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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.
| 22851. |
Which of the following has asymmetric carbon atom? |
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Answer» `CH_(2)Cl-CH_(2)BR` |
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| 22852. |
The vapour pressure of water depends upon: |
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Answer» SURFACE AREA of container |
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| 22853. |
When phenyl magnesium bromide reacts with t-butanol, thhe product would be:- |
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Answer» Benzene |
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| 22855. |
Which of the following substance is used to extinguish fire in substances like oil, fat and petrol ? |
| Answer» Answer :D | |
| 22856. |
Which oneof the following has largest number of isomers ? |
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Answer» `[CR(SCN)_2(CH_3)_4]^(+)` |
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| 22857. |
Which of the following is a female sex hormone: |
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Answer» Estrogen |
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| 22858. |
Which compound does not dissolve in hot, dilute HNO_3 ? |
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Answer» Hgs |
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| 22859. |
The white phosphorous can be changed into ………..by heated it to 420^(@)C in the absence of air aned light. |
| Answer» SOLUTION :RED PHOSPHOROUS | |
| 22860. |
Which of the following reaction occur most rapidly ? |
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Answer» `(CH_(3))_(2)CHOHoverset(HBR)(to)` |
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| 22861. |
Which of the following reactions does NOT lead to formation of ethanol ? |
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Answer» `CH_3COOC_2H_5 overset("HYDROLYSIS")to` |
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| 22862. |
Usually lanthanide forms |
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Answer» ionic compounds |
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| 22863. |
To a 100 mL of 0.1 M weak acid HA solution 22.5 mL of 0.2 M solution of NaOH are added.Now, what volume of 0.1 M NaOH solution be added into above solution, so that pH of resulting solution be 4.7: [Given :(K_b(A^(-))=5xx10^(-10)] |
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Answer» Solution :`HA+NaOHtoNaA+H_2O` `t=0 " " 10 " " 4.5 " " 0 " " 0` t=50% `" " 5 " " 0 " " 5` 5 mL NAOH is required |
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| 22864. |
Which of the following is amphoteric oxide ? Mn_(2)O_(7), CrO_(3), Cr_(2)O_(3), CrO, V_(2)O_(5), V_(2)O_(4) |
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Answer» `V_(2)O_(5), Cr_(2)O_(3)` |
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| 22865. |
When pressure is increases at constant temp volume of gas decreases AB rarr gases, BC rarr vpour +liquid, CD rarr liquid critical point: At this point all the physical propeties of liquid phase will be same as the physical properties in vapour such as, density of liquid = density of vapour T_(c) or critical temp: Temperature abive which a gas can not be liquified P_(c) or critical pressure: minimum pressure which must be applied at critical temp to convert the gas into liqid. V_(c) or critical volume: volume occupied by one mole of gas at T_(c) & P_(c) CRITICAL CONSTANT USING VANDERWAAL EQUATIONS: {:((P+(a)/(V_(m)^(2)))(V_(m)-gb)=RT,rArr,(PV_(m)^(2)+a)(V_(m)-b)=RTV_(m)^(2)),(PV_(m)^(3)+aV_(m)-PbV_(m)^(2) -ab -RTV_(m)^(2) =0, rArr, V_(m)^(3)+V_(m)^(2) (b+(RT)/(P))+(a)/(P)(V)/(m)-(ab)/(P) = 0):} since equation is cubic in V_(m) hence there will be three roots of equation of any temperature and pressure. At critical point all three roots will coincide and will given single value of V_(m) = V_(c) at critical point. Vander Waal equation will be V_(m)^(3) - V_(m)^(2) (b+(RT_(C))/(P_(C))) +(a)/(P_(C)) V_(m) - (ab)/(P_(C)) = 0 ...(i) But at critical point all three roots of the equation should be equal, hence equation should be: V_(m) = V_(c) (V_(m) -V_(c))^(3) = 0 V_(m)^(3) - 3V_(m)^(2) V_(c) +3V_(m) V_(c)^(2) -V_(c)^(3) = 0 ..(2) comparing with equation (1) b +(RT_(c))/(P_(c)) = 3V_(c) ...(i) (a)/(P_(c)) = 3V_(c)^(2) ...(ii) (ab)/(P_(c)) = V_(c)^(3) ..(iii) From (ii) and (iii), V_(c) = 3b From (ii) P_(c) = (a)/(3V_(c)^(2)) substituting P_(c) = (a)/(3(3b)^(2)) = (a)/(27b^(2)) From (i) (RT_(c))/(P_(c)) = 3V_(c) -b = 9b -b = 8b rArr T_(c) = (8a)/(27Rb) At critical point, the slope of PV curve (slope of isotherm) will be zero at all other point slope will be negative zero is the maximum value of slope. ((delP)/(delV_(m)))_(TC) =0 ..(i) (del)/(delV_(m)) ((delP)/(delV_(m)))_(TC) = 0 ...(ii) {Mathematically such points an known as point of inflection (where first two derivatives becimes zero)} using the two T_(c)P_(c) and V_(c) can be calculated by A scientist proposed the following equation of state P = (RT)/(V_(m)) - (B)/(V_(m)^(2)) +(C )/(V_(m)^(3)). If this equation leads to teh critical behaviour then critical temperature is: |
