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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.
| 86401. |
A Wittig reaction with an aldehyde gives |
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Answer» Ketone compound |
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| 86402. |
(A) : With KMnO_(4)//H^(+) , tertiary alcohols are oxidised to ketones . (R) : Tertiary alcohols do not contain alpha- hydrogen . |
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Answer» Both A & R are TRUE , R is the correct explanation of A |
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| 86403. |
(A) : With increase in temperature the condicutivity of metals decreases. (R) : With increase in temperature lattice vibrations increase in metals. |
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Answer» Both (A) and (R) are TRUE and (R) is the correct EXPLANATION of (A) |
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| 86404. |
(A) With HCl ,Fe forms FeCl_(2)andH_(2) (R ) Formed H_(2) prevents the formation of FeCl_(3) |
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Answer» Both (A) and (R ) are TRUE and (R ) is the CORRECT explanation of (A) |
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| 86405. |
(A) With excess HI anisole gives a mixture of iodobenzene and methyl iodide(R) The reaction between anisole and HI takes place via SN^2mechanism. |
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Answer» Both (A) and (R) are TRUE and (R) is the CORRECT EXPLANATION of (A) |
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| 86406. |
(a) With equation, give an example for, (i) Reimer Tiemann reaction(ii) Dehydrogention of a primary alcohol (b) Phenol is an acid but does not react with sodium bicarbonate solution. |
Answer» Solution :(i) When phenol is HEATED with chloroform and sodium hydroxide solution. dium salt of salicyladehyde is obtained which on acidification GIVES salicyladehyde.  (b) Phenol is weaker acid than CARBONIC acid `(H_2 CO_3)`and hence does not LIBERATE from sodium BICARBONATE. |
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| 86407. |
(a) Will pH value of water be same at temperature 25^(@)C and 4^(@)C justify in not more than 2 or 3 sentences. (b) Two students use same solution of ZnSO_(4) and a solution of CuSO_(4). The emf of one cell. Given : 2.303 RT/F=0.06V. |
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Answer» |
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| 86408. |
(A) will not |
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Answer» a.reduce TOLLENS reagent. In `(a), (A)` REDUCES Tollens reagent, since it contains `(-CHO)` group. In `(b), (A)` gives iodoform test, since it contains `(MeCO-)` group. In `(c ) (A)` forms dioxime, since it contains `(-CHO)` and  In `(d), (A)` does not give ceric ammonium nitrate test, since this test is given by alcohols and `(A)` does not contain an ALCOHOLIC group. So the answer is `(d)`. |
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| 86409. |
A will ……..by E_1 reaction . |
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Answer»
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| 86410. |
Why VO_(2)^(+) has lesser oxidizing power than Cr_(2)O_(7)^(2-)? |
| Answer» SOLUTION :i) `CR` is more stable in lower oxidation STATE | |
| 86411. |
(a) Why phenol is more acidic than than ethanol ? (b) Write the mechanism of acid dhydration of ethanol to tield ether: 2CH_(3)CH_(2)OHunderset(413K)overset(H^(+))toCH_(3)CH_(2)OCH_(2)CH_(3) |
Answer» Solution :(a)  Due to resonance, the oxygen atom acquires a partial POSITIVE charge. This weaknes the O-H bond an thus facilitate the release of a proton. Conversely, +I EFFECT of ethyl group increase ELECTRON density which make O-H bond CLEAVAGE difficullt. Hence ethanol less acidic than phenol. (b) `2CH_(3)CH_(2)OHunderset(413K)overset(H^(+))toCH_(3)CH_(2)OCH_(2)CH_(3)` Mechanism: (i) Protonation  (ii) Nucleophilic attack 
