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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.
| 89051. |
A lead storage battery consists of a lead anode and a grid of lead packed with lead dioxide as the cathode. The electrolyte taken is 39% H_(2)SO_(4) by mass having a density of 1.294 g mL^(-1). The battery holds 3.5 L of the acid. During the discharge of the battery, the density H_(2)SO_(4) falls from 1.294 g mL^(-1) to 1.139 g mL^(-1) which is 20% H_(2)SO_(4) by mass Q. The amount of charge in coulombs used up by the battery is nearly |
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Answer» 954180 `=(2xx96500xx9.88786)/(2)=954178` coulombs |
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| 89052. |
(A): Hydrated oxides are usually subjected to roasting (R) : Hydrated oxides on heating loses water and gives oxide |
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Answer» Both A & R are TRUE, R is the correct EXPLANATION of A |
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| 89053. |
A lead storage battery consists of a lead anode and a grid of lead packed with lead dioxide as the cathode. The electrolyte taken is 39% H_(2)SO_(4) by mass having a density of 1.294 g mL^(-1). The battery holds 3.5 L of the acid. During the discharge of the battery, the density H_(2)SO_(4) falls from 1.294 g mL^(-1) to 1.139 g mL^(-1) which is 20% H_(2)SO_(4) by mass Q. Molarity of the solution after the discharge is |
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Answer» 8.136 `THEREFORE`Molarity of the solution`=(8.136" moles")/(3.5L)=2.32` |
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| 89054. |
A hydrated metallic salt(A), light green in colour on earful heating gives a white anhydrous residue(B). (B) is soluble in water and its aqueous solution reacts with No to give a dark brown compound (C). (B) gives a brown residue (D) and a mixture oftwo gases (E) and (F). The gaseous mixture when passed through an acidified KMnO_4solution dischanges the pink colour and when passed through acidified BaCl_2 solution gave a white precipitate. C may be |
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Answer» `FeSO_4.NO` Neutral `FesO_4` gives bown ring test `C= [Fe(H_2O)_5NO] SO_4` `FeSO_4 overset(Delta)rarr underset("(D)")(Fe_2O_3) + underset("(E)")(SO_2) + underset("(F)")(SO_3)` |
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| 89055. |
A lead storage battery consists of a lead anode and a grid of lead packed with lead dioxide as the cathode. The electrolyte taken is 39% H_(2)SO_(4) by mass having a density of 1.294 g mL^(-1). The battery holds 3.5 L of the acid. During the discharge of the battery, the density H_(2)SO_(4) falls from 1.294 g mL^(-1) to 1.139 g mL^(-1) which is 20% H_(2)SO_(4) by mass Q. Moles of sulphuric acid lost during discharge is |
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Answer» 9.88 `=3500mLxx1.294" g "mL^(-1)=4529g` mass of `H_(2)SO_(4)` PRESENT in this solution `=(39)/(100)xx4529g=1766.31g` mass of the solution of after discharge `=3500mLxx1.139" g "mL^(-1)=3986.5g` mass of `H_(2)SO_(4)` present in this soluton, `=(20)/(100)xx3986.5=797.3g` Loss in mass of `H_(2)SO_(4)` during discharge `=1766.31-797.3=969.01g` MOLES of `H_(2)SO_(4)` lost during discharge `=(969.01)/(98)=9.88786` |
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| 89056. |
A hydrated metallic salt(A), light green in colour on earful heating gives a white anhydrous residue(B). (B) is soluble in water and its aqueous solution reacts with No to give a dark brown compound (C). (B) gives a brown residue (D) and a mixture oftwo gases (E) and (F). The gaseous mixture when passed through an acidified KMnO_4solution dischanges the pink colour and when passed through acidified BaCl_2 solution gave a white precipitate. A may be |
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Answer» `FeSO_4.7H_2O` Neutral `FesO_4` gives bown RING TEST `C= [Fe(H_2O)_5NO] SO_4` `FeSO_4 overset(Delta)rarr underset("(D)")(Fe_2O_3) + underset("(E)")(SO_2) + underset("(F)")(SO_3)` |
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| 89057. |
A lead storage battery consists of a lead anode and a grid of lead packed with lead dioxide as the cathode. The electrolyte taken is 39% H_(2)SO_(4) by mass having a density of 1.294 g mL^(-1). The battery holds 3.5 L of the acid. During the discharge of the battery, the density H_(2)SO_(4) falls from 1.294 g mL^(-1) to 1.139 g mL^(-1) which is 20% H_(2)SO_(4) by mass Q. The reacton occurring at the andoe during charging is |
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Answer» `Pb^(2+)+2e^(-)toPb` |
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| 89058. |
