94420 + Interview Questions in Chemistry in Class 12

1.	Conductivity of 0.00241 M acetic acid is 7.896xx10^(-5)" S "cm^(-1). Calculate its molar conductivity and if Lamda_(m)^(@) for acetic acid is 390.5 S cm^(-2)mol^(-1), what is its dissociation constant ?
Answer» Solution :* The CALCULATION of molar conductivity `Lamda_(m)^(c)`: Where, `M=0.00241" "CH_(3)COOH`, `Lamda_(m)^(c)=(kxx1000)/("molarity")""k=7.896xx10^(-5)" S "CM^(-1)` `=(7.896xx10^(-5)(" S "cm^(-1))xx1000(cm^(3)L^(-1)))/(0.00241(mol" "L^(-1)))` * The calculation of dissociation degree `(alpha)` of `CH_(3)COOH`: `therefore alpha=(Lamda_(m)^(c))/(Lamda_(m)^(@)),""alpha=(32.763)/(390.5)=0.0839\|" Where,"alpha="dissociation degree,"Lamda_(m)^(c)=32.763" S "cm^(2)mol^(-1),""Lamda_(m)^(@)=390.5" S "cm^(2)mol^(-1)` * The calculation of dissociation constant `(K_(a))`: The EQUILIBRIUM in acetic acid `(CH_(3)COOH)` solution is as follow: `CH_(3)COOH_((aq))HARR CH_(3)COO_((aq))^(-)+H_((aq))^(+)` `therefore K_(a)=(CalphaxxCalpha)/(C-Calpha)=(Calpha^(2))/((1-alpha))`,`=(0.00241xx(0.0839)^(2))/(1-0.0839),""=(0.00241xx0.0839xx0.0839)/(0.9161)\|" Where, dissociation degree "alpha=0.0839" Molarity "C=0.00241" mol "L^(-1)` `=1.8518xx10^(-5)`

Discussion

2.	Conductivity kappa, equal to _____
Answer» `(1)/(R)(L)/(A)` `(G^(*))/(R)` `Lamda_(m)` `(l)/(A)` Solution : Resistance of any SUBSTANCE is directly proportional to its length and inversely proportional to its cross-sectional area. `R ALPHA(l)/(A)` So, `R=rho(l)/(A)=(l)/(kappaA)` * Where, `rho` is resistance or specific resistance. `therefore rho=(RA)/(l)""therefore (1)/(R)=(1)/(rho)XX(A)/(l)` `(1)/(rho)=kappa""therefore (1)/(R)=kappa((A)/(l))` `therefore kappa=(1)/(R)=kappa((A)/(l))`. . . (A) `(l)/(A)=G^()""therefore kappa=(G^())/(R)` . . . (B) Where `G^(**)=(l)/(A)`=cell constant, `kappa`=conductivity of solution.

Discussion

3.	Conductivity of 0.00241 M acetic acid is 7.896 xx 10^(-5) S cm^(-1) . Calculate its molar conductivity. If Lambda_(m)^(@) for acetic acid is 390.5 S cm^(2) mol^(-1).what is its dissociation constant ?
Answer» SOLUTION :Using the following relation and substituting the values, we get 5) S cm^(-1)) xx 1000 cm^(3) L^(-1))/(0.00241 mol L^(-1)) = 32.76 S cm^(2) mol^(-1)` `alpha = Lambda_(m)^( C)/Lambda_(m)^(@) = (32.76)/(390.5) = 8.4 xx 10^(-2)` DEGREE of dissociation may be obtained as under: `K_(a) -(CALPHA^(2))/(1- alpha) = (0.00241 xx (8.4 xx 10^(-2))^(2))/(1-0.084) = 1.86 xx 10^(-5)`