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| 86412. |
(a) Why is separation of lanthanoid elements difficult ? |
| Answer» SOLUTION :(a) Due to LANTHANIDE contraction, the size of these elements is NEARLY same. | |
| 86413. |
(a) Why isit that onlysulphideoresareconcentratedbyfrothfloatationprocess ?or Why isthefrothfloatationprocessselectedfortheconcentratedofsulphide ores ? (b) Whichofthefollowingores canbeconcentratedbyfrothfloatationmethodandwhy ?Fe_2O _ 3, ZnS, Al_2O _3 |
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Answer» SOLUTION :(a) Thesulphideoreparticlesarepreferentiallywettedby OIL,becomelighterandthusrisetothesurfacealongwiththefrothwhile thegangueparticlesarepreferentiallywettedbywater,becomeheavierandthussettleatthebottomofthetank. Inthisway,SULPHIDE oreparticlesareseparatedand henceconcentratedfromthegangueparticles. (B) Onlysulphideore,i.e., ZNS,isconcentratedbyfrothfloatationprocess. |
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| 86414. |
(a) Why is bithional added to soap ? (b) What is tincture of iodine ? Write its one use. (c) Among the following, which one acts as a food preservative ? Aspartame, Aspirin, Sodium Benzoate, Paracetamol |
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Answer» SOLUTION :(a) Bithional is added to SOAPS to impart antiseptic properties. (b) 2-3 per cent solution of iodine in alochol-water mixture is KNOWN as tincture of iodine. It is applied on WOUNDS as antiseptic. (C) Sodium benzoate acts as food preservative. |
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| 86415. |
(a). Why is beta-D-gllucopyranose the most abundant of naturally occurring aldohexoses? (b). Write the most stable chair conformer for alpha-D- fructopyranose. How does it dirrer from beta-anomer? |
Answer» Solution :All the ring SUBSTITUENTS in the CHAIR CONFORMATION are equatorial. 
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| 86416. |
(a) Why does presence of excess of lithium makes LiCl crystals pink ? (b) A solid with cubic crystal is made of two elements P and Q. Atoms of Q are the corners of the cube and P at the body centre. What is the formula of the compound ? |
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Answer» Solution :(a) When a crystal of LiCl is present in EXCESS of lithium, the lithium atom loses an `e^(-)` to from `Li^(+)` ions. The released `e^(-)` diffuse into the crystal and occupies the anionic sites (F-centre). These `e^(-)` input the pink colour to the crystal. (b) We KNOW that in SIMPLE CUBIC number of ATOMS at corner `(1)/(8) xx 8 = 1` and is also body centre. So the formula of compound is PQ. |
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| 86417. |
(a) Why does PCI_3fumes in moisture ?(b) NO_2dimerises to N_2O_4 ? |
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Answer» Solution :(a)` PCl_3` produces HCl GAS in MOISTURE. Hence `PCI_3`fumes in moist air, `PCl_3 + 3H_2O to H_3PO_3 + HCl` (B) `NO_2`has one unpaired electron on N, ` therefore NO_2`dimerises to PAIR up unpaired electron `O_2N + NO_2 to O_2N - NO_2 " or" N_2O_4` |
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| 86418. |
(a) Why do soaps not work in hard water ? (b) What are the disadvantages of using hard water ? |
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Answer» Solution :(a) Hard water contains calcium and magnesium ions. These ions FORM calcium and magnesium stearate (unsoluble substance) when sodium and potassium soaps are dissolved in hard water. These insoluble substances separate as scum in water and are useless as cleansing agents. (B) (i) Hair washed with hard water looks dull because of sticky precipitate. (ii) Dye does not ABSORB evenly on cloth washed with soap USING hard water because of this sticky and GUMMY mass. |
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| 86419. |