(A) Leaching is the process of converting the ore intooxide andthenreducing it . (R) Leachingis done in a blast furnace. |
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Answer» Both (A) and (R) are true and (R) is the CORRECT EXPLANATION of (A) |
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| 89059. |
(A): Leaching is chemical method used to ore benefication process (R): Leaching is usually performed in metallurgy with the help of a non aqueous solvent |
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Answer» Both A & R are TRUE, R is the CORRECT explanation of A |
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| 89060. |
(A)Leaching is chemical method used to ore benefication process (R)Leaching is usually performed in metallurgy with thehelp of a non-aqueous solvent |
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Answer» Both (A) and (R) are TRUE and (R) is the correct explanation of (A) |
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| 89061. |
(A) Law of conservation of mass is invalid for nuclear fission fusion and disintegration. (R) The law proposes that mass if neither creatednor destroyed in a reaction. |
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Answer» If both (A) and (R) are CORRECT and (R) is the correct EXPLANATION of (A). |
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| 89062. |
A hydrated metallic salt A, light green in colour, on careful heating gives a white anhydrous residue B. Aqueous solution of B reacts with NO to give a dark brown compound C. On strong heating, B decomposes to give a brown residue D and a mixture of two gases E and F. Identify E and F. |
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Answer» `SO_2` and `SO_3` |
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| 89063. |
A layer of chromium metal 0.25 mm thick is to be plated on an auto bumper with a total area 0.32m^(2) from a solution containing CrO_4^(2-) ? What current flow is required for this electroplate if bumper is to be plated in 60s ? The density of churomnium metal is 7.20g / cm^(3) |
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Answer» `4.9 XX 10^(3) A ` ` 6 xx 96500 to 52 gm , Q to 576 gm ` `Q = 6413538.46 = C xx t ,C =( 6413538. 46)/(60) = 1.0689 xx 10^(5) = 10.689 xx 10^(4)` |
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| 89064. |
(A): Last traces of lead from silver is removed by cupellation (R): Lead will be converted to volatile litharge during cupellation, but silver remains. |
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Answer» Both A & R are TRUE, R is the correct explanation of A |
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| 89065. |
A hydrated chloride of metal contains 18.26% metal and 32.42% chloride ion by mass. The specific heat of metal is 0.16 what is hydrated chloride? |
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Answer» |
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| 89066. |
A laser ( light amplification of stimulated emission of radiation ) provides a coherent( in phase ) nearly monochromatic source of high intensty light. Lasers are used in eye surgery, CD // DVD players basic research, etc. some modern eye lasers can be "tuned" to emit a desired wavelength. The following table shows the lambda( nm ) , v( s^(-1))andE ( J) but some of the data is missing : Which of the following statement (s) is // are correct ? ( i) E lies in UV region while G lies in visible region (ii) The energies D and F alone are not sufficient to remove electron from isolated gaseous H-atom. (iii) In the spectrum the increasing order of lambda for all type of lasers is II lt III lt IV lt I (iv) None of the above statement is correct |
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Answer» Only (IV) |
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| 89067. |
A hybrid rocket propellant uses |
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Answer» A liquid oxidizer and a solid FUEL |
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| 89068. |
A laser ( light amplification of stimulated emission of radiation ) provides a coherent( in phase ) nearly monochromatic source of high intensty light. Lasers are used in eye surgery, CD // DVD players basic research, etc. some modern eye lasers can be "tuned" to emit a desired wavelength. The following table shows the lambda( nm ) , v( s^(-1))andE ( J) but some of the data is missing : If an electron absorbs energy equal to D in hydrogen atom in first excited state and after reaching highest excited level it returns to ground level. How many possible lines will be obtained in hydrogen spectrum? |
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Answer» 10 |
