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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 501. |
The cathode reaction during the charging of a lead `-` acid battery leads to theA. Formation of `PbSO_(4)`B. Reduction of `Pb^(2+)` to `Pb`C. Formation of `PbO_(2)`D. Deposition of `Pb` at the anode. |
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Answer» Correct Answer - b Refer section |
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| 502. |
In the lead-acid battery during charging, the cathode reaction isA. formation of `PbO_(2)`B. formation of `PbSO_(4)`C. reduction of `Pb^(2+)` to `Pb`D. decomposition of Pb at the anode |
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Answer» Correct Answer - C During chargin, the reaction that occur are At cathode: `PbSO_(4)(s)+2e^(-)toPb(s)+SO_(4)^(2-)(aq)` At anode: `PbSO_(4)(s)+2H_(2)OtoPbO_(2)(s)+SO_(4)^(2-)(aq)+4H^(+)(aq)+2e^(-)` Thus, `Pb^(2+)` ions of `PbSO_(4)` are reduced to Pb on the cathode while `PbSO_(4)` is oxidized to `PbO_(2)` at anode. (Remember: Anode is the electrode on which oxidation takes place, i.e., loss of electrons and cathode is the electrode on which reduction takes place i.e., gain of electrons further, the electrode which acts as anode during discharge acts as cathode durig charging and vice-versa. |
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| 503. |
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 beA. 0.8B. 0.87C. 0.9D. 0.97 |
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Answer» Correct Answer - D `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%` |
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| 504. |
The resistance of an aqueous solution containing `0.624g` of `CuSO_(4).5H_(2)O` per `100 cm^(3)` of the solution in a conductance cell of cell constant `153.7` per meter iss 520 ohms at 298K. Calculate the moalr conductivity. `(CuSO_(4).5H_(2)O = 249.5)` |
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Answer» Correct Answer - `118.2 mho cm^(2) mol^(-1)` `K=(153.7)/(520)=2.96 xx 10^(-1) "mho m"^(-1)` `=2.96 xx 10^(-3) "mho cm"^(-1)` `lamda_(m)=(K xx 1000)/(c)=(2.96 xx 10^(-3) xx1000)/(2.5 xx 10^(-2))` `1.184 xx10^(2)` `=118.4 "mho cm"^(2) "mol"^(-1)`. |
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| 505. |
How many c.c. of oxygen will be liberated by 2 ampere current flowing for 3 minutes and 13 seconds through acidulated water ?A. 11.2 c.c.B. 33.6 c.c.C. 44.8 c.c.D. 22.4 c.c. |
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Answer» Correct Answer - D `W_(O_(2)) = (E xxi xx t)/(F) = (8 xx 2 xx 193)/(96500) = 0.032` g `(W)/(M) = (V)/(22400) therefore V = (0.032 xx 22400)/(32) = 22.4` c.c. |
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| 506. |
An electric current of 1 amp is passed through acidulated water for 160 minutes and 50 seconds. What is the volume of the hydrogen liberated at the anode (as reduced to NTP)?A. 1.12 litreB. 2.24 litreC. 11.2 litreD. 22.4 litre |
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Answer» Correct Answer - A `V = (l tV_(e))/(96500)` `= (1 xx 9650 xx 11.2)/(96500) = 1.12 "litre"` |
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| 507. |
I. Conductance of electolyte solution increases with temperature II. Resistivity is reciprocal of molar conductivity of electrolyte. III. Cell constant has unit `cm^(-1)`A. if all the statement are correctB. if II and III are correctC. if I and III are correctD. if only II is correct |
| Answer» Correct Answer - C | |
| 508. |
What volume of `O_(2)` at NTP liberated by 5 A current flowing for 193 and through acidulated water ?A. 56 mLB. 112 m LC. 158 mLD. 965 m L |
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Answer» Correct Answer - A `W = ( I xx t xx E)/(F) = (5 xx 193 xx 8)/(96500) = 0.08 g ` At NTP , volume of 32 g of `O_(2)` = 22400 mL `therefore 0.08` g = x ` x = (22400 xx 0.08)/(32) = 56 ` mL |
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| 509. |
