94420 + Interview Questions in Chemistry in Class 12

1.	In a galvanic cell energy changes occursas:
Answer» CHEMICAL ENERGY `rarr` ELECTRICAL energy Electrical energy `rarr` Electrical energy Chemical energy `rarr`Internal energy Internal energy`rarr` ELECTRIC energy Answer :A

Discussion

2.	In a galvanic cell electron flow will be from
Answer» NEGATIVE electrode to POSITIVE electrode Positive electrode to negative electrode there will be no flow of electrons Cathode to anode in the external circuit Solution :`e^(-)` flow anode to cathode

Discussion

3.	In a fuel cell, methanol is used as fuel oxygen gas is used as an oxidizer. The reaction is CH_(3)OH(l)+(3)/(2)O_(2)(g)toCO_(2)(g)+2H_(2)O(l) At 298K, standard Gibbs energies of formation for CH_(3)OH(l),H_(2)O(l) and CO_(2)(g) are -166.2, -237.2 and -394.4 kJ mol^(-1) respectively. If standard enthalpy of combusion of methanol is -726 kJ mol^(-1), efficiency of the fuel cel will be
Answer» 0.8 0.87 0.9 0.97 Solution :`Delta_(R)G=-394.4+2(-237.2)-(-166.2)` `=-702.6" kJ "mol^(-1)` EFFICIENCY`=(Delta_(r)G)/(DELTAH)=(-702.6)/(-726)xx100=97%`

Discussion

4.	In a galvanic cell cathode is positive electrode and anode is negative electrode.
Answer» ANSWER :1

Discussion

5.	In a galvanic cell, after running the cell for sometimes, the concentration of the electrolyte is automatically raised to 3 M HCl. Molar conductivity of the 3 M HCl is about "240 S cm"^(2)" mol"^(-1) and limiting molar conductivity of HCl is about "420 S cm"^(2)" mol"^(-1). If K_(b) of water is "0.52 K kg mol"^(-1), calculate the boiling point of the electrolyte at the end of the experiment.
Answer» 375.6 K 376.3 K 378.1 K 380.3 K Answer :A

Discussion

6.	In a galvanic cell,
Answer» POTENTIAL ENERGY decreases Kinetic energy decreases potential energy CHANGES into ELECTRICAL energy Chemical energy changes into electrical energy Answer :D

Discussion

7.	In a fuel cell methanol is used as fuel and oxygen is used as an oxidiser. The reaction is :CH_(3)OH(l)+(3)/(2)O_(2)(g)rarr CO_(2)(g)+2H_(2)O(l)At 298 K, standard Gibb's energies of formation for CH_(3)OH(l), H_(2)O(l)andCO_(2)(g) are -166.2, - 237.2 and -394.4 kJ/mol respectively. If standard enthalpy of combustion of methanol is - 726 kJ/mol, effeciency of the fuel cell will be :
Answer» `80%` `87%` `90%` `97%` Solution :`CH_(3)OH (l) + (1)/(3)O_(2) (g) rarr CO_(2) (g) + 2H_(2)O (l)` `Delta_(f) H = - 726 kJ//mole` `Delta_(R) G = SIGMA Delta_(f) G_("products") - Sigma Delta_(f) G_("ractants")` `[Delta_(f) G_(CO_(2)) + 2 xx Delta_(f) G_(H_(2)O)] - [Delta_(f) G_(CH_(3)OH) + (3)/(2) Delta_(f) G_(O_(2))]` `= [- 394.4 + 2 xx (- 237.2)] - [- 166.2 + (3)/(2) (0)]` `= - 702.6 kJ//mol` `% efficiency = (Delta_(f) G)/(Delta_(f)H) xx 100 = (- 702.6)/(- 726) xx 100` = 96.77% = 97%

Discussion

8.	In a fuel cell methanol is used as fuel and oxygen gas is used as an oxidizer. The reaction isCH_(3)OH(l)+(3)/(2)O_(2)(g)rarrCO_(2)(g)+2H_(2)O(l) At 298 K standard Gibb's energies of formation for CH_(3)OH(l), H_(2)O(l) and CO_(2)(g) are -166.2, -237.2 and -394.4 kJ mol^(-1) respectively. If standard enthalpy of combustion of methanol is -726 kJ mol^(-1), efficiency of the fuel cell will be
Answer» 0.8 0.87 0.9 0.97 Solution :`CH_(3)OH(l)+(3)/(2)O_(2)(g)rarrCO_(2)(g)+2H_(2)O(l)` `DeltaG_(r)=DeltaG_(F)(CO_(2)(g)+2DeltaG_(f)(H_(2)O(l))-DeltaG_(f)(CH_(3)OH(l))` `-(3)/(2)DeltaG_(f)(O_(2)(g))` `=-394.4+2(-237.2)-(-166.2)-0` `=-394.4-474.4+166.2=-868.8+166.2` `DeltaG_(r)=-702.6 kJ` % EFFICIENCY = `(702.6)/(726)xx100=97%`.

