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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.
| 2051. |
Give two example for gaseous first order reactions. |
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Answer» The following are the examples for gaaseous first order reactions `N_(2)O_(5(g))to N_(2)O_(4(g))+1/2 O_(2(g))` ` SO_(2)Cl_(2(g))to SO_(2(g))+Cl_(2(g))` |
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| 2052. |
Write the Arrhenius equation for the rate constant (k) of a reaction. |
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Answer» Arrhenius equation is `k=Axxe^(-Ea//RT)` k = Rate constant `E_(a)=` activation energy R = gas costant T = Temperature (K) |
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| 2053. |
By how many times the rate constant inhcreases for a rise of reaction temperature by `10^(@)C` ? |
| Answer» For every `10^(@)C` rise of temperature rete constant of chemical reactions may be doubled (some times tripled). | |
| 2054. |
Give two examples for gaseous first order reactions. |
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Answer» The following are the examples for gaaseous first order reactions `N_(2)O_(5(g))to N_(2)O_(4(g))+1/2 O_(2(g))` ` SO_(2)Cl_(2(g))to SO_(2(g))+Cl_(2(g))` |
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| 2055. |
Write the equation useful to calculate half-life `(t_(1//2))` values for zero and first order reactions. |
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Answer» Half life `(t_(1//2))` of zero order reaction At half life `(t_(1//2))[R]=([R]_(0))/(2)` `therefore t_(1//2)=([R_(0)])/(2k)` Half life of a first order reaction `t_(1//2) =(0.693)/(k)` k = rate constant |
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| 2056. |
What are pseudo first order reactions ? Give one example. |
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Answer» First order reactions whose molecularity is more than one ar called pseudo first order reactions `underset("Sucrose")(C_(12)H_(22)O_(11))+H_(2)O overset(H^(+))to underset(" Glucose")(C_(6)H_(12)O_(6))+underset("Fructose")(C_(6)H_(12)O_(6))` Order =1 molecularity =2 |
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| 2057. |
Give the mathematical equation which gives the variation of molar conductivity with `^^_(m)` the molarity (c ) of the solution ? |
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Answer» The mathematical equation which gives the variation of molar conductivity `^^_(m)` with the molarity (c ) of the solutioin is `^^_(m) =(k)/(1000 (lit//cm^(3))xx"molarity(molea/lit)")=(kxx10009(cm^(3)//lit))/("molarity (mole/lit)")` |
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| 2058. |
If equivalent conductance of 1M benzoic aid is 12.8`ohm^(-1)cm^(2)` and 288.42 `ohm^(-1)cm^(2)` respectively. Its egree of dissociation isA. 0.39B. `3.9%`C. 0.0035D. `0.039%` |
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Answer» Correct Answer - B `wedge_(m_(C_(6)H_(5)COOH))^(@)=wedge_((C_(6)H_(5)COO^(-)))^(@)+wedge_((H^(+)))^(@)` `=42+288.42=330.42` `alpha=(wedge_(m)^(@))/(wedge_(m)^(@))=(12.8)/(330.42)=3.9%` |
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| 2059. |
If equivalent conductance of 1 M benzoic acid is 12.8 `ohm^(-1) cm^(2)` and if the conductance of benzoic ion and `H^(+)` ion are 42 and 288.42 `ohm^(-1)` respectively, its degree of dissociation isA. 0.39B. 0.039C. 0.0035D. 0.00039 |