Given the standard potential of the following at `25^(@)C`. `MnO_(2) rarr Mn^(3+), " "lE^(c-)=0.95V` `Mn^(3+) rarr Mn^(2+)," "E^(c-)=1.51V` The standard potential of `MnO_(2)rarr Mn^(2+)` isA. `-0.56V`B. `-2.46V`C. `-1.23V`D. `1.23V` |
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Answer» Correct Answer - d `2e^(-) +MnO_(2) rarr Mn^(2+), E^(c-) =?` `Delta G =-2FE^(c-)` `i. e^(-)+MnO_(2) rarr Mn^(3+), DeltaG_(1)^(c-)=-Fe^(c-)=-0.95F` `ii. e^(-) +Mn^(3+) rarr Mn^(3+), DeltaG_(2)^(c-)=-FE^(c-)=-1.51E^(c-)` Equations `(i) +(ii)` give the desired equation. `:. -2FE^(-) =(-0.95-1.51)F` `E^(c-)1.23V` |
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| 510. |
Find the standard electrode potential of `MnO_(4)^(c-)|MnO_(2).` The standard electrode potential of `MnO_(4)^(c-)|Mn^(2+)=1.51V` and `MnO_(2)|MnO_(2)|Mn^(2+)=1.23V`. |
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Answer» `{:(MnO_(4)^(-)+8H^(+)+5e^(-)rarrMn^(2+)+4H_(2)O,E^(@) =1.51 "volt"),(MnO_(2)+4H^(+)+2e^(-)rarr Mn^(2+) +2H_(2)O , E^(@) = 1.23 "volt"):}/("Subtract" " "MnO_(4)^(-)+4H^(+)+3e^(-)rarrMnO_(2) +2H_(2)O, E^(@) = ?)` `E^(@) = (5 xx 1.51 -2 xx 1.23)/(3) = (7.55 -2.45)/(3) = (5.09)/(3) = 1.70` volt |
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| 511. |
Find the standard electrode potential of `MnO_(4)^(c-)|MnO_(2).` The standard electrode potential of `MnO_(4)^(c-)|Mn^(2+)=1.51V` and `MnO_(2)|MnO_(2)|Mn^(2+)=1.23V`.A. `-0.33V,` the reaction will not occurB. `-0.33V,` the reaction will occurC. `-2.69V,` the reaction will not occurD. `-2.69V,` the reaction will occur |
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Answer» Correct Answer - c `Mn^(2+)+e^(c-) rarr Mn," "E^(c-)=-1.18V` `2Mn^(3+)rarr 2Mn^(3+)+2e^(c-), " "E^(c-)=-1.51V` `ulbar(3Mn^(2+)rarr Mn+2Mn^(3+))` `E_(cell)^(c-)=E_(red)^(c-)+E_(o x i d)^(c-)` `=-1.18+(-1.51)=-2.69V` Negative `EMF` reflects non`-`spontaneous cell reaction. |
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| 512. |
The resistance of a conductivity cell filled with `0.1 M KCl` solution is `100 Omega`. If `R` of the same cell when filled with `0.02 M KCl` solution is `520 Omega`, calculate the conductivity and molar conductivity of `0.02 M KCl` solution. The conductivity of `0.1 M KCl `solution is `1.29 S m^(-1)`.A. `5xx10^(3)`B. `5xx10^(2)`C. `5xx10^(-4)`D. `5xx10^(-3)` |
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Answer» Correct Answer - c `R=(1)/(k)xx(l)/(a) ((l)/(a)=` cell constant `)` For `0.2M` solution `50=(1)/(k)xx(l)/(a)implies50=(1)/(1.4)xx(l)/(a)` `(l)/(a)=70m^(-1)` For `0.5M` solution `280=(1)/(k) xx70` `k=(1)/(4)Sm^(-1)` `^^_(m)((kxx1000)/(M))(10^(-2)m)^(3)` `^^_(m)=(1)/(4)xx((1000)/(M))(10^(-2)m)^(3)` `=(1)/(4)xx(1000)/(0.5)xx10^(-6)` `=500xx10^(-6)=5xx10^(-4)Sm^(2) mol^(-1)` |
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| 513. |
The resistance of a conductivity cell filled with `0.1 M KCl` solution is `100 Omega`. If `R` of the same cell when filled with `0.02 M KCl` solution is `520 Omega`, calculate the conductivity and molar conductivity of `0.02 M KCl` solution. The conductivity of `0.1 M KCl `solution is `1.29 S m^(-1)`.A. `124 xx 10^(-4) S m^2 "mol"^(-1)`B. ` 1240 xx ^(-4) "mol"^(-1)`C. ` 1. 24xx 10^(-4) S m^2 "mol"^(-1)`D. ` 12.4 xx 10^(-4) S m^(2) "mol"^(-1)` |
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Answer» Correct Answer - A `C= 0 . 1 M, R = 100 oh m` `k= 1.29 ohm^(-1) m^(-1)` ` :. K= 1/R xx 1/a` `:. 1/a = k. R= 1.29 = 1.29 xx 100` ohm `C= 0.02 m, R= 530` `k= 1/R xx l/a = 1/(520) xx 129 = 0.248 "ohm"^(-1) m^(-1)` k decreases eith dilution . `:. C = 0.02` and not `0.2` Also `Lambda=k xx 1/(M("in" m//L))` `K+ 1/(M xx 10^(-3) (m//m^3)) =(0.2 48 xx 1)/( 0.02 xx 10^(-3))` ` = 12 .4 xx 10^(-3) Sm^2 "mol"^(-1) = 124 xx 10^(-4) S m^2 "mol"^(-1)`. |
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| 514. |
STATEMENT-1: If SRP of substance is -0.5V, then reduction of substance is possible only in basic medium . SRP of water is -0.8274V and at reduction potential is zero at pH=7A. If both the statements are TRUE and STATEMENTS-2 is the correct explantion of STATEMENTS-4B. If both the statements are TRUE but STATEMENTS-2 is NOT the correct explanation of STATEMENTS-6C. If STATEMENTS-1 is TRUE and STATEMENTS-2 is FALSED. If STATEMENT-1 is FALSE and STATEMENT-2 is TRUE |
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Answer» Correct Answer - C |
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| 515. |
STATEMENT-1: The voltage of mercury cell remains constant for longer period of time. STATEMENT-2: It is because net cell reaction does not involve ions.A. If both the statements are TRUE and STATEMENTS-1 is the correct explantion of STATEMENTS-2B. If both the statements are TRUE but STATEMENTS-1 is NOT the correct explanation of STATEMENTS-2C. If STATEMENTS-1 is TRUE and STATEMENTS-2 is FALSED. If STATEMENT-1 is FALSE and STATEMENT-2 is TRUE |
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Answer» Correct Answer - A |
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| 516. |
At 298 K, The specific conductivity of a saturated solution of silver chloride in water is `2.30xx10^(-5)Scm^(-1)`. Calculate its solubiility in `gL^(-1)` at 298 K. Given `lambda_(m)^(@)(Ag^(+))"and"lambda_(m)^(@)(CI^(-))`are61.9 and 76.3 S `cm^(2)mol^(-1)` respectively. |
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Answer» Correct Answer - `2.382xx10^(-3) gL^(-1)` `Lambda_(m(AgCl))^(oo)=Lambda_(m(Ag^(+)))^(oo)+Lambda_(m(Cl^(-)))^(oo)=(61.9+76.3)" S "mol^(-1)cm^(2)=138.2" S " mol"^(-1)cm^(2)` Solubility (in mol `L^(-1)`) `=(kxx1000" cm"^(3)L^(-1))/(Lambda_(m)^(oo))=((2.3xx10^(-6)" S "cm^(-1))xx(1000 cm^(3)L^(-1)))/((138.2" S mol"^(-1)"cm"^(2)))` `=0.0166xx10^(-3) mol L^(-1)` Solubility (in g`L^(-1)`) `=(0.0166xx10^(-3)" mol "L^(-1))xx("Molar mass of" AgCl)` `=(0.0166xx10^(-3)" mol "L^(-1))xx(143.5" g mo"l^(-1))=2.382xx10^(-3)" g "L^(-1)`. |
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| 517. |
The specific conductivity of a saturated solution of silver chloride is `2.30 xx 10^(-6) m ho cm^(-1) " at " 25^(@)C`. Calculate the solubility of silver chloride at `25^(@)C " if " lamda_(Ag^(+)) = 61.9 m ho cm^(2) mol^(-1) and lamda_(Cl^(-)) = 76.3 m ho cm^(2) mol^(-1)` |
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Answer» Let the solubility of AgCl be S gram mole per litre Dilution `= (1000)/(S)` `Lamda_(AgCl)^(oo) = lamda_(Ag^(+)) + lamda_(Cl^(-))` `= 61.9 + 76.3` `= 138.2 m ho cm^(2) mol^(-1)` Sp. Conductivity `xx` Dilution `= Lamda_(AgCl)^(oo) = 138.2` `2.30 xx 10^(-6) xx (1000)/(S) = 138.2` `S = (2.30 xx 10^(-3))/(138.2) = 1.66 xx 10^(-5)` mol per litre `= 1.66 xx 10^(-5) xx 143.5 g L^(-1) = 2.382 xx 10^(-3) g L^(-1)` |
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| 518. |
50 ml of a buffer of 1 M NH3 and 1 M NH4 + are placed in two volatic cells separately. A current of 3.0 amp is passed through both cells for 10 min. If electrolysis of water takes place as2H2O + O2 + 4e– →4OH– (R.H.S.) 2H2O →4H+ + O2 + 4e– (L.H.S.)then pH of the (1) L.H.S. will increase (2) R.H.S. will increase (3) R.H.S. will decrease (4) Both side will increase |
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Answer» Correct option (2) R.H.S. will increase Explanation: Because of the reactions of electrolysis, [H+] concentration will decrease as a result of which pH will increase. |
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| 519. |
A direct current of 3.0 amp (efficiency 75% ) was passed through 400 ml 0.2 M `Fe_(2)(SO_(4))_(3)` solution for a period of 60 min The resulting solution in cathode chamber was nalaysed by titrating against acidic `KMnO_(4)` solution 20 ml of `KMnO_(4)` required to reach the end point determine the molarity of `KMbO_(2)` solution |
| Answer» Correct Answer - 0.84 M | |
| 520. |
When an aqueous solution of `H_(2)SO_(4)` is electrolysed, the ion discharged at anode isA. `H^(-)`B. `OH^(-)`C. `SO_(4)^(2-)`D. `O^(2-)` |
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Answer» Correct Answer - B `OH^(-),4OH^(-)rarrO_(2)+2H_(2)O+2 e^(-)` |
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| 521. |
When molten lithium chloride (LiCl) is electrolysed, lithium metal is formed at the cathode. If current efficiency is 75% then how many grams of lithium are liberated when 1930 C charge pass through the cell? (Atomic mass of Li=7)A. 0.105B. 0.12C. 0.28D. 0.24 |
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Answer» Correct Answer - A Correct Answer : "A" 0.105
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| 522. |
A `35%` solution of `LiCl` was electrolyzed by using a `2.5 A` current for `0.8 h. ` Assuming the current efficiency of `90%` , find the mass of `LiOH` produced at the end of electrolysis. `(` Atomic mass of `Li=7)` |
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Answer» Number of Faradays `=(90)/(100)((2.5xx0.8xx3600)/(96500))` `=0.067F` `Ew=Mw` of `LiOH=7+16+1=24g mol^(-1)` `(n` factor `=1)` `W_(LiOH)=EwxxEq` `=24 xx 0.067=1.61g` |
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| 523. |
Explain the factors influencing rate of reaction. |
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Answer» Factors affecting rate of a reaction : The rate of a chemical reaction depends on the following. i) Concentration of reactants : The rate of a chemical reaction is directly proporational to the concentration of the reactants. As the concentration of reactants increases the number of molucules increase, collisions between molecules increases as a result number of fruitful collisions increases, hence rate of reaction increases. ii) Nature of reactants :Reactions between ionic compounds are fast when compared to reactions between the covalent compunds. In case of covalent molecules breaking and formation of bonds involved hence the reactions between covalent molecules are very slow. iii) Temperature : As temperature increases, the rate of reaction increases. For every `10^(@)C` rise of temperature the rate of reaction is almost doubled. At high temperatures maximum number of molucules are activated. The collisions between activated molecules are fruitful, hence the rate of reaction increases. iv) Catalyst : In the presence of catalyst also the rate of reaction increases. Catalyst takes the reaction in a new path having low activation energy. |
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| 524. |
Four moles of electrons were transferred from anode to cathode in an experiment on electrolysis of water. The total volume of the tow gases (dry and at `STP)` produced will be approximately (in litres)A. 22.4B. 44.8C. 67.2D. 89.2 |
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Answer» Correct Answer - C From 4 moles of electron wt. of `H_(2)` formed `=4g =44.8`litre at STP wt. of `O_(2)` formed `=4xx8=32g` =22.4 litre at STP Total volume =`44.8+22.4=67.2` litres. |
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| 525. |
What is the EMF of the cell? `Zn(s)|Zn^(2+)+(0.1M)||Sn^(2+)+(0.001M)||Sn(s)`. Given `E^(@)Zn^(2+)//Zn=0.76V,E_(Sn^(@2+)//Sn=-0.14V`A. 0.62VB. 0.56VC. 1.12VD. 0.31V |
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Answer» Correct Answer - B `Zn(s)+Sn^(2+)rarrZn^(2+)(aq)+Sn(s)` `E_("cell")=E_("cell")^(@)-(0.059)/(n)"log"([Zn^(2+)])/([Sn^(2+)])` `=[0.14-(-0.76)]-(0.059)/(2)"log"(0.1)/(0.001)` `=+0.62-(0.059)/(2)"log"100=+0.56V` |
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| 526. |
A concentration cell is shown belowA. `Ag(s)|AgNO_(3)(0.01M)||AgNO_(3)(0.001M)|Ag(s)` The EMF of the cell will b eB. 59.00voltsC. 5.90voltsD. 0.059volts |
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Answer» Correct Answer - C `E.M.F.=(2.303RT)/(nF)"log"(C_(2))/(C_(1))` `=(0.59)/(1)"log"(0.01)/(0.001)=0.059V` |
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| 527. |
Calculate the potential of a `Zn -An^(2+)` electrode in which the molarity of `Zn ^(2+)` is `0.001M.` Given that `E_(Zn^(2+)//Zn)^(0)=-0.76V` `R=0.314JK^(-1) mol ^(-1), F= 96500C mol ^(-1).` |
