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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
If electrolysis of aqueous solution of CuSO_(4) is carried out using graphite electrode, then what is the pH of solution of electrolytic cell ? |
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Answer» pH=14.0 Solution of cell is acidic in nature, solution of `CuSO_(4)` is a salt solution of strong ACID and weak base. And pH of acidic solution is ALWAYS less than 7.0. So `pH lt 7.0` |
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| 2. |
If electrolysis of aqueous CuSO_(4) solution is carried out using Cu- electrons, the reaction taking place at the anode is |
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Answer» `H^(+) + e^(-) rarr H` At cathode , `Cu_((aq))^(2+) + 2e^(-1) rarr Cu_((s))` copper dissolves into the solution from the anode and is deposited at the cathode. |
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| 3. |
If E_((Fe^(2+)|Fe))^(o)=-0.441V and E_((Fe^(3+)|Fe^(2+)))^(o)=0.771V, then what is emf for following reaction ? [Fe+2Fe^(3+) to 3Fe^(2+)]. |
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Answer» 1.653 `Fe^(3+)+e^(-) to Fe^(2+),""E_(2)^(o)=0.771V` Required REACTION : `Fe+2Fe^(3+) to 3Fe^(2+)` `=(Fe to Fe^(2+)+2e^(-))`. . .(oxidation) `+(2Fe^(3+)+2e^(-) to 2Fe^(2+))`. . . (reduction) `E_(CELL)^(o)=E_(Cathode)^(o)-E_("ANODE")^(o)` `=0.771-(-0.441)` `E_(cell)^(o)=1.212V` |
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| 4. |
If E_(Fe^(2+)//Fe)^(o)=-0.441V and E_(Fe^(3+)//Fe^(2+))^(o)=0.771V, the stardard EMF of the reaction Fe+2Fe^(3+)to3Fe^(2+) will be |
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Answer» 1.212V |
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| 5. |
If E_(k) is the average kinetic energy per mole of a gas, then : |
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Answer» <P>`pV = ( 3)/( 2) E_(k)` |
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| 6. |
If E_(Fe^(2+)//Fe)^(@)=0.441 and E_(Fe^(3+)//Fe^(2+))^(@)=0.771V the standard EMF of the reaction Fe+2Fe^(3+)to3Fe^(2+) will be |
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Answer» 1.212 V `E_(cell)^(@)=E_(Fe^(3+)//Fe^(2+))^(@)-E_(Fe^(2+)//Fe)^(@)=0.771V-(-0.441V)=1.212V`. |
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| 7. |
If E_(CIO_(3)^(-)//ClO_(4)^(-))^(@) = -0.36V and E_(ClO_(3)^(-)//ClO_(2)^(-))^(@)= 0.33Vat 300 K . Find the equilibrium constantfor the following reaction . 2ClO_(3)^(-) ClO_(2)^(-)_ClO_(4)^(-) |
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Answer» `(0.04)` the reaction includes to STEPS . `CIO_(3)^(-) ClO_(4)^(-)["OXIDATION"]`,`ClO_(3)^(-)ClO_(2)^(-) ["Reduction"]` |
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| 8. |
If edge length in simple cube, bcc and fcc unit cell is a, then the ratio of atomic radius present in them are ……. |
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Answer» `a/2 : (SQRT3)/2 : (SQRT(2)a)/(2)` |
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| 9. |
If E_(cell)=0.118V for the following reaction, then calculate [H^(+)] and pH at 298K temperature. Pt|H_(2)("1 bar")|H^(+)(10^(-6)M)||H^(+)(xM)|H_(2)("1 bar")|Pt |
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Answer» |
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| 10. |
If E_(cell)^(@) of electrochemical cell=1.1 V for possessing following chemical reaction, then what is E_(cell) ? Zn_((S))+Cu_((aq))^(2+)(0.1M) to Zn^(2+)(1M)+Cu_((S)) |
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Answer» 2.12 V `E_(CELL)=E_(cell)^(@)-(0.0592)/(n)"log"([Zn^(2+)])/([CU^(2+)])` `=1.1-(0.0592)/(2)"log"((1.0)/(0.1))` `=1.1-0.0296log10` `=1.1-0.296=1.0704V~~1.07V` |
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| 11. |
If E_(Cd)^(0) = 0.408 V and E_(Ag)^(0) = -799 V, the emf of the cell Cd| Cd^(2+) || Ag^(+) | Ag is |
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Answer» `-1.207` V |
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| 12. |
If EAN of central metal ion X^(2+) in a non-chelating complex is 34 and atomic number of X is 28. Find out the number of monodentate ligands in compelx. |
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Answer» |
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| 13. |
If EAN of metal is 36 in [M(CO)_(2)(sigma-C_(5)H_(5))(pi-C_(5)H_(5))] Find the atomic number of metal M |
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Answer» `Z implies 26` |
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| 14. |
IfE_(Au^(+)// Au)^(@) is 1.69 V andE_(Au^(3+)//Au)^@ is 1.40 V . Then E_(Au^(+)//Au^(3+))^(@) will be : |
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Answer» 0.19 V (1) `Au^(oplus) + 1E^(-) to Au , DeltaG_1 ^(0) , (2) Au^(+3) + 3E^(-) to Au , DeltaG_2^(0)` (3) `Au^(+3) + 2e^(-) to Au^(+) , DeltaG_(3)^(0) (3)=(2)-(1)` `DeltaG_(3)^(0) = DeltaG_2^(0) -DeltaG_1^(0) , 2E_3^(0) = (3 xx 1.40) - 1.89 , E_(Au)^(+0//Au^(+3))^(0) = -1.255` |
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| 15. |
If EAN of central metal cation M^(2+) in an non-chelating complex is 36 and atomic no. of metal M is 26, then the number of monodentate ligand in this complex are |
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Answer» 5 |
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| 16. |
If E_(1)^(@),E_(2)^(@) and E_(3)^(@) are the standard electrode potential for Fe//Fe^(2+),Fe^(2+),Fe^(2+)//Fe^(3+) and Fe//Fe^(3+) electrodes respectively, derive a relation between E_(1)^(@),E_(2)^(@) and E_(3)^(@). |
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Answer» Solution :`FetoFe^(2+)+2e^(-),DeltaG_(1)^(@)=2E_(1)^(@)F` . . . (i) `Fe^(2+)TOFE^(3+)+E^(-),-DeltaG_(2)^(@)=E_(2)^(@)F` . . (ii) `Fe to Fe^(3+)+3e^(-),-DeltaG_(3)^(@)=3E_(3)^(@)F` . . . (iii) Subracting eqn. (i) from eqn. (iii) , we get `0 to Fe^(3+)-Fe^(2+)+e^(-),-DeltaG_(3)^(@)+DeltaG_(1)^(@)=F(3E_(3)^(@)-2E_(1)^(@))` or `Fe^(2+) to Fe^(3+)+e^(-),-DeltaG_(3)^(@)+DeltaG_(1)^(@)=(3E_(3)^(@)-2E_(1)^(@))F` . . . (iv) Comparing eqn. (ii) and (iv), we get `-DeltaG_(2)^(@)=-DeltaG_(3)^(@)+DeltaG_(1)^(@)`[as eqns. (ii) and (iv) are same] ,brgt i.e., `E_(2)^(@)=F(3E_(3)^(@)-2E_(1)^(@))F` or `E_(2)^(@)=3E_(3)^(@)-2E_(1)^(@)` or `3E_(3)^(@)=E_(2)^(@)+2E_(1)^@` |
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| 17. |
If each orbital can hold a maximum of 3 electrons, the number of elements in 4th periodic table (long form) is. |
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Answer» |
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| 18. |
If E_(1)=0.5V corresponds to Cr^(3)+3e^(-)rarr Cr_((s)) and E_(2)=0.41V corresponds to Cr^(3+)+e rarr Cr^(2+) reactions, calculate the emf (E_(3)) of the reaction Cr^(2+)+e rarr Cr^(2+) reactions, calculate the emf (E_(3)) of the reaction Cr^(2+)+2e rarr Cr_((s)) |
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Answer» Solution :0.655V HINT : `E_(3)=(3E_(1)-E_(2))/(2)` |
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| 19. |
If E_(1)=0.5V corresponds to Cr^(3+)+3e^(-) rarr Cr_((s)) " and " E_(2)=0.41V corresponds to Cr^(3+)+e^(-) rarr Cr^(2+) reactions, calculate the emf (E_(3)) of the reaction Cr^(2+)+2e^(-) rarr Cr_((s)). |
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Answer» Solution :Given : `E_(1)=0.5V` `Cr^(3+)+3E^(-) RARR Cr_((s)) "...(1)"` `E_(2)=0.41V` `Cr^(3+)+e^(-) rarr Cr^(2+) "...(2)"` The REQUIRED reaction is, `Cr^(2+)+2E^(-) rarr Cr_((s))` Then, Formula : `E_(3)=(3E_(1)+E_(2))/2` Solution : `""=(3(0.5)+(0.41))/2=(1.5+0.41)/2` `""=0.955 V` `E_(3)=0.955V` |
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| 20. |
If E_(1 )^(0)is standard electrode potential for Fe//Fe^(+2) ,E_(2)^(0) is for Fe^(+2)//Fe^Fe^(+3) and E_(3)^(0)is for Fe// Fe^(+3)then the relation between E_(1)^(0), E_(2)^(0) and E_(3)^(0)will be : |
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Answer» `E_(3)^(0) = E_(1) ^(0) + E_(2)^(0)` |
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| 21. |
If E^(@) value of given ell is 1.1 V at 298 K temperature, then what is the value of equilibrium constant ? |
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Answer» `10^(-37)` `therefore K=1.65xx10^(37)`. |
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| 22. |