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| 89069. |
A laser ( light amplification of stimulated emission of radiation ) provides a coherent( in phase ) nearly monochromatic source of high intensty light. Lasers are used in eye surgery, CD // DVD players basic research, etc. some modern eye lasers can be "tuned" to emit a desired wavelength. The following table shows the lambda( nm ) , v( s^(-1))andE ( J) but some of the data is missing : The value of A and H will be "........................." and "............................." respectively. |
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Answer» `6329 Å , 6.14 xx 10^(12)` |
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| 89070. |
A human body required 0.01 M activity of a redioactive substance after 24 hours. Half life of the radioactive substance is 6 hours.Theinjectionof maximum activity of the radioactive substance that can be injected is |
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Answer» 0.08 Activity after n half-lives = `([A]_(0))/(2^n)` `:. 0.01 M = ([A_0])/(2^4) " or " [A_0] = 0.1 xx 16= 0.16 M`. |
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| 89071. |
A laser (light amplification by stimulated emission of radiation) provides a coherent (in phase )nearly monochromatic source of high intensityi light. Lasers are used in eye surger, CD/DVD players basic research etc. some miodern eye lasers can be tuned to emit a desired wavelength. The following table shows the lamda(nm),v(s^(-1)) and E(J) but some of the data is missing. Which of the following statement (s) is/are correct? (i) E lies in UV region while G lies in visible region (ii) The energies D and F alone are not sufficietn to remove electron from isolated gaseous H=atom. (iii) In the spectrum the increasing order of lamda, for all type of laser is IIlt III lt IV lt I (iv) None of the above statement is correct |
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Answer» Only (iv) The value of (D) is 2.55 eV which is less than 13.6 eV required to remove a permanently from H-atom (F) is 8.86eV which is also not sufficient `lamda_(1)=6329Å` `lamda_(ii)=4879Å` `lamda_(iii)=1401Å` Hence IV`gt` I `gt` II` gt` III `lamda_(iv)=6630Å` Hence only (i) and (ii) are correct. So, correct option is B |
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| 89072. |
A human body required 0.01M activity of a radioactive substance after 24 hours. Half life of the radioactive substance is 6 hours. The injection of maximum activity of the radioactive substance that can be injected is : |
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Answer» 0.08 `:.` 24 hr =4 half LIFE periods ACTIVITY after n-half lives `=([A]_(0))/(2^(n))` `0.01M=([A]_(0))/(2^(4))` `:.[A]_(0)=0.01xx16=0.16M` |
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| 89073. |
A laser (light amplification by stimulated emission of radiation) provides a coherent (in phase )nearly monochromatic source of high intensityi light. Lasers are used in eye surger, CD/DVD players basic research etc. some miodern eye lasers can be tuned to emit a desired wavelength. The following table shows the lamda(nm),v(s^(-1)) and E(J) but some of the data is missing. The value of A and H will be ............ and .................. respectively. |
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Answer» `6329Å,6.14xx10^(12)` `C=v lamda, lamda =C/v` `lamda=(3xx10^(8)ms^(-1))/(4.74xx10^(14)s^(-1))=6329xx10^(-10)M` `lamda=6329Å=A` For (H)`E=hv , v=E/h` `v=(3xx10^(-9)J)/(6.63xx10^(-34)JS)=4.5xx10^(24)` |
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| 89074. |
A human body excretes certain material through sweating by law similar to radioactivity if technitium is injected in some form in human body the body exretes half the amount in 24 hours A patient is given an injection containing .^(98)Tc The isotope is radioactive with half life of 8 hours The activity just after the injection is 32muCi What will be activity after 48 hrs of the overall excreted material till that time? |
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Answer» `0.125muCi` |
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| 89075. |