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Answer» Given electrode `Zn|Zn^(+2)(0.001m)E_(Zn^(+2)//Zn)^(0) =-0.76V` Nernst equation is `E=E^(0) +(2.303RT)/(nF) log [M^(n+)]` Given `R=8.314J//K.`mole `F = 96500c//`mole `E= E^(0) +(0.059)/(2)log C` `=-0.76+(0.059)/(2)log 0.001` `=-0.76 -(0.059)/(2)xx3` `= -0.76-0.0295xx3` `=-0.76-0.0885 =-0.8485V` |
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| 528. |
The EMF of the following cell is 0.86 volts `Ag|AgNO_(3)(0.0093M)||AgNO_(3)(xM)|Ag`. The value of x will beA. 82.8MB. 2.28MC. 0.228MD. 1.14M |
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Answer» Correct Answer - C `E=(0.059)/(1)"log"(C_(2))/(C_(1))` `0.086=(0.059)/(1)"log"(x)/(0.0093)` `x=0.228M` |
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| 529. |
Given the reaction for the distance of a cobalt-cadmium battery `Co(OH)_(3)+Cd+H_(2)O to Co(OH)_(2)+Cd(OH)_(2)` Which species is oxidised during the discharge of the battery ?A. `Co^(3+)`B. `Co^(2+)`C. CdD. `Cd^(2+)` |
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Answer» Correct Answer - C (c ) Cd is oxidised to `Co^(2+)`. |
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| 530. |
What is the potential for the cell `Cr|Cr^(3+)(0.1M)||Fe^(2+)(0.01M)|Fe` `E^(@)Cr^(3+)// Cr=-0.74V`, `E^(@)Fe^(2+)//Fe=-0.44V`A. `+0.2606V`B. `+0.5212V`C. `+0.1303V`D. `-0.2606V`. |
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Answer» Correct Answer - A Anode half reaction : `CrrarrCu^(3+)+3e^(-)` Cathode half reaction : `Fe^(2+)+ 2Cr^(3+)+6e^(-)` or `Fe^(2)+6e^(-)rarr3Fe` cell reaction `2Cr+3Fe^(2+)rarr2Cr^(2+)+3Fe,n=6` Now `E_("cell")=E_("cell")^(@)-(0.05916)/(6)"log"([Cr^(3+)]^(2))/([Fe^(2+)]^(3))` `=[-0.44-(-0.74)]-(0.05916)/(6)"log"((0.1)^(2))/((0.01)^(3))` `=0.3-0.0394=+0.2606V` |
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| 531. |
Calculate the potential of the following half`-`cells`|` cells`:` `a. Cr|Cr^(3+)(0.1 M)||Fe^(2+)(0.01M)|Fe` Given `: E^(c-)._(Cr^(3+)|Cr)=-0.74V E^(c-)._(Fe^(2+)|Fe)=-0.44V` `b. 6e^(c-)+BrO_(3)^(c-)(aq)+3H_(2)OrarrBr^(c-)(aq)+6OH(aq)` Given `:E^(c-)._((BrO_(3)^(c-)|Br^(c-)))=0.61V,` `[BrO_(3)^(c-)]=2.5xx10^(-3)M,[Br^(c-)]=5.0xx10^(-3)M,pH=9.0` `c. Ag|Ag^(o+)(0.1M)||Cl^(c-)(0.02M)|Cl_(2)(g)(0.5atm)|Pt` Given `E_((Ag^(o+)|Ag))=0.80V,E^(c-)((Cl_(2)|2Cl^(c-))=1.36V` `d. NO_(3) ^(c-)(aq)+2H^(o+)(aq)+e^(-)rarrNO_(2)+H_(2)O` Given `: E^(c-)._(NO_(3)^(c-)|NO_(2)=0.78V` What will be the reductino potential of the half cell in neutral solution ? Assuming all the other species to be at unit concentration. |
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Answer» `a.` Anode reaction `:` `[Cr(s) rarr Cr^(3+)(0.1M)+cancel (3e^(-))]xx2` Cathode reaction `:` `[2e^(-)+Fe^(2+)(0.01M) rarr Fe(s)] x3` Cell reaction `:` `ulbar(2Cr(s)+3Fe^(2+) rarr 2Cr^(3+)+3Fe(s))` `E^(c-)._(cell)=(E^(c-)._(reduction))_(c)-(E^(c-)._(reduction))_(a)` `=-0.44-(-0.74)=0.3V` Using Nernest equation `:n_(cell)=6)` `E_(cell)=E^(c-)._(cell)-(0.059)/(6)log.([Cr^(3+)]^(2))/([Fe^(2+)]^(3))` `=0.3V-(0.06)/(6)log.((0.2)^(2))/((0.01)^(3))` `=0.3V-0.01log.(0.1xx0.1)/(0.01xx0.01xx0.01)` `=0.3V-0.01[log 10^(4)]` `=0.3V-0.01xx4=0.26 V` `b.` Using Nernst equation` : (n_(cell)=6,` take`0.059~~0.06)` `E_(cell)=E^(c-)._(cell)-(0.06)/(6)log .([Br^(c-)][overset(c-)(O)H]^(6))/([BrO_(3)^(c-)])` `(pH=9.0,pOH=14-9=5,[overset(c-)(O)H]=10^(-5)M.` Acitivity of `H_(2)O=1)` `E_((BrO_(3).^(c-)|Br.^(c-)))=E^(c-)._((BrO_(3).^(c-)|Br.^(c-)))` `-(0.06)/(6) log .(5.0xx10^(-3)xx(10^(-5)M)^(6))/(2.5xx10^(-3)M)` `=0.61V-0.01(log2xx10^(-30))` `0.61V-0.01(0.3-30),,,,(log 2 ~~0.3)` `=0.61V-0.01xx(-29.7)` `=0.61V+0.297 V~~0.907V` `c.` Anode reaction `: 2Ag(s) rarr 2 Ag ^(o+)(0.1M)+cancel(2e^(-))` Cathode reaction `:` `Cl_(2)(g)(0.5atm)+cancel(2e^(-))rarr 2Cl^(c-)(0.02M)` Cell reaction `:` `ulbar(2Ag(s)+Cl_(2)(g)(0.5atm)rarr2Ag^(o+)(0.1M)+2Cl^(c-)(0.02M))` `E_(cell)=(E^(c)._(reduction))_(c)-(E^(c-)._(reduction ))_(a)` `=1.36V-0.80V=0.56V` Using Nernst equation `: (n_(cell=2` activity of `Ag=1,` take `0.059~~0.06)` `E_(cell)=E^(c-)._(cell)-(0.06)/(2)log.