a.Why do nucleophiles (Nu^(overset(..)Θ)) add to the (C=C) of alpha-beta-unsaturated carbonyl compounds but not to alkenes ? b.Explain the formation of product in the following reaction: CH_(3)CHunderset((A))= CHCHO overset(overset(Θ)(OD)//D_(2)O)rarr underset((B))underset(("Tetradecuterated compound"))(CD_(3)CH=CD.CHO) c.In the following reaction, three isomers are present in the equilibrium. Which is the most stable ? d.Explain why Michael addition is 3,4-addition. |
Answer» Solution :a.`(Nu^(overset(..)Θ))`adds to the `beta`-C atom to give resonance-stabilised CARBANION enolate.  `Nu^(Θ)` adds to alkene to give localised carbanion (not resonance stabilised), has a very high energy, and is not formed easily.  B. The H atom of Me acide due to delocalisation of negative charge.  Three resonating structure I, II and III with negative charge can accept D form `D_(2)O` to give three products. X, Y, and Z. (I) `overset(D_(2)O)rarr DCH_(2)-UNDERSET((X))CH =CH-CH=O`,(II)`overset(D_(2)O)rarr CH_(2)=underset((Y))CH-overset(D)overset(|)CH-CH=O`  Unstable enol (Z) rearranges to (Y) and (X). When the reaction is performed REPEATEDLY with `O^(Θ)D`, carbanion enolate reforms and the reaction with `D_(2)O` ultimatley produces (B) `(D_(3)C-CH==CD-CH=O)`, tetradeuterated product. c.  d. In Michael reaction, nucleophilic addition reaction at `(C=O)` group of `alpha,beta`-unsaturated carbonyl compounds, GIVES 1,2-addition product (B), which rerranges to unstable intermediate product (C ) and is called 1,4-addition product, that ultimately gives product (D). So the addition in (D) actually is 3,4-and is called Micheel addition. 
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| 86420. |
(a) Why detergents with straight chain of hydrocarbons are prefered over branched chain hydrocarbons? (b) Give one example for detergent with straight chain hydrocarbon. |
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Answer» Solution :(a) As they are biodegradable (b) Sodiumlaurylphate Sodium dodeceyl benzene sulphonate Cetyltrimethyl AMMONIUM bromide (a) Straight chain hydrocarbons are prefered over branched chain hydrocarbons because these are more biodegradable then branched chain hydrocarbons. In branch chainhydrocarbons, TAIL is torque of pollution. (b) Example of straight chain : DETERGENT with HYDROCARBON : `underset("(Lautryl alcohol ethoxylate)")((CH_(3))(CH_(2))_(10)CH_(2)(OCH_(2)CH_(2))_(8).OH)` |
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| 86421. |
a. Why are overline(e) donating groups, such as (-NH_(2)) (-OR), (-R) o- and p- directingtowards SE reaction? Explain by an example. b. Why are overline(e) withdrawing groups, [such as (-NO_(2), (-SO_(3)H),(-CN),(-COOH), etc].m- direection towards SE reaction? explain by example. c. Although halogens are overline(e) withdrawing yet they are o- and p- directing towards SE reaction. Why? |
Answer» Solution :a. Due to `+M` effect (or `oveline(e)` donating effect) of `(-NH_(2))` group, aniline is resonance stablised as shown:  Due to this, the ring is ACTIVATED at `o-` and `p-` positions. Thus, `OVERLINE(e)` density is increased at these positions. HENCE, the incomingelectrophile attacks at these positions giving `o-` and `p-` products. b. Due to `overline(e-)` WITHDRAWING effect of the `(-COOH)` group, benozoic acid is resonance stablised as shown:  due to this, the benzene ring is deactived at `o-` and `p-` positons, thus `overline(e)` density is decreased at these positon. So, comparatively `overline(e)` density at `m-` position is increased, and hence the incomingelectrophileattacks at `m`-postion giving `m-` products, C. Halogens are said to be `pi` donating but `sigma` with-drawing, i.e., they withdraw `overline(e)s` thorugh resonance. The intermediate carbocationformed during `SE` reaction is destablised by the `-1` effect of the halogens.  Resonane stablises the intermediate carbocation and this effect is more prominent at `o-` and `p-` positons, `-I` effect is stronger (due to high `EN` of halogens ) than resonance and causes net `overline(e)`withdrawal, and thus causingnet deactivation. The resonance effect for the attack at `o-` and `p-` positons and hence MAKES thedeactivation lesss for `o-` and `p-` subsitution. |