Discussion

9.	In a fuel cell methanol is used as fuel and oxygen gas is used as an oxidizer. The reaction is CH_(3)OH(l)+3/2 O_(2)(g) rarr CO_(2)(g)+2H_(2)O(l) at 298 K standard Gibb's energies of formation for CH_(3)OH(l), H_(2)O(l) and CO_(2)(g) are -166.2, -237.2 and -394.4" kJ mol"^(-1) respectively. If standard enthalpy of combustion of methanol is -726" kJ mol"^(-1), efficiency of the fuel cell will be -
Answer» `87 %` `90 %` `97 %` `80 %` Solution :`CH_(3)OH(l)+3/2 O_(2) (g) rarr CO_(2)(g)+2H_(2)O(l)` `DeltaG^(@)=DeltaG_(P)^(@)-DeltaG_(R)^(@)` `DeltaG^(@)=[DeltaG_(CO_(2))^(@)+2DeltaG_(H_(2)O)^(@)]-[DeltaG_(CH_(3)OH)^(@)+3/2 DeltaG_(O_(2))^(@)]` `-394.4+2xx(-237.2)-[166.2]` `-868.8-166.2` `G_(1)G^(@)=-702.6" KJ/mol"` EFFICENCY `=(DeltaG^(@))/(DeltaH^(@))xx100=(-702.6 kJ)/(-726 kJ)xx100=96.77%` `[DeltaG_("USEFUL WORK")=DeltaH_("total energy")-TDeltaS_("useless work")]`

Discussion

10.	In a fuel cell methanol is used as fuel and oxygen gas is used as an oxidizer. The reaction is CH_(3)OH(l)+3/2O_(2)(g)toCO_(2)(g)+2H_(2)O(l) At 298 K standard Gibb’s energies of formation for CH_(3)OH(l),H_(2)O(l)andCO_(2)(g)" are "-166.2,-237.2and-394.4kJmol^(-1) respectively. If standard enthalpy of combustion of methanol is -726kJmol^(-1), efficiency of the fuel cell will be -
Answer» 0.87 0.9 0.97 0.8 Answer :C

Discussion

11.	In a fuel cell methanol is used as fuel and oxygen gas is used as an oxidiser. The reaction is CH_(3)OH(l)+3/2O_(2)(g)toCO_(2)(g)+2H_(2)O(l) AT 298 K standard Gibbs energies of formation for CH_(3)OH(l),H_(2)O(l)andCO_(2)(g)" are "-166.2,-237.2and-394.4kJmol^(-1) respectively. If standard enthalpy of combustion of methanol is -726kJmol^(-1), efficiency of the fuel cell will be
Answer» 0.8 0.87 0.9 0.97 Solution :`DeltaG_(F)^(@)(O_(2))=0`. Calculate `DeltaG_(f)^(@)` reaction and then APPLY : Efficiency `(DELTAG^(@))/(DeltaH^(@))xx100`.

Discussion

12.	In a fuel cell hydrogen and oxygen react to produce electricity. In the process hydrogen gas is oxidised at the anode and oxygen at the cathode. If 67.2 litres of H_(2) at STP react in 15 minutes, what is the average current produced? If the entire current is used for electrodeposition of copper from Cu (II) solution, how many grams of copper will be deposited ? Anode reaction : H_(2) + 2OH^(-) rarr 2H_(2)O + 2e^(-) Cathode reaction : O_(2) + H_(2) O + 4e^(-) rarr 4OH^(-)
Answer» SOLUTION :Eq. of HYDROGEN = no. of faradays of ELECTRICITY = eq. of Cu DEPOSITED

Discussion

13.	In a fuel cell H2 and 02 react to produce electricity. In this process 67.2 It of H_2at NTP reacted in 16 min and 5 sec. What is the average current produced.
Answer» 200 AMP 400 amp 500 amp 600 amp Answer :D

