94420 + Interview Questions in Chemistry in Class 12

1.	In a reaction , A+Brarr Product , rate is doubled, and rate increases by a factor of 8 when the concentrations of both the reactants (A and B ) are doubled , rate law for the reaction can be written as -
Answer» rate = k [A] [B] rate `=k [A]^(2)[B]` rate = `k [A][B]^(2)` rate `=k[A]^(2)[B]^(2)` Solution :`A+Brarr` Product `R=k[A]^(alpha)[B]^(BETA)""...[1]` `2r=k[A]^(alpha)[2B]^(beta)""...[2]` `8R=k[2A]^(alpha)[2B]^(beta)""...[3]` Now , by dividing EQUATION (1) by equation (2) we get `(r)/(2r)=(k[A]^(alpha)[B]^(beta))/(k[A]^(alpha)[2B]^(beta))` or, `(1)/(2)=(1)/(2^(beta))` or, `2^(1)=2^(beta) " or, "beta=1` Again by dividing equation (2) by equation (3) we get `(2r)/(8r)=(k[A]^(alpha).[2B]^(beta))/(k[2A]^(alpha).[2B]^(beta))` or, `(1)/(2^(2))=(1)/(2^(alpha)) " or, " 2^(2)=2^(alpha)` or,`alpha=2` So, rate equation will be : `r=k[A]^(2).[B]`

Discussion

2.	In a reaction, A+B toProduct, rate is doubled when the concentration of B is doubled, and rate increases by a factor of 8 when the concentrations of both the reactants (A and B) are doubled, rate law for the reaction can be written as
Answer» `"RATE"=K[A]^(2)[B]` `"rate"=k[A][B]` `"rate"=k[A][B]^(2)` `"rate"=k[A]^(2)[B]^(2)` ANSWER :A

Discussion

3.	In a reaction ,A + B toProduct , rate is doubled when the concentration of B is doubled ,and rate increases by a factor of 8 when the concentrations of both the reactants (A and B) are doubled , rate law for the reaction can be written as
Answer» RATE = `K[A] [B]^(2)` Rate = k `[A]^(2) [B]^(2)` Rate = k [A] [B] Rate= `k [A]^(2) [B]` Solution : That MEANS order of reaction with respect to B is 1 and w.r.t. A is 2 Hence , Rate = `k [A]^(2) [B]^(1)`

Discussion

4.	In a reaction ,A + B to C, the rate expression is R = K[A] [B]^(2) . If the concentration of both the reactants is doubled at constant volume then the rate of reaction will be
Answer» Eight TIMES DOUBLE Quadruple TRIPLE Solution :For the REACTION `[A + B to C]` Rate = R = `K[A][B]^(2)` If [A] = [2A] and [B] = [2B] then new Rate = `R' = K[2A] [2B]^(2)` `(R')/(R) = (K[2A][2B]^(2))/(K[A][B]^(2)) implies (R')/(R) = 8 implies R = ' = 8R`.

Discussion

5.	In a reaction :A+B hArr C+D, the initial concentrations of A and B were 0.9 mol dm^(-3)each.At equilibrium, the concentration of D was found to be 0.6 mol dm^(-3). What is the value of equilibrium constant for the reaction ?
Answer» 3 9 4 8 Answer :C

Discussion

6.	In a reaction, A + 2B hArr 2C, 2.0 mole of A, 3.0 mole of B and 2.0 mole of C are placed in a 2.0 L flask and the equilibrium concentration of C is 0.5 mol/L. The equilibrium constant (K) for the reaction is
Answer» 0.073 0.147 0.05 0.026 Solution : `K_(E )=((0.5)^(2))/((1.25)(2)^(2))=0.05`

Discussion

7.	In a reaction A + 2B to products, the molecularity of the reaction is
Answer» 3 2 1 zero Solution :Molecularity of the REACTION `A + 2 B to P` , is 1 + 2 = 3 , as one molecule of REACTANT A and two MOLECULES of reaction B COLLIDE with each other .

