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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.
| 1. |
An intimate mixture of ferric oxide and aluminium is used as solid fuel in rockets. Calculate the fuel value per `cm^(3)` of the mixture. Heats of formation and densities are as follows: `H_(f(Al_(2)O_(3)))=-399kcal" "mol^(-1),H_(f(Fe_(2)O_(3)))=-199kcal" "mol^(-1)` ltbr. Density of `Fe_(2)O_(3)=5.2g//cm^(3)," Density of "Al=2.7g//cm^(3)` |
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Answer» Correct Answer - `0.9345" k Cal g"^(-1), 3.94" kCal cm"^(-3)` `Fe_(2)O_(3)+2Al rarr Al_(2)O_(3) +2Fe` `DeltaH=-399+199=-200` Total m wt of mixture `=214` `214 g rArr -200 rArr 1 g=-200/214=0.9345" kCal g"^(-1)` Vol. of `Fe_(2)O_(3)=160/5.2 =30.7rArr` Vol. of `Al_(2)O_(3)=54//2.7=20` Total `=30.7+20=50.7 c crArr 50.7 c c=200 rArr 1 c c=200/50.7=3.94" kCal cm"^(-3)` |
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| 2. |
An intimate mixture of `Fe_(2)O_(3)` and `Al` is used in solid fuel rocket. `DeltaH` of `Al_(2)O_(3)` and `Fe_(2)O_(3)` are `399kcal` and `199kcal` respectively. Calculate the fuel value in `kcal g^(-1)` of mixture. |
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Answer» Correct Answer - 1 `2Al+Fe_(2)O_(3) rarr Al_(2)O_(3)+2Fe, DeltaH=?` Given, `2Al+(3)/(2)O_(2)rarr Al_(2)O_(3),DeltaH=-399kcal ...(1)` and `2Fe+(3)/(2)O_(2) rarr Fe_(2)O_(3), DeltaH=-199kcal ...(2)` Subtracting `eq. (2)` from `eq. (10` `underset(Fuel mixture)(2Al+Fe_(2)O_(3))rarr 2Fe=Al_(2)O_(3),DeltaH=-200kcal` `Mol. wt. ` of fuel mixture `=2xx27+2xx56+48=214g` `:. 214g `mixture produce `=200kcal `heat `:. 1 g ` mixture produce `=(200)/(214)=0.9346~~1 kcal g^(-1)` |
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| 3. |
The species which by definition has ZERO standard molar enthyalpy of formation at 298 K isA. `Br_(2)(g)`B. `Cl_(2)(g)`C. `H_(2)O(g)`D. `CH_(4) (g)` |
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Answer» Correct Answer - B Only elementary substance For `Br_(2)(l) Delta_(f) H^(@) =0` |
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| 4. |
In a constant volume calorimeter, 3.5 g of a gas with molecular weight 28 was burnt in excess oxygen at 298.0 K. The temperature process. Given that the heat capacity of the calorimeter is was found to increase from 298.0 K to 298.45 K due to the combustion process. Given that the heat capacity of the calorimeter is `2.5 kJ K^(-1)`, the numerical value for the enthalpy of combustion of the gas in kJ `"mol"^(-1)` is |
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Answer» Correct Answer - `9 kJ mol^(-1)` `q=ms DeltaT=2.5xx0.45=1.125 kJ rArr DeltaE=-1.125/3.5xx28=-9" kJ/mole"` `|DeltaE|=9" kJ/mole"` [Since `DeltaH=DeltaE+Deltan_(g) RT` to determine `DeltaH` from `DeltaE` we must know `Deltan_(g)` (i.e. Chemical reaction or Compound) which is not given therefore having no other option answer is 9] |
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| 5. |
Compute the heat of formation of liquie methyl alcohol is kilojoule per mol using the following data. Heat of vaporisation of liquid methyl alcohol `=38kJ//mol`. Heat of formation of gaseous atoms from the elements in their standard states `:H=218kJ // mol, C=715kJ //mol, O=249kJ //mol.` Average bond energies`:` `C-H415kJ//mol, C-O 356kJ//mol, O-H 463kJ //mol.` |
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Answer» `C_((s))+2H_(2(g))+(1)/(2)O_(2(g)) rarr CH_(3)OH_((l)), DeltaH=?` `DeltaH=` Bond energy data for formation `+` Bond energy data for dissociation `+` Energy released during liquefaction `=-[3C-H+1C-O+1O-H]+[C_(srarrg)+2H-H+(1)/(2)O-O]-[CH_(3)OH_((g))rarr CH_(3)OH_((l))]` `=-[3xx415+356+463]+[715+2xx436+(1)/(2)xx249]-38` `=-266kJ mol^(-1)` |
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| 6. |
The standard heat of formation values of `SF_(6)(g), S(g)`, and `F(g)` are `-1100, 275`, and `80 kJ mol^(-1)`, respectively. Then the average `S-F` bond enegry in `SF_(6)` |
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Answer» Given, `S_((s))+3F_(2(g))rarr SF_(6),DeltaH=-1100kJ ....(1)` `S_((s)) rarr S_((g)),DeltaH=275kJ ...(2)` `(1)/(2)F_(2(g))rarr F_((g)), DeltaH=80kJ ..(3)` To get `SF_(6(g))rarr S_((g))++6F_((g)),,` we can proceed as `eq. (2)+6xxeq.(3)-eq.(1)` `SF_(6(g))rarrS_((g))+6F_((g)),DeltaH=1855kJ` Thus, average bond energy `=(1855)/(6)=309.17` |
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| 7. |
Select the correct option.A. `Delta H_(f) [H (g)]` is equal to `Delta H_("atomisaation")` of `H_(2) (g)`B. `Delta H_(BE) (H - H)` is equal to `Delta H_(f)` ofC. `DeltaH_(BE) (H - H)` is equal to `DeltaH_("atomisation")` of `H_(2) (g)`D. `Delta H_("comustion") [H_(2) (g)]` is equal to `Delta H_(f) [H_(2) (g)]` |
| Answer» Correct Answer - C | |
| 8. |
How much heat will be required at constant pressurent to form `1.28 kg` of `CaC_(2)` from `CaO(s)` & `C(s)`? `{:("Given :",Delta_(f)H^(@) (CaO","s),=,-152 kcal//mol),(,Delta_(f)H^(@) (CaC_(2)","s),=,-14 kcal//mol),(,Delta_(f)H^(@) (CO","g),=,-26 kcal//mol):}`A. `+112` kcalB. `224` kcalC. `3840` kcalD. `2240` kcal |
| Answer» Correct Answer - D | |
| 9. |
At 300 K, the standard enthalpies of formation of `C_(6)H_(5)COOH (s), CO_(2)(g)` & `H_(2)O (l)` are , `-408, -393` & `-286" kJ mol"^(-1)` respectively. Calculate the heat of combustion of benzoic acid at : (i) constant pressure (ii) constant volume. |
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Answer» Correct Answer - (i) -3201 kJ/mol (ii) -3199.75 kJ/mol `C_(6)H_(5)COOH+15/2 O_(2) rarr 7CO_(2)+3H_(2)O(l)` `DeltaH=(7xx-393)+3(-286)-(408)=-3201` `DeltaU=DeltaH-Deltan_(g) RT` `Deltan_(g)=-1/2` `DeltaU=-3199.75 kJ mol^(-1)` |
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| 10. |
Which of the following statement is wrong for `C_((graph ite))rarr C_((diamon d))?DeltaH^(@)=+1.90kJ`A. `DeltaH^(@)``_(f)` for`C_((graph ite))` is zeroB. `DeltaH^(@)``_(f)` for`C_((diamon d))` is `+1.90kJ`C. The enthalpy of formation of element in a form other than its most stable one is non zeroD. None of these |
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Answer» Correct Answer - D The standard heat of formation in most stable state is zero. |
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| 11. |
When a student mixes 50 mL of 1.0 M HCI and 50 mL of 1.0 M NaOH in a coffee cup calorimeter, the temperature of the resultant solution increases from 21.0°C to 27.5°C. Assuming that the calorimeter absorbs only a negligible quantity of heat, that the total volume of the solution is 100 mL, its density 1.0 g mL-1 and that its specific heat is 4.18 J/g, calculate : (a) the heat change during mixing. (b) the enthalpy change for the reaction :HCl(aq ) + NaOH(aq) → NaCl(aq) + H2O |
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Answer» mM of HCI = 50 x 1.0 = 50; mM of NaOH = 50 x 1.0 = 50 Heat changes during mixing = m.s.△T 50 mM of HCI and 50 mM of NaOH = (100 x 1) x 4.18 x 6.5 = 2717 J = 2.72 kJ Further, on mixing 50 mM of HCI and 50 mM of NaOH produces heat = 2.72 kJ 1000 mM of HCI (or 1000 Meq.) and 1000 mM of NaOH produces (acid and bases are monobasic and monoacidic, respectively).