Discussion

4.	Conductivity of 0.00241 M acetic acid is 7.896 xx 10^(-5) S cm^(-1). Calculate its molar conductivity. If Lambda_(m)^(@)for acetic acid is 390.5 S cm^2 mol^(-1) , what is its dissociation constant?
Answer» Solution :`k = 7.896 xx 10^(-5) S cm^(-1)` Molar conductivity can be OBTAINED as under : `Lambda_(m) = (1000 k)/M = (1000 xx 7.896 xx 10^(-5))/0.00241 = (7.896 xx 10^(3))/241 =7896/241 = = 32.76 S cm^(2) mol^(-1)` `ALPHA = (Lambda_(m))/(Lambda_(m)^(@)) =32.76/390.5= 0.0832 = 8.38 %` Use the FOLLOWING relation to calculate DISSOCIATION constant, `K_(a)` `K_(a) = (Calpha^(2))/(1- alpha) =(0.00241 xx (0.084)^(2))/(1-0.084) = 1.86 xx 10^(-5)`

Discussion

5.	Conductivity (kappa), conductance (G) and all constant (G^(@)) are related as___.
Answer» SOLUTION :`kappa=GxxG^(**)`

Discussion

6.	Conductivity is a conductance of ____of the solution and its units are_____
Answer» SOLUTION :`1CM^(3)` or `1M^(3),Omega^(-1)CM^(-1)` or S `m^(-1)`

Discussion

7.	Conductivity increases with the ...................In viscosity .
Answer» SOLUTION :DECREASE

Discussion

8.	Conductivity depends on what ?
Answer» Solution :* Conductivity depends on THREE MAIN FACTOR: (i) Nature of substance (II) Temperature and (iii) PRESSURE

Discussion

9.	Conductivity decreases but molar conductivity increases with ________ in concentration.
Answer» SOLUTION :INCREASE

Discussion

10.	conductivity cell hastwoelectrodes 20mm apart and ofcrosssectionarea 1.8 cm^(2). Find thecellconstant.
Answer» SOLUTION :Distancebetweentwo ELECTRODES = L= 20 mm= 2 cm Crosssectionarea = a= 1.8 `cm^(2)` Cellconstant = b = ? `b= (1)/(a) =(2)/(1.8)= 1.111 cm^(-1)`

Discussion

11.	Conductivity cell consists of two nickel electrodes coated with nickel black.
Answer» ANSWER :FALSE

Discussion

12.	Conductivity cell.
Answer» SOLUTION :

Discussion

13.	Conductanceof 0.1 MKCl (condiuctivity = X ohm^(-1)cm^(-1)) filled in a condictivity cell is Y ohm^(-1). If the conductance of 0.1 Mno OH filled in the csamecell is Z ohm^(-1), them olar conductance of NaOH will be
Answer» `10^(3) (XZ)/(Y)` `10^(4)(XZ)/(Y)` `10(XZ)/(Y)` `0.1 (XZ)/(Y)` Solution :Conductivity (X) = conductance (C ) `XX` CELL constnat `therefore` Cell CONSTANT `=X/Y` Conductivity of `NAOH = X/Y.Z` `Delta M (NaOH) = X/Y . Z xx (1000)/(0.1) =(XZ)/(Y) 10^(4)`

Discussion

14.	Conductance(G)conductivity(K)and molar conductivityLambda_mare terms used in electrolytic conduction.Write any two factors on which conductivity depends on .
Answer» SOLUTION :TEMPERATURE,CONCENTRATION of ELECTROLYTE

Discussion

15.	Conductance of electrolytic solution can be measured with the help of Wheatstone bridge.
Answer» ANSWER :1

Discussion

16.	Conductance(G)conductivity(K)and molar conductivityLambda_mare terms used in electrolytic conduction.How do conductivity and molar conductivity vary with concentration of electrolytic solution?
Answer» SOLUTION :Conductivity(K)of an electrolyte DECREASE in concentration of ELECTROLYTIC solution. Molar conductivity `(^^)`of an electrolyte INCREASES with decreases in concentration of electrolytic solution.

Discussion

17.	Conductance of an electrolytic solution depends :
Answer» NATURE of electrolyte Power of AC source Distance between TWO electrodes None of the above. Answer :A

Discussion

18.	Conductance of a solution of a strong electrolyte is measure of its
Answer» POTENTIAL difference concentration REACTANCE dissociation SOLUTION :Conductance INCREASES with the dissocation.

Discussion

19.	Conductance is directly proportional to area of the vessel and the concentration of solution in itand is inversly proportional to the length of the vessel, then the unit of the constant of proportionality is
Answer» `S. m " MOL"^(-1)` `S. m^(2) " mol"^(-1)` `S^(-2) m^(2) "mol"` `S^(2) m^(2) " mol"^(-2)` ANSWER :B

Discussion

20.	Conducitivity of 2.5 xx 10^(-4) M methonic acid is 5.25 xx 10^(-5) S cm^(-1). Calculate the molar conductivity and degree of dissociation. Given: lambda^(@) (H^(+)) = 349.5 S cm^(2) mol^(-1) and lambda^(@) (HCOO^(-)) = 50.5 S cm^(2) mol^(-1)
Answer» SOLUTION :`Lambda_(HCOOH)^(@)= Lambda_(H^(+))^(@) + Lambda_(HCOO^(-))^(@) =349.5 + 50.5 = 400 S cm^(2) mol^(-1)` `Lambda_(HCOOH) = K xx 1000 cm^(3) L^(-1) "MOLARITY"^(-1)` `=(5.25 xx 10^(-5) S cm^(-1) xx 1000 cm^(3) L^(-1))/(2.5 xx 10^(-4)"mol" L^(-1))= 210 S cm^(2) mol^(-1)` Degree of DISSOCIATION `(ALPHA) = Lambda_(m)/Lambda_(,)^(@) = 210/400 = 0.525` or `52.5%`

Discussion

21.	Condsider the following combination of atomic orbitals : combinding orbitals (internuclear axis ) Combindig orbitals (internuclear axis). (i) s+p_(x)(x)(ii)d_(xy)+d_(xy)(x)(iii)d_(xy)+P_(x)(z)(iv)s+s(z)(v)d_(yz)+d_(yz)(x)(vi)p_(x)+p_(x)(z)(vii)d_(z^(2))+d_(z^(2))(z)(viii)d_(xy)+d_(xy)(z)(ix)p_(x)+p_(x)(y)(x)s+p_(z)(x) Then calculate value of a^(2)+b^(2)+2cd. (where a=sigma M.O.,b=pi M.O. , C=delta M.O., d=non-bonding M.O.)
Answer» ANSWER :`28`

Discussion

22.	Condsider the following combination of atomic orbitals : combinding orbitals (internuclear axis ) Combindig orbitals (internuclear axis). (i) s+p_(x)(x)(ii)d_(xy)+d_(xy)(x)(iii)d_(xy)+P_(x)(z)(iv)s+s(z)(v)d_(yz)+d_(yz)(x)(vi)p_(x)+p_(x)(z)(vii)d_(z^(2))+d_(z^(2))(z)(viii)d_(xy)+d_(xy)(z)(ix)p_(x)+p_(x)(y)(x)s+p_(z)(x) Then calculate value of "a^(2)+b^(2)+2cs". (where a=sigma M.O.,b=pi M.O. , C=delta M.O., d=non-bonding M.O.)
Answer» ANSWER :`28`

Discussion

23.	Conducitivity kappa, is equal to____
Answer» `(1)/(R)(l)/(A)` `(G^())/(R)` `wedge_(m)` `(l)/(A)` Solution :`R=rho.(1)/(A)` or `R=(1)/(KAPPA)(1)/(A)` or `kappa`(conductivity)`=(1)/(R)(1)/(A)`, i.e., (a) As `(1)/(A)=G^(@)` (cell constant), `kappa(1)/(R)G^(@)=(G^())/(R)` , i.e., (B)`.