Discussion

8.	In a reaction, 5g of ethyl acetate is hydrolysed per litre in presence of dil . HCl in 300 minutes. If the reactant to be reduced to 2.5g?
Answer» Solution :ACCORDING to available DATA: a=`22g L^(-1), (a-x)=(22-5)=17 gL^(-1), t=300 min`. For the first order reaction, `k=(2.303)/t log a/(a-x) = 2.303/(300 "min") log (22g L^(-1))/(17 gL^(-1))` `=2.303/(300 "min") log 1.2941 = 2.303/(300 "min") xx 0.1120 = 8.6 xx 10^(-4) min^(-1)`

Discussion

9.	In a reaction, 5g ethyl acetate is hydrolysed per litre in the presence of dil HCl in 300 minutes. If the reaction is of first order and the initial concentration of ethyl acetate is 22g//"L", calculate the rate constant of the reaction.
Answer» SOLUTION :`a=22" G L"^(-1),x=5" g L"^(-1),t=300" MIN".`

Discussion

10.	In a reaction 4 moles of electrons are tranferred to one mole of HNO_(3) when acted as an oxidant .The possible reduction product is
Answer» (1/2) MOLE of `N_(2)` (1/2) mole of `N_(2)O` 1 mole of `NO_(2)` 1 mole of `NH_(3)` Answer :B

Discussion

11.	In a reaction ,2Ato products ,the concentration of A decreases from 0.5 mol L^(-1) to 0.4 mol L^(1) in 10 minutes.Calculate the rate during this interval.
Answer» SOLUTION :Reaction 2A `to` Product, Time =10 minute `THEREFORE Deltat=10 "minute:-0 "minute" =10 "minute"` `Rate=-(1)/(2)(DELTA[A])/(Deltat)` `=-((0.4-0.5)mol L^(-1))/(2XX10 m)` `+(0.1)/(2xx10),ol L^(-1) m^(-1)` `=5.0 xx10^(-3) mol L^(-1) m^(-1)`

Discussion

12.	In a reaction, 2AtoProducts, the concentration of A decreases from 0.5" mol L"^(-1) to 0.4" mol L"^(-1) in 10 minutes. Calculate the rate during this interval.
Answer» Solution :Average rate `=-(1)/(2)(Delta[A])/(DELTAT)=-(1)/(2)([A]_(2)-[A_(1)])/(t_(2)-t_(1))=-(1)/(2)(0.4"M"-0.5" M")/(10" MIN")=-(1)/(2)(-0.1"M")/(10" min")=5XX10^(-3)" M min"^(-1).`

Discussion

13.	In a reaction ,2Ato product, the concentration of A decreases from 0.5 mol L^(-1)to0.4 mol L^(-1)in 10 minutes. Calculate therate during this interval.
Answer» SOLUTION :Average RATE =-1/2(Delta[A])/(Delta)=-1/2([A_2]-[A_1])/(Deltat)=-1/2 ((0.4-0.5))/10` `=-1/2((-0.1)/10)=0.1/20=1/200=0.0`05`

Discussion

14.	In a reaction 2Ararr Products: the concentration of A decreases from 0.5 mol litre^-1 to 0.4 mol litre^-1 in 10 minutes. The rate of reaction during this interval is:
Answer» `0.05M min^-1` `0.005M min^-1` `0.5 M min^-1` `5 M min^-1` ANSWER :B

Discussion

15.	In a reaction 2A toProducts, the concentration of A decreases from 0.5 mol L^(-1) in 10 minutes. Calculate the rate during this interval.
Answer» SOLUTION :Average rate = `(-Delta[A])/(2Deltat)=1/2((0.4-0.5)/10)=5xx10^(-3) M "min"^(-1)`

Discussion

16.	In a reaction, 2A to Products, the concentration of A decreases from 0.5 mol L^(-1) to 0.4 mol L^(-1) in 10 minutes. Calculate the rate during this interval .
Answer» Solution :AVERAGE RATE `=-(1)/(2) (DELTA[A])/(Delta t)= -(1)/(2)([A]_(2)-[A]_(1))/(t_(2)-t_(1))` `= -(1)/(2) xx (0.4M-0.5M)/("10 minutes")= -(1)/(2) xx (-0.1M)/("10 minutes")=5xx10^(-3)" mole minutes"^(-1)`.

