94420 + Interview Questions in Chemistry in Class 12 Page 1793 InterviewSolution

89601.	A gas container observes Maxwell distribution of speeds. If the number of molecules between the speed 5 and 5.1 km sec^(-1) at 25^(@)C be 'n', what would be the number of molecules between this range of speed if the total number of molecules in the vessel are doubled ?
Answer» `N^(2)` NATURE of the GAS `n//2` 2n Answer :D

Discussion

89602.	A gas consists of 5 molecules with a velocity of 5ms^(-1) , 10 molecules with a veclocity of 3ms^(-1) and 6 molecules with a velocity of 6ms^(-1). The ratio of average velocity and rms velocity of the molecles is 9.5xx10^(-x) Where x is……….
Answer» Solution :Average VELOCITY `=(5xx5+10xx3+6xx6)/21=4.33` RMS velocity `=SQRT((5xx25+10xx9+6xx36)/21)=4.53` `(V_(av))/(V_("rms"))=0.955=9.5xx10^(-1)` `x=1`

Discussion

89603.	A gas Cl_(2) at 1 atm is bubble through a solution containing a mixture of 1 M Br^(-1) and 1 M F^(-1) at 25^(@)C if the reduction potential is FgtClgtBr, then:
Answer» Cl will oxidise BR and not F Cl will oxidise F and not Br Cl will oxidise both Br and F Cl will REDUCE both Br and F. Answer :A

Discussion

89604.	A given gas cannot be liquefied if its temperature is
Answer» FORCES of ATTRACTION are zero under ordinary conditions Froces of attraction are LOW under ordinary conditions Forces of attraction are eitherhigh or low under ordinary conditions FORCE of attraction are high under ordinary conditions. Answer :A

Discussion

89605.	A gas can be liquefied if:
Answer» forces of ATTRACTION are low under ORDINARY conditions forces of attraction are HIGH under ordinary conditions forces of attraction are ZERO under ordinary conditions forces of attraction either high or low under ordinary conditions Answer :A::B::D

Discussion

89606.	A gas can be liquefied at temperature T and pressure P provided
Answer» <P>`T=T_(c)` and `Plt P_(C)` `TltT_(c)` and `PgtP_(c)` `TgtT_(C)` and `PgtP_(C)` `TgtT_(C)` and `PltP_(C)` Solution :The two very important CONDITIONS for liquefying a gas are 1. Temperature should be LOWER than the critical temperature `(TltT_(c))` 2. PRESSURE should be greater than the critical pressure `(PgtP_(C))`.

Discussion

89607.	A gas bulb of 1-litre capacity contains 2 xx 10^21molecules of nitrogen exerting a pressure of 7.57 xx 10^3Nm^(-2) . Calculate the root-mean-square (rms) speed and the temperature of the gas molecules. If the ratio of the most probable speed to the root-mean-square speed is 0.82, calculate the most probable speed for these molecules at this temperature.
Answer» Solution :Use `C = sqrt((3pV)/(MM)) , V = 10^(-3) m^3 , m = (28 xx 10^(-3))/(Av. Const)` `494.18ms^(-1) , 274.15K, 405.22 ms^(-1)`

Discussion

89608.	A gasbelievedto bethecauseofexplosionin coalmines is :
Answer» `CH_(4)` `C_(2) H_(6)` `CH_(3)H_(8)` `C_(4)H+_(10)` ANSWER :A

Discussion

89609.	A gas at low temperature does not react with the most of compounds. It is almost inert and is used to create inert atmosphere in bulbs. The combustion of thisgas exceptionally and endothermic reaction based on the given information, we can conclude that the gas is
Answer» oxygen nitrogen carbon mono-oxide hydroge Answer :B

Discussion

89610.	A gas at a pressure of 5 atm is heated from 0^(@)C to 546^(@)C and simultaneously compressed to one third of its original volume. Hence final pressure is:
Answer» 10atm 30atm 45atm 5atm Solution :`(PV)/(T)=` CONSTANT `(5xxV)/(273)=(ptxx(V)/(3))/(819)` `rArrP_(r)=45atm`

