (or) hydrogent because the standar free energy of formation`(DeltaG^(@))` of`Cu_(2)S` is greater then those of`CS_(2)` and `H_(2)S` `{:(2Cu_(2)S+Crarr4Cu+CS_(2)),(Cu_(2)S+H_(2)rarrCu+H_(2)S):}}" Not fessible"` `rarr` The `DeltaG^(@)` of copper oxide is less than that of`CO_(2)` `therefore` The sulphide ore is firts converted to oxide by roasting and then reduced . `2Cu_(2)S+3O_(2)underset("Heat")overset("Roasting")rarr2Cu_(2)+2SO_(2)` `2Cu_(2)+O+Crarr4Cu+CO_(2)["feasible"].` The following are the conditions that Al might be expected to reduce MgO. The Equations for the formation of two oxides are `(4)/(3)Al_((s))+O_(2(g))rarr(2)/(3)Al_(2)O_(3(s))` `2Mg_((s))+O_(2(g))rarr2MgO_((s))` From the Ellingham diagram the two curves of these oxides formation intersect each otherrat a certain point. The corresponding value of`DeltaG^(@)` becomes zero for the reduction of MgO by aluminium metal, `2MgO_((s))+(4)/(3)hArr2Mg_((s))+(2)/(3)Al_(2)O_(3_(s))` From the above information the reduction of MgO by Al metal cannot occur below this temperature(1665 K) instead ,Mg can reduce`Al_(2)O_(3)` to al below 1665K. `rarr` Al-Metal can reduce MgO to Mg above 1665 K because`DeltaG^(@)` for `Al_(2)O_(3)`is less compared to that of MgO. `3MgO_((s))+2Al_((s))overset(T gt 1665K)rarrAl_(2)O_(3(s))+2Mg_((s))`

Discussion

45.	The chlorination of toluene in presence of ferric chloride gives predominantly
Answer» BENZYL chloride m-chlorotoluene. benzal chloride o-and p-chlorotoluene Answer :D

Discussion

46.	The chlorine atom in ClF_(5) involves
Answer» `SP^(3)` HYBRIDISATION `sp^(3)d` hybridisation `sp^(3)d^(2)` hybridisation `sp^(3)d^(3)` hybridisation. Answer :C

Discussion

47.	The chlorination of methane is an example of :
Answer» ELIMINATION reaction substitution reaction ADDITION reaction oxidation reaction Solution :CHLORINATION of METHANE is an example of substitution reaction.

Discussion

48.	The chlorination of alkane involves
Answer» CL FREE RADICALS `Cl^+` SPECIES `Cl^-` species `CH_3` free radicals Answer :B

Discussion

49.	The chloride that turns black on addition of NH_4OH is :
Answer» AGCL `PbCl_2` `Hg_2Cl_2` Both (B) and ( C ) ANSWER :C

Discussion

50.	The chloride that can be extracted with ether:
Answer» NaCl LiCl `BaCl_2` `CaCl_2` ANSWER :B

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

The co-ordination number and oxidation number of X in the compound[X(SO_(4))(NH_(3))_(5)] will be :

The co-ordination number and oxidation number of X in the following compound [X(SO_4)(NH_3)_5]C1 will be:

The co-ordination compounds with bidentate ligands are called as

The co-ordination compound of which one of the following compositions will produce two equivalents of AgCl on reaction with aqueous silver nitrate solution

The CN of Ni^(+2)=4 NiCl_(2)+KCN_("excess")toA (cyano complex) NiCl_(2)+conc.KCl_("(excess)")toB (Chloro complex) The IUPAC name of A and B are

The CN of Ni^(+2)=4 NiCl_(2)+KCN_("excess")toA (cyano complex) NiCl_(2)+conc.KCl_("(excess)")toB (Chloro complex) The hybridisation of A and B are

The C.N. of a M^(2+) in MX_(2) is 8. Hence, C.N. of X^(– ) is -

The CN of Ni^(+2)=4 NiCl_(2)+KCN_("excess")toA (cyano complex) NiCl_(2)+conc.KCl_("(excess)")toB (Chloro complex) Predict the magnetic nature of A and B

The CN of copper in complex formed by adding excess of NH_(3) to CuSO_(4) is

The ClO radicals decay by second-order reaction. If the initial concentration is2.5 xx 10^(-5) "mole" dm^(-3), calculate its (a) first half-life, (b) second half-life, and (c) concentration of ClO radical after 4 min. The rate constant of the process is 2.25 xx 10^7 dm^3 "mol"^(-1) s^(-1)

The calomel present in an electrode is :

The cleavage of carbon-oxygen bond in ether takes place in ….

The cleavage of an ethyl methyl ether with cold hydrogen iodide will give

The cleaning action of soap and detergent in water is due to the formation of

The cleaning action of soap or detergent is due to

The cleavage of an aryl-alkyl ether with hydrogen halide will give :

The clean water should have BOD value

The Clausius-Clapeyron equation may be given as:

The classification of drugs in the following manner is most conveniant for doctors

The class of drugs used for the treatment of stress is

The class of drugs used for the treatment of stree is

The class of drugs used for the treatment of cut or wound is :

The class of chemical compounds used for the treatment of stress are called

The class of compounds that are reduced to primary alcohols and also respond to Fehling.s solution are known as:

The claisen schmidt condensation product of benzaldehyde and acetone is

The cis form of the complex [RhCl_(3)(py)_(3)] is called …… isomer whereas its trants form is called …… isomer.

The circulation of blood in human body supplies O_2 and releases CO_2 The concentration of O_2 and CO_2 is variable but on the average, 100 mL blood contains 0.02 g of O_2 and 0.08 g CO_2 . The volume of O_2 and CO_2 at 1 atm and body temperature 37^@C , assuming 10 litre blood in human body is :

The Cl-C-Cl angle in 1,1,2,2-tetra-chloroethene and tetrachloromethane respectively will be about:

The chromyl chloride testresponds pourly with the chlorides of Pb,Ag so Sn but failwith the chlorides of

Chromatographic technique was first introduced by ___________.

The choice of a reducing agent in a particular case depends the thermodynamic factors. How far do you agree with this statement? Support your opinion with examples.

The choice of areducing agent in the extraction of a particular case depends on thermodynamic factor. Explain.

The choice of a reducing agent in a particular case depends on thermodynamic factor. How far do you agree with this statement? Support your opinion with example.

Thechoiceof areducingagentin aparticularcasedepends on thermodynamicfactor. How far do youagreewiththisstatement ?Supportyouropinionwithtwoexamples.

The choice of a reducing agent in a particular case depends on thermodynamic factor. How far do you agree with this statement ? Support your opinion with two examples.

The choice of a reducing agent in a particular case depends on thermodynamic factor. How far do you agree with this statement? Support your opinion with two examples.

The choice of a reducing agent in a particular case depends on thermo -dynamic factor. Explain with two examples.

The chlorination of toluene in presence of ferric chloride gives predominantly

The chlorine atom in ClF_(5) involves

The chlorination of methane is an example of :

The chlorination of alkane involves

The chloride that turns black on addition of NH_4OH is :

The chloride that can be extracted with ether: