94420 + Interview Questions in Chemistry in Class 12 Page 1796 InterviewSolution

89751.	A first order reaction is 25% complete in one hour . At the end of two hours the extent of reaction is :
Answer» 0.5 0.3125 0.4375 0.75 Solution :(C ) `k = (2.303)/(t) "LOG"([A]_(0))/([A])` `k = (2.303)/(1) "log" ([A]_(0))/(0.75[A]_(0))` `= (2.303)/(1) "log" 1.33` `=(2.303)/(1) xx0.12876` Now `t = (2.303)/(k) "log" ([A]_(0))/([A])` `2 = (2.303)/(0.2876)"log" ([A]_(0))/([A])` log `([A]_(0))/([A])= (2xx0.2876)/(2.303)=0.2497 ` or `([A])/([A]_(0))=1.77 ` `([A]_(0))/([A])= (1)/(1.77) == 56 .35 % ` `:. ` Extent of REACTION `= 100 -56.3 = 43. 75% `

Discussion

89752.	A first order reaction is 25% complete in 30 minutes. Calculate the specific reaction rate.
Answer» SOLUTION :`K=2.303/309"LOG"[A]_0/(75/100[A]_0)=2.303/30"log"4/3=(2.303xx0.1250)/30=0.0096 min^-1=9.6xx10^-3min^-1` `t_(1/2)=0.693/9.6xx10^3=6930/96=72.18min`.`

Discussion

89753.	A first order reaction is 25% complete in 40 minutes. Calculate the value of rate constant. In what time will the reaction be 80% completed?
Answer» Solution :Apply the reaction `k=2.303/t log ([R]_0)/([R])` 25% complete reactions means, `[R]=0.75[R]_0 and t=40 MINUTES` Substituting the values in the equation above, we have `k=2.303/(40 minutes)log""([R])_0/(0.7[R]_0)=2.303/(40 minutes)[log4-log 3]` or `k=2.303/(40 minutes) (0.6021-0.4771)` `or k=2.303/(40 minutes) TIMES 0.125=0.007197` WRITING the first order equation again and subsituting new values, we have `k=2.303/tlog""([R]_0)/([R])` 80% complete reaction means `[R]=0.2[R]_0` Substituting the values in the equation above, we have `0.007197=2.303/tlog""[R]_0/(0.2[R]_0)` or `t=2.303/0.007197 timeslog 5=2.303/0.007197 times 0.699` or `t=223.68 minutes`

Discussion

89754.	A first order reaction is 20% completed in 10 minutes . Calculate the time taken for the reaction to go to 80% completion.
Answer» Solution :Applying the first ORDER equation, `k=(2.303)/tlog.([R]_0)/([R])` At t = 10 min , R100 -20 `:. k = (2303)/t log10100/((100-20))` `:. k = (2303)/t LOG.(100)/(80)` `= 0.0223"min"^(-1)`

Discussion

89755.	A first order reaction is 20% completed in 10 minutes. Calculate the time taken for the reaction to go to 80% completion.
Answer» SOLUTION :Applying the FIRST order equation, `k = (2.303)/(t) "log" ([R]_(0))/([R])` At t= 10 min, R= 100-20 `:. K = (2303)/(t) "log"_(10) (100)/((100-20))` `t= (2303)/(10) "log"_(10) (100)/(80)` `=0.0223 "min"^(-1)`

Discussion

89756.	A first order reaction is 20% complete in 20 minutes. Calculate the time taken for the reaction to go to 80% completion.
Answer» SOLUTION :144.3 MINUTES

Discussion

89757.	A first order reaction is 20% complete in 10 minutes. Calculate time taken for reaction to go to 80% .
Answer» SOLUTION :`K=(2.303)/(t)log_(10).(100)/((100-20))` `=(2.303)/(10)log_(10).(100)/(80)=0.223minute^(-1)` Again applying first ORDER equation `t=(2.303)/(K)log_(10).(100)/((100-80))` `=(2.303)/(0.0233)log_(10).(100)/(20)=72.18min`.