A large numberof monomers (Simple molecules) combine together to form a larger molecule ( macro molecule) called as polymer. Each polymer is made up of a repeating structural unit. A polymer is said to be homopolymer if the structural unit is derived from one type of monomer molecules. if the repeating structural unit of a polymer is derived from more than one different types of monomers, the polymer is said to a copolymer. The homopolymers as well as copolymers may be formed by addition or condensation reactions. Alekenes and dienes polymerize by additions (Chain growth) mechanism involving carbocations, carbanions or free radical intermeidates. Dienes (Chloroprene) polymerise by 1, 4 addition mechanism to give cis or trans polymers. Natural rubber is, however, cis-polyisoprene. Natural rubber is quite soft and tacky but these properties can be improved by a process called vulcanization. In contrast, bifunctional monomer molecules undergo condensation or step-growth polymerization. Polymers which can be heated and reshpaed as many times as desired are called thermoplastics (polyethene, polystyrene, PVC, teflon etc). While those which can be heated onoy once to give a particular shape are called thermosetting polymers (bakelite, melmac etc.) Which is not a macro molecule? |
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Answer» DNA |
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| 89076. |
A large numberof monomers (Simple molecules) combine together to form a larger molecule ( macro molecule) called as polymer. Each polymer is made up of a repeating structural unit. A polymer is said to be homopolymer if the structural unit is derived from one type of monomer molecules. if the repeating structural unit of a polymer is derived from more than one different types of monomers, the polymer is said to a copolymer. The homopolymers as well as copolymers may be formed by addition or condensation reactions. Alekenes and dienes polymerize by additions (Chain growth) mechanism involving carbocations, carbanions or free radical intermeidates. Dienes (Chloroprene) polymerise by 1, 4 addition mechanism to give cis or trans polymers. Natural rubber is, however, cis-polyisoprene. Natural rubber is quite soft and tacky but these properties can be improved by a process called vulcanization. In contrast, bifunctional monomer molecules undergo condensation or step-growth polymerization. Polymers which can be heated and reshpaed as many times as desired are called thermoplastics (polyethene, polystyrene, PVC, teflon etc). While those which can be heated onoy once to give a particular shape are called thermosetting polymers (bakelite, melmac etc.) Which of the following sets contain only copolymers? |
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Answer» SBR, glyptal, nylon-66 |
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| 89077. |
A human body excretes certain material through sweating by law similar to radioactivity if technitium is injected in some form in human body the body exretes half the amount in 24 hours A patient is given an injection containing .^(98)Tc The isotope is radioactive with half life of 8 hours The activity just after the injection is 32muCi How much time will elapse before the activity of patient falls to 16 muCi? |
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Answer» 3.8 hrs |
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| 89078. |
A laser (light amplification by stimulated emission of radiation) provides a coherent (in phase )nearly monochromatic source of high intensityi light. Lasers are used in eye surger, CD/DVD players basic research etc. some miodern eye lasers can be tuned to emit a desired wavelength. The following table shows the lamda(nm),v(s^(-1)) and E(J) but some of the data is missing. If an electrons absorbs energy equalto D in hydrogen atom in first excited state and after reaching highest excited level it returns to ground level. How many possible lines will be obtained in hydrogen spectrum? |
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Answer» 10 `=(6.63xx10^(-34)xx3xx10^(8))/(4.879xx10^(-7))` (D)`=4.1xx10^(-19)` In EV `4.1xx10^(-19)=2.55eV` The e is already present at first excited level. So, by absorbign this much energy e can move up to FOURTH level and when it returns to ground level total lines will be `(n(n-1))/2=6` |
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| 89079. |
(a) How would you account for the following: (i) Actinoid contraction is greater than lanthanoid contraction. (ii) Transition metals form coloured compounds. (b) Complete the following equation : 2MnO_(4)^(-)+6H^(+)+5NO_(2)^(-)to |
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Answer» Solution :(a) (i) Actinoid contraction is greater than lanthanoid contraction. Contraction in the size of the atom and ions is greater in actinoids COMPARED to LANTHANOIDS because of poor shielding by 5f electrons. (ii) When an electron from a lower energy d-orbital is excited to a higher energy d-orbital, the energy of excitation corresponds to the frequency of LIGHT absorbed. This frequency lies in the visible range. The COLOUR observed corresponds to the COMPLEMENTARY colour of the light absorbed. (b) The completed equation is : `2MnO_(4)^(-)+6H^(+)+5NO_(2)^(-)to2Mn^(2+)+5NO_(3)^(-)+3H_(2)O` |