Discussion

89611.	A gas at low temperature does not react with most of the compounds. It is almost inert and is used to create inert atmosphere in bulbs. The combustion of this gas is exceptionally an endothermic reaction. Based on the given information, we can conclude that the gas is :
Answer» oxygen nitrogen carbon monoxide hydrogen Answer :B

Discussion

89612.	A gas at 10^(@)C occupies a volume of 283ml. If it is heated to 20^(@)C keeping the pressure constant, the new volume will be
Answer» 293 ml 283ml 566ml 586ml Solution :`V_(1) = 283 ml`, `T_(1) = 273+ 10 = 283 K` `V_(2) = ?` `T_(2) = 273 + 20 = 293K` At constant P, `( V_(1))/( T_(1)) = ( V_(2))/( T_(2))` or `V_(2) = ( V_(1)T_(2))/( T_(1))` `V_(2) = ( 283 xx 293)/( 283) = 293ml`

Discussion

89613.	A gas at 1 atm is bubbled through a solution containing 1MY^(-)and1MZ^(-) " at " 25^(@)C.Ifthe reduction potential of ZgtYgtZ,then
Answer» Ywill oxidize X and not Z Y will oxidise Z and not X Y will oxidize both x and Z Y will REDUCE both X and Z. Solution :SINCE the reduction potential of Y is greater than that of X but LOWER thanthat of Z. THEREFORE , Y will oxidize X but BOT Z.

Discussion

89614.	A gas absorbs a photon of 355nm and emits at two wavelengths. If one of the emissions is at 680 nm. The other is at:
Answer» `743 nm 518nm 1035 nm 325 nm Solution :`(HC)/(LAMDA)=(hc)/(lamda_(1))+(hc)/(lamda_(2))` or `1/(lamda)=1/(lamda_(1))+1/(lamda_(2))` `1/355=1/680+1/(lamda_(2))` or `1/(lamda_(2))=1/355-1/680=(680-355)/(355xx680)` `1/(lamda_(2))=1.346xx10^(-3):.lamda_(2)=743nm`

Discussion

89615.	A gas absorbs 35.8 cal of heat and underrgoes an expansion from 1L to 1.5 L against the external pressure of 1.2 atm. The change in the internal energy of the system is which of the following? [Use 1 atm =1.01325 xx 10^(5) Nm^(-2), 1 cal = 4.18 J)
Answer» `-210.68` J `+210.68` J `-89.12` J `+88.849` ANSWER :D

Discussion

89616.	A gas absorbs 120 J of heat and expands against an external pressure of 1.10 atm from a volume of 0.52 to 2.0 L. The change in internal energy is (1 L atm = 101.3 J) :
Answer» `-167.1 J` `-47.1 J` `-287.1 J` 287.1 J ANSWER :B

Discussion

89617.	A gas ‘X' is passed through water to form a saturated solution. The aqueous solution on treatment with silver nitrate gives a white precipitate. The saturated aqueous solution also dissolves magnesium ribbon with evolution of a colourless gas 'Y’: Identify ‘X' and 'Y’.
Answer» `X=CO_(2)`,`Y=Cl_(2)` `X=Cl_(2)`,`Y=CO_(2)` `X=Cl_(2)`,`Y=H_(2)` `X=H_(2)`,`Y=Cl_(2)` Solution :`Cl_(2)+H_(2)toHOCl+HCl` `2HCL+"Mg"toMgCl_(2)+H_(2)UARR` `Hence,(C( is the correct answer

Discussion

89618.	A galvanic is set up from a zinc bar weighing 50 g and 1.0 litre, 1.0 M CuSO_(4) solution. How long would the cell run assuing it delivers a steady current of 1.0 ampere?
Answer» Solution :We have to calculate the TIME in which all Cu or Zn will be consumed on producing a steady current of 1 ampere. `ZntoZn^(2+)+2e^(-)` `underline(Cu^(2+)+2e^(-)TOCU"")` `Zn+Cu^(2+)toZn^(2+)+Cu` 1L, 1M `CuSO_(4)` CONTAIN `CuSO_(4)=1` mole=1 mole of `Cu^(2+)` `Zn=50g=(50)/(65.4)`mole `lt1` Thus, Zn will be consumed first completely `ZntoZn^(2+)+2e^(-)` 1 mole Zn (65.4g) requires charge=`2xx96500C` `therefore50g` Zn will require charge`=(2xx96500)/(65.4)xx50C=147554C` Q=Ixxtthereforet=(Q)/(I)=(147554)/(1)sec=41` hrs