Discussion

89758.	A firstorderreactionis 20% complete in 10 min. How long it takes to complete 80% ?
Answer» Solution :Applying first ORDER integral rate equation, equation, rate constant K is gives as, `K=(2.303)/(t)"log"(a)/(a-x)RARR(2.303)/(10)"log"(100)/(100-20)=0.0223" MIN"^(-1)` TIME required for `80%` completion of the first order reaction, `t_(0.8)` `t_(0.8)=(2.303)/(k)"log"(100)/(!00-80)=(2.303)/(0.0223)"log"(100)/(20)=72.2` min

Discussion

89759.	A first order reaction in 50% completed in 30 minutes at 27^@C and in 10 minutes at 47^@C . Calculate the reaction rate constant at 27^@C and the energy of activation of eh reaction in kJ mol^(-1)
Answer» SOLUTION :For a first order reaction `k=(0.693)/(t_(1//2))` At `27^@C` `k_(27^@C)=(0.693)/30=0.0231"min"^(-1)` At `47^@C` `k_(47^@C)=(0.693)/10=0.0693"min"^(-1)` Now applying the following equation: `log.k_2/k_1=""/(2.303R)[(T_2-T_1)/(T_1T_2)]` or `log_(10).(0.0231)/(0.0693)=(-E_a)/(2303xx8.314).((320-300)/(320xx300))` or `-log_(10)0.3333=(E_a)/(19.1471)xx20/96000` `E_a=(19.1471xx96000)/20xxlog0.3333` `=-91906xx(-0.4772)` `=43857"J MOL"^(-1)` `=43.857"KJ mol"^(-1)`

Discussion

89760.	A first order reaction has rate constant of k=1.15xx10^(-3)s^(-1). How long will it take for 6 g of reactant to reduce to 3 g ?
Answer» ANSWER :`896.6" s"`

Discussion

89761.	A first order reaction has rate constant 1.15xx10^(-3) s^(-1) How long will 5 g of this reactant take to reduce to 3g?
Answer» Solution :SUPPOSE molecular MASS of reactant =M g `mol^(-1)` Rate constant K=`1.15xx10^(-3)` INITIAL mass =5 g `therefore"Initial mole"=(5g)/(M g mol^(-1))` `therefore [R]_(0)=(5)/(M)mol` Final mass =3g `therefore` Final mole =`(3g)/(MG mol^(-1))` `therefore [R]_(t)=(3)/(M) mol` Change in time `Deltat=(?)` For first order reaction , Rate constant k`=(2.303)/(t)` log `([R]_(0))/([R]_(t))` `therefore t=(2.303)/(k)` log `([R]_(0))/([R]_(t))` `=(2.303)/(1.15xx10^(-3)s^(-1))` log `((5)/(M))/((3)/(M))` `(2.303)/(1.15xx10^(-3)s^(-1))` log `(5)/(3)` `(2.303)/(1.15xx10^(-3)s^(-1))xx0.2219=444.3s`

Discussion

89762.	A firstorderreactionhasactivationenergy22.8 kJat 300 K . Ifthe rateconstantof thereation is1.42 xx 10^(2) min^(-1) Calculatethe frequencyfactor .
Answer» SOLUTION :A=2.2 `XX 10^(4) s^(-1)`

Discussion

89763.	A first order reaction has a specific rreaction rate of 10^(-2)s^(-1). How much time will it take for 20 g of the reactant of reduce to 5 g ?
Answer» 238.6 second 138.6 second 346.5 second 693.0 second Solution :Given, ` k = 10^(-2) s^(-1).` HENCE, `t_(1//2) = (0.693)/(k) = (0.693)/(10^(-2))` second `20 g to 5 g ` takes TWO half LIVES `( 20g to 10g to 5 g)` Hence, `t =2 xx (0.693)/(10^(-2)) = 138*6` second

Discussion

89764.	A first order reaction has a specific reaction rate of 10^(-2)"sec"^(-1). How much time will it take for 20 g of the reactant to reduce to 5g ?
Answer» `693.0"SEC"` `238.6"sec"` `138.6"sec"` `346.5"sec"` ANSWER :A