Discussion

14.	In a flask, the weight ratio of CH_(4) (e) and SO_(2) (g) at 298 K and I bar is 1 : 2. The ratio of the number of molecules of SO_(2) (g) and CH_(4) (g) is
Answer» `1:4` `4:1` `1:2` `2:1` SOLUTION :Let mass of `CH_(4)(G) = 1g` Number of moles of `CH_(4)(nCH_(4)) =1/16` Number of MOLECULES of `CH_(4)(g) = 1/16 xx N_(A)` Let the mass of `SO_(2)(g) = 2g` Number of moles of `SO_(2)(g) (n_(SO_(2))) = 2/64` Number of molecules of `SO_(2)(g) = 2/64 xx N_(A)` Ratio of number of molecules of `SO_(2)` (e) and number of molecules of `CH_(4)`(g) `=2/64 xx N_(A) : 1/16 xx N_(A) rArr 1/32 : 1/16 rArr 1:2`

Discussion

15.	In a first reaction at 27^(@) C and 47^(@)C 50% reaction complete in 30 min.and 10 min.Calculate the energy of activation.
Answer» SOLUTION :43857 J `MOL^(-1)`

Discussion

16.	In a flask colourless N_2O_4 is in equilibrium with brown colourless NO_2. At equilibrium when the flask is heated at 100^Oc the brown colour deepens and on cooling it becomes less coloured. The change in enthalpy DaltaH, for the system is:
Answer» Negative Positive Zero Undefined Answer :B

Discussion

17.	In a fission reaction the nucleus of an element
Answer» Loses only some elementary NUCLEAR PARTICLES from another nuclear Captures some elementary nuclear particles from another nucleus Breaks up into several SMALLER nuclei Breaks up into two smaller nuclei with the LOSS of some elementary nuclear particles Solution :The definition of nuclear fission

Discussion

18.	In a first order reaction x rarry , if k is the rate constant and the initial concentration of the reactant x is 0.1 M , then , the half life is ...........
Answer» `((log2)/K)` `((0.693)/((0.1)k)` `((ln2)/k)` none of these SOLUTION :`k=(1/t)LN(([A_0])/([A]))` `[A_0]=0.1,[A]=0.05` `k=(1/t_(1//2))ln((0.1)/(0.05))` `k=(1/(t_1//2))ln(2)impliest_(1//2)=(ln(2))/k`

Discussion

19.	In a first order reaction the time reaquired for the concentration to decrease from 6 moles to 3 moles is 40 minutes in such a reaction what time will be taken for the conversion reactant from 12 moles to 6 moled?
Answer» 20 minutes 40 minutes 80 minutes 160 minutes Answer :B

Discussion

20.	In a first order reaction ,the concentration of the reactant,decreases from 0.8 M to 0.4 M in 15 minutes .The time taken for the concentration to change from 0.1 M to 0.025 M is……
Answer» 7.5 minutes 15 minutes 30 minutes 60 minutes Answer :C

Discussion

21.	In a first order reaction , the concentration of the reactant is reduced to 1/8 of the initial concentration in 75 minutes at 298 K . What is the half-life period of the reaction in minutes
Answer» 50 15 45 25 Solution :Let a = 1 , a-x = 1/8 , t = 75 min For first order reaction `k = (2.303)/(t)` LOG `(a)/(a-x)` `k = (2.303)/(75)` log `(1)/(1//8) = (2.303 xx 0.903)/(75) "min"^(-1)` For first order reaction `t_(1//2) = (0.693)/(k) = (0.693 xx 75)/(2.303 xx 0.903) = 25` min .

Discussion

22.	In a first order reaction, the concentration of the reactant is reduced from 0.6 mol L^(-1) to 0.2 mol L^(-1) in 5 minutes. Calculate the rate constant of the reaction.
Answer» Solution :Substituting the values in the first ORDER EQUATION `k=(2.303)/(t)"LOG"([R]_(0))/([R])` `k=(2.303)/(5"MIN")xx"log"(0.6)/(0.2) or k=(2.303)/(5)xx log3` or `k=(2.303)/(5)xx0.4771=0.2198`

Discussion

23.	In a first order reaction, the concentration of the . reactant is decreased from 1.0M to 0.25M in 20 minute. The rate constant of the reaction would
Answer» `10 MI n^(-1)` `6.93 mi n^(-1)` `0.693mi n^(-1)` `0.0693 mi n^(-1)` SOLUTION :`t= 2.303/klog. (a)/((a-x))` `:. K=2.303/20log. (1)/0.25` `- 0.06931 MIN^(-1)`