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Answer» Correct Answer - B `(^^_(C_(6)H_(5)COOH)^(@))=lambda_(C_(6)H_(5)COO^(-))^(@)+lambda_(H^(+))^(@)` `=42+288.42=330.42` `alpha=(^^_( m)^(@))/(^^_( m)^(@))=(1 2.8)/(330.42)= 0.03 9=3.0%` |
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| 2060. |
If `Zn^(2+)//Zn` electrode is diluted 100 times, then the charge in reduction potential isA. increase of 59mVB. decrease of 59mVC. increase of 25.5mVD. decrease of 2.95V |
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Answer» Correct Answer - B `E_(Zn^(2+)//Zn)=E_(Zn^(2+)//Zn)^(@)-(0.059)/(2)"log"(1)/(|Zn^(2+)|)` `E_(Zn^(2+)//Zn)=E_(Zn^(3+)//Zn)+0.0295"log"C` When solution is diluted 100 times. `E_(Zn^(2+)//Zn)=E_(Zn^(2+)//Zn)^(@)+0.0295"log"(C)/(100)` `=E_(Zn^(2+)//Zn)^(@)+0.0295"log" C-0.0295xx2` `=E_(Zn^(2+)//Zn)-0.059` `E_(Zn^(2+)//Zn)-E_(Zn^(3+)//Zn)=-0.059V=-59mV` |
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| 2061. |
If the `Zn^(2+)//Zn` electrode is diluted to 100 times then the change in e.m.f.A. Increase of 59 mVB. Decrease of 59 mVC. Increase of 29.5 mVD. Decrease of 29.5 mV |
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Answer» Correct Answer - A `E_(cell)=(0.059)/(n)log((1)/(C))=-(0.059)/(2)log((1)/(100))` `=-(0.059)/(2)(-2)=0.059V=59mV`. (increase) |
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| 2062. |
The voltage of the cell `Zn(s)|Zn(CN)_(4)^(2-) (0.450M),CN^(-) (2.65 xx 10^(-3)M)||Zn^(2+)(3.84 xx 10^(-4)M)|Zn(s)` is `E =+0.099V`. Calculate the constant `K_(f)` for `Zn^(2+) +4CN^(-) rarr Zn(CN)_(4)^(2-)`, the only `Zn^(2+) +CN^(-)` complex reaction of importance, |
| Answer» Correct Answer - `5.4 xx 10^(16)` | |
| 2063. |
The voltage of the cell `Pb(s)|PbSO_(4)(s)|NaHSO_(4)(0.600M)||Pb^(2+)(2.50 xx 10^(-5)M)|Pb(s)` is `E = +0.061V`. Calculate `K_(2) = [H^(+)] [SO_(4)^(2-)]//[HSO_(4)^(-)]` the dissociation constant for `HSO_(4)^(-)`. Given `Pb(s) +SO_(4)^(2-) rarr PbSO_(4) +2e^(-) (E^(@) = 0.356V),E^(@) (Pb^(2+)//Pb) =- 0.126V`. |
| Answer» Correct Answer - `1.06 xx 10^(-2)` | |
| 2064. |
Accidentally chewing on a stray fragmet of aluminium foil can causes a sharp tooth path if the aluminium comes in contact with an amalgam filing. The filing, an alloy of silver, tin and mercury, acts as the cathode of a tiny galvanic cell, the aluminium behaves as the anode, and salivas serves as the electrolyte. when the aluminium and the filling come in contact, and electric current passage from the aluminium to the filling which is sensed by a nerve in the tooth. Aluminium is oxidized at the anode, and `O_(2)` gas is reduced to water at the cathode. `E_(AI^(3+)//AI)^(@) =- 1.66 E_(O_(2)H^(+)//H_(2)O)6^(@) = 1.23 V` The standard reduction potential of the reaction, `H_(2)O +e^(-) rarr (1)/(2)H_(2)+OH^(-)` at `298K` is:A. `E^(@) =(RT)/(2F)In K_(w)`B. `E^(@) =(RT)/(F)In ]P_(H_(2))]^(1//2) [OH^(-)]`C. `E^(@) =(RT)/(F)In ([P_(H_(2))]^(1//2))/([H^(+)])`D. `E^(@) =(RT)/(F)In K_(w)` |
| Answer» Correct Answer - A | |
| 2065. |