Discussion

89619.	A galvix cell is constructed using the redox reaction, (1)/(2)H_(2_(g))+AgCl_((s))hArrH_((aq))^(+)+Cl_((aq))^(-)+Ag_((s)) It is represented as
Answer» A) `Pt\|H_(2(G))\|HCl_(("soln"))\|\|AgNO_(3" (soln)")\|Ag` B) `Ag\|AgCl_((s))\|KCl_(("soln"))\|\|HCl_(("soln")),H_(2(g))\|Pt` C) `Pt\|H_(2(g))\|KCl_(("soln"))\|AgCl_((s))\|Ag` D) `Pt\|H_(2(g)),HCl_(("soln"))\|\|AgCl_((s))\|Ag` Solution :`Pt\|H_(2)(g)\|HCl_(("soln"))\|\|AgCl(s)\|Ag` `H_(2)` undergoes OXIDATION and AgCl `(Ag^(+))` undergoes reduction.

Discussion

89620.	A galvanic cell is set up from a zinc bar weighing 50g and 1.0 litre, 1.0M, CuSO_(4) solution. How long would the cell run, assuming it delivers a steady current of 1.0 ampere
Answer» 48hrs 41hrs 21hrs 1hr Answer :C

Discussion

89621.	A galvanic cell was operated under almost ideally reversible conditions at a current of 10^(-16) amp. How long would it take to deliver 1 mole of electrons ?
Answer» `9.65 XX 10^(20)s` 96,500 s `9.65 xx 10^(-12) s` NONE of these Answer :A

Discussion

89622.	A galvanic cell is constructed with Ag//Ag^(+) as one elecrode and Fe^(2+)//Fe^(3+) as the second electrode. Calculate the concentration of Ag^(+) ions at which the E.M.F. of the cell will be zero at equimolar concentrations of Fe^(2+) and Fe^(3+) ions. Given E_(Ag^(+)//Ag)^(@)=0.80V,E_(Fe^(3+)//Fe^(2+))^(@)=0.77V
Answer» Solution :Given electrode potential values show that EMF of the CELL will be positive only if reduction occurs at silver electrode. Therefore, the cell REACTION will be `Fe^(2+)+Ag^(+)toFe^(3+)+Ag` Hence, the cell may be represented as: `Fe^(2+)\|Fe^^(3+)\|\|Ag^(+)\|Ag` `thereforeE_(cell)^(@)=E_(Ag^(+)//Ag)^(@)-E_(Fe^(3+)//Fe^(@+))^(@)=0.80-0.77=0.03` V Applying nernst equation to the above reaction, `E_(Cell)=E_(Cell)^(@)-(0.0591)/(1)"log"([Fe^(3+)][Ag])/([Fe^(2+)][Ag^(+)])` But `[Fe^(2+)]=[Fe^(3+)]` (given) and [Ag]=1. `thereforeE_(cell)=E_(Cell)^(@)-0.0591"log"(1)/([Ag^(+)]) therefore0=0.03+0.0591log[Ag^(+)]` or `log[Ag^(+)]=-(0.03)/(0.0591)=-0.5076=1.4924` or `[Ag^(+)]`=antilog `overline(1).4924=3.1xx10^(-1)=0.31M`

Discussion

89623.	A galvanic cell is constructed of two hydrogen electrodes, one immersed in a solution with H^+ at 1M and the other in 1M KOH. Calculate E_(cell). If 1M KOH solution is replaced by 1M NH_3, will E_(cell) be higher or lower than in 1M KOH?
Answer» SOLUTION :0.83 V LOWER

Discussion

89624.	A galvanic cell is composed of two hydrogen electodes, one of which is a standard one . In which of the following solutins the other electrode be immersed to get maximum emf:
Answer» `0.1MHCI` `0.1MCH_3COOH` `0.1 MH_3PO_4` `0.1MH_2SO_4` ANSWER :D

Discussion

89625.	A galvanic cell is composed of two hydrogen electrodes one of which is a standard one. In which of the following solutions should the other electrode be immersed to get maximum emf ?
Answer» 0.1 M HCl `0.1 M CH_(3)COOH` `0.1 M H_(3)PO_(4)` `0.1 M H_(2)SO_(4)` Solution :(d) `0.1 M H_(3)PO_(3)` will liberate maximum `H^(+)` ions in solution.