Discussion

24.	In a first order reaction, the concentration of the reactant is decreased from 1.0 M to 0.25 M in 20 minute. The rate constant of the reaction would be
Answer» `10min^-1` `6.931min^-1` `0.6931min^-1` `0.06931min^-1` ANSWER :D

Discussion

25.	In a first order reaction the concentration of the reactant decreases from 800 mol/ dm^(3) to 50 mol/ dm^(3) in 2xx10^(4) sec . The rate constant of the reaction in sec^(-1) is :
Answer» `2XX10^(4)` `3.45xx10^(-5)` `1.386xx10^(-4)` `2xx10^(-4)` Solution :(C ) k = `(2.303)/(t) "log" ([A]_(0))/([A])` `k = (2.303)/(2xx10^(4))"log" (800)/(50) = 1.38 xx10^(-4) s^(-1)`

Discussion

26.	In a first order reaction the a/(a - x) was found tu be 8 after 10 minute. The rate constant is
Answer» `(2.303xx3log2)//10` `(2.303xx21og3)//10` `10 XX 2.303 xx 2 log 3` `10 xx 2.303 xx 3 log 2` SOLUTION :`t= 2.303/k log. (a)/((a-x))` Thus, `k= 2.303/10log8 = 2.303/10log2^3` `= (2.303xx 3log2)/10`

Discussion

27.	In a first order reaction the concentration of the reactant decreases from 0.8 to 0.4 M in 15 minutes . The time taken for the concentration to change from 0.1 M to 0.025 M is :
Answer» a) 7.5 MINS B) 15 mins C) 30 mins d) 60 mins Solution :(C ) `t_(1//2) ` is 15 MIN `:.` Conc . Reduces from 0.1 to0.025 M in `2t_(1//2)` i.e 30 min

Discussion

28.	In a first order reaction, the concentration of the reactant, decreases from 0.8 M to 0.4 M in 15 minutes. The time taken for the concentration to hange from 0.1 M to 0.025 M is
Answer» 30 minutes 60 minutes 7.5 minutes 15 minutes Solution :`0.8Moverset(15"MIN")(to)0.4M ,0.1overset(15)(to)0.05overset(15)(to)0.025`

Discussion

29.	In a first order reaction, the concentration of the reactant, decrease from 0.8M to 0.4M in 15minutes. The time taken for the concentration to changes from 0.1M to 0.025M is
Answer» 7.5 minutes 15 minutes 30 minutes 60 minutes Solution :The `t_((1)/(2))` is 15 minuts. To fall the concentrationn from 0.1 to 0.025 we need TWO HALF LIVES i.e., 30 minutes.

Discussion

30.	In a first order reaction, the (a)/(a-x) was found to be 10 min. The rate constant is
Answer» `((2.303xx3log2))/(10)` `((2.303xx2log3))/(10)` `10xx2.303xx2log3` `10xx2.303xx3log2` ANSWER :A

Discussion

31.	In a first order reaction the (a)/((a-x)) value was found to be 10 after 15 minutes . The rate constant of the reaction is
Answer» `0.153 "MIN"^(-1)` `0.135 "min"^(-1)` `0.315 "min"^(-1)` `0.351 "min"^(-1)` Solution :`K = (2.303)/(t)` LOG `(a)/(a-x)` `k = (2.303)/(15)` log 10 = `0.153 "min"^(-1)`

Discussion

32.	In a first order reaction , on plotting a graph between t_(1//2) and concentration of reactant
Answer» A curve is obtained A STRAIGHT line is obtained having slope `45^@` A straight line is parallel to TIME axis A straight line is parallel to the concentrationaxis Answer :D

Discussion

33.	In a first-order reaction of the type: A(g)to2B(g), the initial and final pressures are P_(1) and p respectively. The rate constant can be expressed by
Answer» <P>`k=1/t"LN"(p_(1))/(2p_(1)-p)` `k=1/t"ln"(p_(1))/(p_(1)-p)` `k=1/t"ln"(p_(1))/(p-p_(1))` `k=1/t"ln"(p_(1))/(p)` ANSWER :A