Accidentally chewing on a stray fragmet of aluminium foil can causes a sharp tooth path if the aluminium comes in contact with an amalgam filing. The filing, an alloy of silver, tin and mercury, acts as the cathode of a tiny galvanic cell, the aluminium behaves as the anode, and salivas serves as the electrolyte. when the aluminium and teh filling come in contact, and electric current passage from the aluminium to the filling which is sensed by a nerve in the tooth. Aluminium is oxidized at the anode, and `O_(2)` gas is reduced to water at the cathode. `E_(AI^(3+)//AI)^(@) =- 1.66 E_(O_(2)H^(+)//H_(2)O)6^(@) = 1.23 V` Net reaction taking place when amalgam is in contact with aluminium foil:A. `AI + O_(2) +OH^(-) rarr AI(OH^(-)) +H_(2)O`B. `4AI +3O_(2)+12H^(+) rarr 4AI^(3+) +6H_(2)O`C. `4AI +3O_(2) rarr 4AI_(2)O_(3)`D. `2H_(2)+O_(2) rarr 2H_(2)O` |
| Answer» Correct Answer - A | |
| 2066. |
Calculate the minimum mass of NaOH required to be added in RHS to consume all the `H^(+)` present in RHS of the cell of emf + 0.701 volt at `25^(@)C` before its use. Also report the emf of the cell after addition of NaOH. `Zn|underset("0.1 M")(Zn^(2+))||underset("1 litre")(HCl)|underset("1 atm")(Pt (H_(2)g)), E_(Zn//Zn^(2+))^(@)=0.760 V` |
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Answer» The cell reaction is, `Zn+2H^(+) rarr Zn^(2+) +H_(2)` Applying Nernst equation, `E_(cell)=E^(@)-0.0591/2"log"([Zn^(2+)])/([H^(+)]^(2))` `0.701=0.760-0.0591/2"log" ([Zn^(2+)])/([H^(+)]^(2))` So, `"log"([Zn^(2+)])/([H^(+)]^(2))=(0.0591xx2)/0.0591=2` or `([Zn^(2+)])/([H^(+)]^(2))=10^(2)` `[H^(+)]^(2)=0.1/10^(2)=10^(-3)` `[H^(+)]=0.0316" mol L"^(-1)` Thus, 0.0316 mol/litre of NaOH is required to neutralise `H^(+)` ions. Mass of NaOH = `0.0316xx` Mol. mass of NaOH `=0.0316xx400=1.264 g` After addition of NaOH, the solution becomes neutral, i.e., the concentration of `H^(+)` ions in cathodic solution becomes `10^(-7)`. Applying again Nernst equation, `E_(cell)=E_(cell)^(@)-0.0591/2"log"([Zn^(2+)])/([H^(+)]^(2))` `=0.760-0.0591/2"log"0.1/((10^(-7))^(2))=0.3759` volt |
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| 2067. |
The emf of cell is 1.3 volt. The positive electrode has potential of 0.5 volt. The potential of negative electrode is :(a) 0.8 V (b) -0.8 V (c) 1.8 V (d) – 1.8 V |
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Answer» Option : (b) -0.8 V |
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| 2068. |
Which of the following species gains electrons more easily?(a) Na+(b) H+ (c) Mg+ (d) Hg+ |
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Answer» Option : (b) H+ |
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| 2069. |
What would be the electrode potential of a silver electrode dipped in a saturated solution of `AgCl` in contact with `0.1 M KCl` solution at `25^(@)C` ? `E^(c-)._(Ag^(o+)|Ag)=0.799 V ` `K_(sp)` of `AgCl=1xx10^(-10)` |
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Answer» `Ag|AgCl,0.1 M KCl` `E=E^(c-)-(0.059)/(1) log .(1)/([Ag^(o+)])` `[Cl^(c-)]=0.1` `[Ag^(o+)]=(K_(sp))/([Cl^(c-)])=(10^(-10))/(0.1)=10^(-9)` `E=0.799-(0.059)/(1) log .(1)/([10^(-9)])` `=0.799-0.059xx9=0.799-0.531=0.268V` |
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| 2070. |
The electrode potential of a silver electrode dipped in 0.1 M AgNO3 solution at 298 K is (E0red of Ag = 0.80 volt)(a) 0.0741 V (b) 0.0591 V (c) 0.741 V (d) 0.859 V |
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Answer» Option : (c) 0.741 V |
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| 2071. |
Select the correct relations :A. `-DeltaS =((delE)/(delT))_(P) xx nF`B. `((delE)/(delT))_P = (DeltaG - DeltaH)/(T)`C. `DeltaS = ((delE)/(deltT))_(P) xx nF`D. `((delE)/(delT))_(P) = ((delDeltaS)/(delT))_(P)` |