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Answer» `(8B)/(27RC)` `(delp)/(delV_(m)) = 0 RARR -(RT_(C))/(V_(m)^(2)) +(2B)/(V_(m)^(3)) - (3C)/(V_(m)^(4)) = 0 rArr -RT_(c)+ (2B)/(V_(m)) -(3C)/(V_(m)^(2)) = 0 rArr RT_(c)V_(m)^(3) - 2BV_(m) +3C = 0` as equation will have repeated root then `D = 0 rArr T_(c) = (B^(2))/(3RC)` |
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| 22866. |
When pressure is increases at constant temp volume of gas decreases AB rarr gases, BC rarr vpour +liquid, CD rarr liquid critical point: At this point all the physical propeties of liquid phase will be same as the physical properties in vapour such as, density of liquid = density of vapour T_(c) or critical temp: Temperature abive which a gas can not be liquified P_(c) or critical pressure: minimum pressure which must be applied at critical temp to convert the gas into liqid. V_(c) or critical volume: volume occupied by one mole of gas at T_(c) & P_(c) CRITICAL CONSTANT USING VANDERWAAL EQUATIONS: {:((P+(a)/(V_(m)^(2)))(V_(m)-gb)=RT,rArr,(PV_(m)^(2)+a)(V_(m)-b)=RTV_(m)^(2)),(PV_(m)^(3)+aV_(m)-PbV_(m)^(2) -ab -RTV_(m)^(2) =0, rArr, V_(m)^(3)+V_(m)^(2) (b+(RT)/(P))+(a)/(P)(V)/(m)-(ab)/(P) = 0):} since equation is cubic in V_(m) hence there will be three roots of equation of any temperature and pressure. At critical point all three roots will coincide and will given single value of V_(m) = V_(c) at critical point. Vander Waal equation will be V_(m)^(3) - V_(m)^(2) (b+(RT_(C))/(P_(C))) +(a)/(P_(C)) V_(m) - (ab)/(P_(C)) = 0 ...(i) But at critical point all three roots of the equation should be equal, hence equation should be: V_(m) = V_(c) (V_(m) -V_(c))^(3) = 0 V_(m)^(3) - 3V_(m)^(2) V_(c) +3V_(m) V_(c)^(2) -V_(c)^(3) = 0 ..(2) comparing with equation (1) b +(RT_(c))/(P_(c)) = 3V_(c) ...(i) (a)/(P_(c)) = 3V_(c)^(2) ...(ii) (ab)/(P_(c)) = V_(c)^(3) ..(iii) From (ii) and (iii), V_(c) = 3b From (ii) P_(c) = (a)/(3V_(c)^(2)) substituting P_(c) = (a)/(3(3b)^(2)) = (a)/(27b^(2)) From (i) (RT_(c))/(P_(c)) = 3V_(c) -b = 9b -b = 8b rArr T_(c) = (8a)/(27Rb) At critical point, the slope of PV curve (slope of isotherm) will be zero at all other point slope will be negative zero is the maximum value of slope. ((delP)/(delV_(m)))_(TC) =0 ..(i) (del)/(delV_(m)) ((delP)/(delV_(m)))_(TC) = 0 ...(ii) {Mathematically such points an known as point of inflection (where first two derivatives becimes zero)} using the two T_(c)P_(c) and V_(c) can be calculated by If the critical constants for a hypotentical gas are V_(c) = 150 cm^(3) mol^(-1). P_(c) = 50 atm and T_(c) = 300K. then the radius of the molecule is: [Take R = (1)/(12)Ltr atm mol^(-1)K^(-1)] |
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Answer» <P>`((75)/(2piN_(A)))^(1//3)` `P_(c) = (a)/(27b^(2)), T_(c) = (8a)/(27Rb)` `(P_(c))/(T_(c)) = (R )/(8B) rArr b = (300 xx 1//12)/(8 xx 50) = (1)/(16)` `4 xx (4)/(3) pi r^(3). N_(A) = (1)/(16) rArr r = ((3)/(256piNa))^(1//2)` |
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| 22867. |
When carbon dioxide is passed through an ethereal solution of CH_(3)MgBr and the product is treated with mineral acid , we get , |
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Answer» ETHANAL |
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| 22868. |
When pressure is increases at constant temp volume of gas decreases AB rarr gases, BC rarr vpour +liquid, CD rarr liquid critical point: At this point all the physical propeties of liquid phase will be same as the physical properties in vapour such as, density of liquid = density of vapour T_(c) or critical temp: Temperature abive which a gas can not be liquified P_(c) or critical pressure: minimum pressure which must be applied at critical temp to convert the gas into liqid. V_(c) or critical volume: volume occupied by one mole of gas at T_(c) & P_(c) CRITICAL CONSTANT USING VANDERWAAL EQUATIONS: {:((P+(a)/(V_(m)^(2)))(V_(m)-gb)=RT,rArr,(PV_(m)^(2)+a)(V_(m)-b)=RTV_(m)^(2)),(PV_(m)^(3)+aV_(m)-PbV_(m)^(2) -ab -RTV_(m)^(2) =0, rArr, V_(m)^(3)+V_(m)^(2) (b+(RT)/(P))+(a)/(P)(V)/(m)-(ab)/(P) = 0):} since equation is cubic in V_(m) hence there will be three roots of equation of any temperature and pressure. At critical point all three roots will coincide and will given single value of V_(m) = V_(c) at critical point. Vander Waal equation will be V_(m)^(3) - V_(m)^(2) (b+(RT_(C))/(P_(C))) +(a)/(P_(C)) V_(m) - (ab)/(P_(C)) = 0 ...