Discussion

17.	In a reaction , 2A rarr products , the concentration of A decreases from 0.5 "mol L"^(-1) to 0.4 "mol L"^(-1) in 10 minutes . Calculate the rate during this interval.
Answer» SOLUTION :AVERAGE rate `=1/2(Delta[A])/(Deltat)=1/2([A]_2-[A]_1)/(t_2-t_1)` `=-1/2((0.4M-0.5M)/(10MIN))=-1/2((-0.1M)/(10min))=5xx10^(-3) "M MIN"^(-1)`

Discussion

18.	In a reaction, 2A rarr products, the concentration of A decreases from 0.5 "mol L"^(-1) to 0.4 "mol L"^(-1) in 10 minutes. Calculate the rate during this interval ?
Answer» Solution :Average rate `=-(1)/(2)(Delta [A])/(Delta L)` `=-(1)/(2)([A]_(2)-[A]_(1))/(t_(2)-t_(1))` `=-(1)/(2)((0.4-0.5)/(10))` `=-(1)/(2)((-0.1)/(10))` `= 0.005 "MOL L"^(-1)"MIN"^(-1)` `= 5xx10^(-3)"M min"^(-1)`.

Discussion

19.	In a reaction 2A product, the concentration of A decreases from 0.5 mol L-1 to 0.4 mol L-1 in 10 minutes. Calculate the rate during this interval.
Answer» Solution :AVERAGE RATE= `-1/2 (Delta[A])/(Delta t)=-1/2[([A]_3-[A]_1)/(Deltat_2-t_1)]= -1/2[(0.4 - 0.5)/10]=0.005 MOL L^-1 mi N^(-1) = 5 times 10^-3 mol L^-1 mi n^(-1)`

Discussion

20.	In a reaction2A+btoA_(2)B the reaction A will disappear at
Answer» HALF the rate at which B disappears the same rate at which B disappears the same rate at which `A_(2)B` is formed TWICE the rate at which B disappears ANSWER :D

Discussion

21.	In a radioactive decay, an emitted electron comes from
Answer» Nucleus of the atom Inner ORBITAL of the atom Outermost orbit of the atom Orbit having principal QUANTUM NUMBER one Solution :In nucleus electrons formed by the following decay `._(0)n^(1) rarr ._(+1)p^(1) + ._(-1)e^(0)`

Discussion

22.	In a quantitative determination of iron in an ore, an analyst converted 0.42 g of the ore into its ferrous form . This required 42.00 mL of 0.1 N solution of KMnO_(4) for titration. How many moles of KMnO_(4) were used for titration ? (Fe =56)
Answer» Solution :In this problem `FE^(2+)` is oxidised to `Fe^(3+)` by `KMnO_(4)` Moles of `KMnO_(4) =("equivalents")/("factor relating mo.wt and eq.wt")""...(Eqn.6ii)` ` = (0.0042)/5 = 0.00084` mole . Note : Thus we see how the equations 1 to 7 have been put into USE in the PROBLEMS given above . However , application of these euations and the ability to use them come only from PRACTICE. The STUDENTS are advised to apply the said rules in as many problems as possible ] .

Discussion

23.	In a radioactive decay, an emitted electron comes from :
Answer» inner ORBITAL of the atom outermost orbital of the atom nucleus of the atom orbit having principal QUANTUM number one. Answer :C

Discussion

24.	In a quantitative determination of iron in an ore, an analyst converted 0.42 g of the ore into its ferrous form . This required 42.00 mL of 0.1 N solution of KMnO_(4) for titration. What is the molarity of KMnO_(4) solution used ?
Answer» Solution :In this problem `Fe^(2+)` is oxidised to `Fe^(3+)` by `KMnO_(4)` Molarity of `KMnO_(4) = ("normality")/("factor relating mol.wt and eq.wt")` i.e., change in ON ....(Eqn .6i) ` =(0.1)/5 = 0.02 M ` `{UNDERSET(+7)(MnO_(4)^(-))to underset(+2) (MN^(2+))}`