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| 89080. |
A large numberof monomers (Simple molecules) combine together to form a larger molecule ( macro molecule) called as polymer. Each polymer is made up of a repeating structural unit. A polymer is said to be homopolymer if the structural unit is derived from one type of monomer molecules. if the repeating structural unit of a polymer is derived from more than one different types of monomers, the polymer is said to a copolymer. The homopolymers as well as copolymers may be formed by addition or condensation reactions. Alekenes and dienes polymerize by additions (Chain growth) mechanism involving carbocations, carbanions or free radical intermeidates. Dienes (Chloroprene) polymerise by 1, 4 addition mechanism to give cis or trans polymers. Natural rubber is, however, cis-polyisoprene. Natural rubber is quite soft and tacky but these properties can be improved by a process called vulcanization. In contrast, bifunctional monomer molecules undergo condensation or step-growth polymerization. Polymers which can be heated and reshpaed as many times as desired are called thermoplastics (polyethene, polystyrene, PVC, teflon etc). While those which can be heated onoy once to give a particular shape are called thermosetting polymers (bakelite, melmac etc.) Which of the following sets contains only thermoplasters? |
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Answer» Polythene, Bakelite, Nylon-6 |
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| 89081. |
(a) How would you account for the following: (i) Highest fluoride of Mn is MnF, whereas the highest oxide is Mn_(2)O_(7). (ii) Transition metals and their compounds show catalytic properties. (b) Complete the following equation: 3MnO_(4)^(2-)+4H^(+)to |
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Answer» Solution :(a) (i) The ability of oxygen to form MULTIPLE bonds to METALS explains why the highest oxide of Mn is `Mn_(2)O_(7)` while the highest fluoride is `MnF_(4)`. (ii) Catalytic properties of transition metals and their compounds is due to their ability to adopt multiple oxidation STATES and to form COMPLEXES. (b) The completed EQUATION is : `3MnO_(4)^(2-)+4H^(+)to2MnO_(4)^(-)+MnO_(2)+2H_(2)O` |
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| 89082. |
A large number of simple molecules called monomers combine together to form a macromolecule called a polymer. Each polymer has a repeating structural unit. If the repeating structural unit is derived from one type of monomer molecules. The polymer is said to be a homopolymer and if it is derived from two or more different types of monomer molecules , the polymer is said to be a copolmer. Both homopolymers and copolymers may be formed either by addition or condensation reactions. Alkenes and dienes polymerize by addition (chain growth) mechanism involving carbocations , carbanions or free radical intermediates. Dienes (chloroprene, isoprene, etc.) polymerize by 1, 4-addition mechanism to give cis-or trans-polymers. Natural rubber is quite soft and tacky but these properties can be improved by a process called vulcanization. In contrast , bifunctional monomer molecules undergo , condensation or step-growth polymerization. Polymers which can be heated and reshaped as many times as desired are called thermoplastics , (polythene, polystyrene, PVC, teflon, etc.) while those which can be heated only once to give a particular shape are called thermosetting polymers (bakelite, melmac, etc.). Which of the following sets contains only thermoplastics ? |
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Answer» POLYTHENE, BAKELITE, Nylon 6 |
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| 89083. |
(a) How would you account for the following : (i) Actinoid contraction is greater than lanthanoid contraction (ii) Transition metals from coloured compound. (b) Complete the following equation. 2MnO_(4)^(-)+ 6H^(+) + 5NO_(2)^(-) rarr |
| Answer» SOLUTION :(B)`2MnO_(4)^(-) + 6H^(+) + 5NO_(2)^(-) RARR 2Mn^(2+) + 5NO_(3)^(-) + 3H_(2)O` | |
| 89084. |