Discussion

89765.	A first orderreactionhasa specific reaction rate of 10^(-2) S^(-1). How much timewill it take for 20 gof thereactanttoreduceto 5 g?
Answer» 238.6s 138.6s 346.5s 693.0s Solution :for a first orderreaction rate constant`(K) =(2303)/(t) log (a) /(a-x)` wherea initial CONCENTRATION `a-x=`concentration after time't' .t= timein sec given a=20 g,a-x =5g.k =`10^(-2)` `thereforet=(2303)/(10^(-2))log (20)/(5) ` `=1386s` ALTERNATIVELY , Half - lifeforthefirstorderreaction `(t_(1//2))/(2)=(0.693)/(10^(-2))=69.3 s` twohalf -livesarerequiredfor thereductionof 20 g of REACTANT into5g `20 g OVERSET( t_(1//2)) to 10g overset( t_(1//2))to 5g.` `therefore"totaltime "=2t_(1//2)=2xx69.3=138.6s`

Discussion

89766.	A first order reaction has a specific reaction rate of 10^(-2) sec^(-1) . How much time will it take for 20 g of the reactant to reduce to 5 g
Answer» 138.6 sec 346. 5 sec 693.0 sec 238 . 6 sec Solution :HALF life of a FIRST order reaction `t_(1//2) = (0.693)/(K) = (0.693)/(10^(-2)) = 69.3` sec Method - 1 Total time = `2t_(1//2) = 2 xx 69.3= 138.6` sec Method-2 `t = (2.303)/(K)` log `([A]_(0))/([A]_(t))` `t = (2.303)/(10^(-2)) "log" (20)/(5) implies t = 138.6` sec

Discussion

89767.	A first order reaction has a rate constant of 0.0051" min"^(-1). If we begin with 0.10 M concentration of the reactant, what concentration of the reactant will remain in the solution after 3 hours ?
Answer» Solution :`k=(2.303)/(t)log""([A]_(0))/([A]):.0.0051" MIN"^(-1)=(2.303)/(3xx60" min")log""(0.10)/([A])` or `log""(0.10)/([A])=0.3986" or "(0.10)/([A])="ANTILOG "(0.3986)=2.5" or "[A]=(0.10)/(2.5)=0.04" M"`

Discussion

89768.	A first order reaction has a rate constant 1.15xx10^(-3)s^(-1). How long will 5 g of this reactant take to reduce to 3 g?
Answer» Solution :GIVEN : `[A]_(0)=5 g, [A]=3 g, k=1.15xx10^(-3)s^(-1)`. Applying first order KINETICS equation and substituting the values, we get `t=(2.303)/(t)"log"([A]_(0))/([A])=(2.303)/(1.15xx10^(-3)s^(-1))"log"(5g)/(3G)=2.00xx10^(3) (log 1.667)s` `=2.0xx10^(3) xx 0.2219s=443.8s`

Discussion

89769.	A first order reaction has a rate constant , 1.15xx10^(-3)s^(-1). How long will 5 g of this reactant take to reduce to 3 g ?
Answer» SOLUTION :Here , `[R]_0 = 5 G , [R]3 g` `k = 1.15 xx10^(-3)s^(-1)`. As the reaction is of first order, `k = (2.303)/(t)log.([R]_0)/([R])` `t = (2.303)/(overset(..)uoverset(..)uoverset(..)uxx-3-1)log.(5g)/(3g)=2.00xx10^3(log1.667)s ` `2.0xx10^3xx0.2219s = 443.85=444s`

Discussion

89770.	A first order reaction has a rate constant 1.15 xx 10^(-3)s^(-1). How long will 5g of this reacant take to reduce to 3g ?
Answer» SOLUTION :Data, `[R]_(0)(5)/(M)' [R] (3)/(M)` M is molar MASS of the REACTANT ` k = 1.15 xx 10^(-3)s^(-1)` Formula: `k = (2.303)/(t) log ""([R]_(0))/([R])"" `t = (2.303)/(1.15 xx 10^(-3)) log ""((5)/(M))/((3)/(M))` ` t = (2.303)/(1.15 xx 10^(-3)) xx 0.2218 "" t 444 s.`

Discussion

89771.	A first order reaction has a rate constant 1.15xx10^(-3)s^(-1). How long will 5 g of this reactant take to reduce to 3 g ?
Answer» Solution :Here, `[A]_(0)=5G.[A]=3g,k=1.15xx10^(-3)s^(-1).` As the REACTION is of 1st ORDER, `t=(2.303)/(t)log""([A]_(0))/([A])=(2.303)/(1.15xx10^(-3)s^(-1))log""(5g)/(3g)=2.00xx10^(3)(log1.667)s` `=2.0xx10^(3)xx0.2219s=443.8s~~444s.`