Discussion

89626.	A galvanic cell has electrical potential of 1.1 V. if an opposing potential of 1.1 V is applied to this cell, what will happen to the cell reaction and current flowing through the cell?
Answer» SOLUTION :No CURRENT will FLOW and the REACTION STOPS.

Discussion

89627.	A galvanic cell has electrical potential of 1.1 V. If an opposing potential of 1.1 V is applied to this cell, what will happen to the cell reaction and the current flowing through the cell ?
Answer» SOLUTION : The REACTION STOPS and no current FLOWS through the circuit.

Discussion

89628.	A galvanic cell has electrical potential of 1.1 V. If an opposing potential of 1.1 V is applied to this cell, what will happen to the cell reaction and current flowing through the cell ?
Answer» Solution :* When `E_(CELL)^(Theta)=1.1V and E_(EXT)^(Theta)=1.1V` pass in opposite direction, `E_(cell)^(Theta)=E_(ext)^(Theta)` * Then cell reaction stops MEANS, there is no CHEMICAL reaction in cell and no CURRENT is pass through the cell.

Discussion

89629.	A galvanic cell has electrical potential of 1.1 V . If an opposing potential of 1.1 Vis applied to this cell, what will happen to the cell reaction and current flowing through the cell ?
Answer» Solution :when an oppsoing POTENTIAL of 1.1 V is APPLIED to a galvanic cell having ELECTRICAL potential of 1.1 V then cell REACTION stops completely and no CURRENT will flow through the cell.

Discussion

89630.	A galvanic cell has E_(cell)^(@)=1.5" V". If an opposing potential of 1.5" V" is applied to cell, what will happen to the cell reaction and current flowing through the cell ?
Answer» SOLUTION :On applying opposing potential, reaction CONTINUES to take PALCE till the MAGNITUDE of opposing e.m.f. reaches the value of 1.1" V". After this , reaction STOPS and no further reaction will take place.

Discussion

89631.	A galvanic cell consists of a metallic zinc plate immersed in 0.1 M Zn(NO_(3))_(2) solution. Calculate the emf of the cell at 25^(@)C. Write the chemical equations for the electrode reactions and represent the cell. (Given : E_(Zn^(2+)//Zn)^(@) = - 0.76 volt and E_(Pb^(2+)//Pb)^(@) = - 0.13 volt)
Answer» Answer :`ZN rarr Zn^(2+) + 2e^(-),""Pb^(2+)+2e^(-) rarr Pb` `Zn\|underset(0.1 M)(Zn^(2+))\|\|underset(0.02 M)(Pb^(2+))\|Pb` `E_(cell)^(@)=0.63" VOLT"E_(cell)=0.609" volt"`

Discussion

89632.	A galvanic cell consists of a metallic zinc plate immersed in 0.1 M Zn (NO_(3))_(2) solution and metallic plate of lead in 0.02M Pb (NO_(3))_(2) solution.Calculte the emf of the cell. Write the chemical equation for the electrode reactions and represent the cell. (Given E_(Zn^(2+)//Zn)^(@)=-0.76V,E_(Pb^(2+),Pb)^(@)=-0.13V)
Answer» ANSWER :0.6094V, `Zn+PB^(2+)TOZN^(2+)+Pb;Zn\|Zn^(2+)(0.1M)\|\|Pb^(2+)(0.02M)\|Pb`

Discussion

89633.	A galvanic cell consists of
Answer» CADMIUM cell two HALF cells three half cells lead accumulator Answer :B

Discussion

89634.	A galvanic cell after use is recharged by passing current through it. What type of cell is it? Give an example.
Answer» SOLUTION :SECONDARY CELL , Ex: Lead STORAGE battery.