A large number of simple molecules called monomers combine together to form a macromolecule called a polymer. Each polymer has a repeating structural unit. If the repeating structural unit is derived from one type of monomer molecules. The polymer is said to be a homopolymer and if it is derived from two or more different types of monomer molecules , the polymer is said to be a copolmer. Both homopolymers and copolymers may be formed either by addition or condensation reactions. Alkenes and dienes polymerize by addition (chain growth) mechanism involving carbocations , carbanions or free radical intermediates. Dienes (chloroprene, isoprene, etc.) polymerize by 1, 4-addition mechanism to give cis-or trans-polymers. Natural rubber is quite soft and tacky but these properties can be improved by a process called vulcanization. In contrast , bifunctional monomer molecules undergo , condensation or step-growth polymerization. Polymers which can be heated and reshaped as many times as desired are called thermoplastics , (polythene, polystyrene, PVC, teflon, etc.) while those which can be heated only once to give a particular shape are called thermosetting polymers (bakelite, melmac, etc.). Which is not a macromolecule ? |
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Answer» DNA |
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| 89085. |
A large number of simple molecules called monomers combine together to form a macromolecule called a polymer. Each polymer has a repeating structural unit. If the repeating structural unit is derived from one type of monomer molecules. The polymer is said to be a homopolymer and if it is derived from two or more different types of monomer molecules , the polymer is said to be a copolmer. Both homopolymers and copolymers may be formed either by addition or condensation reactions. Alkenes and dienes polymerize by addition (chain growth) mechanism involving carbocations , carbanions or free radical intermediates. Dienes (chloroprene, isoprene, etc.) polymerize by 1, 4-addition mechanism to give cis-or trans-polymers. Natural rubber is quite soft and tacky but these properties can be improved by a process called vulcanization. In contrast , bifunctional monomer molecules undergo , condensation or step-growth polymerization. Polymers which can be heated and reshaped as many times as desired are called thermoplastics , (polythene, polystyrene, PVC, teflon, etc.) while those which can be heated only once to give a particular shape are called thermosetting polymers (bakelite, melmac, etc.). Which of the following sets contain only copolymers ? |
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Answer» SBR, GLYPAL, Nylon6,6 |
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| 89086. |
(a) How will you prepare : (i) K_(2)MnO_(4) from MnO_(2) ? (ii) Na_(2)Cr_(2)O_(7) from Na_(2)CrO_(4) ? (b) Account for the following: (i) Mn^(2+) is more stable than Fe^(2+) towards oxidation to +3 state. (ii) The enthalpy of atomisation is lowest for Zn is 3d series of the transition elements. (iii) Actinoid elements show wide range of oxidation states. |
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Answer» Solution :(a) (i) `K_(2)MnO_(4)` from `MnO_(2)` `MnO_(2)` is fused with potassium hydroxide and an oxidant agent like `KNO_(3)`. `2MnO_(2)+4KOH+O_(2)to2K_(2)MnO_(4)+2H_(2)O` (ii) `Na_(2)Cr_(2)O_(7)` from `Na_(2)CrO_(4)` Yellow solution of sodium chromate is filtered and acidified with sulphuric acid to give a solution from which orange sodium dichromate can be crystallised. `2Na_(2)CrO_(4)+2H^(+)to Na_(2)Cr_(2)O_(7)+2Na^(+)+H_(2)O` (b) (i) `Mn^(2+)` is more stable than `Fe^(2+)` towards oxidation to +3 state. `Mn^(2+)` has d-orbital configurations as `3d^(5)` and `Mn^(3+)` has the configuration `3d^(4). 3d^(5)` (being half-filled) is a stable configuration compared to `3d^(4)`. Hence, `Mn^(2+)` is more stable towards oxidation to +3 state. `Fe^(2+)` has d-orbital configuration as `3d^(6)` and `Fe^(3+)` has the more stable configuration `3d^(5)`. Hence, `Fe^(2+)` has a greater tendency towards oxidation to +3 state. (ii) Enthalpy of atomisation depends upon the number of unpaired electrons. Since there are no unpaired electrons in the ATOM of zinc in the 3d series, it has the lowest enthalpy of atomisation. (iii) Actinoid elements SHOW a wide range of oxidation STATES. This is due to the fact that 5f, 6d and 7s levels are of comparable energies. Electrons from any of these levels could be lost giving rise to the POSSIBILITY of greater number of oxidation states. |
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| 89087. |