Discussion

89772.	A first order reaction has a rate constant of 1.15xx10^(-3) s^(-1). How long will 5 g of this reactant take to reduce to 3 g ?
Answer» SOLUTION :t = 444 SECONDS

Discussion

89773.	A first order reaction has a half life period of 69.3 sec. At 0.10mol litre^-1 reactant concentration, rate will be:
Answer» `10^-4Msec^-1` `10^-3Msec^-1` `10^-1Msec^-1` `6.93X10^-1Msec^-1` ANSWER :B

Discussion

89774.	A first order reaction completes 25% of the reaction in 100 mins. What are the rate constant and half life values of the reactions ?
Answer» SOLUTION :`k=2.8773xx10^(-3)"MIN"^(-1), t^(1//2)=240.85" MINS"`

Discussion

89775.	A first order reaction completes 25% of the reaction in 100 mins. What are the rate constant and half life values of the reaction?
Answer» SOLUTION :`k=2.8773xx10^(-3)"MIN"^(-1), t^(1//2)=240.85" MINS"`

Discussion

89776.	A first order reaction ArarrBrequires activation energy of 70"kJ.mol"^(-1)When a 20% solutions of A was kept at25^(@)Cfor 20 min, 25% decomposition took place. What will be the percentage decomposition in the same time in a 30% solution maintained at 40^(@)C? Assume that activation energy remains constant in this range of temperature.
Answer» Solution :From the Arrhenius equation, `logk=logA-(E_(a))/(2.303RT)` or, `LOG.(k_(2))/(k_(1))=(E_(a))/(2.303R)[(1)/(T_(1))-(1)/(T_(2))]` `thereforelog.(k_(2))/(k_(1))=(70xx10^(3))/(2.303xx8.314)[(1)/(298)-(1)/(313)] " or, " (k_(2))/(k_(1))=3.872"" ...[1]` For a FIRST order reaction, `k=(2.303)/(t)log.(a)/(a-X)` Given x = 0.25a, `thereforea-x=(a-0.25a)=0.75a " at " t=20min` `thereforek_(1)=(2.303)/(20)log.(a)/(0.75a)=0.014386"min"^(-1)""...[2]` Putting the value of `k_(1)`into equation [1] , we have `k_(2)=3.872xx0.014386=0.05571"min"^(-1)` For a first order reaction, `k_(2)=(2.303)/(t)log.(a)/(a-x)` or,`0.05571=(2.303)/(20)log.(a)/(a-x)` [x = decrease in concentration of the reactant ] or, `log.(a)/(a-x)=(0.05571xx20)/(2.303)` or, `log.(a)/(a-x)=0.48381 " or, " x=0.6717a` Hence , the percentage DECOMPOSITION`=(0.6717a)/(a)xx100` `=67.17%`

Discussion

89777.	A first-order reaction Ato B , requires activation energy of 70 kJ "mole"^(-1) . When a 20% solution of A was kept at 25^@C for 20 minutes, 25% decomposition took place. What will be the per cent decomposition in the same time in a 30% solution maintained at 40^@C ? Assume that activation energy remains constant in this range of temperature.
Answer» SOLUTION : First calculate `k_25`, then `k_49`. Again find out per cent DECOMPOSITION at `40^@C` 67.17%

Discussion

89778.	A first order reaction, A to B , requires activation energy of 70 kJ mol^(-1). When a 20%solution of A was kept at 25^(@)C for 20 minutes , 25% decomposition took place. What will be the per cent decomposition in the same time in a 30% solution maintained at 40^(@)C ? Assume that activation energy remains constant in this range of temperature.
Answer» ANSWER :`67.21%`

Discussion

89779.	A first order reaction, 3A (g) to 2B(g) + 3C (g) Starting with pure A the pressure developed after 4 min and infinite time is 4.5 atm and 5 atm respectively then calculate time (in minutes) required for 87.5% decomposition of A .
Answer» ANSWER :6

Discussion

89780.	A first order nuclear reaction is half completed in 45 minutes. How long does it need 99.9% of the reaction to be completed
Answer» 5 hours 7.5 hours 10 hours 20 hours Solution :`K = (0.693)/(0.75 HR) = (2.303)/(t) "log" (a)/(a - 0.999a)` `= (2.303)/(t) log 10^(3) = 7.5 hr`