| Answer» Correct Answer - B::C | |
| 2072. |
Which of the following rection(s) is (are) involved in the rustin of iron ?A. `Fe_((s)) rarr Fe_((aq.))^(2+) + 2e`B. `O_(2(g)) + 4H_((aq.))^(+) + 4e rarr 2H_(2)O_(l)`C. `2Fe_((s)) + O_(2(g)) + 4H_((aq.))^(+) rarr 2Fe_((aq.))^(2+) + 2H_(2)O_((l))`D. `Fe^(2+) + 2e rarr Fe_((s))` |
| Answer» Correct Answer - A::B::C | |
| 2073. |
Which are true for a standard hydrogen electrode ?A. The hydrogen ion concentration is `1 M`B. Temperature is `25^(@)C`C. Pressure of hydrogen is `1` atmD. It contains a metallic conductor which does not adsorb hydrogen. |
| Answer» Correct Answer - A::B::C | |
| 2074. |
One faraday is the amount of charge:A. that liberates `1 g` equivalent of a metal form its solutionB. `96515` coulombC. that liberates `31.78 g` of `Cu`D. that liberates `1//2 g` atom of `Cu` |
| Answer» Correct Answer - A::B::C::D | |
| 2075. |
If ∧m and ∧0 are the molar conductivities of a weak electrolyte at concentration C and at zero concentration, then the dissociation constant Ka is given by :(a) Ka = \(\frac{∧^2_m\times C}{∧_0-∧_m}\)(b) Ka = \(\frac{∧^2_m\times C}{∧_0(∧_0-∧_m)}\) (c) Ka = \(\frac{∧^2_0\times C}{(∧_0-∧_m)}\) (d) Ka = \(\frac{∧_m\times C}{∧_0(∧_0-∧_m)}\) |
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Answer» Option : (b) Ka = \(\frac{∧^2_m\times C}{∧_0(∧_0-∧_m)}\) |
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| 2076. |
What weight of copper will be deposited by passing 2 Faradays of electricity through a cupric salt? (atomic mass = 63.5)(a) 63.5 g (b) 31.75 g (c) 127 g (d) 12.7 g |
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Answer» Option : (a) 63.5 g |
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| 2077. |
Coulomb is the quantity of charge defined as:A. One ampere of current passing for `1` sec1B. The charge which deposits `0.001118 g` of `Ag` on cathodeC. The charge which deposits electrochemical equivalence of metalD. `1//2` ampere current for two second |
| Answer» Correct Answer - A::B::C | |
| 2078. |
On the electrolysis of a `10^(-6) M HCl` solution:A. `O_(2)` gas is product at the anodeB. `Cl_(2)` gas is producted at the anodeC. `H_(2)` gas is produced at the cathodeD. `O_(2)` gas is produced at the cathode |
| Answer» Correct Answer - A::C | |
| 2079. |
On electrolusis, which of the following arragements will lead tio oxygen being evolved at the anode ?A. Dilute `H_(2)SO_(4)` with copper electrodesB. Fused `Na_(2)OH_(4)` with an iron cathode and a nickel anodeC. Aqueous `AgNO_(3)` solution with platinum electrodesD. Dilute `H_(2)SO_(4)` with platinum electrodes |
| Answer» Correct Answer - B::C::D | |
| 2080. |
What would be the product of electrolusis if molten ` ICI_3` is electrlysed ?A. ` I_2` is produced at cathode and ` CI_2` is proluced at anodeB. ` I_2` is produced at cathode and ` I_2` is proluced at anode C. Botg ` I_2` and ` VI_2` are liberated at both electrodesD. ` ICI_2` is produced at cathode and `ICI_4` is produced at anode |
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Answer» Correct Answer - C In the molten states `ICI_3` ionises as folws `1ICI_(3) underset(larr)(rarr)IC_(2)^(+)l_(4)^(-)`. Hence both `I_2` and `CI_2` are produced at both the elecrtrodes. |
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| 2081. |