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| 86422. |
(a) Why are Mn^(2+) compounds more stable than Fe^(2+) towards oxidation to their + 3 state? (b) What are interstitial compounds ? Why are such compounds well-known for transition metals ? (c) Write electronic configuration of Pm^(3+). |
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Answer» Solution :(a)`Mn^(2+) (3d^5 )-` exactlyhalffilledandectrastableand hashigh3rdionisationenthalpy`FE^(2+ ) ( 3d ^6 )` can loseelectron easilytoget`Fe^(3+)(3d^5)` i.e.,extrastableconfiguration (b )In suchcompounds the small atomssuchas `h,c ` B,Netcoccupyinterstitialsites in theirlatticeseg. TiC`TiH_2,Fe_3H`steeletcsuchcompoundsTransitionmetalsformsuchcompoundsbecausetheycantakeup smallatomsof C , h,B,N etcin theirinerstitialsitesin theirlattices (c ) ` PM^(3-1) (61-3=58e)` `1s^2 2s^2 2p^6 3s^2 3d^63d^(10) 4s^24p^(10)5s^2 5p^6 4f^4 "in" [Xe ]_54 4F^4 ` |
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| 86423. |
(a) Why are Mn^(2+) compounds more stable than Fe^(2+) towards oxidation to their +3 state? (b) What are interstitial compounds ? Why are such compounds well-known for transition metals? (c) Write electronic configuration of Pm^(3+). |
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Answer» Solution :(a)`Mn^(2+) (3d^(5))` - exactly half FILLED and extra stable and has HIGH 3rd ionisation enthalpy `FE^(2+) (3d^(6))` can lose electron easily to get `Fe^(3+) (3d^(5))` i.& extra stable configuration. (b) In such compounds, the small atoms such as H, C, B, N etc. occupy interstitial sites in their lattices eg. Tic, `TiH_(2), Fe_(3)H` Steel etc. Such compounds are called interstitial compounds. Transition metals form such compounds because they can take up small atoms of C, H, B, Netc. in their interstitial sites in their lattices. (c)`Pm^(3-1)(61-3=58e)` `1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(10)4s^(2)4p^(6)4d^(10)5s^(2)5p^(6)4f^(4)` in `[Xe]_(54)4f^(4)` |
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| 86424. |
(a) Why are detergents called soapless soap ? (b) Why do soaps not work in hard water ? |
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Answer» SOLUTION :(a) Detergents are called soapless soap because they have SIMILAR CLEANING ACTION with soap but they do not contain sodium or potassium salts of fatty acids. (b) Hard water contain calcium and magnesium salts which combine with soaps to FORM corresponding salts. These salts are not water soluble and get separated as curdy white precipitate resulting in wastage of soap. |
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| 86425. |
(a). Why are alkyl halides insoluble in water? (b). Why is butan-1-ol optically inactive but butane-2 ol is optically active in nature? (c). Although chlorine is an electron withdrawing group yet it is ortho para directing in electrophilic aromatic substitution reaction. Why? |
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Answer» Solution :(a). Athough ALKYL halides contain a polar `C-X` bond, they are not involved in any intermolecular HYDROGEN BONDING with water. THEREFORE, these are water insoluble. (B). Butan-2-ol has a chiral carbon while it is absent in butan-1-ol. (c). N/A |
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| 86426. |
(a) Why are alkyl halides insoluble in water ? (b) Why is Butan-1- ol optically inactive but Butan-2-ol is optically actove ? (c ) Although chlorine is an electron-withdrawing group, yet it is ortho-para-directing in electrophilic aromatic substitution reactions. Why? |