Discussion

89781.	A first order gas reaction A_(2)B_(2)(g) to 2A (g) +2B(g) at the temperature 400^(@)C has the rate constantk=2.0xx10^(-4)s^(-1). What percentage of A_(2)B_(2) is decomposed on heatingfor 900 seconds?
Answer» Solution :For the first ORDER reaction, use the relation `k=(2.303)/(t)"log"([R]_(0))/([R])` Given that `k=2.0xx10^(-4)s^(-1) and t =900s` Substituting the values in the first order equation, we have `2.0xx10^(-4) =(2.303)/(900) "log" ([R]_(0))/([R])` or `"log"([R]_(0))/([R])=(2.0xx10^(-4)xx900)/(2.303)=0.0781` TAKING antilogarithm of the above `([R]_(0))/([R])=1.197 or ([R])/([R]_(0))=(1)/(1.197)=0.8354` or Fraction decomposed `= 1-0.8354=0.1646` Percentageof `A_(2)B_(2)` decomposed `=0.1646xx100=16.46`.

Discussion

89782.	A firstordergas -phasereaction hasan energyofactivationof 240 kJ mol^(-1). Iffrequencyfactorof thereactionis 1.6 xx 10^(13) s^(-1) , calculateitsrateconstant at 600 K .
Answer» Solution :Given :ENERGYOF activation` = E_(a) 240 kJ mol^(-1) = 240 XX 10^(3) J mol^(-1)` Frequencyfactor = A = `1.6 xx 10^(13) s^(-1)` Temperature`= T= 600 K ` Rateconstant= k = ? By Arrheniusequation, `k= A xx e^(-E_(a)//RT) :." In" k = " In" A - (E_(a))/(RT)` `:. 2.303log _(10) k= 2.303 log_(10)A- (E_(a))/(RT)` `log_(10)k =log_(10) A - (E_(a))/(2.303 RT) = log_(10)1.6 xx 10^(13)- (240 xx 10^(3))/(2.303 xx 8.314 xx 600)` `= 13. 204- 20.9= - 7.696` `:. k= AL-7 .696= ALbar(8).304 = 2.01 xx 10^(-8) s^(-1)`

Discussion

89783.	(A) First ionisation potential of helium is highest (R) The electron gain enthalpy of helium is taken as zero
Answer» Both (A) and (R) are TRUE and (R) is the correct explanation of (A) Both (A) and (R) are true and (R) is not the correct explanation of (A) (A) is true but (R) is false Both (A) and (R) are false Answer :B

Discussion

89784.	A first order decomposition reaction takes 40 minutes for 30% decomposition. Calculate its t_(1//2) value.
Answer» Solution :`K=(2.303)/(t)log""(a)/(a-x)=(2.303)/(40" MIN")log""(a)/(a-0.30a)=0.0089" min"^(-1),t_(1//2)=(0.693)/(k)=(0.693)/(0.0089" min"^(-1))=77.86" min."`

Discussion

89785.	A firework gave bright crimson light. It is probably a salt of:
Answer» Ca Sr Ba Mg Answer :B

Discussion

89786.	A fire of lithium, sodium and potassium can be extinguished by
Answer» `H_(2)O` Nitrogen `CO_(2)` Asbestoes BLANKET Solution :Carbon dioxide does not help in BURNING , also it forms carbonate with ALKALI METALS.

Discussion

89787.	A finely divided state of catalyst is more efficient because in this state
Answer» more ACTIVE CENTRES are formed more surface area is a available more energy is sotred in the catalyst none of the above Answer :A::B

Discussion

89788.	(a) Find the value of deltaG^circ at 25^circ for the following electrochemical cell. Cu\|Cu^2+ 9(1M) Ag ( 1M) Ag Write the equation of anodic and cathodic reaction occur during rusting of iron. (i) At Anode :- 2Fe to 2Fe^2+ +4e^- (ii) At Cathod :- O_2+4H ^+ 4e ^- to 2H_2O.
Answer» SOLUTION :` E_(CELL)^E_(cell)^CIRC` = `E^circ R-E_(l)^circ = 0.34= 0.46V` `Delta G^circ` =`nF E^circ =-2xx 96,487 xx 1.14 = -2 xx 96,487 xx 0.46 = -88768.04j`. (i) At Anode :`- 2Fe to 2Fe^2+ +4E` (ii) At Cathod :- `O_2 + 4H^+ 4e to 2H_2O`.