Discussion

89635.	(A) Galactose is the C_4epimer of glucose. (R) Glucose and galactose differ in configura- tion at C_4.
Answer» Both A & R are TRUE and R is the CORRECT EXPLANATION of A Both A & R are true, but R is not the correct explanation of A A is true, R is FALSE A is false, R is true Answer :A

Discussion

89636.	A g of N_(2) and 1g of O_(2) are put in a two-litre flask at 27^(@)C. Calculate partial pressure of each gas, the total pressure and the composition of the mixture in mole percentage.
Answer» SOLUTION :0.44 ATM, 0.82 atm, 53.3%, 46.7%

Discussion

89637.	A : Fullerene is the purest allotrope of carbon . R : They have smooth structure without danging bonds .
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 :A

Discussion

89638.	(A) Fuelinn aeroplane has a high percentagefo highlybranched chain alkanes . (R ) Octanenumberofbranchedalkanes is lessthanthatof straightnumberalaknes .
Answer» if both(A) and (R ) are correct and ( R) is thecorrect explanationof the(A) IFBOTH(A) and (R ) are correct but ( R) is not thecorrectexplanationof the(A) if (A) is correct but ( R) is INCORRECT is (A)in incorrect but( R)is correct Answer :C

Discussion

89639.	A fuel oil contains significant quantity of Sulphur. When the oil is burnt, the Sulphur is oxidized to SO_2. In a city 465 tonnes of SO_2 are emitted by power plants each day. If 50% of SO_2 comes from the combustion of fuel oil that contains 3% Sulphur by mass, how many tonnes of oil are burnt per day?
Answer» 3875 tonnes 3108 tonnes 4365 tonnes 2176 tonnes Answer :A

Discussion

89640.	A fuelhavinghighoctanenumbercontainsmainly :
Answer» ALKANES straightchainalkanes BRANCHED CHAIN hydrocarbons areanes Answer :C

Discussion

89641.	A fuel cell is feed with 8 g of H_(2) and 40 g of O_(2) . How long would the fuel cell run to give a current of 5A?
Answer» 1608 minutes 8041 minutes 3217 minutes 12867 minutes Solution :`H_(2)+(1)/(2)O_(2)RARR H_(2)O` `2gH_(2)` REQUIRES 16g O So, `8hH_(2)` will require = 64 g `O_(2)` But OXYGEN given is 40 g, so oxygen is a limiting reagent. So, `((40)/(32))xx4xx96500=5xx"t(sec)"` `t=(40xx4xx96500)/(32xx5xx60)="1608 min"`

Discussion

89642.	A fuel contains 25% n-heptane and 75% iso-octane. Its octane number is :
Answer» 50 75 100 25 Answer :B

Discussion

89643.	A fuel cell is a cell that is continously supplied with an oxidant and a reductant so that if can deliver a current indefinitely. Fuel cells offer the possibility of achieving high thermodynamic efficiency in the conversion of Gibbs energy into mechanical work.Internal combustion engines at best convert only the fraction (T_2-T_1)//T_2 of the heat of combustion into mechanical work. While the thermodynamic efficiency of the fuel cell is given by, eta=(DeltaG)/(DeltaH), where DeltaG is the Gibbs energy change for the cell reaction and DeltaH is the enthalpy change of the cell reaction.A hydrogen-oxygen fuel cell may have an acidic or alkaline electrolyte. Pt\|H_2(g)\|H^(+)(aq.)\|\|H_2O(l)\|O_(2)(g)\|Pt , (2.303 RT)/F=0.06 The above fuel cell is used to produce constant current supply under constant temperature & 30 atm constant total pressure conditions in a cylinder.If 10 moles H_2 and 5 moles of O_2 were taken initially. Rate of consumption of O_2 is 10 milli moles per minute. The half-cell reactions are 1/2O_2(g)+2H^+(aq)+2e^(-)toH_2O(l) E^(@)=1.246 V 2H^+(aq)+2e^(-) to H_2(g) E^(@)=0 To maximize the power per unit mass of an electrochemical cell, the electronic and electrolytic resistances of the cell must be minimized.Since fused salts have lower electolytic resistances than aqueous solutions, high-temperature electrochemical cells are of special interest for practical applications.The above fuel cell is used completely as an electrolytic cell with Cu voltameter of resistance 26.94 Omega using Pt electrodes. Initially Cu voltameter contains 1 litre solution of 0.05 M CuSO_4 [H^(+)] in solution after electrolysis (Assuming no changes in volume of solution.)
Answer» 0.015 M 0.03 M 0.025 M 0.01 M Solution :`i=V/R=1.3/26.94=0.04825 A` TIME of ELECTROLYSIS `=5/(10xx10^(-3))=500` MINUTES. `Cu^(2+)(aq)+H_2OtoCu^(+)(aq)+1/2O_2+2H^(+)(aq)` `:.` MOLES of `H^+` formed `=(ixxt)/F=(0.04825xx500xx60)/96500=0.015 M`