(i) But at critical point all three roots of the equation should be equal, hence equation should be: V_(m) = V_(c) (V_(m) -V_(c))^(3) = 0 V_(m)^(3) - 3V_(m)^(2) V_(c) +3V_(m) V_(c)^(2) -V_(c)^(3) = 0 ..(2) comparing with equation (1) b +(RT_(c))/(P_(c)) = 3V_(c) ...(i) (a)/(P_(c)) = 3V_(c)^(2) ...(ii) (ab)/(P_(c)) = V_(c)^(3) ..(iii) From (ii) and (iii), V_(c) = 3b From (ii) P_(c) = (a)/(3V_(c)^(2)) substituting P_(c) = (a)/(3(3b)^(2)) = (a)/(27b^(2)) From (i) (RT_(c))/(P_(c)) = 3V_(c) -b = 9b -b = 8b rArr T_(c) = (8a)/(27Rb) At critical point, the slope of PV curve (slope of isotherm) will be zero at all other point slope will be negative zero is the maximum value of slope. ((delP)/(delV_(m)))_(TC) =0 ..(i) (del)/(delV_(m)) ((delP)/(delV_(m)))_(TC) = 0 ...(ii) {Mathematically such points an known as point of inflection (where first two derivatives becimes zero)} using the two T_(c)P_(c) and V_(c) can be calculated by Identify the wrong statement related to the above graph: |
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Answer» between `50K` and `150K` temperature and pressure ranging from 10 atm to 20 atm matter may have LIQUID state.  At 100 K and pressure below 20 atm it may have liquid or gaseous state depending on the pressure. |
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| 22869. |
Which of the following acids will give maximum yield of alkyl chloride in Huns diecker reaction |
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Answer» `CH_3-CH_2-CH_2-COOH` |
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| 22870. |
Which of the following statements is not ture about amophous solids ? |
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Answer» On HEATING they may become CRYSTALLINE at certain temperature. |
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| 22871. |
What is meant by sacrifical anode ? |
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Answer» |
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| 22872. |
Which aldehyde, does not react with Fehling's solution ? |
| Answer» SOLUTION :BENZALDEHYDE | |
| 22873. |
Which of the following substrates is/are more reactive than ethyl bromide for S_(N)2 reaction? |
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Answer» `BR-CH_(2)-Br` |
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| 22874. |
What are the used of acetyl chloride? |
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Answer» Solution :(i) ACETYL CHLORIDE is used as acetylating agent in organic analysis (II) It is used in the DETECTION and estimation of `- OH, – NH_2`groups in organic COMPOUNDS. |
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| 22875. |
What is obtained, when propene is treated with N-bromosuccinimide |
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Answer» `CH_(3)-underset(BR)underset(|)(C)=CH_(2)` 
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| 22876. |
Which of the following has the smallest number of molecules? |
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Answer» `22.4xx10^(3)ML "of" CO_(2)` gas |
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| 22877. |
The reagent used in Clemmensen reduction is |
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Answer» CONC. `H_(2)SO_(4)` |
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| 22878. |
What is the basic difference between the electron gain enthalpy and electropositivity? |
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Answer» Solution :The amount of energy released when an electron is ADDED to a NEUTRAL gaseous isolated atom is called electron gain enthalpy. `X_(g) + e^(-) rarr X_(g)^(-) , Delta H = -ve` Electropositivity is the tendency to lose electron. It is directly related to the METALLIC character. More the ELECTROPOSITIVE character of an electron and thus the element is more metallic in nature. If electon gain enthalpy is more, the element has more tendency to gain electron. so that element will have non-metallic character. Elements having more electropositive character can act as strong reducing agents. while the elements having more electron gain enthalpies will act as OXIDISING character. |
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| 22879. |
The standard reduction electrode potentials of your metals A,B, C and D are -3.65,-1.68,-0.80 and +0.86.The highest chemical activity will be exhibited by : |
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Answer» A |
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| 22880. |
Which of the followingis/are neurologically active drug/s . |
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Answer» Aspirin |