Discussion

25.	In a given reaction _ZX^A rightarrow _(Z+1)Y^A rightarrow_(Z-1)K^(A-4) rightarrow _(Z-1)K^(A-4) Radioactive radiations are emitted in the sequence of
Answer» `ALPHA,BETA,gamma` `beta,alpha,gamma` `gamma,alpha,beta` `beta,gamma,alpha` Solution :The COMPLETE sequence is : `_(Z)^(A)Poverset(-beta)to""_(Z+1)^(A)Qoverset(-alpha)to""_(Z-1)^(A-4)Roverset(gamma)to""_(Z-1)^(A-4)S`

Discussion

26.	In a quantitative determination of iron in an ore, an analyst converted 0.42 g of the ore into its ferrous form . This required 42.00 mL of 0.1 N solution of KMnO_(4) for titration. What is the percentage of iron in the ore?
Answer» SOLUTION :In this PROBLEM `Fe^(2+)` is oxidised to `Fe^(3+)` by `KMnO_(4)` ` % " of iron" = (0.2352)/(0.42) xx 100 = 56.00 % `

Discussion

27.	In a quantitative determination of iron in an ore, an analyst converted 0.42 g of the ore into its ferrous form . This required 42.00 mL of 0.1 N solution of KMnO_(4) for titration. How many grams of iron were present in the sample ?
Answer» Solution :In this problem `Fe^(2+)` is oxidised to `Fe^(3+)` by `KMnO_(4)` Wt.of IRON = equivalent `xx` eq.wt ...(Eqn.4) ` = 0.0042 xx 56 =0.2352 g` `{Fe^(2+)toFe^(3+):. "of Fe"=("at.wt")/("change in ON")=56/1 = 56}`.

Discussion

28.	In a quantitative determination of iron in an ore, an analyst converted 0.42 g of the ore into its ferrous form . This required 42.00 mL of 0.1 N solution of KMnO_(4) for titration.How many equivalents of iron were present in the sample of the ore taken for analysis ?
Answer» Solution :In this PROBLEM `FE^(2+)` is oxidised to `Fe^(3+)` by `KMnO_(4)` Equivalent of Fe present in the SAMPLE. ` = ("m.e of"KMnO_(4)"solution")/1000` ...(eqn.3) ` =(4.2)/1000 = 0.0042 `

Discussion

29.	In a quantitative determination of iron in an ore, an analyst converted 0.42 g of the ore into its ferrous form . This required 42.00 mL of 0.1 N solution of KMnO_(4) for titration. How many milliequivalents of KMnO_(4) does 42.00 mL of 0.1N solution represent ?
Answer» SOLUTION :In this PROBLEM `Fe^(2+)` is oxidised to `Fe^(3+)` by `KMnO_(4)` (i) m.e of `KMnO_(4)` solution =`0.1 xx42 =4.2 "" ….(Eqn 1)`

Discussion

30.	In a pseudo first order reaction of hydrolysis of an ester in H_2O, the following results were obtained : Calculate the average rate of reaction between the time interval 30 to60 sec.
Answer» Solution :AVERAGE RATE during 30-60 SEC. `=(0.17-0.31)/(60-30)=4.67xx10^(-3) "MOL L"^(-1) "sec"^(-1)`

Discussion

31.	In a pseudo first order reaction of hydrolysis of an ester in H_2O, the following results were obtained : Calculate the pseudo first order rate constant for the hydrolysis of ester.
Answer» Solution :`K_30=2.303/t "log"([A]_0)/([A])=2.303/30 "log"0.55/0.31` `K_60=2.303/60 "log"0.55/0.17` `K_90=2.303/90"log"0.55/0.085` AVERAGE `K=1.98xx10^(-2) "sec"^(-1)`