([Ag^(o+)]^(2)[Cl^(c-)]^(2))/((p_(Cl_(2)(g))))` `E_(cell)=0.56V-0.03 (log.((0.1)^(2)(0.2)^(2))/(0.5 atm))` `=0.56V-0.03(log 8xx10^(-4))` `=0.56V-0.03(log 2^(3)+log 10^(-4))` `=0.56V-0.03(3 log 2 - 4) " "(log 2 ~~ 0.3)` `=0.56V-0.03(3xx0.3-4)` `=0.56 V+0.03xx3.1V` `=0.653V` `d. pH` of natural solution `=7:. [H^(o+)]=10^(-7)M` `n_(cell)=1,` concentration of `[NO_(2)]=[H_(2)O]=[NO_(3).^(c-)]=1` `E_((NO_(3).^(c-)|NO_(2)))=E^(c-)._((NO_(3).^(c-)|NO_(2)))-(0.059)/(1)log.([NO_(2)][H_(2)O])/([NO_(3).^(c-)][H^(o+)]^(2))` `=0.78V-0.059log.(1)/((10^(-7))^(2))` `=0.78V-0.059[log10^(14)]` `=0.78V-0.059xx14` `=0.78 V-0.059xx14` ltbr. `=-0.046V` |
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| 532. |
Calculate the emf of the cell at `25^(@)C` `Cr|Cr^(3+)(0.1M)||Fe ^(2+)(0.01M) |Fe,` given that `E_(Cr^(3+)//Cr )^(0)=-0.74V and E_(Fe ^(2+)//Fe )^(0)=-0.44V` |
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Answer» Given cell is `Ce|Cr _((0.1M))^(+3)||Fe_((0,01M))^(+2)|Fe` `E^(0) of Cr ^(+3)//Cr =-0.74V` `E^(0)of F^(+2)//Fe =-0.44V` `E_(Cr^(3)//C)=E^(0) +(0.059)/(3)log _(10) [Cr^(+3)]` `=-0.74+(0.054)/(3)log 0.1` `=-0.76V` `E_(Fe^(+2)//Fe) =-0.44+(0.059)/(2)log 0.01` `=-0.44-0.059=-0.499V` EMF of cells `=E_(RHS) -E_(LHS)` `=(-0.499) -(-0.76)=0.261V` |
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| 533. |
Calculate the EMF of the cell containing chromium and cadmium electrodes (Given `E_(Cr^(3+)//Cr)^(@)=-0.74V,E_(Cd^(2+)//Cd)^(@)=-0.40V`) |
| Answer» Correct Answer - 0.34V | |
| 534. |
given `E_(S_(2)O_(8)^(2-)//SO_(4)^(2-)^(@)=2.05V` `E_(Br_(2)//Br^(-))^(@)=1.40V` `E_(Au^(3+)//Au)^(@)=1.10V` ,brgt `E_(O_(2)//H_(2)O)^(@)=1.20V` Which of the following is the strongest oxidizing agent ?A. `Cr^(3+)`B. `Mn^(2+)`C. `MnO_(4)^(c-)`D. `Cl^(c-)` |
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Answer» Correct Answer - c As per data mentioned,` MnO_(4)^(c-)` is strongest oxidizing agent as it has maximum standard reduction potential value. |
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| 535. |
The standard reduction potential at `25^(@)C` of the reaction `2H_(2)O+2e^(-)hArrH_(2)+2overset(Θ)(O)H` is `-0.8277V`. Calculate the equilibrium constant for the reaction. `2H_(2)OhArrH_(3)O^(o+)+overset(Θ)(O)H` at `25^(@)C` . |
| Answer» Correct Answer - `1.04xx10^(-14)` | |
| 536. |
`E^(c-)` for `Cr^(3+)+3e^(-) rarr Cr `and `Cr^(3+)+e^(-) rarr Cr^(2+)` are `-0.74 V` and `-0.40V`, respectively, `E^(c-)` for the reaction is `Cr^(+2)+2e^(-) rarr Cr`A. `-0.91V`B. `+0.91V`C. `-1.14V`D. `+0.34V` |
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Answer» Correct Answer - a `Cr^(3+)+3e^(-) rarr Cr,` `Cr^(3+)+e^(-) rarr Cr^(2+), ` `ulbar(Cr^(2+)+2e^(-) rarr Cr,)` `E^(c-)._(3)=(n_(1)E^(c-)._(1)-nE^(c-)._(2))/(n_(3))` `=([3xx(-0.74)]-[(-1)xx(-0.40)])/(2)=-0.91V` |
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| 537. |
For the half-cell raction, `2H_(2)O+2e^(-)toH_(2)+2OH^(-),E^(@)=-0.8277V` at 298K. Autoprotolysis constant of water calculted from this value will beA. `1xx10^(-10)`B. `1xx10^(-12)`C. `1xx10^(-13)`D. `1xx10^(-14)` |
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Answer» Correct Answer - D Autoprotolysis constant of `H_(2)O=[H^(+)][OH^(-)]=K_(w)` Redn. Half reaction: `2H_(2)O+2e^(-)toH_(2)+2OH^(-),E^(@)=-0.8277V` `underset("Ox. Half reaction:"H_(2)to2H^(+)+2e^(-),E^(@)=0.0)` Cell reaction: `2H_(2)OhArr2H^(+)+2OH^(-),E^(@)=-0.8277V` `E=E^(@)-(0.0591)/(2)"log"[H^(+)]^(2)[OH^(-)]^(2)` At equilibrium, E=0. Hence, `E^(@)=(0.0591)/(2)log[H^(+)]^(2)[OH^(-)]^(2)` `therefore-0.8277=0.0591log[H^(+)][OH^(-)]^(2)` `=0.0591logK_(w)` or `logK_(w)=14` or `K_(w)=10^(-14)`. |
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| 538. |