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| 22881. |
Which one of the following will have a meso-isomer also? |
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Answer» 2-Chlorobutane |
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| 22882. |
The standard EMF for the cell reaction , Zn + Cu^(2+) to Cu + Zn^(2+)is 1.1 volt at25^@ C . The EMF for the cell reaction , when 0.1 M Cu^(2+) and 0.1 M Zn^(2+) solutions are used , at25^@ Cis |
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Answer» 1.10 V |
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| 22883. |
Which monomer is used in preparation of orlon ? |
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Answer» `CF_(2)=CF_(2)` |
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| 22884. |
Which one of the following is used as an emulsifier? |
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Answer» SODIUM META sulphite |
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| 22885. |
Which of the following enzymes are used to convert starch into alcohol : |
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Answer» MALTOSE, DIASTASE |
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| 22886. |
Which of the following is simple ether : |
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Answer» `C_6H_5OCH_3` |
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| 22887. |
Which of the following statements are incorrect |
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Answer» The net increase in entropy of the system is ZERO in any reversible cyclic PROCESS * |
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| 22888. |
Which of the following natural product is not a polymer: |
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Answer» DNA |
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| 22889. |
Unlike dry cell, the mercury cell has a constant cell potential throughout its useful life. Why? |
| Answer» Solution :OVERALL REACTION does not involve any ion whose concentration may CHANGE. Hence, it GIVES CONSTANT potential. | |
| 22890. |
Which of the following is used in electro-plating process ? |
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Answer» AGCL SOLID |
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| 22891. |
Which of the following is arranged in order of increasing melting point? |
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Answer» `ZnltCultNiltFe` |
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| 22892. |
What is denaturation of proteins. |
| Answer» SOLUTION :On HEATING a PROTEIN it loseits PROPERTY it is calleddenatuartion of PROTEINS. | |
| 22893. |
Which major product is formed in the reaction, RCN +H_2 O_2 overset(OH^(-) )to |
| Answer» SOLUTION :`RCONH_2` | |
| 22894. |
What are partial and total vapour pressures at 25^(@)C above the solution having equal numbers of molecules of benzene and toluene ? The vapour pressures of benzene and toluene at this temperature are 95.1 and 28.4mmHg respectively. What is the composition of the vapour in equilibrium with benzene toluene solution ? |
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Answer» |
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| 22895. |
Which is used as substitute for platinum in jewellery? |
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Answer» ROLLED gold |
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| 22896. |
Which is not a mineral of aluminium: |
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Answer» Anhydrite |
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| 22897. |
van't Hoff factor is the ratio of ……………..molecular mass to ……………molecular mass. |
| Answer» SOLUTION :CALCULATED, OBSERVED | |
| 22898. |
For the reaction 2H(g)rarr H_(2)(g), the sign of DeltaH and DeltaS respectively are : |
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Answer» Solution :(i) The given reaction, `2H_((g))rarrH_(2(g))` is the FORMATION of `H_(2(g))` from FREE atoms. (ii) In the bond formation, energy is released equal and opposite to bond enthalpy of `H_(2) ` i.e., `-DeltaH_(H-H)^(@)` . Hence the reaction is exothermic and `DeltaAH` is negative. (iii) Since two H atoms FORM one `H_(2)` molecule, `Deltan=1-2=-1` and disorder is DECREASED. Hence ENTROPY change `Delta S lt O`. |
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| 22899. |
Which of the following has highest paired 4f electrons ? |
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Answer» `Yb ^(3+)` |
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| 22900. |
Which of the following has the highest reducing power? |
| Answer» Solution :`I_(2)` is the weakest oxidising AGENT and hence HI has the HIGHEST reducing power. | |