Discussion

32.	In a pseudo first order hydrolysis of ester in water the following results were obtained : {:(t//s,0,30,60,90),("[Ester]M",0.55,0.31,0.17,0.085):} Calculate the pseudo first order rate constant for the hydrolysis of ester.
Answer» Solution :`k = 1.98 XX 10^(-2) SEC^(-1)`

Discussion

33.	In a pseudo first order hydrolysis of ester in water the following results were obtained : {:(t//s,0,30,60,90),("[Ester]M",0.55,0.31,0.17,0.085):} Calculate the average rate of reaction between the time interval 30 to 60 seconds.
Answer» SOLUTION :`4.67 xx 10^(-3)MS^(-1)`

Discussion

34.	In a pseudo first order hydrolysis of an ester in water, the following results were obtained : {:("t/s",,,0,,,30,,,60,,,90),(["Ester"]//"mol L"^(-1),,,0.55,,,0.31,,,0.17,,,0.085):} (i) Calculate the average rate of reaction between the time interval 30 to 60 seconds. (ii) Calculate the pseudo first order rate constant for the hydrolysis of ester.
Answer» Solution :(i) AVERAGE rate during the interval 30-60 sec. `=(C_(2)-C_(1))/(t_(2)-t_(1))=(0.31-0.17)/(60-30)=(0.14)/(30)" MOL L"^(-1)s^(-1)=4.67xx10^(-3)" mol L"^(-1)s^(-1)` (II) `k'=(2.303)/(t)log""([A_(0)])/([A])" in which "[A_(0)]=0.55" M"` `t=30" sec",k'=(2.303)/(30s)log""(0.55)/(0.31)=1.91xx10^(-2)s^(-1)` `t=60" sec",k'=(2.303)/(60s)log""(0.55)/(0.17)=1.96xx10^(-2)s^(-1)` `t=90" sec",k'=(2.303)/(90s)log""(0.55)/(0.085)=2.07xx10^(-2)s^(-1)` Average `k'=(1.91+1.96+2.07)/(3)xx10^(-2)=1.98xx10^(-2)s^(-1).`

Discussion

35.	In a pseudo first order hydrolysis of ester in water, the following results were obtained. i) Calculate the average rate of reaction between the time interval 30 to 60 seconds. ii) Calcualte the pseudo first order rate constant for the hydrolysis of ester.
Answer» SOLUTION :Average rate of reaction = `("Change in molar concentration")/("Change in time INTERVAL")` `=(0.17 - 0.31)/(60-30) = (-0.14M)/(30s) = -4.67 xx 10^(-3)MS^(-1)` the rate constant, `K=(2.303)/t log[A_(0)]/[A], [A_(0)]=0.55 M` For t=30s , `k_(1)=2.303/(30s) log[0.55]/[0.31]= 2.303/(30s) xx 0.249 = 1.91 xx 10^(2)s^(-1)`. For t=60s, `k_(2) = 2.303/(60s) log [0.55]/[0.31]=2.303/(60s)xx0.5099 = 1.96 xx 10^(-2)s^(-1)` For t=60s , `k_(2)=2.303/(60s), k_(2) = 2.303/(60s)log[0.55]/[0.17]=2.303/(60s) xx 0.5099 = 1.96 xx 10^(-2)s^(-1)`. For t=90 s, `k_(3)=2.303/(90S)log[0.55]/[0.085]=2.303/(90s) xx 0.811 = 2.07 xx 10^(-2)s^(-1)`. Average rate constant (k) = `(k_(1)+k_(2)+k_(3))/(3) = (1.91 + 1.96+2.07)/(3) xx 10^(-2)s^(-1)` `=1.98 xx 10^(-2)s^(-1)`.