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| 12. |
Which plot represents for an exothermic reaction ?A. B. C. D. |
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Answer» Correct Answer - A `DeltaH=H_(p)-H_(R)` Thus, `DeltaH` is negative because `H_(p)ltH_(R)` |
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| 13. |
Heat of neutralisation of `NaOH` and `HCl` is `-57.46 kJ //` equivalent. The heat of ionisation of water in `kJ //mol` is `:`A. `-57.46`B. `+57.46`C. `-114.92`D. `+114.92` |
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Answer» Correct Answer - B `H^(+)+OH^(-)rarr H_(2)O, DeltaH=-57.46kJ //eq.` `(` heat of neutralisation `)` `H_(2)O rarr H^(+)+OH^(-), DeltaH=57.46kJ//eq. or kJ //mol` |
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| 14. |
Statement `:` Heat of neutration for `HF ` is `-68.552kJ //eq.` whereas for `HCl` it is `-57.26kJ //eq.` Explanation `:` The acid `HF` is weak acid. |
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Answer» Correct Answer - a No doubt `HF` is weak acid but the higher values are due to extensive hydration of `F^(-)` ion being smallest anion `(` only `H^(-)` is smaller than `F^(-))`. |
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| 15. |
On the basis of the following thermochemical data `(Delta f G^(@)H_((aq))^(+)=0)` `H_(2)O(l) rarr H^(+)(aq)+OH^(-)(aq), DeltaH=57.32 kJ` `H_(2)(g)+1/2 O_(2)(g) rarr H_(2)O(l), DeltaH=-286.20 kJ` The value of enthalpy of formation of `OH^(-)` ion at `25^(@)C` is -A. `-228.88 kJ`B. `+ 228.88 kJ`C. `-343.52 kJ`D. `-22.88 kJ` |
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Answer» Correct Answer - A `H_(2)O(l) rarr H^(+) (aq) +OH^(-)(aq), DeltaH=57.32 kJ" "`....(i) `H_(2)(g) +1/2 O_(2)(g) rarr H_(2)O(l), DeltaH=-286.20 KJ" "`.....(ii) For the reaction `H_(2)(g)+1/2 O_(2) (g) rarr H^(+)(aq) +OH^(-)(aq)" "DeltaH=?` By adding equation (i) and (ii) `H_(2)(g)+1/2 O_(2)(g) rarr H^(+) (aq) +OH^(-) (aq)" "DeltaH=57.32 kJ -286.20 kJ` `DeltaH=-280.88" kJ/mol"` |
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| 16. |
Heat of neutralization `(DeltaH)` of `NH_(4)OH` and `HF` are `-51.5` and `-68.6 kJ` respectively. Calculate their heat of dissociation? (i) `HCl (aq)+NaOH(aq) rarr NaCl (aq)+H_(2)O, " "DeltaH= -57.3 kJ` (ii) `HCl (aq)+underset("(weak base)")(NH_(4)OH (aq)) rarr NH_(4)Cl (aq)+H_(2)O, " "DeltaH= -51.5 kJ` |
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Answer» `:.` The heat of dissociation of `NH_(4)OH`, `DeltaH=-51.5-(-57.3)=5.8 kJ` Similarly we have `HF(aq) +NaOH(aq) rarr NaF(aq)+H_(2)O, DeltaH=-68.6 kJ` `:.` The heat of diffociation of `HF`, `DeltaH =-68.6-(-57.3)=-11.3 kJ` |
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| 17. |
Calculate the heat of neutralization by mixing `200mL` of `0.1 MH_(2)SO_(4)` and `200mL ` of `0.2 MKOH `if heat generated by the mixing is `2.3 kJ`. |
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Answer» `Meq. `of `H_(2)SO_(4)=200xx0.1xx2=40` `(H_(2)SO_(4)` is dibasic `)` `Meq. `of `KOH=200xx0.2=40` `:. 40 Meq. Of H_(2)SO_(4)` and `40Meq.` of `KOH` on mixing gives heat `=2.3kJ` `:. 1000meq. ` of `H_(20SO_(4)` and `1000Meq.` of `KOH` on mixing gives heat `=(2.3xx1000)/(40)=57.5kJ` |
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| 18. |
For the reaction `CO (g) + (1)/(2) O_(2) (g) rarr CO_(2) (g)` using data given in table find out incorrect statement (s) among the following `{:(,Delta H_(f)^(@) (kJ//"mole"),S^(@) (J//K"mole"),),(CO (g),-110,+197,),(O_(2) (g),0,+205,),(CO_(2) (g),-395,+213,):}` Assume vibration modes of motion do not contribute to heat capacity at low temperature.A. `Delta H^(@) gt Delta U^(@)` for the reaction at 298 KB. In standard state conditin, the rection `CO (g) + (1)/(2) O_(2) (g) rarr CO_(2) (g)` attain equilibrium at very hight temperature.C. At low temperature `(d(Delta H)^(@))/(dT) = - ve`D. In a `CO, O_(2)` fuel cell electrical energy obtained by cell `gt |Delta H_("combustion")^(@) [CO (g)]|` |
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Answer» Correct Answer - A `DeltaH^(@)-DeltaU^(@)=Deltan_(g)RT` since `Deltan_(g)=-ve` `rArr DeltaH^(@)-DeltaU^(@) lt 0` `DeltaH^(@)=-ve` `DeltaS^(@)=-ve" "," "C_(p)=3/2 R+R" "," "=5/2 R` `Delta_(r)C_(p)=-ve` |
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| 19. |
The following sequence of reaction occurs in commercial production of aqueous of aqueous nitric acid. `{:(4 NH_(3) (g) + 5O_(2) (g) rarr 4NO (g) + 6 H_(2) O (l),Delta H = - 904 kJ.....(1),),(2 NO (g) + O_(2) (g) rarr 2NO_(2) (g) ,Delta H = - 112 kJ.....(2),),(3 NO_(2) (g) + H_(2) O (l) rarr 2HNO_(3) (aq) + NO (g),Delta H = - 140 kJ.....(3),):}` Determine the total heat liberated (in kJ/mol) at constant pressure for the production of exactly 1 mole of aqueous nitric acid from `NH_(3)` by this process.A. 986B. 493C. 246.5D. none of these |
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Answer» Correct Answer - B 1 mole of `HNO_(3)=3/2` moles of `NO_(2) rarr 3/2` mole of `NO rarr 3/2` mole of `NH_(3)` `(3/2xx1/4)(904)-(3/2xx1/2)(112)-(3/2xx1/3)(140)=493" kJ/mol"` Heat liberated `=493" kJ/mol"` |
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| 20. |
If `Ag_(2)O (s)` is ecposed to atmosphere having pressure 1 atm and temperature 1 atm and temperature `27^(@)C`. Under these conditions comment whether it will dissociate spontaneously or not. `2Ag_(2)O (s) hArr 3Ag(s)+O_(2)(g)` `{:("Given :",,DeltaH_(f)^(@)" (kJ/mol)",DeltaS^(@)" (J/K mol) at "27^(@)C),(,Ag(s),0,42.0),(,Ag_(2)O(s),-30,121.0),(,O_(2)(g),0,204.0):}` (Air consist of `20% O_(2)` by volume) Take `: R=8.3 JK^(-1) mol^(-1)` |
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Answer» `DeltaG^(@)=DeltaH^(@)-T DeltaS^(@)` `DeltaH^(@)=DeltaH_(f)^(@)("product")-DeltaH_(f)^(@) ("reactants")` `=2xx30=60 kJ` `DeltaS^(@)=204+4(42)-2(121)=+130` `DeltaG^(@)=DeltaH^(@)-T DeltaS^(@)=60000-300xx130` `DeltaG^(@)=21000 J=-RT ln K` `log K=-(21000/(300xx8.3xx2.3))` `K_(p)=2.15xx10^(-4)` atm The dissociation of `Ag_(2)O` is nonspontaneous at `27^(@)C` |
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| 21. |
Calculate `Delta G` (in kJ) for the reaction at 300 K, `H_(2)(g)+Cl_(2)(g) rarr 2HCl (g)` Given at 300 K, `BE_(H-H) = 435 kJ mol^(-1), BE_(Cl-Cl) = 240 kJ mol^(-1), BE_(HCl) = 430 kJ mol^(-1)` Entropies of `H_(2), Cl_(2)` and HCl are 131, 223 and `187 JK^(-1) mol^(-1)` respectively.A. 191B. 291C. `-191`D. None of these |
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Answer» Correct Answer - C `DeltaH=(435+240-2xx430)=- 185" kJ mol"^(-1)` `DeltaS=(2xx187-(131+223))/(1000)=20/1000"kJ mol"^(-1)` `DeltaG=DeltaH-TDeltaS` `=-185-(300xx20)/(1000)=-185-6=-191" kJ mol"^(-1)` |
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| 22. |
Statement`:` Heat of combustion are always exothermic. Explanation `:` Combustio of `N_(2)` to give `NO` is exothermic. |
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Answer» Correct Answer - a `(1)/(2) N_(2)+(1)/(2)O_(2) rarr NO,DeltaH=+ve.` It is due to high bond energy of `N_(20`. Note this equation does not represent heat of combustion of `N_(2)(` Partial combustion `)`. |
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| 23. |
Heat of neuralisation is amount of heat evolved or absorbed when `1g-` equivalent of an acid reacts with `1g-` equivalent of a base in dilute solution . If weak acid or weak bae are neutralised, the heat released during neutralisation is somewhat lesser than `-13.7kcal` or `-57.27 kJ` . Het of neutralisation is also referred as heat of formation of water from `H^(+)` and `OH^(-)` ions `i.e., H^(+)+OH^(-) rarr H_(2)O, DeltaH=-13.7kcal. ` `200mL` of `0.1M NaOH` is mixed with `100mL` of `0.1M H_(2)SO_(4)` in 1 experiment. In `II` experiment `100mL` of `0.1M NaOH` is mixed with `50mL` of `0.1M H_(2)SO_(4)`. Select the correct statements: `(1)` heat liberated in each of the two reactions is `274cal`. `(2)` heat liberated in `I` is `274 cal` and in `II` is `137 cal`. ,brgt `(3)` temperature rise is `I` reaction is equal to the temperature rise in `II`. `(4)` temperature rise in `I` reaction is equal to the temperaure rise in `II`A. `1,3`B. `2,4`C. `2,3`D. `1,4` |
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Answer» Correct Answer - b `1 Meq.` of `NaOH=20, Meq . Of H_(2)SO_(4)=20` ltbr. `II Meq. Of NaOH=10, Meq. ofH_(2)SO_(4)=10` In `I,DeltaH=-(13.7xx10^(3)xx20)/(1000)` `=274 cal. `taken by `300mL, ` In `II: DeltaH=-(13.7xx10^(3)xx10)/(1000)` `=137cal. `taken by `150 mL` |
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| 24. |
