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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.
| 86251. |
(a)Aluminium-Galvanising metals (b)Zinc-Cooking vessels (c )Iron-Cutlery (d)Copper-Dental fillings |
| Answer» SOLUTION :(C )Iron-Cutlery | |
| 86252. |
(a)Alluminium-design of aeroplane (b)Zinc oxide-Cosmetics (c )Zinc sulphite-X-ray screens (d)Iron-Artifici al limb joints |
| Answer» SOLUTION :(d)Iron-Artificial LIMB JOINTS | |
| 86253. |
(a)Aluminium-Corundum (b)Limonite-Iron (c )Galena-Lead (d)Tin-Siderite |
| Answer» SOLUTION :(d)Tin-Siderite | |
| 86254. |
What is B ? |
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Answer»
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| 86255. |
(AAK_MCP_36_NEET_CHE_E36_015_Q01)The reagents, A and B are respectively |
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Answer» `H_2/Pt`, `NaBH_4` |
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| 86256. |
(AAK_MCP_36_NEET_CHE_E36_014_Q01) Major product P in the above reaction is |
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Answer» (`AAK_MCP_36_NEET_CHE_E36_014_A01`) |
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| 86257. |
(AAK_MCP_35_NEET_CHE_E35_024_Q01)Total number of isomers (including stereo isomers) of B obtained in the above reaction are |
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Answer» 2 |
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| 86258. |
(A)Adsorption is a surface phenomenon.(R ) During adsortion, residual force of surface decreases |
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Answer» Both (A) and (R ) are TRUE and (R ) is the CORRECT EXPLANATION of (A) |
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| 86259. |
a.(A) + KBr rarryellow ppt . (B) (A) + "conc". H_(2)SO_(4) overset(Delta) to brown vapours intensifield with cu- turnings. (B) dissolves in lypo forming a soluble complex (C ) what are (A),(B) and (C ) and explain their reactions ? b. SO_(3)^(2-) and SO_(4)^(2-) both give while ppt with BaCI_(2) solution .How is SO_(3)^(2-) detected in presence of SO_(4)^(2-)? c. Na_(2)B_(4)O_(7).10 H_(2)O +"conc".H_(2)SO_(4) overset(Delta) to (A) overset(CH_(2)OH Delta) to (B) idenity (A) and (B) d. (A) + dii H_(2)SO_(4) overset(Delta) to gas (B) Gas (B)turns K_(2)Cr_(2)O_(7) //H^(Theta) solutiongreen Aq solution of (A) + BaCI_(2) rarrwhile ppt .(C ) Filltrate after removing (C) + Br_(2) waterrarr while ppt dissolve in qammon ium acetate solution Example. |
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Answer» Solution :(A) `AgNO_(3) (B) AGBR (C ) Na_(2)[Ag(S_(2)O_(3))_(2)]` b. `BaSO_(4)` is insoluble in cone `HCI` while `BaSO_(3)` remainsoluble Soparate `HaSO_(4)` and add `Hr_(2)` WATER into filrate `So_(3)^(2-)` is axidised to `SO_(4)^(2-)` and given white ppt of `BaSO_(4)` `Br_(2) + H_(2)O rarr 2HBr + O` `BaSO_(3) + O rarr BaSO_(4) darr` c. (A) `H_(2)BO_(3)` (B) `(CH_(3)O)_(3)B` burns with green edges flame d. (A) `SO_(3)^(2-) + SO_(4)^(2-)`(B) :`SO_(2)` (C ) `BaSO_(4)` e. `2CH_(3)COONH_(4) + UNDERSET("White ppt".)(PbSO_(4) rarr) underset("Soluble")((CH_(3)COO)_(2)) Pb + (NH_(4))_(2)SO_(4)` |
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| 86260. |
(A)A colloidal sol of Fe(OH)_(3) formed by peptisation carries positive charge. (R ) During formation of Fe(OH)_(3) sol , electrons are lost by the particles. |
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Answer» Both (A) and (R ) are TRUE and (R ) is the correct explanation of (A) |
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| 86261. |
(A)A catalyst does not alter the equilibriumconstant of a reaction (R ) The catalyst forms a complex with the reactants and provides an alternate path with lower energy of activation for the reaction. The forward and the backward reactions are affected to the same extent. |
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Answer» Both (A) and (R ) are TRUE and (R ) is the CORRECT EXPLANATION of (A) |