Discussion

89644.	A fuel cell is a cell that is continously supplied with an oxidant and a reductant so that if can deliver a current indefinitely. Fuel cells offer the possibility of achieving high thermodynamic efficiency in the conversion of Gibbs energy into mechanical work.Internal combustion engines at best convert only the fraction (T_2-T_1)//T_2 of the heat of combustion into mechanical work. While the thermodynamic efficiency of the fuel cell is given by, eta=(DeltaG)/(DeltaH), where DeltaG is the Gibbs energy change for the cell reaction and DeltaH is the enthalpy change of the cell reaction.A hydrogen-oxygen fuel cell may have an acidic or alkaline electrolyte. Pt\|H_2(g)\|H^(+)(aq.)\|\|H_2O(l)\|O_(2)(g)\|Pt , (2.303 RT)/F=0.06 The above fuel cell is used to produce constant current supply under constant temperature & 30 atm constant total pressure conditions in a cylinder.If 10 moles H_2 and 5 moles of O_2 were taken initially. Rate of consumption of O_2 is 10 milli moles per minute. The half-cell reactions are 1/2O_2(g)+2H^+(aq)+2e^(-)toH_2O(l) E^(@)=1.246 V 2H^+(aq)+2e^(-) to H_2(g) E^(@)=0 To maximize the power per unit mass of an electrochemical cell, the electronic and electrolytic resistances of the cell must be minimized.Since fused salts have lower electolytic resistances than aqueous solutions, high-temperature electrochemical cells are of special interest for practical applications. If lambda_m^(oo)(Cu^(2+))=0.01S m^(2) "mole"^(-1),lambda_m^(oo)(H^(+))=0.035S m^(2) "mole"^(-1) and lambda_m^(oo)(SO_(4)^(2-))=0.016S m^(2) "mole"^(-1),specific conductivity of resulting solution left in copper voltameter after above electrolysis is
Answer» 2.57S `m^(-1)` 1.75S `m^(-1)` 1.525S `m^(-1)` 2.25 S `m^(-1)` Solution :In RESULTING solution, `[Cu^(2+)]=0.05-0.0075=0.0425 M, [H^+]=0.015 M and [SO_4^(2-)]=0.05 M` `K=K_(Cu^(2+))+K_(H^+)+K_(SO_4^(2-))=(lambda_(Cu^(2+))^ooxx[Cu^(2+)])/1000+(lambda_(H^+)^(oo)xx[H^+])/1000+(lambda_(SO_4^(2-))^(oo)xx[SO_4^(2-)])/1000` `(0.01xx10^4xx0.0425)/1000+(0.035xx10^4xx0.015)/1000+(0.016xx10^4xx0.05)/1000=0.00425+0.00525+0.008=0.0175 S cm^(-1)` `=1.75 Sm^(-1)`

Discussion

89645.	A fuel cell is
Answer» the voltaic CELLS in which discontinuous supply of fuels are send at anode to give oxidation the voltaic cell in which fuels such as : `CH_(4) , H_(2) , CO` are USED up at anode it involves the reactions of `H_(2) - O_(2)` FUEL cell such as Cathode : `2 H_(2) + 4OH^(-) to 4 H_(2) O(l) + 4e` Anode : `O_(2) + 2 H_(2)O (l) + 4e to 4 OH^(-)` the efficiency of a `H_(2) - O_(2)` fuel cell is very low SOLUTION :`CH_(4) , H_(2) , CO` are used fuel .