The heat of dissociation of `H_(20` is `435kJ mol^(-1)`. If `C_((s)) rarr C_((g)), DeltaH=720kJ mol^(-1)`, calculate the bond energy per mole of the `C-H` bond in `CH_(4)` molecule . `(DeltaH_(f) fo r CH_(4)=-75kJ mol^(-1))` |
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Answer» Correct Answer - `416.25kJ mol^(-1);` |
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| 25. |
When `100 c.c.` of a mixture of methane `(CH_(4))` and ethylene `(C_(2)H_(4))` was exploded with an excess of oxygen , the volume of carbon dioxide produced `(` measured at the same temperature and pressure `)` was `160 c.c.` Calculate the heat evolved when `22.4` litre of the mixture of methane and ethylene `(` measured at `N.T.P. )` is completely oxidised to carbon dioxide and water at constant volume. `CH_(4)+2O_(2)=CO_(2)+2H_(2)O+212000cal,` `C_(2)H_(4)+3O_(2)=2CO_(2)+2H_(2)O+333000 cal.` |
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Answer» Correct Answer - `284.60kcal;` |
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| 26. |
A person takes 15 breaths per minute. The volume of air inhaled in each breath is 448 ml and contains 21% of oxygen by volume. The exhaled air contains 16% of oxygen by volume. If all the oxygen is used in the combustion of sucrose, how much of the Sucrose is burnt in the body per day & how much heat is evolved. `Delta H_(com)` of sucrose is `= 6000 kJ mol^(-1)`. Take temperature to be 300 K throughout. [Assuming `V_("inhaled air") = V_("exhalded air"), P_(atm) = 1atm`] |
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Answer» Correct Answer - 9.822 MJ/day evolved, 560 gm Oxygen inhaled `=21/100xx448=94.08` Exhaled air `=16/100xx448=71.68` Oxygen used `=22.4 ml` (In one breath) Total oxygen in a day `=22.4 mL` (in one breath) Total oxygen used in a day `=22.4xx15xx60xx24=483840 mL =483.84 L` Apply `PV=nRT` `n=19.644` Moles of sucrode used `=1/12xx19.644=1.637` Total energy evolved `=1.637xx6000=9822 kJ` |
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| 27. |
`C_(p)` and `C_(v)` denote the molar specific heat capacities of a gas at constant pressure and volume respectively. Then `:`A. `C_(p)+C_(v)` is larger for a diatomic ideal gas than for a monoatomic ideal gas.B. `C_(p)-C_(v)` is larger for a diatomic ideal gas ghtan for a monoatomic ideal gas.C. `C_(p).C_(v)` is larger for a diatomic ideal gas than for monoatomic ideal gas.D. `C_(p)//C_(v0` is larger for a diatomic ideal gas than for a monoatomic ideal gas. |
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Answer» Correct Answer - A::C |
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| 28. |
Calculate the heat produced when 3.785 litre of actane `(C_(8)H_(18))` reacts with oxygen to form `CO` & water vapour at `25^(@)C`. The density of octane is 0.7025 gm/ml. enthalpy of combustion of `C_(8)H_(18)` is `-1302.7` k Cal/mol. `{:(DeltaH_(f)^(@) CO_(2)(g)=-94.05" k Cal mol"^(-1),,,DeltaH_(f)^(@) CO(g)=-26.41" k Cal mol"^(-1),),(DeltaH_(f)^(@) H_(2)O(l)=-68.32" k Cal mol"^(-1),,,DeltaH_(f)^(@) H_(2)O (g)=-57.79" k Cal mol"^(-1)):}` |
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Answer» Correct Answer - 15.5528 M Cal `C_(8)H_(18)+25/2 O_(2) rarr 8 CO_(2) +9H_(2)O` `DeltaH_(f)^(@)C_(8)H_(18) =-64.58` `C_(8)H_(18) +17/2 O_(2) rarr 8 CO+9 H_(2)O` `DeltaH=-666.73` `DeltaH` for `3.785` litre of octane `=(666.73xx3.785xx0.7025xx10000)/(114)` `15552.8" kCal"=15.5528" M kCal"` |
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| 29. |
The enthalpy of formation of `C_(2)H_(5)OH(l)` is `-66 k Cal//mol`. The enthalpy of combustion of `CH_(3)-O-CH_(3)` is `-348 k Cal//mol`. Given that the enthalpies of formation of `CO_(2)(g)` and `H_(2)O(l)` are `-94 k Cal//mol` & `-68 k Cal//mol` respectively, calculate `Delta H` for the isomerisation of ethanol to methoxymethane. All data are all `25^(@)C`. |
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Answer» Correct Answer - `22" kCal mol"^(-1)` `C_(2)H_(5)OH rarr CH_(3)COCH_(3)` `DeltaH=66+348-188-204=22" kCal mol"^(-1)` |
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| 30. |