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| 86262. |
A5.25% solution of a substance is isotonic with a 1.5% solution of urea (molar mass = "60 g mol"^(-1)) in the same solvent. If the densities of both the solutions are assumed to be equal to "1.0 g cm"^(-3), molar mass of the substance will be |
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Answer» `"210.0 g MOL"^(-1)` Molar conc. Of the SUBSTANCE = Molar conc. Of urea `(52.5gL^(-1))/(M)=(15gL^(-1))/("60 g mol"^(-1))` `("100 g solution = 100 mL as d = 1 g mL"^(-1))` `therefore""M=(52.5xx60)/(15)="210 g mol"^(-1)` |
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| 86263. |
A^(3+) overset(-1.2V) to A^(2+) overset(-0.56 V) to A^(1+) Where -1.2V and -0.56 V represent the reduction potentials of A^(3+) and A^(2+). Which of the following is true. |
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Answer» The reaction `2A^(2+) to A^(3+) + A^(+1)` is spontaneous |
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| 86264. |
A_2 (g)+ B_2(g) hArr 2AB(g). The equilibrium constant of above reaction at 100^@C is 50. If a one litre falsk containing one mole of A_2 is connected to a two litre flask containing two moles of B_2, how many moles of AB will be formed at 373 K? |
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Answer» 3.5 |
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| 86265. |
A_(2)B is anideal gas, which decompase according f to theequation : A_(2) B toA_(2)+ (1)/(2)B_(2) . Atstart , the initial pressureis 100 mm Hgand after 5 minutes , the pressureis 120 mm Hg. Whatis the averagerateof decompositionof A_(2)B ? AssumeT and V are constant . |
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Answer» Solution :The decomposition reaction of gaseous `A_(2)B` is given as `A_(2)B to A_(2)+(1)/(2)B_(2)` `100 0 0` at start of the reaction `100-2x 2x X` after the reaction `100-2x+2x+x=100+x=120"MM"` `x=20mm "or" 2x=40mm` The decrease in pressure of reactant SUBSTANCE, `A_(2)B` in `5` min is `40` mm. The rate of decomposition of `A_(2)B=(40)/(5)` `8 "mm min"^(-1)=0.133"mm" s^(-1)` |
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| 86266. |
A_(1)g of an element gives A_(2)g of its oxide. The equivalent mass of the element is: |
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Answer» `(A^(2)-A_(1))/(A_(1))XX8` |
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| 86267. |
A_(1)g of an element gives A_(2)g of its chloride, the equivalent mass of the element is: |
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Answer» `(A_(1))/(A_(2)-A_(1))xx35.5` |
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| 86268. |
A_(2) + 2B_(2) to A_(2)B_(4) and (3)/(2) A_(2) + 2B_(2) to A_(3) B_(4) Two substance A_(2) & B_(2) are allowed to react completely to from A_(2)B_(4) & A_(3)B_(4) mixture of leaving none of the reactants. Using this information calculate the composition of final mixture when mentioned amount of A_(2) & B_(2) are taken. (i) If (5)/(4) moles of A_(2) & 2 moles of B_(2) is taken (ii) If (7)/(2) moles of A_(2) & 6 moles of B_(2) is taken (iii) If 4 moles of A_(2) & 6 moles of B_(2) is taken |
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Answer» (ii) `[A_(3)B_(4)=1A_(2)B_(4)=2]` (III) `[A_(3)B_(4)=2A_(2)B_(4)=1]` |
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| 86269. |
A1_(2)O_(3) is reduced by electrolysis at low potential and high current. If 4.0 xx 10^(4) amperes of current is passed through molten A1_(2)O_(3) for 6 hrs, what mass of aluminium is produced? (Assume 100% current efficiency) |
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Answer» `9.0xx10^(3)` G |
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| 86270. |
A_1 and A_2 are two ores of metal M.A_1 on calcination gives black precipitate, CO_2 and water. Identify A_1 and A_2. |