Discussion

89646.	A fuel cell generates a standard electrode potential of 0.7 V, involving 2 electrons in its cell reaction. Calculate the standard free energy change for the reaction. "Given F= 96487 C mol"^(-1).
Answer» Solution :STANDARD free energy CHANGE for the reaction `DeltaG^(@)=-nFE^(@)` `"Given : "E^(@)=0.7V, n=2, F="96500 C MOL"^(-1)` `DeltaG^(@)=-2xx96500xx0.7V` `=-135100 J` `=-135kJ`

Discussion

89647.	A fuel cell is :
Answer» The voltic CELLS in which continuous supply of fuels are SEND at ANODE to give oxidation The voltaic cell in which fuels such as,`CH_4, H_2, CO` are used up at anode It involves the reaction of`H_2 -O_2` fuel cell such as : Anode : `2H_2 +4OH^- rarr4H_2O(l)+4e` Cathode:`O_2 +2H_2O(l) + 4e rarr4OH^-` All Answer :D

Discussion

89648.	A fuel cell involves combustion of butane at 1 atm and 298 K C_(4)H_(10)(g)+(13)/(2)O_(2)(g)to4CO_(2)(g)+5H_(2)O(l),DeltaG^(@)=-2746" kJ "mol^(-1) The E_(cell)^(@) will be
Answer» 0.545 V 1.09 V 0.922 V 0.755 V Solution :`overset(-10)(C_(4))overset(+10)(H_(10))(g)+(13)/(2)O_(2)(g)to4overset(+4)(C)overset(-4)(O_(2))+5H_(2)O(L)` Change in oxidation number of carbon `=4(+4)-(-10)=26` or `(13)/(2)O_(2)^(@)+26e^(-)to13O^(2-)` `(8O^(2-)` in `4CO_(2)` and `5O^(2-)` in `5H_(2)O)` Thus, CELL reaction involves 26 ELECTRONS, i.e., n=26 `E_(cell)^(@)=(DeltaG^(@))/(nF)=(-(-2746)xx1000)/(26xx96500)=+1.09V`

Discussion

89649.	A fuelcell develops an electrical potential from the combustion of butane at1 bar and 298 K C_4H_(10(g)) + 6.5O_(2(g)) to 4CO_(2(g)) + 5H_2O_((l)) , DeltaG^(0) = -2746 kUJ// mol
Answer» 4.74 V 0.547 V 4.37 V 1.07 V Solution :`C_4H_(10(g)) + 6.5O_(2(g)) to 4O_(2(g)) + 5H_2O_((l))` ` N = [4-(2.5)] xx 4 ` carbon atoms , n = 26 `DeltaG = -nFE^(0) - 2746 = -2.6 xx 96,500 xx E , E =(-2746)/(26 xx 96,500) = 1.07V`

Discussion

89650.	Afualhasthesameknockingpropertyas amixture of 70%iso - octane(2,2,4 -trimethylpentane ) and30% n-heptaneby volume, theoctanenumberof afuel is :
Answer» 70 40 100 50 Answer :A

Discussion

Explore topic-wise InterviewSolutions in .

A gas container observes Maxwell distribution of speeds. If the number of molecules between the speed 5 and 5.1 km sec^(-1) at 25^(@)C be 'n', what would be the number of molecules between this range of speed if the total number of molecules in the vessel are doubled ?

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A gas at 1 atm is bubbled through a solution containing 1MY^(-)and1MZ^(-) " at " 25^(@)C.Ifthe reduction potential of ZgtYgtZ,then

A gas absorbs a photon of 355nm and emits at two wavelengths. If one of the emissions is at 680 nm. The other is at:

A gas absorbs 35.8 cal of heat and underrgoes an expansion from 1L to 1.5 L against the external pressure of 1.2 atm. The change in the internal energy of the system is which of the following? [Use 1 atm =1.01325 xx 10^(5) Nm^(-2), 1 cal = 4.18 J)

A gas absorbs 120 J of heat and expands against an external pressure of 1.10 atm from a volume of 0.52 to 2.0 L. The change in internal energy is (1 L atm = 101.3 J) :

A gas ‘X' is passed through water to form a saturated solution. The aqueous solution on treatment with silver nitrate gives a white precipitate. The saturated aqueous solution also dissolves magnesium ribbon with evolution of a colourless gas 'Y’: Identify ‘X' and 'Y’.

A galvanic is set up from a zinc bar weighing 50 g and 1.0 litre, 1.0 M CuSO_(4) solution. How long would the cell run assuing it delivers a steady current of 1.0 ampere?