Discussion

89789.	A film of pyridine filled a rectangular wire loop in which one side could be moved. Given that the wire loop is 8.53 cm wide and that a force of 6.48 xx 10^(-3) N is needed to move the side, determine the value of the surface tension. What is the work necessary to stretch the film a distance of 0.1 cm?
Answer» Solution :The FORCE is related to the surface tension y and the width `l` by: `F=2l gamma` `THEREFORE gamma =F/(2l) = (6.48 XX 10^(-3))/(2 xx 8.53 xx 10^(-2))` `=3.79 xx 10^(-2) NM^(-1)` As `gamma =("work done" (W))/("area (A)")` `w = gamma xx A = 2(3.79 xx 10^(-2)) xx (8.53 xx 10^(-2)) xx (0.1 xx 10^(-2))` `=6.5 xx 10^(-6)` J

Discussion

89790.	A few students of class XII were playing in the stadium under the supervision of their coach. All of a sudden, Nitin fell down. The coach immediately rushed towards Nitin and examined the injury. He opend up his sporting kit and pulled out a small plastic pack. He squeezed hard the pack and placed it on the injured portion of his skin. Nitin got some conform and relief. Later coach took Nitin to hospital for further treatment. Read the above passage and answer the following questions : (i) What values were expressed by coach towards Nitin ? (ii) Name the chemical contained in the plastic pack. What is this pack called and how does it work ? (iii) Name two other uses of the chemical contained in the small plastic pack.
Answer» Solution :(i) The coach expressed the human values and helped NITIN who was in pain. He provided the first aid and took him to the hospital for further treatment. (ii) The chemical contained in the small plastic pack is ammonium nitrate and the small pack is called the instant cold pack. An instant cold plastic pack has two compartments , one contains solid ammonium nitrate `(NH_(4)NO_(3))` and the other contains water. When the plastic pack is squeezed hard, the wall between the two compartment breaks. The `NH_(4)NO_(3)` mixes with water, endothermic reaction occurs and the temperature drops. The cold pack is then placed on the injured tissues. It takes out the heat from the injured tissues, reduces SWELLING and bleeding by decreasing the capillary size of the blood vessels under the bruised skin. As a result, the person gets relief from the pain. (III) Ammonium nitrate is a USEFUL fertilizer because plants usually take up nitrogen in the form of nitrate and ammonium ions. Ammonium nitrate mixed with fuel oil is one of the most deadly explosives used for coal mining and demolition of buildings. During explosion, oxidation-reduction reaction occurs , `NH_(4)^(+)` ion is the reducing agent while `NO_(3)^(-)` ion is the oxidising agent. As a result, fuel oil `(C_(17)H_(36))` gets oxidized and a large volume of gases is produced which causes explosion : `52 NH_(4)NO_(3)(s) + C_(17)H_(36)(l) rarr 52 N_(2)(g) + 17 CO_(2)(g) + 122 H_(2)O(g)` But unfortunately, in the recent years, this explosive has been used in terrorist attacks in India and around the world.

Discussion

89791.	A few drops of conc. H_2 SO_4 is added to the solution of FeSO_4 in water to
Answer» Convert `Fe^(3+)` ions (if any) into `Fe^(2+)` ions prevent salt hydrolysis Both (A) and (B) None of these SOLUTION :A few drops of CONC. `H_(2)SO_(4)` added to a solution of `FeSO_(4)[ "or" Al_(2)(SO_(4))_(3)]` a salt hydrolysis and thus helps in getting a clear solution.

Discussion

89792.	A fertilizer contains20% nitrogen by mass. To proivided a fruit tree with an equivalent of 1 kg of nitrogen, the quantity of fertilizer required is
Answer» 20 kg 0.20 kg 0.05 kg 5 kg Solution :Mass % = `("(mass of nitrogen )")/("(mass of FERTILISER)")XX100` `RARR =(1 kg)/x xx100 rArr x=100/20=5 kg `

Discussion

89793.	A fertile soil is likely to have a pH of :
Answer» 3 9 `6-7` 14 Answer :C

Discussion

89794.	A ferromagnetic substance becomes the permanent magnet when it is placed in the magnetic field because
Answer» all the domains get ORIENTED in the direction of MAGNETIC field all the domains get oriented in the direction OPPOSITE to the direction of magnetic field domains get oriented randomly domains are not AFFECTED by magnetic field Solution :Ferromagnetic substances can be magnetised PERMANENTLY.