Discussion

36.	In a pseudo first order hydrolysis of an ester in water, and following results were obtained: (i) Calculate the average rate of reaction between the time interval 30 to 60 seconds. (ii) Calculate the pseudo first order rate constant for the hydrolysis of ester.
Answer» SOLUTION :(i) Average rate during the INTERVAL `30-60` seconds `=(C_(2)-C_(1))/(t_(2)-t_(1))=(0.31-0.17)/(60-30)=(0.14)/(30)" MOL L"^(-1)s^(-1)=4.67xx10^(-3)" mol L"^(-1)s^(-1)` (ii) `k=(2.303)/(t)"log"([A_(0)])/([A])` in which `[A_(0)]=0.55M` At t = 30 seconds, `k=(2.303)/(30s)"log"(0.55)/(0.31)=1.91xx10^(-2)s^(-1)` At t = 60 seconds,`k=(2.303)/(60S)"log" (0.55)/(0.17)=1.96xx10^(-2)s^(-1)` At t = 90 seconds, `k=(2.303)/(90s)"log"(0.55)/(0.085)=2.07xx10^(-2)s^(-1)` Average `k=(1.91+1.96+2.07)/(3)xx10^(-2)=1.98xx10^(-2)s^(-1)`.

Discussion

37.	In a protein, variousaminoacidsliked togetherby ………
Answer» Peptidebond Dativebond `ALPHA` - GLYCOSIDIC BOND `BETA` - Glycosidicbond ANSWER :A

Discussion

38.	In a protein molecule various amino acids are linked together by :
Answer» `beta`-glycosidic BOND PEPTIDE bond dative bond `ALPHA`-glycosidic bond Solution :peptide bond

Discussion

39.	In a protein ,the different type of attractions that exist are(A) H bonding(B) hydrophobic(C ) ionic(D) covalent
Answer» B, C and D only A, C and D only A, B and C only A, B, C and D ANSWER :C

Discussion

40.	In a protein molecule various amino acids are linked together by
Answer» peptide BOND dative bond `ALPHA`-GLYCOSIDIC bond `BETA`-glycosidic bond Answer :A

Discussion

41.	In a protein molecule various amino acids are linked together by-
Answer» `ALPHA`-GLYCOSIDIC BOND `BETA`- glycosidic acid peptide bond dative bond Answer :C

Discussion

42.	In a process for producing acetic acid, oxygen gas is bubbledinto acetaldehyde containing manganesed (II) acetate(catalyst under pressure at 60^(@) C . 2 CH_(3) CHO + O_(2) to 2CH_(3) COOHIn a laboratorytest of thisreaction20 g of CH_(3) CHO and 10 g of O_(2) were putinto a reaction vessel . (a)Howmanygrams of CH_(3) COOH can be produced ?(b) Howmany grams of the excess reactant remain after thereaction is complete ?
Answer» Solution :`{:(" "(20//40)""(10//32)""0". . . .initial mole" ),(2 CH_(3)CHO""+O_(2)""to ""2 CH_(3) CO OH),( ""0""(0.3125-0.2272)""0.4545". . . . final mole" ),(""=0.0853):}` (a) `CH_(3) CHO`is the limitingreagent and hence amount of `CH_(3) C O O H` produced ` = 0 . 4545 xx 60 G`= 27 . 27 g (b)Amount of `O_(2)" LEFT " = 0 . 853 xx 32 ` = 2 . 73 g

Discussion

43.	In a protein molecule amino acids are linked together by:
Answer» PEPTIDE BOND DATIVE bond GLYCOSIDIC bond phospodiestes bond ANSWER :A

Discussion

44.	In a primitive cubic lattice, the percentage of void volume is
Answer» 52.36% 25.95% 74.05% 47.64% Solution :In simple CUBIC UNIT cell, volume occupied by SPHERES = 52.36% Empty or void volume = 100 - 52.36 = 47.64%

Discussion

45.	In a polluted lake, the index of pollution is
Answer» BOD and Daphnia DO and Artemisia Presence of frogs Diatoms larvae Answer :A

Discussion

46.	In a polymer sample 30% of molecules have a molecular mass 20,000 , 40 % have 30,000 and the rest 60,000 . What is the weight average molecular mass of the polymer
Answer» 40300 30600 43333 50400 Answer :C