Discussion

24.	Conditions of standard state used in thermochemistry is
Answer» `0^(@)C` and 1atm `20^(@)C` and 1 atm `25^(@)C` and 1 atm 0 K and 1 atm ANSWER :C

Discussion

25.	Condition for maximum yield of C_2H_5Cl is
Answer» `C_2H_6"(excess)"+Cl_2overset"UV light"to ` `C_2H_6+Cl_2 underset"ROOM TEMP."overset"dark"to` `C_2H_6+Cl_2 "(excess)"overset"UV light"to` `C_2H_6+Cl_2 overset"UV light"to` Solution :`C_2H_6"(excess)"+Cl_2overset"U.V. light"to underset"(MAJOR PRODUCT )"underset"Ethyl chloride"(C_2H_5Cl)+HCl`

Discussion

26.	Conditions for the formation of NH_(3) in Haber's process are
Answer» HIGH temperature, low PRESSURE Optimum temperature, high pressure High temperature, high pressure Optimum temperature, low pressure ANSWER :2

Discussion

27.	Condisder the followingreactionsat 1100^(@)C (i)2C+O_(2) rarr 2CO, triangle G^(@)=-460 kJ mol^(-1) (ii) 2Zn+O_(2) rarr 2 Zno, triangle G^(@)=-360 kj mol^(-1) Basedon the observation selectthe correct alternate
Answer» zinccan beoxidised by co zinc oxidecan be reduced by carbon both a and B noneof the above Solution :`ZnO+Crarr Zn +CO TRIANGLE G^(@) =- ve` Hencethisis sponstaneous REACTION

Discussion

28.	Condensation product of the benzaldehyde and acetophenone in alkaline medium at 293 K is
Answer» `C_(6)H_(5)-CH_(2)-CO-CH_(2)-C_(6)H_(5)` `C_(6)H_(5)-CO-CH_(2)-C_(6)H_(4)-CH_(3)` `C_(6)H_(5)-CH=CH-CO-C_(6)H_(5)` `C_(6)H_(5)-CH_(2)-C_(6)H_(5)-CO-CH_(3)` Answer :C

Discussion

29.	Condensation polymers are formed from monomers
Answer» which have bifunctional groups which havemultiple `(C=C)` or ` (C-=N) or (C=C)`BONDS in which elimination can take PLACE in which addition can take place Solution :Condensationpolymersare formed from MONOMERS which havebifunctional GROUP , e.g. Nylon-6,6 is a condensation polymer . `nNH_(2)(CH_(2))_(6)NH_(2)+n(CH_(2))_(4)(COOH)_(2)to(--underset("Nylon-66")(NH(CH_(2))_(6))NHCO(CH_(2))_(4)CO--)_(n)+nH_(2)O`

Discussion

30.	Condensation polymer.
Answer» Solution :The polymers formed by the repeated condensation REACTION betweem TWO bifunctional or trifunctional monomer UNITS whichresults in ELIMINATION of small molecules like water or alcohol are CALLED condensation polymers. Example : Nylon - 6,6.

Discussion

31.	Condensation of ethyl amine with acctaldehyde gives a compound A which upon reduction with hydrogen and nickel gives
Answer» diethyl amine ethyl methyl amine triethyl amine dimethyl amine. Solution :`CH_(3)CH_(2)NH_(2)+O = CHCH_(3)underset(-H_(2)O)(RARR)` `CH_(3)CH_(2)N=CHCH_(3)OVERSET(H_(2)//Ni)(rarr)CH_(3)CH_(2)underset("Diethylamine")(-NH-CH_(2))CH_(3)`

Discussion

32.	Condensation of secondarynitroalkanewith aldehyde procedsvia ____ ion
Answer» `C^(+)` `C^(-)` `N^(+)` `N^(-)` SOLUTION :CARBANION FORMATION

Discussion

33.	Condensation of acetone in conc. H_(2)SO_(4) and trimerisation of propyen by passing through hot copper tube give same compound X. The compound X is
Answer» Phorone Mesitylene Mesityl oxide Mixture of A and B. Solution :

Discussion

34.	Condensation of caprolactam gives:
Answer» Nylon-6,6 Nylon-6 Nitrile rubber Nylon -6,10 Answer :B

Discussion

35.	Condensation homopolymers among the following are : SBR , polyester , bakelite , nylon 6 , PVC , starch , nylon 6,6 glyptal , natural rubber.
Answer» SOLUTION :TWO . NYLON -6, STARCH.