In passing `3F` of electricity through three electrolytic cells connect in series containing `Ag^(o+),Ca^(2+),` and `Al^(3+)` ions, respectively. The molar ratio in which the three metal ions are liberated at the electrodes isA. `1:2:3`B. `2:3:1`C. `6:3:2`D. `3:4:2` |
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Answer» Correct Answer - c `1Eq` of `Ag^(o+):1Eq` of `Ca^(2+):1Eq` of `Al^(3+)` `1 mol ofAg^(o+),(1)/(2)molof Ca^(2+),(1)/(3)mol of Al^(3+)` `:. mol ` ratio i s`1:(1)/(2):(1)/(3)=6:3:2` |
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| 2082. |
What would be the product of electrolusis if molten ` ICI_3` is electrlysed ?A. ` I_2` is produced at cathode and ` CI_2` is proluced at anodeB. ` I_2` is produced at cathode and ` I_2` is proluced at anodeC. Botg ` I_2` and ` VI_2` are liberated at both electrodesD. ` ICI_2` is produced at cathode and `ICI_4` is produced at anode |
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Answer» Correct Answer - C In the molten states `ICI_3` ionises as folws `1ICI_(3) underset(larr)(rarr)IC_(2)^(+)l_(4)^(-)`. Hence both `I_2` and `CI_2` are produced at both the elecrtrodes. |
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| 2083. |
`1` mol each of `AG AgBO_3, CuSO_4` and `AICI_3` is electrolysed. The number of faradays rewuired is in the ration of:A. ` 1:1:1`B. `1:2:3:`C. `3:2:1`D. `1:3:1` |
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Answer» Correct Answer - B `Ag^(+) + e^(-) rarr Ag` ` Cu^(2+) + 2e^(-) rarr Cu` ` Al^(1+) +3e^(-) rarr Al` Thus, munber of faradauys required by `1` mol each ` =1 :2: 3.` |
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| 2084. |
When electrolysis of silver nitrate solution is carried out using silver electrodes, wich of the following reaction occurs at the anode?A. `Ag to Ag^(+) + e^(-)`B. `Ag^(+) + e^(-) to Ag`C. `2 H_(2)O to 4 H^(+) + O_(2) + 4 e^(-)`D. `4 OH^(-) to 2H_(2) + O_(2) + 4 e^(-)` |
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Answer» Correct Answer - A At the anode , Ag is oxidized most easily (or `NO_(3)^(-)` ions attack Ag electrode to convert it into `Ag^(+)` ions ) . |
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| 2085. |
By passageof 1 F of electricityA. 1 mol of Cu is depositedB. 0.5 mol of Mgis depositedC. 9g of Al is depositedD. 5.6 L of `O_(2)` gas evolved at anode |
| Answer» Correct Answer - B::C::D | |
| 2086. |
It is not adviasable to:A. Stir sugar solution with a steel spoonB. Stiir copper sulphate solution with a silver spoonC. Stir copper sulphate solution with a zinc spoonD. Stir silver nitrate solution with a copper spoon |
| Answer» Correct Answer - C::D | |
| 2087. |
Why does alkaline medium inhibit the rusting of iron ? |
| Answer» `OH^(-)` ions released in the alkaline medium combine with the `H^(+)` ions available from `H_(2)CO_(3)` `(CO_(2)` and `H_(2)O` to form `H_(2)O`. Since `H^(+)` ions take part in the redox reaction responsible for the rusting of iron, the removal of these ions checks the rusting of iron. | |
| 2088. |
Reactions taking place in a fuel cell are:A. `O_(2)(g)+2H_(2)O(l)+4e^(-)to4OH^(-)(aq)` at the cathodeB. Reaction in (a) at the cathodeC. `2H_(2)(g)+4OH^(-)(aq)to4H_(2)O(l)+4e^(-)` at the anodeD. Reaction in (c) at the cathode |
| Answer» Correct Answer - A::C | |
| 2089. |