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Answer» Solution :`A_1` is malachite `CuCO_3.Cu(OH)_3` `CuCO_3Cu(OH)_2overset(DELTA)toCuO+CO_2uarr+H_2O` `CuO` is the black solid. `CuCO_3.Cu(OH)_2overset(dil.HCl)toCuCl_2+CO_2+H_2O` `CuCl_2overset(KI)toCu_2I_2+KCl+I_2` `A_2` is COPPER glance `Cu_2S`, a sulphide ore. `Cu_2Sunderset(Delta)overset(O_2)toCu_2O+SO_2` `Cu_2S+Cu_2OtoCu+SO_2` The GAS `SO_2` gives GREEN colour with acidified `K_2Cr_2O_7.K_2Cr_2O_7+H_2SO_4+3SO_4toK_2SO_4+Cr_2(SO_4)_3+4H_2O` |
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| 86271. |
A_(1) & A_(2) are two ores of metal M. Ar on calcination gives black precipitate, CO_(2)& water. |
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Answer» SOLUTION :`CuCO_(3), CU(OH)_(2), Cu_(2)S, Cu, CUO, Cu_(2)l_(2), SO_(2)` `A_(1) = Cu(OH)_(2), CuCO_(3)` `A_(2)= Cu_(2)S` `Cu(OH)_(2) CuCO_(3)OVERSET(cabination)to 2CuO+CO_(2)+overset(H_(2)O)(Black 0Solid)` `Cu(OH)_(2)CuCO_(3)overset(dillHCl)toCUCl_(2)+CO_(2)+3H_(2)O` `2CU_(2)Cl_+4Klto2Cu_(2)I_(2)+I_(2)+4KCI` `underset(A_(2))(2Cu_(2)S)+3O_(2)to2CU_(2)O+2SO_(2)` Cu_(2)S+3O_(2)to6Cu+SO_(2)` |
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| 86272. |
A_(1) and A_(2) are two ores of metal M. A_(1) on calcination gives black precipitate, CO_(2) and water. Identify A_(1) and A_(2). |
| Answer» | |
| 86273. |
'1^@ alcohol on dehydrogenation with a heavy metal catalyst (Cu or Ag) gives a Ketone. |
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Answer» |
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| 86274. |
A0.1M sodium acetate solution was prepared. The K_(h)=5.6xx10^(-10) |
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Answer» The degree of hydrolysis is `7.48xx10^(-5)` `K_(h)=((0.1h)(0.1h))/(0.1(1-h))=0.1h^(2)` `implies5.6xx10^(-10)=0.1h^(2)impliesh=7.48xx10^(-5)` `[thereforeltltltlt1]` `[OH^(-)]=ch=7.48xx10^(-5)xx10^(-1)=7.48xx10^(-6)implies` `[H^(+)]=(K_(w))/([OH^(-)])=(10^(-14))/(7.48xx10^(-6))=1.33xx10^(-9)`implies pH=8.8` approx Hence choices (a), (c) and (d) are correct while (b) is wrong. |
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| 86275. |
A0.020 m solution of each of the following compounds is prepared. Which solution would you expect to freeze at -0.149 ""^@C ? |
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Answer» `[Co(en)_2Cl_2]CL` `DeltaT_(f) = i xx K_(f) xx m` So,`T_(f-)^(@) - T_(f) =i xx K_(f) xx m` 0.149 ` = ixx 1.86 xx 0.020` Hence , i=4 The compound having the VALUE of i=4 will have the freezing POINT `-0.149^(@)C` i.e , `[Cr(NH_3)_6]Cl_(3)`. |
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| 86276. |
(A) ZSM-5 is used as a catalyst is petrochemical industries .(R ) Zeolites are three dimensional network silicates in which some silicon atoms are replaced by aluminium atom. |
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Answer» Both (A) and (R ) are true and (R ) is the correct EXPLANATION of (A) |
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| 86277. |
(A): Zr and Hf have nearly equal atomic radii. (R): Zr and Hf belong to the same group. |
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Answer» Both (A) and (R) are TRUE and (R) is the CORRECT EXPLANATION of (A) |
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| 86278. |
A Zn rod weighing 25g was kept in 100 mL of 1M CuSO_(4) solution. After a certain time the molarity of Cu^(2+) in solution was 0.8. What was molartiyof SO_(4)^(2-)? Whatwas the weight of Zn rod after cleaning ? ("At. Weight of "Zn = 65.4.) |
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Answer» |
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| 86279. |
(A): Zn(II) compounds are diamagnetic. (R): Zn(II) has all its electrons paired. |
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Answer» Both (A) and (R) are TRUE and (R) is the CORRECT EXPLANATION of (A) |
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| 86280. |