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Answer» SOLUTION :(a) For an alkyl halide to dissolve in WATER, energy is required to overcome the attractions between alkyl halide molecules and break the hydrogen bonds in water molecules. Less energy is released when NEW attractions are set up between haloalkane and water molecules. These are not as strong as the original hydrogen bonds in water. Therefore alkyl halides are INSOLUBLE in water. (b) `underset("Achira(symmetric)")underset("Butan-1-ol")(CH_(3)CH_(2)CH_(2)-overset(H)overset(|)underset(H)underset(|)C-OH)` `underset("CHIRAL (asymmetric)")underset("Butan-2-ol")(CH_(3)CH_(2)-overset(H)overset(|)underset(OH)underset(|)C-CH_(3))` Butan-1-ol is an achiral compound. Therefore it is optically inactive. Butan-2-ol is a chiral compound. Therefore it is optically active. (c ) The lone pairs of electrons on chlorine, take part in resonance with the benzene ring making ortho and para positions negative. Therefore atom it is ortho and para directing in electrophilic aromatic substitution. |
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| 86427. |
Why 3d-series of elements acts as good catalyst? |
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Answer» Solution :(a) Transition metals and their compounds exhibit catalytic activity due to (i) Ability to adopt to multiple oxidation STATES (ii) Transition metals can form complexes with reactant molecules using vacant d-orbitals. This lowers the energy of activation of the reaction and increases RATE of the reaction. (iii) Due to the presence of large SURFACE area in finely DIVIDED state. (B) Due to `d^(0)` configuration of `Ti^(4+)`, it is colourless where `Cr^(3+)` has `d^3` configuration facilitating d-d transition. |
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| 86428. |
A white, water insoluble solid A turns yellow on heating and becomes white on cooling. Solid A gives a clear solution B when treated with HCl solution and C when treated with NaOH solution. When H_2 S is passed through B nothing is obtained. However if B is made neutral, H_2 S caused the precipitation of white compound D. Identify C. |
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Answer» `NaAlO_2` |
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| 86429. |
(A) White suspension of lead (II) sulphate turns blackish on passing ozone though it. (R ) Ozone oxidises SO _(4) ^(2-)to S _(2) O _(8) ^(2-) ion. |
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Answer» Both (A) and (R ) are true and (R ) is the CORRECT explanation of (A) |
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| 86430. |
Assertion (A): White tin is an example of tetragonal system.Reasoning (R): For a tetragonal system a = b = c and a = p = y * 90^@. |
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Answer» Both (A) and (R) are true and (R) is the CORRECT explanation of (A) |
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| 86431. |
A white substance, when strongly heated, leaves a residue which is yellow while hot and white when cold, and gives a gas that turns baryta water milky and accidified dichromate paper green. The substance can be |
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Answer» `ZnCO_(3)` |
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| 86432. |
A white substance was alkaline in solution. Which of the following substances could it be |
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Answer» `Fe_(2)O_(3)` |
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| 86433. |
A white substance (A) reacts with dilute H_(2)SO_(4) to produce a colourless gas (B) and a colourless solution ( C).The reaction between (B) and acidified K_(2)Cr_(2)O_(7) solution produces a green solution and a slightly coloured precipitate (D).The substance (D)burns in air to produce a gas (E) which reacts with (B) to yield (D) and a colourless liquid. Anhydrous copper sulphate is turned blue on addition of this colouless liquid.Addition of aqueous NH_(3) or NaOH to ( C) produces first a white precipitate which dissolves in the excess of the respective reagent to produce a clear solution in each case. Identify (A),(B),( C),(D) and (E). |