In which case `(E_(cell)-E_(Cell)^(@))` is zero:A. `Cu|Cu^(2+)(0.01M)||Ag^(+)(0.1M)|Ag`B. `Pt(H_(2))|pH=1||Zn^(2+)(0.01M)|Zn`C. `Pt(H_(2))|pH=1||Zn^(2+)(1M)|Zn`D. `Pt(H_(2))|H^(+)(0.01M)||Zn^(2+)(0.1M)|Zn` |
| Answer» Correct Answer - A::B | |
| 2090. |
Rust is a mixture ofA. `Fe_(2)O_(3) . xH_(2)O`B. `FeO`C. `Fe(OH)_(3)`D. `Fe` |
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Answer» Correct Answer - A Rust is hydrated ferric oxide i.e., `Fe_(2)O_(3) . xH_(2)O` . |
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| 2091. |
The formation of rust on the surface or iron occurs through the reactions:A. `Fe(s)toFe^(2+)(aq)+2e^(-)` at anodeB. `O_(2)(g)+4H^(+)(aq)+4r^(-)to2H_(2)O(l)` at cathodeC. `4Fe^(2+)(aq)+O_(2)(g)+4H_(2)O(l)to2Fe_(2)O_(3)(s)+8H^(+)`D. `Fe_(2)O_(3)(s)+xH_(2)O(l)toFe_(2)O_(3).xH_(2)O` |
| Answer» Correct Answer - A::B::C::D | |
| 2092. |
Write the chemical formula of rust. |
| Answer» The chemical formula of rust is : `Fe_(2)O_(3).xH_(2)O` | |
| 2093. |
Assertion `(A) :` Galvanized iron does not rust. Reason `(R): Zn` has a more negative electrode potential than `Fe`.A. If both assertion and reason are correct and reason is correct explanation for assertion.B. If both assertion and reason are correct but reason is not correct explanation for assertion.C. If assertion is correct but reason is incorrect.D. If assertion and reason both are incorrect. |
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Answer» Correct Answer - A (a) Reason is the correct explanation for assertion. |
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| 2094. |
Why does not iron rust even if zinc coating is broken in a galvanised iron pipe ? |
| Answer» Zinc being more reactive than iron will sacrifice itself for the sake of iron. As a result, zinc will participate in the corrosion while iron will remain intact. For more details, consult Section 23. | |
| 2095. |
Select the wrong relation `(s)`.A. `DeltaS=((delE)/(delT))_(P)xxnF`B. `-DeltaS=((delE)/(delT))_(P)xxnF`C. `((delE)/(delT))_(P)=((delDeltaS)/(delT))`D. `((delE)/(delT))_(P)=(DeltaH+nEF)/(T)` |
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Answer» Correct Answer - a,d `DeltaG=DeltaH-TDeltaS` and `DeltaG=DeltaH+T((delDeltaG)/(delT))_(P)` `:. (del(DeltaG)/(delT))_(P)=(DeltaG-DeltaH)/(T)=-(TDeltaS)/(T)=-DeltaS` `:. DeltaS=+nF((delE)/(delT))_(P)` Also, `-nFE=DeltaH+Txx(-nF)((delE)/(delT))_(P)` `:. ((delE)/(delT))_(P)=(DeltaH+nEF)/(T)` |
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| 2096. |
When `4.0 A` of current is passed through a `1.0L , 0.10M Fe^(3+)(aq)` solution for `1.0 ` hour, it is partly reduced to `Fe(s)` and partly of `Fe^(2+)(aq)`. The correct statements `(s)` is `(` are `):`A. `0.10 mol` of electrons are required to convert all `Fe^(3+)` to `Fe^(2+)`B. `0.025 mol` of `Fe(s)` will be deposited.C. `0.075 mol` of iron remains as `Fe^(2+)`.D. `0.050 mol` of iron remains as `Fe^(2+)` |
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Answer» Correct Answer - a,b,c Number of Faradays `=(4xx1xx3600)/(96500)=0.15` Initially, moles of `Fe^(3+)=0.1xx1 =0.1` First, `Fe^(3+)` will get reduced to `Fe^(2+)` . `Fe^(3+)+e^(-) rarr Fe^(2+)` `1F-=1 mol Fe^(3+)` deposited `implies0.15 F -=0.15 mol Fe^(3+)` deposited `gt Fe^(3+) ` available. Thus, `1 mol Fe^(3+)-=1F` `implies 0.1 mol Fe^(3+)-=0.1 F` electricity is used `-= 0.1 mol Fe^(2+)` produced `implies0.15 -0.1 =0.05 F ` electricity left for the reduction of `Fe^(2+)` . `Fe^(2+) + 2e^(-) rarr Fe` `2F-=1 mol Fe^(2+)` `implies0.05 F -=(0.05)/(2) =0.25 mol Fe^(2+)` reduced `-= 0.025 mol Fe` deposited . `implies Fe^(2+)` left `=0.1-0.025=0.075 mol` |