Discussion

34.	In a first-order reaction, it takes the reactant 40.5 minutes to be 25% decomposed. Find the rate constant of the reaction.
Answer» Solution :SINCE the reactant is 25% DECOMPOSED in 40.5 min, the remaining reactants is `75%," i.e."(a-x)=0.75" at "40.5" min."` `k=(2.303)/(t)LOG""(a)/(a-x)=(2.303)/("40.5 min")log""(a)/(0.75a)` `=0.05686log" 1.33 min"^(-1)` `=0.05686xx0.1249" min"^(-1)` `=7.1xx10^(-3)" min"^(-1)`

Discussion

35.	In a first order reactiona//(a-x) was found to be 8 after l0minute. The rate constant is
Answer» `(2.303xx3 LOG2)//10` `(2.303xx2log3)//10` `10xx2.303xx2log3` `10xx2.303xx3log2` ANSWER :A

Discussion

36.	In a first order reaction ArarrBif k is rate constant and initial concentration of the reactant A is 0.5 M then the half - life is :
Answer» `("LOG"2)/(k) ` `("log"2)/(ksqrt(0.5))` `(In2)/(k)` `(0.693)/(0.5 k)` Solution :(C ) `t_(1//2) ` for a FIRST ORDER reaction is : `t_(t//2) = (0.693)/(k) "or " = (In2)/(k)`

Discussion

37.	In a first-order reactionA to pif it takes 20 minutes to bring about decompositionof 30% of the initial substance, calculate the time to decompose (i) 60% of it, and (ii) all of it.
Answer» SOLUTION :51.3 MIN, INFINITE

Discussion

38.	In a first - order reaction A to B, if K is the rate constant and initial concentration of the reactant is 0.5 M, then half-life is
Answer» `("In"2)/(K)` `("In"2)/(Ksqrt(0.5))` `(log_(10)2)/(K)` `(0.693)/(0.5K)` ANSWER :A

Discussion

39.	In a first order reaction A rarr products , 60% of the given sample of A decomposes in 40 min . What is the halg life of the reaction .
Answer» Solution :`k=(2.303)/tloga/(a-x)` `k=(2.303)/(40min)log(100)/((100-60))impliesk=(2.303)/40log(100/40)` `k=0.05575(0.3979)impliesk=0.02287"MIN"^(-1)` `t_(1//2)=(0.6932)/k=(0.6932)/(0.02287)` `t_(1//2)=30.31min`.

Discussion

40.	In a first order reaction A to B , if k is rate constant an initial concentration of the reactant A is 0.5 M then the half - life is
Answer» `(0.693)/(0.5 K)` `(LOG 2 )/(K)` `(log 2 )/(K SQRT(0.5))` `("In"2)/(K)` Answer :B

Discussion

41.	In a first order reaction A rarr product, 80% of the given sample of compound decomposes in 40 min. What is thehalf life period of the reaction ?
Answer» Solution :GIVEN, 80% of sample DECOMPOSES Time = 40 MIN. `t_(1//2)` (half life = ? ) Consider, initial concentration of sample is 100. `:.[A]_(o)=100` `:.[A]_(t)=100-80` `=20 ` Final concentration `:.k=(2.303)/(t)log_(10).([A]_(o))/([A]_(t))` `=(2.303)/(40)log_(10).(100)/(20)` `=(2.303)/(40) log_(10)5` `=(2.303)/(40)xx0.6989` `k=0.04024` `:.` Half life `t_(1//2)=(0.693)/(k)` `=(0.693)/(0.04024)=17.22` Half-life is `17.22` min.

Discussion

42.	In a first order reaction, A+B to , If K is rate constant and initial concentration of the reactant A is 0.5M, then the half-life is
Answer» (`log`2)/(k) (LOG2)/(ksqrt(0.5)) `(IN2)/(k)` `(0.693)/(0.5k)` Solution :For a first order reaction `k=(2.303)/(t)"log"(a)/(a-x)` When `t=t_((1)/(2))` `k=(2.303)/(t_((1)/(2)))"log"(a)/(a-a//2)` or, `t_((1)/(2))=(2.303)/(k)"log"_(10)2=(In2)/(k)`

Discussion

43.	In a first order reaction, 20% reaction is completed in 24 minutes.The percentage of reactat remaining after 48 minutes is
Answer» 60 64 81 80 Answer :B

Discussion

44.	In a fcc lattice, the number of neighbours for a given lattice point is ............
Answer» 12 8 6 14 ANSWER :A

Discussion

45.	In a first order reaction, 0.16 moles of reactant decreases to 0.02 moles in 144 minutes, its half life is
Answer» 24 MIN 12 min 72 min 48 min Answer :D