A galvix cell is constructed using the redox reaction, (1)/(2)H_(2_(g))+AgCl_((s))hArrH_((aq))^(+)+Cl_((aq))^(-)+Ag_((s)) It is represented as

A galvanic cell is set up from a zinc bar weighing 50g and 1.0 litre, 1.0M, CuSO_(4) solution. How long would the cell run, assuming it delivers a steady current of 1.0 ampere

A galvanic cell was operated under almost ideally reversible conditions at a current of 10^(-16) amp. How long would it take to deliver 1 mole of electrons ?

A galvanic cell is constructed of two hydrogen electrodes, one immersed in a solution with H^+ at 1M and the other in 1M KOH. Calculate E_(cell). If 1M KOH solution is replaced by 1M NH_3, will E_(cell) be higher or lower than in 1M KOH?

A galvanic cell is composed of two hydrogen electodes, one of which is a standard one . In which of the following solutins the other electrode be immersed to get maximum emf:

A galvanic cell is composed of two hydrogen electrodes one of which is a standard one. In which of the following solutions should the other electrode be immersed to get maximum emf ?

A galvanic cell has electrical potential of 1.1 V. if an opposing potential of 1.1 V is applied to this cell, what will happen to the cell reaction and current flowing through the cell?

A galvanic cell has electrical potential of 1.1 V. If an opposing potential of 1.1 V is applied to this cell, what will happen to the cell reaction and the current flowing through the cell ?

A galvanic cell has electrical potential of 1.1 V. If an opposing potential of 1.1 V is applied to this cell, what will happen to the cell reaction and current flowing through the cell ?

A galvanic cell has electrical potential of 1.1 V . If an opposing potential of 1.1 Vis applied to this cell, what will happen to the cell reaction and current flowing through the cell ?

A galvanic cell has E_(cell)^(@)=1.5" V". If an opposing potential of 1.5" V" is applied to cell, what will happen to the cell reaction and current flowing through the cell ?

A galvanic cell consists of a metallic zinc plate immersed in 0.1 M Zn(NO_(3))_(2) solution. Calculate the emf of the cell at 25^(@)C. Write the chemical equations for the electrode reactions and represent the cell. (Given : E_(Zn^(2+)//Zn)^(@) = - 0.76 volt and E_(Pb^(2+)//Pb)^(@) = - 0.13 volt)

A galvanic cell consists of

A galvanic cell after use is recharged by passing current through it. What type of cell is it? Give an example.

(A) Galactose is the C_4epimer of glucose. (R) Glucose and galactose differ in configura- tion at C_4.

A g of N_(2) and 1g of O_(2) are put in a two-litre flask at 27^(@)C. Calculate partial pressure of each gas, the total pressure and the composition of the mixture in mole percentage.

A : Fullerene is the purest allotrope of carbon . R : They have smooth structure without danging bonds .

(A) Fuelinn aeroplane has a high percentagefo highlybranched chain alkanes . (R ) Octanenumberofbranchedalkanes is lessthanthatof straightnumberalaknes .

A fuelhavinghighoctanenumbercontainsmainly :

A fuel cell is feed with 8 g of H_(2) and 40 g of O_(2) . How long would the fuel cell run to give a current of 5A?

A fuel contains 25% n-heptane and 75% iso-octane. Its octane number is :

A fuel cell is

A fuel cell generates a standard electrode potential of 0.7 V, involving 2 electrons in its cell reaction. Calculate the standard free energy change for the reaction. "Given F= 96487 C mol"^(-1).

A fuel cell is :

A fuel cell involves combustion of butane at 1 atm and 298 K C_(4)H_(10)(g)+(13)/(2)O_(2)(g)to4CO_(2)(g)+5H_(2)O(l),DeltaG^(@)=-2746" kJ "mol^(-1) The E_(cell)^(@) will be

A fuelcell develops an electrical potential from the combustion of butane at1 bar and 298 K C_4H_(10(g)) + 6.5O_(2(g)) to 4CO_(2(g)) + 5H_2O_((l)) , DeltaG^(0) = -2746 kUJ// mol

Afualhasthesameknockingpropertyas amixture of 70%iso - octane(2,2,4 -trimethylpentane ) and30% n-heptaneby volume, theoctanenumberof afuel is :