The molar heat capacities at constant pressure (assumed constant with respect to temperature) of A, B and C are in ratio of `1.5 : 3.0 : 2.5`. If enthalpy change for the exothermic reaction `A+2B rarr 3C` at `300 K` and `310 K` is `DeltaH_(1)` and `DeltaH_(2)` respectively thenA. `DeltaH_(1) gt DeltaH_(2)`B. `Delta_(1) lt DeltaH_(2)`C. `DeltaH_(1)=DeltaH_(2)`D. If `T_(2) gt T_(1)` then `DeltaH_(2) gt DeltaH_(1)` 7 if `T_(2) lt T_(1)` then `DeltaH_(2) lt DeltaH_(1)` |
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Answer» Correct Answer - C `Delta_(r)C_(p)=3xx2.5-1xx1.5-2xx3rArr 0," ":. " "DeltaH_(1)=DeltaH_(2)` |
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| 31. |
Calculate the enthalpy of combustion of benzene `(l)` on the basis of the following data: a. Resonance energy of benzene `(l) =- 152 kJ// mol` b. Enthalpy of hydrogenation of cyclohexene `(l) =- 119 kJ//mol` c. `Delta_(f)H^(Theta)C_(6)H_(12)(l) =- 156 kJ mol^(-1)` d. `Delta_(f)H^(Theta) of H_(2)O(l) =- 285.8 kJ mol^(-1)` e. `Delta_(f)H^(Theta)of CO_(2)(g) =- 393.5 kJ mol^(-1)` |
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Answer» Correct Answer - `DeltaH_(C)^(@) ("benzene")=-3267.4" kJ mol"^(-1)` Resonance energy = observed energy - calculated energy Calculate energy `=-3419.4` Observed `=-3419.4-(-152)=-3267.4" kJ mol"^(-1)` |
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| 32. |
The enthalpy of formation of ethane, ethylene and benzene from the gaseous atoms are -2839.2, -2275.2 and `-5506" kJ mol"^(-1)` respectively. Calculate the resonance energy of benzene. The bond enthalpy of `C-H` bond is given as equal to +410.87 kJ/mol. |
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Answer» Correct Answer - `-23.68` kJ/mol (i) `2C+6H rarr CH_(3)-CH_(3)" "DeltaH=-2839.2` `-2839.2=0-[(410.87xx6)]+C-C]` `C-C=373.98` (ii) `2C+4H rarr C_(2)H_(4)` `-22275.2=0-[1643.48+C=C]` `(C = C)=631.72` (iii) `6C+6H square rarr C_(6)H_(6)` `DeltaH_("calculayed")=5482.32` Resonance energy `=5482.32-5506=-23.68" kJ mol"^(-1)` |
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| 33. |
the enthalpy of combustion of acetylene is -312 kCal per mole. If enthalpy of formation of `CO_(2)` & `H_(2)O` are `-94.38` & `-68.38` kCal per mole respectively, calculate `C equiv C` bond enthalpy. given that enthalpy of atomisation of C is 150 kCal per mole and `H-H` bond enthlpy and `C-H` bond enthalpy are 103 kcal per mole and 93.64 kCal per mole respectively. |
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Answer» Correct Answer - `E_(C equiv C)= 160.86` kCal/mol Enthalpy of formation of `C_(2)H_(2)=54.92` `2 C(s)+H_(2) rarr HC equiv C-H` `54.92=403-187.28-x" "[x=B.E." of "C equiv C]` `x=160.86" kCal/mol"` |
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| 34. |
The enthalpy of formation of enthane and benzene from the gaseous atoms are `-2839.2` and `-5506 kJ//mol` respectively. Bond enthalpy of `C = C` bond is Given: Resonance energy of benzene `=-23.68" kJ/mol"` Bond enthalpy of `C-H` bond `=411.0" kJ/mol"`A. 373.98 kJ/molB. 632.24 kJ/molC. 647.5 kJ/molD. 1896.72 kJ/mol |
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Answer» Correct Answer - B `-2839.2=0-[6xx411+DeltaH_(C-C)]` `DeltaH_(C-C)=373.2" KJ/mol"` `6C(g) +6H(g) rarr C_(6)H_(6)(g)` `DeltaH_("theoritical")=0-[6xx411+3xx373.2+3x]` `R.E.="Actual"-"Theoritical"` `-23.68=-5506+[6xx411+3xx373.2+3x]` `3x=5506-23.68-2466-1119.6` `x=632.24" KJ/mol "`or`" "DeltaH_(C-C)=632.24" KJ/mol"` |
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| 35. |
Using the given data calculate enthalpy of formation of acetone (g). [All value in `"KJ mol"^(-1)`] bond enthalpy of : `{:(C-H=413.4,,,C-C=347.0,,,(C=O)=728.0,,),((O=O)=495.0,,,H-H=435.8,,,Delta_("sub")H" of "C=718.4.,):}` |
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Answer» Correct Answer - `-192.3" kJ mol"^(-1)` `3C+3H_(2)+1/2 O_(2) rarr CH_(3) -underset(O)underset("||")(C)-CH_(3)` Calculate `DeltaH` `DeltaH=-192.3" kJ mol"^(-1)` |
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| 36. |
Using the data provided, calculate the multiple bond energy `(kJ mol^(-1))` of a `C-=C` bond in `C_(2)H_(2)`. That energy is `(` take the bond energy of a `X-H` bond as `350kJ mol^(-1))`. `2C_((s))+H_(2(g))rarr C_(2)H_(2(g)), Delta=225kJ mol^(-1)` `2C_((g))rarr 2C_(g)), DeltaH=1410kJ mol^(-1)` `H_(2(g)) rarr 2H_((g)),DeltaH=330kJ mol^(-1)`A. 1165B. 837C. 865D. 815 |