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Answer» `Cr_(2)O_(7)^(2-) + 3H_(2)S+8H^(+) to 2Cr^(3+) ("green solution")+3S DARR +7H_(2)O` `S+O_(2) to SO_(2) , 2H_(2)S +SO_(2) to 2H_(2)O+3S` `ZN^(2+)+2OH^(-) toZn(OH)_(2) darr` (white) `Zn^(2+)+2NH_(3)+ H_(2)O hArrZn(OH)_(2) darr("white")+NH_(4)^(+)` `Zn(OH)_(2) darr +4NH_(3) HARR [Zn(NH_(3))_(4)]^(2+) +2OH^(-)` |
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| 86434. |
A white substance having alkaline nature in solution: |
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Answer» `Fe_2O_3` |
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| 86435. |
A white substance a reacts with dilute H_(2)SO_(4) to produce a colourless gas B and a colourless solution C. The reaction between B and acidified K_(2)Cr_(2)O_(7) solution produces a green solution and a slightly coloured precipitate D. The substance D burns in air to produce a gas E which reacts with B to yield D and a colourless liquid. Anhydrous copper sulphate is turned blue on addition of this colourless liquid. Additio of aqueous NH_(3) or NaOH to C produces first a precipitate, produce a clear solution in each case. Identify A, B, C, D and E. Write the equations of the reactions involved. |
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| 86436. |
Awhite substance Areactswithdilute H_(2)SO_(4) isproduce acolourless B and acidified K_(2)Cr_(2)O_(7) solutionproduces a greensolutionand a slighlycolouredprecipitate D .Thesubstance B toburnsin airtoproducea gas Ewhich reactswith B ito yiedD and a colouless liquid .Anhyhdrous coppersulphateis turned blue on additionof thiscolourlessliquid additionod aqueous NH_(3) or NaOHto Cproducefirsta precipitatewhichdissolve in theexcess of the respective reagenttoproduce a clearsolutionin eachcaseidentify A,B,C, and E. Writetheequationof thereactioninvolved. |
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Answer» Solution :`A = ZnS,B = H_(2)S ,C = ZnSO_(4),D =- S , E = SO_(2)` A white substance A reactswith dilute `H_(2)SO_(4)` to produce a colourless gas .B and a colourless solution C .The reactionbetween B and acidified `K_(2)Cr_(2)O_(7)` solution produces a green solutionand a slightlycoloured precipitate D.Thisprecipitatemust be sul,phate , so gas B must be hydrogensulphide ,Sulpbuur in airto produce a gas sulphur dioxidewhich reactswithhydrogen sulphide to yield sulphuragainand a colourless LIQUID water .Anhydrous coppersulphateis turnedblue on addition of water a precipitatewhichdissolve a clearsolutionin excesssolutionof therespectiveregaoin to produce a clearsolutionin eachcase .Thissuggesist that Cmust be sulphateofeitherzine oraluminium so, A may be zine sulphide . `ZnS +H_(2)SO_(4) rarr H_(2)S + ZnSO_(4)` `Cr_(2)O_(7)^(2-) + 4H^(o+) ("From" H_(2)SO_(4)) + 3H_(2)S rarrunderset("Green solution")(2Cr^(3+)) +7H_(2)O +3S darr` `S + SO_(2) rarr SO_(2)` `SO_(2) +2H_(2)S rarr 2H_(2)O +3S` `ZnSO_(4) +2NaOH rarr underset("White")(Zn(OH)_(2)darr) +Na_(2)SO_(4)` `Zn(OH)_(2) +2NaOH rarr Na_(2)ZnO_(2) +2H_(2)O` |
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| 86437. |
A white substance (A) on heating with excess of dil HCl gave an offensive smelling gas (B) and a solution (C ). Solution (C ) on treatment with aqueous NH_(3) did not give any precipitate but on treatment with NaOH solution gave a precipitate (D) which dissolves in excess of NaOH solution. (A) on strong heating in air gave a strong smelling gas (E) and a solide (F). Solid (F) dissolved completely in HCl and the solution gave a precipitate with BaCl_(2), in acid solution. Identify A to F and write the chemical equation for various reactions involved. |