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| 2097. |
A cell, as shown below, consists of there compartments separated by porous pots. The first contains a cobalt eletrode in `5.0L` of `0.10M Co(NO_(3)),` the second contains `5.0L` of `0.10M KNO_(3),` the third contains an `Ag` electrode in `5.0L` of `0.10M AgNO_(3)`. Assuming that current with in the cell iis carried equally by the negative and positive ions by passign `0.1F` of electricity. Given `: Co^(2+)+2e^(-) rarr Co" "E^(c-)=-0.28V` Ag^(o+)+e^(-) rarr Ag" "E^(c-)=0.80V` `2Ag^(o+)+Corarr Co^(2+)+2Ag" "E^(c-)._(cell)=1.08V` The final concentration of `K^(o+)` in `I,II, `and `III` compartment isA. `0.090,0.0,0.0100M`B. `0.0,0.0100,0.090M`C. `0.0,0.090,0.0100M`D. `0.100,0.90,0.0M` |
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Answer» Correct Answer - c `:. [K^(o+)]` in `I,II,` and `III` compartment is `:` `(0.0,0.090,0.0100)M` |
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| 2098. |
A cell, as shown below, consists of there compartments separated by porous pots. The first contains a cobalt eletrode in `5.0L` of `0.10M Co(NO_(3)),` the second contains `5.0L` of `0.10M KNO_(3),` the third contains an `Ag` electrode in `5.0L` of `0.10M AgNO_(3)`. Assuming that current with in the cell iis carried equally by the negative and positive ions by passign `0.1F` of electricity. Given `: Co^(2+)+2e^(-) rarr Co" "E^(c-)=-0.28V` Ag^(o+)+e^(-) rarr Ag" "E^(c-)=0.80V` `2Ag^(o+)+Corarr Co^(2+)+2Ag" "E^(c-)._(cell)=1.08V` The final concentration of `NO_(3)^(c-)` in `I,II,` and `III` compartment isA. `0.100,0..210,0.0900M`B. `0.210,0.100,0.0900M`C. `0.0900,0.210,0.100M`D. `0.900,0.100,0.210M` |
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Answer» Correct Answer - b `:. [NO^(c-)._(3)` in `I,II,` and `III` compartment is `:` `(0.210,0.100,0.0900)M` |
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| 2099. |
A cell, as shown below, consists of there compartments separated by porous pots. The first contains a cobalt eletrode in `5.0L` of `0.10M Co(NO_(3)),` the second contains `5.0L` of `0.10M KNO_(3),` the third contains an `Ag` electrode in `5.0L` of `0.10M AgNO_(3)`. Assuming that current with in the cell iis carried equally by the negative and positive ions by passign `0.1F` of electricity. Given `: Co^(2+)+2e^(-) rarr Co" "E^(c-)=-0.28V` Ag^(o+)+e^(-) rarr Ag" "E^(c-)=0.80V` `2Ag^(o+)+Corarr Co^(2+)+2Ag" "E^(c-)._(cell)=1.08V` The final concentration of `Co^(2+)` in `I, II, `and `III` compartment isA. `0.105,0.005,0.0M`B. `0.005,0.105,0.0M`C. `0.105,0.0,0.005M`D. 0.0,0.005,0.105M` |
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Answer» Correct Answer - a `:. [Co^(2+)]` in `I,II,` and `III` compartment is `:` `(0.105,0.005,0.0)M` |
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| 2100. |
The equillibrium constant for thefollowing general reaction is `10^(30)`. Calculate `E^(@)` for the cell at 298 K. `2X_(2)(s)+3Y^(2+)(aq)to2X_(2)^(3+)(aq)+3Y(s)`A. `+0.105V`B. `+0.2955 V`C. 0.0985 VD. `-0.2955` V |
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Answer» Correct Answer - B |
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