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Answer» Correct Answer - D Ethyne molecule breaks as `C_(2)H_(2(g))rarr 2C_((g))+2H_((g)) ….(i)` Given equations are `2C_((s))+H_(2(g))rarr H-C-=C-H_((g)) , DeltaH=225kJ mol^(-1) ....(ii)` `2C_((s)) rarr 2C_((g)), DeltaH=1410kJ mol^(-1) ...(iii)` `H_(2(g))rarr 2H_((g)), DeltaH=330kJ mol^(-1) ...(iv)` `:. 225=2xxDeltaH_(su b)C_((s))+B.E._((H_(2)))-2B.E._(C-H)-1xxB.E._(c-=C)` `225=2xx705+330-2xx350-B.E._(C-=C)` `225=1410+330-700-B.E._(C-=C)` `:. B.E._(C-=C)=815kJ mol^(-1)` |
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| 37. |
A system is provided `50xx10^(3)J` energy and work done on the system is `100J.` The change in internal energy is `:`A. `50kJ`B. `50.1kJ`C. `150J`D. `50J` |
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Answer» Correct Answer - B `q=DeltaU+(-W)` or `DeltaU=q+W` Here `W` is work done on the system `=50xx10^(3)+100=50100J` `=50.1kJ` |
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| 38. |
Which fuel provides the highest calorific values ?A. CharcoalB. KeroseneC. WoodD. Dung |
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Answer» Correct Answer - B An experiment fact. |
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| 39. |
The melting point of a certain substance is `70^(@)C` and its normal boiling point is `450^(@)C`. Its enthalpy of fusion is `30.0cal//g`, enthalpy of vaporisation is `45.0 cal//g` and its heat capacity is `0.215 cal g^(-1)K^(-1)`. Calculate the heat required to convert `100g` of substance from solid state at `70^(@)C` to vapour state at `450^(@)C`. |
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Answer» `underset(70^(@)C)(Soli d) overset(DeltaH_(1))rarrunderset(70^(@)C)(Liqui d)overset(DeltaH_(2))rarr underset(450^(@)C)(Liqui d) overset(DeltaH_(3))rarr underset(450^(@)C)(Vapour)` `DeltaH_(Total)=DletaH_(1)+DeltaH_(2)+DeltaH_(3)` `=30xx100+100xx0.215xx380+45xx100` `3000+8170+4500` `=15670cal=15.67kcal` |
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| 40. |
Calculate enthalpy change of the following reaction `:` `H_(2)C=CH_(2(g))+H_(2(g)) rarr H_(3)C-CH_(3(g))` The bond energy of `C-H,C-C,C=C,H-H` are `414,347,615` and `435k J mol^(-1)` respectively. |
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Answer» `DeltaH_(Reaction)=` Bond energy data for the formation of bond `+` Bond energy data for the dissociation of bond `=-[1-(C-C)+6(C-H)]+[1-(C=C)+4(C-H)+1-H-H)]` `=-1(C-C)-2(C-H)+1(C=C)+1(H-H)` `=-347-2xx414+615+435=-125kJ` |
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| 41. |
Calculate enthalpy change of the following reaction `:` `H_(2)C=CH_(2(g))+H_(2(g)) rarr H_(3)C-CH_(3(g))` The bond energy of `C-H,C-C,C=C,H-H` are `414,347,615` and `435k J mol^(-1)` respectively.A. `+125kJ`B. `-125kJ`C. `+250kJ`D. `-250kJ` |
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Answer» Correct Answer - B `DeltaH=-[e_(C-C)+6xxe_(C-H)]+[e_(C=C)+4xxe_(C-H)+e_(H-H)]` `=-[347+6xx414]+[615+4xx414+435]` `=-125kJ` |
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| 42. |
The enthalpies of combustion of carbon and carbond monoxide are `-393.5` and `-283kJ mol^(-1)` respectively . The enthalpy of formation of carbond monoxide per mole is `:`A. `-110.5kJ`B. `676.5kJ`C. `-676.5kJ`D. `110.5kJ` |
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Answer» Correct Answer - A `C+O_(2)rarr CO_(2),DeltaH=-393.5kJ` `ul ( {: (CO,+(1)/(2)O_(2),rarrr,CO_(2),,DeltaH=-283kJ),(-,-,,-," +"):} )` `C+(1)/(2)O_(2) rarr CO,DeltaH=-110.5kJ` |
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| 43. |
The heat of atomisation of `PH_(3(g))` is `228kcal mol^(-1)` and that of `P_(2)H_(4)` is `355kcal mol^(-1)`. Calculate the average bond energy of `P-P` bond.A. 102B. 51C. 26D. 204 |
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Answer» Correct Answer - B `PH_(3) rarr P+3H,DeltaH=228` `H_(2)P-PH_(2) rarr 2P+4H,DeltaH=355` `e_(P-P)+4xxe_(P-H)=355` `e_(P-P)+4xx(228)/(3)=355` `:. e_(P-P)=51` |
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| 44. |
The heat of atomisation of `PH_(3)(g)` and `P_(2)H_(4)(g)` are `954 kJ mol^(-1)` and `1485 kJ mol^(-1)` respectively. The `P-P` bond energy in `kJ mol^(-1)` is |