If E^0 of calomel electrode (C^(o-)|Hg_2Cl_2|Pt|] is 0.27V, its potential whenKCl=0.01 M would be- |
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Answer» 0.25V |
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| 23. |
If eis thecharge of an electron in esu m is the mass in grams and v the voltage 'h' is the planck constant in erg sec then the wavelength of the electron in cm is |
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Answer» `H//sqrt(Ve)` i.e., 1/2 m `upsilon^2=eV` or `upsilon=sqrt((2eV)/m)` `therefore lambda=h/(m upsilon)=h/(m((2eV)/m)^(1//2))=h/sqrt(2mVe)` |
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| 25. |
If (dx)/(dt)=k[H_(3)O^(+)]^(n) and rate becomes 100 times when pH changes form 2 to 1. Hence order of reaction is |
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Answer» 1 |
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| 26. |
If during a reacton ,in presence of enzyme.heat is evolved or adsorbed ,then what change we can obseve in it? |
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Answer» Increases |
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| 27. |
If D_(T) and D_(0) are the theoretical and observed vapour densities at a definite temperature and alpha bethe degree of dissociation of a substance. Then, α in the terms of D_(0),D_(T) and n (number of moles of product formed fro 1 mole reactant) is calculated by the formula:- |
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Answer» `ALPHA=(D_(0)-D_(T))/((1-n)D_(T))` `(D_(T))/(D_(0))=(1+alpha(n-1))/(1)` `alpha=(D_(T)-D_(0))/((n-1)D_(0))`. |
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| 28. |
If doubling the concentration of a reaction A increases the rate 4 times and tripling the concentration of A increases the rate 9 times , the rate is proportional to |
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Answer» Concentration of A |
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| 29. |
If dissociation for reaction, PCl_(5)hArrPCl_(3)+Cl_(2) is 20% at 1 atm, pressure. Calculate K_(c) |
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Answer» `0.04` `=(0.2xx0.2)/(0.8)=(0.04)/(0.8)=0.05` |
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| 30. |
If distance between two Pt electrode is 2 cm, cross intercept of 4.0 cm^(2) and resistance is 25Omega, then find out the molar conductivity of 0.5 M solution. |
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Answer» |
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| 31. |
If dispersed phase is gas and dispersion medium is liquid then the type of colloid is… |
| Answer» Solution :Foam | |
| 32. |
If dispersed phase is a liquid and the dispersion medium is a solid the colloid is known as : |
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Answer» a sol |
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| 33. |
If dispersed phase and dispersion medium both are solid, this type of sol is called ......... |
| Answer» SOLUTION :Solid sol | |
| 34. |
If density of some lake water is 1.25 g m L^(-1)and contains 92 g of Na^+ ions per kg of water, calculate the molality of Na^+ ion in the lake. |
| Answer» Solution :MOLALITY of `Na^+` IONS in the lake = `(92 //23)/(1 ) = 4`. | |
| 35. |
If density of a gaseous mixture of dinitrogen tetroxide (N_(2)O_(4)) and nitrogen dioxide (NO)_(2) is 2.5 gm/L at 127^(@)C and 1 atm pressure. [R=0.08 atm lit/mole-k] K_(p) for N_(2)O_(4)(g)hArr2NO_(2) is: |
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Answer» `0.90` 0.74 atm 0.26atm `k_(p)=((0.26)^(2))/(0.74)=0.091` `k_(p)=0.09` |
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| 36. |
If densities of two gases are in the ratio 1 : 2 and their temperatures are in the ratio 2 : 1, then the ratio of their respective molar mass at certain pressure is: |
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Answer» `1:1` `(M_(1))/(M_(2))=(d_(1)T_(1))/(d_(2)T_(2))=(1xx2)/(2xx1)=1/1` |
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| 37. |
If density of a gaseous mixture of dinitrogen tetroxide (N_(2)O_(4)) and nitrogen dioxide (NO)_(2) is 2.5 gm/L at 127^(@)C and 1 atm pressure. [R=0.08 atm lit/mole-k] Partial pressure of N_(2)O_(4) is: |