(A) Zn,Cd^(-) and Hgof group 12 are transition metals (R) Partially filled d sublevel is present in the atomic state as well as in their common oxidation state. |
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Answer» Both (A) and (R) are TRUE and (R) is the CORRECT EXPLANATION of (A) |
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| 86281. |
(A) Zn^(+2) is diamagnetic (R) The electrons are lost from 4s orbital to form Zn^(+2) |
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Answer» Both (A) and (R) are TRUE and (R) is the CORRECT EXPLANATION of (A) |
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| 86282. |
A zinc rod is placed in 250 mL of 1 M CuSO_4 solution for such a period that the molarity of Cu^(2+) becomes 0.85 M. The molarity of SO_4^(2-) at this stage is : |
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Answer» 1 M |
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| 86283. |
A zinc rod is placed in 0.095 M zinc chloride solution at 25^(@)C. emf of this half cell is -0.79V. Calculate E_(Zn^(2+)//Zn)^(@). |
| Answer» SOLUTION :`-0.76V` | |
| 86284. |
A zinc rod is placed in 0.095 M zinc chloride solution at 25^(@)C. EmF of this half cell is -0.79V. Calculate E^(@)""_(Zn2+//Zn). |
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Answer» Solution :Given : E = -0.79 V n = 2 `[ZN^(+)]=0.095 M` Formula : `E=E^(@)-0.0591/n log Zn^(2+)` Solution : `E^(0)""_(Zn^(2+)//Zn)=E+0.0591/2logZn^(2+)` `E^(0)""_(Zn^(2+)//Zn)=-0.76+0.0591/2log0.095` `""=-0.79+0.02889=-0.76V` `""E^(@)=-0.76V`. |
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| 86285. |
A zinc rod is dipped in 0.1 M solution of ZnSO_(4). The salt is 95% dissociated at this dilution at 298K. Calculation the electrode potential (E_(Zn^(2+)//Zn)^(@)=-0.76V) |
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Answer» Solution :The electrode reaction written asreduction reaction is: `ZN^(2+)+2e^(-)Zn(n=2)` Applying NERNST equation, we GET `E_(Zn^(2+)//Zn)=E_(Zn^(2+)//Zn)^(@)-(0.0591)/(2)"log"(1)/([Zn^(2+)])` As 0.1 M `ZnSO_(4)` solution is 95% dissociated, this means that in the solution, `[Zn^(2+)]=(95)/(100)xx0.01M=0.095M` `thereforeE_(Zn^(2+)//Zn)=-0.76-(0.0591)/(2)"log"(1)/(0.095)=0.76-0.02955(log1000-log95)` `=0.76-0.02955(3-1.9777)=-0.76-0.03021=-0.79021` volt |
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| 86286. |
(A) Zinc oxide is yellow coloured when hot(R) Zinc oxide has metal excess defect due to anionic vacancies |
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Answer» Both (A) and (R) are true and (R) is the correct explanation of (A) |
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| 86287. |
Azero order reaction X rarr Product, with an initial concentration 0.02 M has a half life of 10 min. If one starts with concentration 0.04 M, then the half life is |
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Answer» 10 s |
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| 86288. |
(A): Zinc does not show characteristic properties of transition metals (R): The outer most shell of zinc is completely filled |
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Answer» Both A & R are TRUE, R is the correct EXPLANATION of A |
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| 86289. |
(A): Zinc can be refined from lead by distillation (R): Vapour pressure of lead is more compared to that of zinc |
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Answer» Both A & R are true, R is the CORRECT EXPLANATION of A |
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| 86290. |
A zero order reaction is one whose rate is independent of…. |
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Answer» TEMPERATURE of the REACTANTS |
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| 86291. |
A zero order reaction is 20% complete in 20 minutes . Calculate the value of the rate constant . In what time will the reaction be 80% complete ? |