A large number of simple molecules called monomers combine together to form a macromolecule called a polymer. Each polymer has a repeating structural unit. If the repeating structural unit is derived from one type of monomer molecules. The polymer is said to be a homopolymer and if it is derived from two or more different types of monomer molecules , the polymer is said to be a copolmer. Both homopolymers and copolymers may be formed either by addition or condensation reactions. Alkenes and dienes polymerize by addition (chain growth) mechanism involving carbocations , carbanions or free radical intermediates. Dienes (chloroprene, isoprene, etc.) polymerize by 1, 4-addition mechanism to give cis-or trans-polymers. Natural rubber is quite soft and tacky but these properties can be improved by a process called vulcanization. In contrast , bifunctional monomer molecules undergo , condensation or step-growth polymerization. Polymers which can be heated and reshaped as many times as desired are called thermoplastics , (polythene, polystyrene, PVC, teflon, etc.) while those which can be heated only once to give a particular shape are called thermosetting polymers (bakelite, melmac, etc.). Teflon, styron and neoprene are all |
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Answer» copolymers |
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| 89088. |
(a) How will you prepare nitric acid (b) Discuss the structure of nitric acid and give its important uses . |
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Answer» Solution :(a) By heating potassium NITRATE with conc. `H_2SO_4` . `KNO_3 + H_2SO_4 to KHSO_4 + HNO_3` (b) Structure of NITRIC acid  Uses of nitric acid (i) it is used as a reagent in laboratory. (II) it is used as a oxidising agent (iii) it is used in the MANUFACTUREOF FERTILIZERS |
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| 89089. |
(a) How will you prepare benzoic acid (i) from ethyl benzene (ii) by using Grignard reagent? (b) How is benzoic acid converted to (i) Benzyl alcohol (ii) Benzyl ethanoate ? |
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Answer» |
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| 89090. |
A large number of simple molecules called monomers combine together to form a macromolecule called a polymer. Each polymer has a repeating structural unit. If the repeating structural unit is derived from one type of monomer molecules. The polymer is said to be a homopolymer and if it is derived from two or more different types of monomer molecules , the polymer is said to be a copolmer. Both homopolymers and copolymers may be formed either by addition or condensation reactions. Alkenes and dienes polymerize by addition (chain growth) mechanism involving carbocations , carbanions or free radical intermediates. Dienes (chloroprene, isoprene, etc.) polymerize by 1, 4-addition mechanism to give cis-or trans-polymers. Natural rubber is quite soft and tacky but these properties can be improved by a process called vulcanization. In contrast , bifunctional monomer molecules undergo , condensation or step-growth polymerization. Polymers which can be heated and reshaped as many times as desired are called thermoplastics , (polythene, polystyrene, PVC, teflon, etc.) while those which can be heated only once to give a particular shape are called thermosetting polymers (bakelite, melmac, etc.). Chloroprene is the monomer unit in : |
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Answer» POLYSTYRENE |
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| 89091. |
(a) How will you distinguish between the following pairs of compunds: (i) Aniline are Ethanamine (ii) Aniline and N-methylaniline (b) Arrange the following compounds in decreasing order of their boiling points: Butanol, Butanamine, Butane |
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Answer» SOLUTION :(a) (i) Aniline and Ethanamine Ethanamine on treatment with `NaNO_(2)//HCl` evolves `N_(2)` gas and ethanol is formed. Aniline does not give this test, no `N_(2)` is evolved. However aniline reacts with `NaNO_(2)//HCl` at `0-5^(@)C` to form benzene diazonium chloride. (II) Aniline and N-methylaniline Aniline and N-methylaniline can be DISTINGUISHED by performing carbylamine test. Aniline on heating with chloroform and potassium hydroxide evolved carbylamine which has a foul SMELL. N-methylaniline does not give this test. (b) Decreasing order of boiling points: Butanol > BUTANAMINE > Butane |
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| 89092. |
A large number of molecular rearrangements involve anionotropic migration of a group to an adjacent atom with incomplete octet. The migration group may either detach from the migration origin and then form bond with the migration terminus or may remain partly bonded to both migration origin and terminus to form a bridged ion intermediate. similar to the electron deficent carbon such as carbocation and carbene intermediates, electron-deficient nitrogen species are known in which the nitrogen bears a positive charge or the nitrogen has an open sextet of electrons, i.e., a nitrene. Q. Which of the following is stable product for given reaction? HO-CH_(2)-CH_(2)-CH_(2)-CH_(2)-OH overset(H^(+))to ? |