Discussion

47.	In a photoelectric effect, the energy of the photon striking a metallic surface is 5.6xx10^(-19)J. The kinetic energy of the ejected electron is 12.0xx10^(-20)J .The work function is
Answer» `6.4 XX 10^(-19) J` `6.8 xx 10^(-19) J` `4.4 xx 10^(-19) J` `6.4 xx 10^(-24) J` ANSWER :C

Discussion

48.	In a pesudo first order hydrolysis of ester in water, the following results were obtained. {:("1",0,30,60,90),("[Ester]mol L"^(-1),0.55,0.31,0.17,0.085):}Calculate the pseudo first order rate constant for the hydrolysis of ester.
Answer» Solution :For a PSEUDO FIRST order reaction, `k=(2.303)/(t)"log"([R]_(0))/([R])` For, `t = 303` `k_(1)=(2.303)/(t)"log"(0.55)/(0.31)` `=1.911xx10^(-2)s^(-1)` For, `t = 603` `k_(2)=(2.303)/(60)"log"(0.55)/(0.17)` `=1.957xx10^(-2)s^(-1)`, For, `t = 90 s` `k_(3)=(2.303)/(90)"log"(0.55)/(0.085)` `=2.075xx10^(-2)s^(-1)`. Then, average rate constant, `k=(k_(1)+k_(2)+k_(3))/(3)` `=((1.911xx10^(-2))+(1.957xx10^(-2))(2.075xx10^(-2)))/(3)` `= 1.98xx10^(-2)s^(-1)`.

Discussion

49.	In a pesudo first order hydrolysis of ester in water, the following results were obtained. {:("1",0,30,60,90),("[Ester]mol L"^(-1),0.55,0.31,0.17,0.085):} Calculate the average rate of reaction between the time interval 30 to 60 seconds.
Answer» Solution :AVERAGE rate of reaction between the time interval, 30 to 60 SECONDS `=("d[ESTER]")/(dt)` `=(0.31-0.17)/(60-30)=(0.14)/(30)` `=4.67xx10^(-3)"mol L"^(-1)s^(-1)`.

Discussion

50.	In a periodic table the basic character of oxides
Answer» increases from LEFT to right and decreases from TOP to bottom decreases from right to left and increase from top to bottom decreases from left to right and increases from top to bottom decreases from left to right and increases from bottom to top. Solution :As the electronegativity of element increases, ACIDIC character of OXIDES increases. So, in a GROUP, basic nature increases on moving down and decreases along a period.

Discussion

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In a reaction ,2Ato product, the concentration of A decreases from 0.5 mol L^(-1)to0.4 mol L^(-1)in 10 minutes. Calculate therate during this interval.

In a reaction 2Ararr Products: the concentration of A decreases from 0.5 mol litre^-1 to 0.4 mol litre^-1 in 10 minutes. The rate of reaction during this interval is:

In a reaction 2A toProducts, the concentration of A decreases from 0.5 mol L^(-1) in 10 minutes. Calculate the rate during this interval.

In a reaction, 2A to Products, the concentration of A decreases from 0.5 mol L^(-1) to 0.4 mol L^(-1) in 10 minutes. Calculate the rate during this interval .

In a reaction , 2A rarr products , the concentration of A decreases from 0.5 "mol L"^(-1) to 0.4 "mol L"^(-1) in 10 minutes . Calculate the rate during this interval.

In a reaction, 2A rarr products, the concentration of A decreases from 0.5 "mol L"^(-1) to 0.4 "mol L"^(-1) in 10 minutes. Calculate the rate during this interval ?

In a reaction 2A product, the concentration of A decreases from 0.5 mol L-1 to 0.4 mol L-1 in 10 minutes. Calculate the rate during this interval.

In a reaction2A+btoA_(2)B the reaction A will disappear at

In a radioactive decay, an emitted electron comes from

In a quantitative determination of iron in an ore, an analyst converted 0.42 g of the ore into its ferrous form . This required 42.00 mL of 0.1 N solution of KMnO_(4) for titration. How many moles of KMnO_(4) were used for titration ? (Fe =56)

In a radioactive decay, an emitted electron comes from :

In a quantitative determination of iron in an ore, an analyst converted 0.42 g of the ore into its ferrous form . This required 42.00 mL of 0.1 N solution of KMnO_(4) for titration. What is the molarity of KMnO_(4) solution used ?