Discussion

36.	Concrete is a mixture of:
Answer» Cement, lime and water Cement, sand and water Cement, sand, GRAVEL and water Cement, SLAKED lime and water Answer :C

Discussion

37.	Concider the following statements, (I) As the oxidation number of a metal increases, ionic character decreases.
Answer» I,II,III and IV only II II and III II and IV Solution :As the oxidation NUMBER of a metal increases, ionic character decreases. In CASE of `Mn,Mn_(2)O_(7)` is a covalent green oil which possess perdominantly acidic character and it gives `HMnO_(4).V_(2)O_(5)` is amphoteric, it gives `VO_(4)^(3-)` as WELL as `VO_(2)^(+)` salts. CrO is basic while `Cr_(2)O_(3)` is amphoteric.

Discussion

38.	Conc.HNO_3can be stored in container of :
Answer» Cu Sn Zn Al Answer :D

Discussion

39.	Conc.HCl reacts rapidly with :
Answer» `CH_3CH_2OH` `(CH_3)_2CHOH` `(CH_3)_3COH` All Answer :C

Discussion

40.	Concentration terms such as mass percentage, ppm, mole fraction and molality are independent of temperature, however molarity is a function of temperature. Explain.
Answer» SOLUTION : Molarity of a solution is defined as the number of moles of solute dissolved in one LITRE of solution. SINCE volume depends on temperature and UNDERGOES a change with change in temperature, the molarity will change with change in temperature. On the other hand, mass does not change with change in temperature. Mass percentage ppm, MOLE fraction and molality involve only mass and remain unchanged by changing temperature.

Discussion

41.	Concentration terms such as mass percentage, ppm mole fraction and molality are independent of temperature, however molarity is a function of temperature. Explain.
Answer» Solution :Molarity of a solution is a weight by VOLUME relationship to represent its strength and defined as .the number of moles of solute dissolved in one litre of solution.. Since volume depends on temperature and undergoes a change with change in temperature, the molarity will also change with change in temperature. On the other hand, the other concentration terms such as mass percentage, oom, mole fraction and molality are based upon mass by mass relationship of solute and SOLVENT present in a binary solution. Mass does not change with change in temeprature, as a result these concentration terms remain UNCHANGED with variation of temperature. According to the DEFINITION of all these terms, mass of the solvent used for making the solution is related to the mass of solute.

Discussion

42.	{:("Concentration process","Name of process"),("A) Hydrogen on finely","1) Physisorption"),("divided palladium","2) Chemisorption"),("B) Hydrogen on nickel","3) Occlusion"),("C) Hydrogen on charcoal","4) Desorption"):} The correct match is
Answer» `{:(A,B,C),(1,2,3):}` `{:(A,B,C),(3,1,4):}` `{:(A,B,C),(3,2,1):}` `{:(A,B,C),(4,3,2):}` ANSWER :C

Discussion

43.	Concentration process
Answer» ANSWER :C

Discussion

44.	…….concentration phenol solution acts as antiseptic.
Answer» `2 %` `1%` `0.2%` `0.1 %` SOLUTION :`0.2%`

Discussion

45.	Concentration of the reactant in first order is reduced to (1)/(e^(2)) after
Answer» ONE NATURAL LIFE TIME two natural life time THREE natural life time four natural life time Answer :B

Discussion

46.	Concentration of the ore is done :
Answer» To MAKE the ore porus To make the ore fit for electrolysis To DRIVE off the volatile IMPURITIES To free it from as MANY as impurities as possible . Answer :D

Discussion

47.	Concentration of the Ag^+ions in a saturated solution of Ag_2 C_2 O_4 is 2.24xx10^-4 mol L^-1 solubility product of Ag_2 C_2 O_4 is
Answer» `2.42xx10^-8 mol^3 L^-3` `2.66xx10^-12 mol^3 L^-3` `4.5xx10^-11 mol^3 L^-3` `5.619xx10^-12 mol^3 L^-3` ANSWER :D