`E^(@) ` of two reactions are given below : `Cr^(3+) + 3e rarr Cr, E^(@) = -0.74 V` `OCl^(-) + H_(2)O + 2e rarr Cl^(-) + 2OH^(-), E^(@) = 0.94 V` What will be the `E^(@)` for ? `3OCl^(-) + 2Cr + 3H_(2)O rarr 2Cr^(3+) + 3Cl^(-) + 6OH^(-)`A. `1.68 V`B. `0.20 V`C. `-1.68 V`D. `-0.20 V` |
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Answer» Correct Answer - A `E^(@) = E_(OP_(Cr))^(@) + E_(RP_(OCl^(-)))^(@)` `= 0.74 + 0.94 = 1.68 V` |
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| 539. |
During the electrolysis of aqueous `CuSO_(4)` solutions using `Pt` elecrodesA. `Cu` is deposited at the anodeB. `O_(2)(g)` is liberated at cathodeC. `Cu` is deposited at the cathodeD. `H_(2)(g)` is liberated at the cathode |
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Answer» Correct Answer - C The species we should consider for half reactions are `Cu^(2+)(aq.), SO_(4)^(2-)(aq.)`, and `H_(2)O` Possible cathode half-reactions are `Cu^(2+)(aq.)+2e^(-) rarr Cu(s), E^(@) = 0.34 V` `2H_(2)O(l)+2e^(-) rarr H_(2)(g)+2OH^(-)(aq.): E^(@) = -0.83 V` Because the copper electrode potential is much larger than the reduction potential of water, `Cu^(2+)` is reduced and `Cu` is deposited at the cathode. `2SO_(4)^(2-)(aq.) rarr S_(2)O_(8)^(2-)(aq.)+2e^(-) E^(@) = 2.01 V` `2H_(2)O(l) rarr O_(2)(g)+4H^(+)(aq.)+4e^(-), E^(@) = 1.23V` Because the second electrode potential is lower, `H_(2)O` is oxidized and `O_(2)(g)` is liberated at the anode. However, if electrolysis is carried out using copper electrodes then another possible anode reaction is `Cu(s) rarr Cu^(2+)(aq.)+2e^(-) E^(@) = 0.34 V` Thus, copper anode will dissolve or `Cu` anode will loose mass. |
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| 540. |
Electrolysis of molten magnesium chloride results in the formation of 0.4 g magnesium. Calculate the volume of chlorine liberated during the reaction at STP. When an electrolytic cell consisting of molten `AlCl_(3)` and molten `NaCl` are connected in series with the above cell, calculate the amount of aluminium and sodium deposited at cathode. |
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Answer» `MgCl_(2) rarr Mg^(+2) + 2C^(-)` For the deposition for 12 g of Mg (equivalent weight of magnesium) one faraday is required `:.` 12 g requires 1 faraday 0.4 g requires ? `= (0.14)/(12) = 0.03` Faraday 1 Faraday gives 11.2 L of `Cl_(2)` at STP, therefore, 0.03 faraday gives ? `= 0.336 1 L " of " Cl_(2)` at STP 1 faraday required to deposit 9 g of Al `:.` 0.03 faraday required to deposit ? = 0.27 g 1 faraday required to deposit 23 g of Na `:.` 0.03 faraday required to deposit ? `= 0.03 xx 23 = 0.69 g` |
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| 541. |
What would be the magnitude of `EMF` of the following cell`:` at `25^(@)C` ? The ionization constant of acetic acid, `K_(a)~ 10^(-5)`, while that of formic acid, `K_(a)~10^(-4)`A. `0.0295V`B. `0.059V`C. `-0.059V`D. `-0.0295V` |
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Answer» Correct Answer - b At anode `:` `0.1M` acetic acid `+0.1M` sodium acetate `implies` Acidic buffer. `:. pH_(a)=pK_(a)+log [(sal t)/(Aci d)]" "(pK_(a(CH_(3)COOH))=5)` `=5+log((0.1)/(0.1))=5` At cathode `:` `0.1M` formic acid `+0.1M` sodium formate `implies` Acidic buffer. `:.pH_(c)=pK_(a)+log((0.1)/(0.1))" "[pK_(a(HCOOH))=4]` `pH_(c)=4` `E_(cell)=-0.059(pH_(c)-pH_(a))=-0.059*4-5)=0.059V` |
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| 542. |
The standard reduction potential at `25^(@)C` of the reaction `2H_(2)O+2e^(-) rarr H_(2)+2OH^(-)` is -0.8277 volt. Calculate the equuilibrium constant for the reaction, `2H_(2)O hArr H_(3)O^(+)+OH^(-)` at `25^(@)C` |
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Answer» Correct Answer - `~~10^(14)` `H_(2)O+e^(-) rarr ½H_(2)+OH^(-) ("Cathode"), E^(@)=-0.8277` volt `H_(2)O + ½H_(2) rarr H_(3)O^(+)+e^(-) " (Anode)", E^(@)=0` `E^(@)` for the cell `=-0.8277` volt Apply now `E^(@)=.0591/n [log K]" "(n=1)` |
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| 543. |