Discussion

89795.	A ferromagnetic substance becomes a permanent magnet when it is placed in a magnetic field because ....
Answer» all the domains GET ORIENTED in the direction of magnetic FIELD. all the domains get oriented in the direction opposite to the direction of magnetic field. domains get oriented randomly. domains are not affected by magnetic field. ANSWER :A

Discussion

89796.	A ferromagnetic substance becomes permanent magnet when it is placed in a magnetic field because ………………
Answer» DOMAINS get oriented randomly. Domains are not AFFECTED by MAGNETIC field. All the domains get oriented in the DIRECTION opposite to the direction of magnetic field. All the domains get oriented in the direction of magnetic field. Answer :D

Discussion

89797.	A: Ferrocene may prepared by the reaction of Grignard reagent ferrous chloride R: Ferrocene is so and it bonded complex
Answer» If both Assertion & REASON are true and the reason is the correct explanation of the assertion, then MARK (1) If both Assertion & Reason are true but the reason is not the correct explanation of the assertion, then mark (2). ) If Assertion is true statement but Reason is false, then mark (3). If both Assertion and Reason are false STATEMENTS, then mark (4) ANSWER :C

Discussion

89798.	(a) Ferric oxide crystallizes in a hexagonal close - packed array of oxide (O^(2-)) ions with two out of every three octahedral holes occupied by iron ions. What is the formula of ferric oxide ?(b) In cadium iodide every alternate octahedral hole in an HCP array of iodide I^(-) ions is occupied by a cadmium ion. What is the formula of cadium iodide.
Answer» Answer :(a) `Fe_(4)O_(6)` or `Fe_(2)O_(3)`,(b) `CaCl_(3)` or `Cdl_(2)`

Discussion

89799.	(A): [Fe(edta)] complex is octahedral in shape (R): edta ligand is hexadentate and forms six bonds with the metal atom undergoing d^2sp^3 hybridization
Answer» Both A & R are TRUE, R is the correct EXPLANATION of A Both A & R are true, R is not correct explanation of A A is true, R is FALSE A is false, R is true Answer :A

Discussion

89800.	(A): FeO is basic in character. (R): Oxides of transition metals are basic when metal is in lower oxidation state.
Answer» Both (A) and (R) are TRUE and (R) is the CORRECT EXPLANATION of (A) Both (A) and (R) are true and (R) is not the correct explanation of (A) (A) is true but (R) is FALSE Both (A) and (R) are false Solution :Both (A) and (R) are true and (R) is the correct explanation of (A)

Discussion

Explore topic-wise InterviewSolutions in .

A first order reaction is 25% complete in one hour . At the end of two hours the extent of reaction is :

A first order reaction is 25% complete in 30 minutes. Calculate the specific reaction rate.

A first order reaction is 25% complete in 40 minutes. Calculate the value of rate constant. In what time will the reaction be 80% completed?

A first order reaction is 20% completed in 10 minutes . Calculate the time taken for the reaction to go to 80% completion.

A first order reaction is 20% completed in 10 minutes. Calculate the time taken for the reaction to go to 80% completion.

A first order reaction is 20% complete in 20 minutes. Calculate the time taken for the reaction to go to 80% completion.

A first order reaction is 20% complete in 10 minutes. Calculate time taken for reaction to go to 80% .

A firstorderreactionis 20% complete in 10 min. How long it takes to complete 80% ?

A first order reaction in 50% completed in 30 minutes at 27^@C and in 10 minutes at 47^@C . Calculate the reaction rate constant at 27^@C and the energy of activation of eh reaction in kJ mol^(-1)

A first order reaction has rate constant of k=1.15xx10^(-3)s^(-1). How long will it take for 6 g of reactant to reduce to 3 g ?

A first order reaction has rate constant 1.15xx10^(-3) s^(-1) How long will 5 g of this reactant take to reduce to 3g?

A firstorderreactionhasactivationenergy22.8 kJat 300 K . Ifthe rateconstantof thereation is1.42 xx 10^(2) min^(-1) Calculatethe frequencyfactor .