Discussion

48.	Concentration of the Ag^(+) ions in a saturaed solution of Ag_(2)C_(2)O_(4) is 2.2 xx 10^(-4) mol L^(-1). Solubility product of Ag_(2)C_(2)O_(4) is :
Answer» `2.66 xx 10^(-12)` `4.5 xx 10^(-11)` `5.3 xx 10^(-12)` `2.42 xx 10^(-8)` Solution :`Ag_(2)C_(2)O_(4) HARR 2Ag^(+) + C_(2)O_(4)^(2-)` `2.2 xx 10^(-4)M 1.1 xx 10^(-4) M` `K_(SP) = [Ag^(+)]^(2)[C_(2)O_(4)^(2-)]` `= [2.2 xx 10^(-4)]^(2).[1.1 xx 10^(-4)]` `K_(sp) = 5.3 xx 10^(-12)`.

Discussion

49.	Concentration of the Ag^+ ion in a saturated solution of Ag_2C_2O_4 is 2.24 times 10^-4 mol L^-1 solubility product of Ag_2C_2O_4 is
Answer» `2.42 TIMES 10^-8 MOL^3 L^-3` `2.66 times 10^-12 mol^3 L^-3` `4.5 times 10^-11 mol^3 L^-3` `5.619 times 10^-12 mol^3 L^-3` Solution :`5.619 times 10^-12 mol^3 L^-3` `Ag_2C_2O_4leftrightarrow2Ag^++C_2O_4^(2-)` `[C_2O_4^(2-)]=(2.24 times 10^-4)/2MOL L^-1` `=1.12 times 10^-4 mol L^-1` `K_(SP)=[Ag^+]^2[C_2O_4^(2-)]` `=(2.24 times 10^-4 mol L^-1)^2 (1.12 times 10^-4 mol L^-1)` `=5.619 times 10^-12 mol^3 L^-3`

Discussion

50.	Concentration of SO_(4)^(2-) ions in a solution prepared by mixing 100 ml 0.1 M BaCl_(2) and 400 ml 0.2 M Na_(2)SO_(4).
Answer» `(8)/(50)M` `(7)/(50)M` `(7)/(40)M` `(1)/(25) M` Solution :`{:(BaCl_(2),+,Na_(2) SO_(4),rarr,BaSO_(4),+,2NaCl,),(0.01 "mole",,0.08 "mole",,,,,),(-,,0.07 "mole",,UNDERSET(PPt darr)(0.01 "mole"),,0.01 "mole",):}` `{:(Na_(2)SO_(4),rarr,2Na^(+),+,SO_(4)^(2-),),(0.07 "mole",,-,,-,),(-,,0.14 "mole",,0.07 "mole",):}` `[SO_(4)^(2-)] = (n_(SO_(4)^(2-)))/(V_(1) + V_(2)) = (0.07)/(0.5) = (7)/(50) M`

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

Conductivity of 0.00241 M acetic acid is 7.896xx10^(-5)" S "cm^(-1). Calculate its molar conductivity and if Lamda_(m)^(@) for acetic acid is 390.5 S cm^(-2)mol^(-1), what is its dissociation constant ?

Conductivity kappa, equal to _____

Conductivity of 0.00241 M acetic acid is 7.896 xx 10^(-5) S cm^(-1) . Calculate its molar conductivity. If Lambda_(m)^(@) for acetic acid is 390.5 S cm^(2) mol^(-1).what is its dissociation constant ?

Conductivity of 0.00241 M acetic acid is 7.896 xx 10^(-5) S cm^(-1). Calculate its molar conductivity. If Lambda_(m)^(@)for acetic acid is 390.5 S cm^2 mol^(-1) , what is its dissociation constant?

Conductivity (kappa), conductance (G) and all constant (G^(@)) are related as___.

Conductivity is a conductance of ____of the solution and its units are_____

Conductivity increases with the ...................In viscosity .

Conductivity depends on what ?

Conductivity decreases but molar conductivity increases with ________ in concentration.

conductivity cell hastwoelectrodes 20mm apart and ofcrosssectionarea 1.8 cm^(2). Find thecellconstant.

Conductivity cell consists of two nickel electrodes coated with nickel black.

Conductivity cell.