Discussion

46.	In a fcc lattice, atom A occupies the corner positionsand atom B occupies the face centred position. If one atom of B is missing from one of the face centred points, the formula of the compound is
Answer» `A_2B` `AB_2` `A_2B_3` `A_2B_5` SOLUTION :`{:("ATOM A","Atom B"),(Z=8xx1/8=1 , Z=5xx1/2=5/2):}` `rArr A:B=1:5/2 rArr` 2:5 So, formula of COMPOUND will be `A_2B_5`

Discussion

47.	Element 'A' and 'B' form a compound with cubic structure in which 'A' atoms are at the corners of the cube and 'B' atoms at the face centres. What is the formula of the compound ?
Answer» Solution :Number of A type ATOMS in the UNIT cell, `8 xx 1/8 =1` Number of B type atoms in the unit cell, `6 xx 1/2 =3` Hence the formula is `AB_3`

Discussion

48.	Ina fcc arrangement, the corner atoms are A type and those at face centres are B type. What is the simplest formula of the compound ?
Answer» SOLUTION :Number of A type atoms in the UNIT cell, `8 xx 1/8 =1` Number of B type atoms in the unit cell, `6 xx 1/2 =3` Hence the FORMULA is `AB_3`

Discussion

49.	In a factory, 40 kg of calcium is produced in 2 hours. If capacity of flow of current is 50%, then how much aluminium can be obtained by passing same current for 2 hours ?
Answer» 22 kg 18 kg 9 kg 27 kg Solution :`(W_(Ca))/(E_(Ca))=(W_(Al))/(E_(Al))` `therefore (40)/(20)=(W_(Al))/(9)` `therefore W_(Al)=18KG`.

Discussion

50.	In a face centred unit cell (fcc), the number of atoms present is :
Answer» 4 2 3 5 Answer :A

Discussion

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In a first order reaction ,the concentration of the reactant,decreases from 0.8 M to 0.4 M in 15 minutes .The time taken for the concentration to change from 0.1 M to 0.025 M is……

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In a first order reaction, the concentration of the reactant is reduced from 0.6 mol L^(-1) to 0.2 mol L^(-1) in 5 minutes. Calculate the rate constant of the reaction.

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In a first order reaction, the concentration of the reactant is decreased from 1.0 M to 0.25 M in 20 minute. The rate constant of the reaction would be

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In a first order reaction the a/(a - x) was found tu be 8 after 10 minute. The rate constant is

In a first order reaction the concentration of the reactant decreases from 0.8 to 0.4 M in 15 minutes . The time taken for the concentration to change from 0.1 M to 0.025 M is :

In a first order reaction, the concentration of the reactant, decreases from 0.8 M to 0.4 M in 15 minutes. The time taken for the concentration to hange from 0.1 M to 0.025 M is

In a first order reaction, the concentration of the reactant, decrease from 0.8M to 0.4M in 15minutes. The time taken for the concentration to changes from 0.1M to 0.025M is

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In a first order reactiona//(a-x) was found to be 8 after l0minute. The rate constant is

In a first order reaction ArarrBif k is rate constant and initial concentration of the reactant A is 0.5 M then the half - life is :

In a first-order reactionA to pif it takes 20 minutes to bring about decompositionof 30% of the initial substance, calculate the time to decompose (i) 60% of it, and (ii) all of it.

In a first - order reaction A to B, if K is the rate constant and initial concentration of the reactant is 0.5 M, then half-life is

In a first order reaction A rarr products , 60% of the given sample of A decomposes in 40 min . What is the halg life of the reaction .

In a first order reaction A to B , if k is rate constant an initial concentration of the reactant A is 0.5 M then the half - life is

In a first order reaction A rarr product, 80% of the given sample of compound decomposes in 40 min. What is thehalf life period of the reaction ?

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In a fcc lattice, atom A occupies the corner positionsand atom B occupies the face centred position. If one atom of B is missing from one of the face centred points, the formula of the compound is

Element 'A' and 'B' form a compound with cubic structure in which 'A' atoms are at the corners of the cube and 'B' atoms at the face centres. What is the formula of the compound ?

Ina fcc arrangement, the corner atoms are A type and those at face centres are B type. What is the simplest formula of the compound ?

In a factory, 40 kg of calcium is produced in 2 hours. If capacity of flow of current is 50%, then how much aluminium can be obtained by passing same current for 2 hours ?

In a face centred unit cell (fcc), the number of atoms present is :