In a given reaction _ZX^A rightarrow _(Z+1)Y^A rightarrow_(Z-1)K^(A-4) rightarrow _(Z-1)K^(A-4) Radioactive radiations are emitted in the sequence of

In a quantitative determination of iron in an ore, an analyst converted 0.42 g of the ore into its ferrous form . This required 42.00 mL of 0.1 N solution of KMnO_(4) for titration. What is the percentage of iron in the ore?

In a quantitative determination of iron in an ore, an analyst converted 0.42 g of the ore into its ferrous form . This required 42.00 mL of 0.1 N solution of KMnO_(4) for titration. How many grams of iron were present in the sample ?

In a quantitative determination of iron in an ore, an analyst converted 0.42 g of the ore into its ferrous form . This required 42.00 mL of 0.1 N solution of KMnO_(4) for titration.How many equivalents of iron were present in the sample of the ore taken for analysis ?

In a quantitative determination of iron in an ore, an analyst converted 0.42 g of the ore into its ferrous form . This required 42.00 mL of 0.1 N solution of KMnO_(4) for titration. How many milliequivalents of KMnO_(4) does 42.00 mL of 0.1N solution represent ?

In a pseudo first order reaction of hydrolysis of an ester in H_2O, the following results were obtained : Calculate the average rate of reaction between the time interval 30 to60 sec.

In a pseudo first order reaction of hydrolysis of an ester in H_2O, the following results were obtained : Calculate the pseudo first order rate constant for the hydrolysis of ester.

In a pseudo first order hydrolysis of ester in water the following results were obtained : {:(t//s,0,30,60,90),("[Ester]M",0.55,0.31,0.17,0.085):} Calculate the pseudo first order rate constant for the hydrolysis of ester.

In a pseudo first order hydrolysis of ester in water the following results were obtained : {:(t//s,0,30,60,90),("[Ester]M",0.55,0.31,0.17,0.085):} Calculate the average rate of reaction between the time interval 30 to 60 seconds.

In a pseudo first order hydrolysis of ester in water, the following results were obtained. i) Calculate the average rate of reaction between the time interval 30 to 60 seconds. ii) Calcualte the pseudo first order rate constant for the hydrolysis of ester.

In a pseudo first order hydrolysis of an ester in water, and following results were obtained: (i) Calculate the average rate of reaction between the time interval 30 to 60 seconds. (ii) Calculate the pseudo first order rate constant for the hydrolysis of ester.

In a protein, variousaminoacidsliked togetherby ………

In a protein molecule various amino acids are linked together by :

In a protein ,the different type of attractions that exist are(A) H bonding(B) hydrophobic(C ) ionic(D) covalent

In a protein molecule various amino acids are linked together by

In a protein molecule various amino acids are linked together by-

In a protein molecule amino acids are linked together by:

In a primitive cubic lattice, the percentage of void volume is

In a polluted lake, the index of pollution is

In a polymer sample 30% of molecules have a molecular mass 20,000 , 40 % have 30,000 and the rest 60,000 . What is the weight average molecular mass of the polymer

In a photoelectric effect, the energy of the photon striking a metallic surface is 5.6xx10^(-19)J. The kinetic energy of the ejected electron is 12.0xx10^(-20)J .The work function is

In a pesudo first order hydrolysis of ester in water, the following results were obtained. {:("1",0,30,60,90),("[Ester]mol L"^(-1),0.55,0.31,0.17,0.085):}Calculate the pseudo first order rate constant for the hydrolysis of ester.

In a pesudo first order hydrolysis of ester in water, the following results were obtained. {:("1",0,30,60,90),("[Ester]mol L"^(-1),0.55,0.31,0.17,0.085):} Calculate the average rate of reaction between the time interval 30 to 60 seconds.

In a periodic table the basic character of oxides