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Answer» `HO-CH_(2)-CH_(2)-CH_(2)-CH=CH_(2)` |
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| 89093. |
(a) How will you distinguish between Pentan-2-one and Pentan-3-one with the help of Iodoform test ? (b) How will you bring about following conversions ? (i) Benzoic acid to m-Nitrobenzyl alcohol. (ii) Benzaldehyde to Benzophenone. (iii) Benzoic acid to Benzamide. |
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Answer» Solution :(a) Pentan-2-one gives iodoform TEST i.e. YELLOW ppt on WARMING with `I_(2)` and NaOH but pentan-3-one does not. `CH_(3)-overset(O)overset(||)C-CH_(2)CH_(2)-CH_(3)+3I_(2)+4NaOH overset(Delta)to CH_(3)CH_(2)CH_(2)COONa+underset("Yellow ppt.")(CHI_(3))+3NaI+3H_(2)O` `CH_(3)-CH_(2)-overset(O)overset(||)C-CH_(2)-CH_(3)+I_(2)+NaOH overset(Delta)to "No reaction"` (b)  
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| 89094. |
(a) How will you convert: (i) Aniline into fluorobenzene. (ii) Benzamide into Benzylamine . (iii)Ethanamineto N, N-Diethylethanamine. (b) Write the structures of A and B in the following : CH_(3)CH_(2)CN overset(OH^(+))underset("partial hydrolysis")to A overset(NaOH+Br_(2))to B (ii) CH_(3)CH_(2)Broverset((i)KCN)underset((ii)LiAIH_(4))toA overset(HNO_(2))underset(0@C)toB |
Answer» SOLUTION : (i)Aniline into fluorobenzene  (ii) benzamide into benzylamine  (iii) Ethanamine to N, N-Diethylethnamine `CH_(3)CH_(2)NH_(2)overset(C_(2)H_(5)Cl)toC_(2)H_(5)-oversetoverset(H)(|)N-C_(2)H_(5)overset(C_(2)H_(5)Cl)toC_(2)H_(5)-undersetunderset(C_(2)H_(5))(|)N-C_(2)H_(5)` (b) (i) STURCTURES of A and B are given as under : A: `CH_(3)CH_(2)CONH_(2)` B: `CH_(3) CH_(2)NH_(2)` (ii) STRUCTURE of A and B are given as under : A: `CH_(3)CH_(2)CH_(2)NH_(2)` B: `CH_(3)CH_(2)CH_(2)OH` |
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| 89095. |
A large number of antibiotics have been isolated from : |
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Answer» BACTERIA actinomycetes |
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| 89096. |
(a) How will you convert : (i)Benzoyl chloride to benzaldehyde (ii) Propanone to 2-propanol ? (b) Name the structure of the product formed when formaldehyde reacts with semicarbazide (NH_(2)CONHNH_(2)). |
| Answer» SOLUTION :(B) SEMICARBAZONE. | |
| 89097. |
(a) How will you convert an alkyl halide to a primary amine whose molecule has one more carbon atom than the used alkyl halide molecule ? (b) Why are amines more basic than the comparable alcohols ? |
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Answer» Solution :(a) The conversion can be carried out as under (ascent of series) : `""UNDERSET("Chloromethane")(CH_(3)Cl)+KCN rarr underset("Ethane nitrile")(CH_(3)C equiv N) underset(4[H])overset(Na//C_(2)H_(5)OH)(rarr) underset("ETHANAMINE")(CH_(3)CH_(2)NH_(2))` (b) Oxygen is more electronegative than nitrogen. Therefore, oxygen has less tendency to donate the electrons COMPARED to nitrogen. That is why AMINES containing N are more basic than alcohols containing oxygen. |
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| 89098. |
A large nuber of antibotics have been isolated from |
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Answer» BACTERIA actinomycetes |
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| 89099. |
a. How will you convert an alkyl halide to a primary amine whose molecule has one carbon atom more than the used alkyl halide molecule? b. Why are amines more basic than the comparable alcohols. |
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Answer» Solution :a. `underset("CHLOROMETHANE")(CH_(3)Cl)+KCNtounderset("Ethananitrile")(CH_(3)C-=N) underset(4[H])overset(Na//C_(2)H_(5)OH)(to) underset("Ethanamine")(CH_(3)CH_(2)NH_(2))` B. Due less electronegativity of oxygen atom as compared to nitrogen, AMINES are more basic than alcohols. |
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| 89100. |
A large irregularly-shaped closed tank is first evacuated and then connected to a 50-litre cylinder containing compressed nitrogen gas. The gas pressure in the cylinder, originally at 21.5 atm, falls to 1.55 atm after it is connected to the evacuated tank. Calculate the volume of the tank. |
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Answer» Solution :Let the volume of the TANK be V LITRES. AS the no. of moles of `N_(2)` before and after connecting it to the tank will be same, `(21.5 xx 50)/(RT) = (1.55 xx (50 + V))/(RT) ""[n = (pV)/(RT)]` `(("moles of " N_(2) " before"),("CONNECTION")) (("moles of " N_(2) " after"),("connection"))` `therefore V = 643.5` litres |
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