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Answer» SOLUTION :(i) A = 100% , x = 20% ,Therefore , a -x = 100 - 20 = 80 For the zero order reaction `k=(x/t)impliesk=(20/20)=1` Rate constant for reaction = 1 (II) To calculate the time for 80% of COMPLETION k = 1 , = 10 , x = 80% , t = ? Therefore , `t = (x/k)=(80/1)=80min ` |
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| 86292. |
A zero order reaction is one : |
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Answer» whose rate is not AFFECTED by CONCENTRATION |
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| 86293. |
A zero order reaction is one |
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Answer» In which reactants do not react |
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| 86294. |
(A) Zeolites are good shape-selective catalysts (R ) Zeolites have honeycomb - like structures |
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Answer» Both (A) and (R ) are TRUE and (R ) is the CORRECT explanation of (A) |
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| 86295. |
A zeo order reaction X rarrProduct, with an initial concentration 0.02 M has a half life of 10 min . If one starts with concentration 0.04 M , then the half life is .... |
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Answer» 10 s for n = 0 `t_(1//2)=1/(2k[A_0]^(-1))` `t_(1//2)=([A_0])/(2k)` `t_(1//2)=prop[A_0]""....(1) ` Given `[A_0]=0.02M , t_(1//2)=10 min ` `[A_0]=0.04 M , t_(1//2)`= ? Substitute in (1) ` 10 min prop 0.02 M ""...(2)` `t_(1//2) prop 0.4 M ""...(3)` Dividing Eq. (3) by Eq . (2) we get , `implies (t^(1//2))/(10 min )=(0.04M)/(0.02M)` `t_(1//2)=2xx10 ` min = 20 min . |
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| 86296. |
(a) You are provided with the following four reagents LiAlH_(4), I_(2)//NaOH, NaHSO_(3) and Schiff's reagent. Which two reagents can be sued to distinguish between the compound in each of the following pairs: (i) CH_(3)CHO and CH_(3)COCH_(3) (ii) CH_(3)CHO and C_(6)CH_(5)CHO (iii) C_(6)H_(5)COCH_(3) and C_(6)H_(5)COC_(6)H_(5) (b) Write a note on Cannizzaro reaction |
| Answer» Solution :For answer, CONSULT solved EXAMPLE 5. | |
| 86297. |
(a) You have the following substances : NH_(3), O_(2), Pt and H_(2)O. Write equations for the preparation of N_(2)O from these substances. (b) Considering the fact that N_(2) makes up about 79% of the atmosphere, why don't animals use the more abundant N_(2) instead of O_(2) for biological reactions. |
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Answer» Solution :(a) `{:(4NH_(3)(g)+5O_(2)(g)underset(1100 K)OVERSET(Pt)(RARR)4NO (g) + 6 H_(2)O(l),,,2NO (g)+O_(2)(g) rarr 2NO_(2)(g)),(3NO_(2)(g)+H_(2)O(l) rarr 2HNO_(3)(aq)+NO(g),,,NH_(3)(g)+HNO_(3)(aq) rarr NH_(4)NO_(3) (aq)),(NH_(4)NO_(3)(aq)underset("at room temperature")overset("Vacuum evaporation")rarr NH_(4)NO_(3)(s),,,NH_(4)NO_(3)(s)overset(523 K)rarr N_(2)O(g) + 2H_(2)O (l)):}` (b) Animals need large AMOUNT of energy to move around and maintain the body temperature. Therefore, to obtain the required energy, it is much easier for them to break weaker double bond (493.4 kJ `mol^(-)`)of `O_(2)` than breaking the much stronger triple bond (941.4 kJ `mol^(-)`) of `N_(2)`. |
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| 86298. |
A yellow product formed when an organic compound C_(3)H_(8)O is warmed with aqueous solution of sodium carbonate and iodine solution. The product is ….. |
| Answer» Solution :Iodoform | |
| 86299. |
A yellow precipitate will be obtained ifAgNO_3 is added to a solution of : |
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Answer» `KIO_3` |
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| 86300. |
(A) Xenonmonoxytetrafluoride has square planar structure. (R) Xenon in XeOF_4 undergoes dsp^2hybridisation. |
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Answer» Both (A) and (R) are TRUE and (R) is the CORRECT explanation of (A) |
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