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Answer» Solution :Solution C gives ppt. with NaOH solution which is soluble in excess of NaOH solution hence the cation should be of the amphoteric metal like Zn or Al. Again solid F is soluble in HCL and gives white precipitate with `BaCl_(2)`. Therefore ANION must be `SO_(4)^(2-)` ion. Nowthe A gives offensive SMELLING gas hence A may be ZnS or `Al_(2)S`. But `Al_(20S_(3)` on hetaing in air does not form `Al_(2)(SO_(4))_(3)` . Chemical reactions are as follows `:` `underset((A))(ZnS)+2HClrarr underset((C ))(ZnCl_(2))+underset((B))(H_(2)S uarr)` `underset((C ))(ZnCl_(2))+2NaOH rarr underset((D))(Zn(OH)_(2)darr)+2NaCl` `Zn(OH)_(2)+2NaOH rarr NA_(2)ZnO_(2)+2H_(2)O` `underset((A))(2ZnS)+3O_(2)rarr underset((F))(2ZnO)+underset((E))(2SO_(2))` `underset((A))(ZnS)+2O_(2)rarrZnSO_(4)(F)` `ZnSO_(4)+BaCl_(2)rarrBaSO_(4)darr +ZnCl_(2)` |
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| 86438. |
A white sublimable solid, when boiled with a NaOH solution, give a colour less gas thal turns Nessler's Teagent brown. The solid on being heated with solid K_(2)Cr_(2)O_(7) and conc. H_(2)SO_(4), gives red brown vapours. The white solid can be |
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Answer» `NH_(4)Cl` `K_(2)Cr_(2)O_(7)+H_(2)SO_(4)` with `Cl^(-)` gives `CrO_(2)Cl_(2)` and with `Br^(-)` gives `Br_(2)` both are red |
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| 86439. |
A white solid (X) gives golden yellow flame on platinum wire. A piece of red litmus paper turns white when it is dipped into a freshly made aqueous solution of white solid. The formula of (X) is |
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Answer» `Na_(2)O` |
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| 86440. |
A white solid reacts with dilute HCI to give colourless gas that de-colourises aqueous bromine. The solid is likely to be. |
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Answer» SODIUM carbonate |
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| 86441. |
A white solid reacts with dil. HCL to give colourless gas that decolourise aqueous bromine. The solid is most likely to be: |
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Answer» SODIUM carbonate |
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| 86442. |
A white solidis either Na_(2)O or Na_(2)O_(2). A piece of red litmus paperturn when in is freshlymade aqueoussolutionof thewhitesolid. I Identifythe substance and explain withbalancedequation ii. Explainwhatwouldhappen to the red litmusif thewhitesolid werethe othere compound |
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Answer» Solution :`Na_(2)O_(2) rarr 2H_(2)O rarr 2NAOH + H_(2)O_(2)` `H_(2)O_(2) rarr H_(2)O + [O]` `H_(2)O_(2)` TURN thecolourof redlithium paper to white due to its bleanchingaction by OXIDATION .Therefore the givensubstance is `Na_(2)O_(2)` |
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| 86443. |
A white solid forms Rinmann's green in the charcoal cavity test in an oxidising flame. On treatment with dilute H_(2)SO_(4) this solid produces a gas that turns an acidified dichromate paper green and lead acetate paper black. The white solid is |
| Answer» Solution : ZnS | |
| 86444. |
A white solid A reacts with dilute H_(2)SO_(4) to produce a colourless gas B and a colourless solution C. the reaction between B and acidified dichromate yields a green solution and a slightly coloured precipitate D. the substance D, when burnt in air, gives a gas E which reactswith B to yield D and a colourlesss liquid. anhydrous copper sulphate turns blue with this colourless liquid. the addition of aqueous NH_(3) or NaOH toC produces a precipitate that dissolves in an excess of the reagent to form a clear solution. Q. Suppose the solution obtained by the treatment off the solution C with an excess of NaOH is acidified with acetic acid and the gas B is passed through it. which of the following will obtained? |
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Answer» A colourless solution `H_(2)SO_(4)` suggets that the substance is a zinc salt. thus, A is most likely ZnS. let us go through the reactions now. 