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Answer» Correct Answer - 213 kJ/mol `PH_(3) rarr P+3H" "DeltaH=954` `P-H` bond enthalpy `=954/3=318 kJ` `H-underset(H)underset(|)(P)-underset(H)underset(|)(P)-H rarr 2P+4H" "DeltaH=1485` `1485=P-P+(4xx318)` `P-P=213" kJ mol"^(-1)` |
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| 45. |
The heat of atomisation of `PH_(3(g))` is `228kcal mol^(-1)` and that of `P_(2)H_(2)` is `355kcal mol^(-1)`. Calculate the average bond energy of `P-P` bond. |
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Answer» `PH_(3(g))rarr P_((g))+3H_((g)), DeltaH=228kcal//mol` `H_(2)P=PH_(2)rarr 2P_((g))+4H_((g)),DeltaH=355kcal//mol` `:. DeltaH=e_(P-P)+4e_(P-H)` or `355-e_(P-P)+4xx(228)/(3)( :. e_(P-H)=(228)/(3))` `:. e_(P-P)=51kcal//mol` |
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| 46. |
Standard molar enthalpy of formation of `CO_(2)` is equal to `:`A. zeroB. the standard molar enthalpy of combustionC. the sum of standard molar enthalpies of formation of `CO` and `O_(2)`D. the standard molar enthalpy of combustion of carbon `( ` graphite `)` |
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Answer» Correct Answer - D Standard molar heat enthalpy `(H^(@))` of a compound is equal to its standard heat of formation from most stable states of initial components. |
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| 47. |
Calculate the standard enthalpy of combustion of `CH_(4)`, it standard enthalpies of formation of `CH_(4(g)),H_(2)O_((l)),` and `CO_(2(g))` are `-74.81kJ mol^(-1), -285.83kJ mol^(-1)` and `-393.51kJ mol^(-1)` respectively. |
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Answer» The desired equation is `:` `CH_(4)+2O_(2)rarr CO_(2)+2H_(2)O, DeltaH_(1)^(@)=? …(1)` Given that `:` `C_((s))+2H_(2) rarr CH_(4(g)),` `Delta_(2)^(@)=-74.81kJ …(2)` `H_(2(g))+(1)/(2)O_(2(g))rarr H_(2)O_((l)),` `DeltaH_(3)^(@)=-285.83kJ ...(3)` `C_((s))+O_(2(g)) rarr CO_(2(g)),` `DeltaH_(4)^(@)=-393.51kJ ....(4)` The obtain `eq. (1)` following mathematical operation must be performed. `2xx(3)+(4)-(1)` `2H_(2(g))+O_(2(g))+C_((s))+O_(2(g))-C_((s))-2H_(2) rarr 2H_(2)O_((l))+CO_(2(g))-CH_(4(g))` or `CH_(4(g))+2O_(2(g)) rarr CO_(2(g))+2H_(2)O_((l)) ...(i)` Same mathematical treatment is done with `DeltaH` values `DeltaH_(1)^(@)=[2xxDeltaH_(3)^(@)+DeltaH_(4)^(@)]-[DeltaH_(2)^(@)]` `=[2xx(-285.83)+(-393.5)]-[-74.81]` Thus, enthalpy of combustion of `CH_(4)` `(i.e., DeltaH_(1)^(@)=-890.36kJ)` |
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| 48. |
Calculate standard enthalpies of formation of carbon-di -sulphide `(l)`. Given the standard enthylpy of combustion of carbon (s), sulphur (s) & carbon-di-sulphide `(l)` are :- `393.3, -293.72` and `-1108.76 kJ mol^(-1)` respectively. |
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Answer» Correct Answer - 128.02 kJ `C+2S rarr CS_(2)` `DeltaH=-393.3-587.44+1108.76=128.02 kJ` |
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| 49. |
`Delta_(f)^(@)` for `CO_(2(g)), CO_((g))` and `H_(2)O_((g))` are `-393.5,-110.5` and `-241.8kJ mol^(-1)` respectively. The standard enthalpy change `(` in `kJ)` for given reaction is `:` `CO_(2(g))+H_(2(g)) rarr CO_((g))+H_(2)O_((g))`A. `+524.1`B. `+41.2`C. `-262.5`D. `-41.2` |
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Answer» Correct Answer - B Given, `C+O_(2) rarr CO_(2), DeltaH^(@)=-393.5kJ ….(1)` `C+(1)/(2)O_(2) rarr CO, DeltaH^(@)=-110.5kJ ….(5)` `ul (H_(2)+(1)/(2)O_(2)rarr H_(2)O, DeltaH^(@)=-241.8kJ ....(3))` `CO_(2)+H_(2) rarr CO+H_(2)O,DeltaH^(@)=+41.2` By `(2)+(3)-(1)` `DeltaH^(@)=+41.2` |
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| 50. |
From the following data at `25^(@)C`, Calculate the standard enthalpy of formation of `FeO(s)` and of `Fe_(2)O_(3) (s)`. `{:(,"Reaction",Delta_(r)H^(@)" (kJ/mole)"),((1),Fe_(2)O_(3)(s)+3C ("graphite") rarr 2Fe(s) +3CO(g),492.6),((2),FeO(s)+C("graphite") rarr Fe(s)+CO(g),155.8),((3),C("graphite")+O_(2) (g) rarr CO_(2)(g),-393.51),((4),CO(g)+1//2O_(2)(g) rarr CO_(2)(g),-282.98):}` |
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Answer» Correct Answer - `-266.3` kJ/mol and `-824.2` kJ/mol (i) `Fe+1/2 O_(2) rarr FeO` `DeltaH=-155.8-393.51+282.98=-266.3" kJ mol"^(-1)` (ii) `2Fe+3CO rarr Fe_(2)O_(3)+3C` `DeltaH=-492.6+848.94-1180.53=824.19" kJ mol"^(-1)` |
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