Electrolysis of aqueous solutions of which of the following substances results in only the decomposition of water ?A. Potassium chlorideB. Zinc sulphateC. Potassium hydroxideD. Sodium phosphate |
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Answer» Correct Answer - c,d Electrolysis of `ZnSO_(4)` is done for the purification of `Zn` using `Al` cathode on which it gets reduced `(` factual statement `).` |
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| 544. |
On passing `0.5 mool` of electrons through `CuSO_(4)` and `Hg_(2)(NO_(3))_(2)` solutions in series using inert electrodesA. `0.5 mol ` of `Cu` is depositedB. `0.5 mol` of `Hg` is depositedC. `0.125 mol` of `O_(2)` is producedD. `0.5 mol ` of `O_(2)` is produced |
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Answer» Correct Answer - b,c `Cathode : [{:(Cu^(2+),+,H^(o+),,Hg_(2)^(2+)),(Hg_(2).^(2+),+,2e^(-),rarr,2Hg),(implies0.5F,,=, 0.5,mol e Hg):}` `Anode : [{:(SO_(4)^(2-),,overset(o+)(O)H,,NO_(3)^(c-),),(2overset(o+)(O)H,,rarr,O_(2),2H_(2)O,4e^(-)),(implies4F,,-=,1molimplies, 0.5F,-=(1)/(8)mol e O_(2)):}` |
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| 545. |
A dilute aqueous solution of `CuSo_4` is electrolysed using platinum electrods. The products at the anode and cathode are:A. `O_2,H_2`B. `H_2,O_2`C. `O_2,Cu`D. `S_2O_8^(2-),H_2` |
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Answer» Correct Answer - C |
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| 546. |
The standard reduction potential at `25^(@)C` for the reaction, volt. The equilibrium constant for the reaction `:` isA. `10^(-12)`B. `10^(-14)`C. `10^(-11)`D. `10^(-11)` |
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Answer» Correct Answer - b Anode`: H_(2)implies2H^(o+)+2e^(-)," "E^(c-)=0V` Cathode`: 2H_(2)O+2e^(-) rarr H_(2)+2overset(c-)(O)H,E^(c-)=-0.8277V` `2H_(2)OhArr2H^(o+)2overset(Θ)(O)H` Use `:=(E^(c-)._(cell)xxn_(cell))/(0.059)=logK_(w)` |
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| 547. |
When an aqueous solution of `CaCl_(2)` is electrolyzed using inert electrods, which of the following is `(` are `)` true ?A. Calcium deposites on cathode.B. Calcium deposits an anodeC. Chloride is liberated on anodeD. Calcium hydroxide precipitates near cathode on prolonged hydrolysis |
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Answer» Correct Answer - c,d `Cathode : [{:((X)Ca^(2+),+,2e^(-),rarr,Ca),(or,,,,),((sqrt())2H_(2)O,+,2e^(-), rarr,H_(2)O+2overset(o+)(O)H):}` `Anode : [{:((X)4overset(o+)(O)H,rarr,O_(2)+2H_(2)O,+,4e^(-)),(or,,,,),((sqrt())2Cl^(c-),rarr,Cl_(2), +,2e^(-)):}` |
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| 548. |
The standard electrode of a metal ion `(Ag|Ag^(o+))` and metal `-` insoluble salt anion `(Ag|AgCl|Cl^(c-))` are related asA. `E_(Ag^+"|"Ag)^(@)=E_(Cl(-)"|"AgCl"|"Ag)^(@)+(RT)/FInK_(sp)`B. `E_(Cl(-)"|"AgCl"|"Ag)^(@)=`E_(Ag^+"|"Ag)^(@)+(RT)/FInK_(sp)`C. `E_(Cl(-)"|"AgCl"|"Ag)^(@)=`E_(Ag^+"|"Ag)^(@)-(RT)/FIn([Cl^-])/K_(sp)`D. `E_(Cl(-)"|"AgCl"|"Ag)^(@)=`E_(Ag^+"|"Ag)^(@)-(RT)/FIn([Cl^-])/K_(sp)` |
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Answer» Correct Answer - B |
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| 549. |
A galvanic cell is composed of two hydrogen electrods, one of which is a standard one. In which of the following solutions should the other electrode be immersed to get maximum e.m.f:A. `0.1 M HCl`B. `0.1 M CH_(3)COOH`C. `0.1 M H_(3) PO_(4)`D. `0.1 M H_(2)SO_4` |
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Answer» Correct Answer - D `H_(2)SO_(4)` will furnish maximum `H^(+)` . |
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| 550. |
A galvanic cell is composed of two hydrogen electrods, one of which is a standard one. In which of the following solutions should the other electrode be immersed to get maximum e.m.f:A. 0.1M HClB. 0.1 M `H_2SO_4`C. 0.1 M `NH_4OH`D. 0.01 M HCOOH |
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Answer» Correct Answer - C |
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