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Answer» 0.62atm `a_(0)-a_(0)aloha 2a_(0)alpha` `M_(avg)=(a_(0)xxM.Wr.N_(2)O_(4))/(a_(0)(1+alpha))` `PM=dRT` `1xxM_(avg)=2.5xx0.08xx400` `M_(avg)=(92)/(1+alpha)=80` `P_(N_(2)O_(4))=(1-alpha)/(1+alpha)xxP_(1)` `1+alpha=(92)/(80)` `P_(N_(2)O_(4))=(0.85)/(1.15)xx10.74` atm `alphah=0.15` `P_(NO_(2))=1-P_(N_(2)O_(4)` `1-0.74=0.26` atm |
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| 38. |
if density of an ideal gas when plotted against pressure exerted, shows the above variation at 273K. Find the molar mass (in gm/mol) the gas? |
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Answer» 19.4 `d=P((M)/(RT))` SLOPE `=TAN 60^(@)=(M)/(RT)` `M=sqrt(3)xx0.0821xx273approx39gm` |
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| 39. |
If DeltaT_(f) is the depression in freezing point for the electrolyte and DeltaT_(f)^(@) for the non-electrolyte of the same concentration, then Van't Hoff factor (i) is |
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Answer» `DeltaT_(F)XX DeltaT_(f)^(@)` |
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| 40. |
If DeltaH_(f(C_(2)H_(6)))^(0)(g) = -85 KJH mol^(-1) , DeltaH_(f(C_(3)H_(8)))^(0) (g) = -104 KJ mol^(-1), DeltaH^(0)for C (s) rarr C(g) is 718 KJ mol^(-1) and heat of formation of H-atom is 218 KJmol^(-1) then : |
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Answer» `DeltaH_(C-C) = 345` KJ Also, `DeltaH_(f(C_(2)H_(6)))^(0)(G) = 2xx 718 + 3xx436 - a - 6b"" (DeltaH_(C-C) = a, DeltaH_(C-H) = B)` `a+ 6b = 2829"""…."(i)` `3C + 4H_(2) rarrC_(2)H_(8)` `DeltaH_(f(C_(3)H_(8)))^(0) (g) = 3XX 718 + 4 xx 436 - 2a - 8b` `2a + 8b = 4002"""....."(ii)` `a = 345"&" b=414 KJ.` |
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| 41. |
If DeltaH_("vaporisation") of substance X (l) (molar mass : 30 g/mol) is 300 J/g at it's boiling point 300 K, then molar entropy change for revesible condensation process is |
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Answer» 30 J/mol K |
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| 42. |
If Delta_(t) represents CFSE in a tetrahedral crystal field t_(2) ….. The bari centre and e lies …… the bari centr. |
| Answer» SOLUTION :`0.4 Delta_(t)` above, `0.6 Delta_(t)` below | |
| 43. |
If DeltaH_f^@" for " H_2O and H_2O_2 " are " -280 and -188 kJ mol^(-1), the enthalpy change for the reaction, 2 H_2O_2 (l) to 2H_2O (l) + O_2(g) is : |
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Answer» `-196 KJ mol^(-1)` |
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| 44. |
If DeltaH is the enthalpy change and DeltaE the change in internal energy accompanying a gaseous reaction then |
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Answer» `DeltaH` is always less than `DeltaE` |
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| 45. |
If DeltaH is the change in enthalpy and DeltaE the change in internal energy accompanying a gaseous reaction |
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Answer» `DeltaH` is ALWAYS GREATER than `DeltaE` |
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| 46. |
If DeltaH is the change in enthalpy and DeltaE the change in internal energy for a gaseous reaction then |
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Answer» `DELTAH ` is ALWAYS GREATER than `DeltaE` |
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| 47. |
If DeltaH is the change in enthylpy and DeltaU, the change in internal energy accompanying a gaseous reactant then |
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Answer» `DeltaH` is always greater than `DELTAE` |
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| 48. |
If DeltaG^(@)[HI(g)=-1.7kJ], the equilibrium constant for the reaction 2HI(g)hArr H_(2)(g)+I_(2)(g) at 25^(@)C is |
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Answer» 24 |
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| 49. |
If DeltaG^@ (HI, g) is 1.7 kJ, what is the equilibrium constant at 20^@C for 2HI(g) hArr H_2(g)+ I_2(g) |
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Answer» 24 |
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