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| 86445. |
A white solid A reacts with dilute H_(2)SO_(4) to produce a colourless gas B and a colourless solution C. the reaction between B and acidified dichromate yields a green solution and a slightly coloured precipitate D. the substance D, when burnt in air, gives a gas E which reactswith B to yield D and a colourlesss liquid. anhydrous copper sulphate turns blue with this colourless liquid. the addition of aqueous NH_(3) or NaOH toC produces a precipitate that dissolves in an excess of the reagent to form a clear solution. Q. Which of the following reactions are relevant to the action of NH_(3) or NaOH solution on C? |
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Answer» `Zn(OH)_(2)+4NH_(3) to [Zn(NH_(3))_(4)]^(2+)+2OH^(-)` `H_(2)SO_(4)` suggets that the substance is a zinc salt. thus, A is most likely ZNS. let us go through the REACTIONS now. 
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| 86446. |
A white solid A reacts with dilute H_(2)SO_(4) to produce a colourless gas B and a colourless solution C. the reaction between B and acidified dichromate yields a green solution and a slightly coloured precipitate D. the substance D, when burnt in air, gives a gas E which reactswith B to yield D and a colourlesss liquid. anhydrous copper sulphate turns blue with this colourless liquid. the addition of aqueous NH_(3) or NaOH toC produces a precipitate that dissolves in an excess of the reagent to form a clear solution. Q. What would appear if the gas B were passed through an aqueous solution solution of Pb(NO_(3))_(2) ? |
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Answer» A white precipitate solution in hot dilute `HNO_(3)` `H_(2)SO_(4)` suggets that the substance is a zinc salt. thus, A is most likely ZnS. LET us go through the reactions now. 
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| 86447. |
A white solid A reacts with dilute H_(2)SO_(4) to produce a colourless gas B and a colourless solution C. the reaction between B and acidified dichromate yields a green solution and a slightly coloured precipitate D. the substance D, when burnt in air, gives a gas E which reactswith B to yield D and a colourlesss liquid. anhydrous copper sulphate turns blue with this colourless liquid. the addition of aqueous NH_(3) or NaOH toC produces a precipitate that dissolves in an excess of the reagent to form a clear solution. Q. What would happen if the gas E were passed through an acidified KMnO_(4) solution? |
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Answer» Bleaching of the permangnate solution without any precipitation `H_(2)SO_(4)` suggets that the substance is a zinc salt. thus, A is most likely ZnS. let us go through the reactions now. 
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| 86448. |
A white solid A reacts with dilute H_(2)SO_(4) to produce a colourless gas B and a colourless solution C. the reaction between B and acidified dichromate yields a green solution and a slightly coloured precipitate D. the substance D, when burnt in air, gives a gas E which reactswith B to yield D and a colourlesss liquid. anhydrous copper sulphate turns blue with this colourless liquid. the addition of aqueous NH_(3) or NaOH toC produces a precipitate that dissolves in an excess of the reagent to form a clear solution. Q. Which of the following gases are B and E respectively? |
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Answer» `CO_(2) and SO_(2)` `H_(2)SO_(4)` suggets that the substance is a zinc salt. thus, A is most likely ZnS. let us go through the REACTIONS now. 
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| 86449. |
A white solid 'A' on heating gives off a gas which turns lime water milky. The residue is yellow when hot but turns white on cooling. This solid 'A' is |
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Answer» ZINC sulphide `(Hot)underset("YELLOW")(ZnO) HARR underset("white")(ZnO)(co ol)` |
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| 86450. |
A white sodium salt dissolves readily in water to give a solution which is neutral to litmus. When silver nitrate solution is added to the solution, a white precipitate is obtained which does not dissolve in dil. HNO_3. The anion could be |
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Answer» `CO_(3)^(2-)` |
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