Conductanceof 0.1 MKCl (condiuctivity = X ohm^(-1)cm^(-1)) filled in a condictivity cell is Y ohm^(-1). If the conductance of 0.1 Mno OH filled in the csamecell is Z ohm^(-1), them olar conductance of NaOH will be

Conductance(G)conductivity(K)and molar conductivityLambda_mare terms used in electrolytic conduction.Write any two factors on which conductivity depends on .

Conductance of electrolytic solution can be measured with the help of Wheatstone bridge.

Conductance(G)conductivity(K)and molar conductivityLambda_mare terms used in electrolytic conduction.How do conductivity and molar conductivity vary with concentration of electrolytic solution?

Conductance of an electrolytic solution depends :

Conductance of a solution of a strong electrolyte is measure of its

Conductance is directly proportional to area of the vessel and the concentration of solution in itand is inversly proportional to the length of the vessel, then the unit of the constant of proportionality is

Conducitivity of 2.5 xx 10^(-4) M methonic acid is 5.25 xx 10^(-5) S cm^(-1). Calculate the molar conductivity and degree of dissociation. Given: lambda^(@) (H^(+)) = 349.5 S cm^(2) mol^(-1) and lambda^(@) (HCOO^(-)) = 50.5 S cm^(2) mol^(-1)

Conducitivity kappa, is equal to____

Conditions of standard state used in thermochemistry is

Condition for maximum yield of C_2H_5Cl is

Conditions for the formation of NH_(3) in Haber's process are

Condisder the followingreactionsat 1100^(@)C (i)2C+O_(2) rarr 2CO, triangle G^(@)=-460 kJ mol^(-1) (ii) 2Zn+O_(2) rarr 2 Zno, triangle G^(@)=-360 kj mol^(-1) Basedon the observation selectthe correct alternate

Condensation product of the benzaldehyde and acetophenone in alkaline medium at 293 K is

Condensation polymers are formed from monomers

Condensation polymer.

Condensation of ethyl amine with acctaldehyde gives a compound A which upon reduction with hydrogen and nickel gives

Condensation of secondarynitroalkanewith aldehyde procedsvia ____ ion

Condensation of acetone in conc. H_(2)SO_(4) and trimerisation of propyen by passing through hot copper tube give same compound X. The compound X is

Condensation of caprolactam gives:

Condensation homopolymers among the following are : SBR , polyester , bakelite , nylon 6 , PVC , starch , nylon 6,6 glyptal , natural rubber.

Concrete is a mixture of:

Concider the following statements, (I) As the oxidation number of a metal increases, ionic character decreases.

Conc.HNO_3can be stored in container of :

Conc.HCl reacts rapidly with :

Concentration terms such as mass percentage, ppm, mole fraction and molality are independent of temperature, however molarity is a function of temperature. Explain.

Concentration terms such as mass percentage, ppm mole fraction and molality are independent of temperature, however molarity is a function of temperature. Explain.

{:("Concentration process","Name of process"),("A) Hydrogen on finely","1) Physisorption"),("divided palladium","2) Chemisorption"),("B) Hydrogen on nickel","3) Occlusion"),("C) Hydrogen on charcoal","4) Desorption"):} The correct match is

Concentration process

…….concentration phenol solution acts as antiseptic.

Concentration of the reactant in first order is reduced to (1)/(e^(2)) after

Concentration of the ore is done :

Concentration of the Ag^+ions in a saturated solution of Ag_2 C_2 O_4 is 2.24xx10^-4 mol L^-1 solubility product of Ag_2 C_2 O_4 is

Concentration of the Ag^(+) ions in a saturaed solution of Ag_(2)C_(2)O_(4) is 2.2 xx 10^(-4) mol L^(-1). Solubility product of Ag_(2)C_(2)O_(4) is :

Concentration of the Ag^+ ion in a saturated solution of Ag_2C_2O_4 is 2.24 times 10^-4 mol L^-1 solubility product of Ag_2C_2O_4 is

Concentration of SO_(4)^(2-) ions in a solution prepared by mixing 100 ml 0.1 M BaCl_(2) and 400 ml 0.2 M Na_(2)SO_(4).

Conductivity is a conductance of of the solution and its units are_