A first order reaction has a specific rreaction rate of 10^(-2)s^(-1). How much time will it take for 20 g of the reactant of reduce to 5 g ?

A first order reaction has a specific reaction rate of 10^(-2)"sec"^(-1). How much time will it take for 20 g of the reactant to reduce to 5g ?

A first orderreactionhasa specific reaction rate of 10^(-2) S^(-1). How much timewill it take for 20 gof thereactanttoreduceto 5 g?

A first order reaction has a specific reaction rate of 10^(-2) sec^(-1) . How much time will it take for 20 g of the reactant to reduce to 5 g

A first order reaction has a rate constant of 0.0051" min"^(-1). If we begin with 0.10 M concentration of the reactant, what concentration of the reactant will remain in the solution after 3 hours ?

A first order reaction has a rate constant 1.15xx10^(-3)s^(-1). How long will 5 g of this reactant take to reduce to 3 g?

A first order reaction has a rate constant , 1.15xx10^(-3)s^(-1). How long will 5 g of this reactant take to reduce to 3 g ?

A first order reaction has a rate constant 1.15 xx 10^(-3)s^(-1). How long will 5g of this reacant take to reduce to 3g ?

A first order reaction has a rate constant 1.15xx10^(-3)s^(-1). How long will 5 g of this reactant take to reduce to 3 g ?

A first order reaction has a rate constant of 1.15xx10^(-3) s^(-1). How long will 5 g of this reactant take to reduce to 3 g ?

A first order reaction has a half life period of 69.3 sec. At 0.10mol litre^-1 reactant concentration, rate will be:

A first order reaction completes 25% of the reaction in 100 mins. What are the rate constant and half life values of the reactions ?

A first order reaction completes 25% of the reaction in 100 mins. What are the rate constant and half life values of the reaction?

A first order reaction, 3A (g) to 2B(g) + 3C (g) Starting with pure A the pressure developed after 4 min and infinite time is 4.5 atm and 5 atm respectively then calculate time (in minutes) required for 87.5% decomposition of A .

A first order nuclear reaction is half completed in 45 minutes. How long does it need 99.9% of the reaction to be completed

A first order gas reaction A_(2)B_(2)(g) to 2A (g) +2B(g) at the temperature 400^(@)C has the rate constantk=2.0xx10^(-4)s^(-1). What percentage of A_(2)B_(2) is decomposed on heatingfor 900 seconds?

A firstordergas -phasereaction hasan energyofactivationof 240 kJ mol^(-1). Iffrequencyfactorof thereactionis 1.6 xx 10^(13) s^(-1) , calculateitsrateconstant at 600 K .

(A) First ionisation potential of helium is highest (R) The electron gain enthalpy of helium is taken as zero

A first order decomposition reaction takes 40 minutes for 30% decomposition. Calculate its t_(1//2) value.

A firework gave bright crimson light. It is probably a salt of:

A fire of lithium, sodium and potassium can be extinguished by

A finely divided state of catalyst is more efficient because in this state

(a) Find the value of deltaG^circ at 25^circ for the following electrochemical cell. Cu|Cu^2+ 9(1M) Ag ( 1M) Ag Write the equation of anodic and cathodic reaction occur during rusting of iron. (i) At Anode :- 2Fe to 2Fe^2+ +4e^- (ii) At Cathod :- O_2+4H ^+ 4e ^- to 2H_2O.

A few drops of conc. H_2 SO_4 is added to the solution of FeSO_4 in water to

A fertilizer contains20% nitrogen by mass. To proivided a fruit tree with an equivalent of 1 kg of nitrogen, the quantity of fertilizer required is

A fertile soil is likely to have a pH of :

A ferromagnetic substance becomes the permanent magnet when it is placed in the magnetic field because

A ferromagnetic substance becomes a permanent magnet when it is placed in a magnetic field because ....

A ferromagnetic substance becomes permanent magnet when it is placed in a magnetic field because ………………

A: Ferrocene may prepared by the reaction of Grignard reagent ferrous chloride R: Ferrocene is so and it bonded complex

(A): [Fe(edta)] complex is octahedral in shape (R): edta ligand is hexadentate and forms six bonds with the metal atom undergoing d^2sp^3 hybridization

(A): FeO is basic in character. (R): Oxides of transition metals are basic when metal is in lower oxidation state.