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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.
| 50051. |
A hydrocarbon contains 85.7% carbon. If 42 mg of the compound contain 3.01 xx 10^20 molecules, find the molecular formula of the compound. |
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| 50052. |
A hydrocarbon containing two double bonds on reductive ozonolysis gave glyoxal, ethanal and propanone . Give the structure of the hydrocarbon along with its IUPAC name . |
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Answer» Solution :To write the structure of the products of OZONOLYSIS with their carbonyl GROUPS facing each other. `underset"ETHANAL"(CH_3-oversetoverset(H)(|)C=O) "" underset"Glyoxal"(O=oversetoverset(H)|C-oversetoversetH|C=O)"" underset"Propanone"(O=oversetoverset(CH_3)|C-CH_3` Step 2 . To write the structure of the hydrocarbon Remove oxygen atoms from each of the three carbonyl COMPOUNDS and connect them by double bonds , we have, `underset"2-Methylhexa-2,4-diene"(underset6CH_3-underset5oversetoversetH|C=underset4oversetoversetH|C-underset3oversetoversetH|C=underset2oversetoverset(CH_3)|C-underset1CH_3)` Thus, the GIVEN hydrocarbon is 2-methylhexa-2,4-diene. |
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| 50053. |
A hydrocarbon C,H, (A) reacts wit A reacts with HBr to form compound (B). Compound (B) rencts with aqueous potassium hydroxide to give (C) of molecular formula C_(3)H_(6)O to give (C) of molecular formula C.H.O. What are (A) Ans. (B) and (C). Explain the reactions |
Answer» SOLUTION : 1. The hydrocarbon withmolecular formula ` C_3H_6` (A) is definedas propene , `CH_3-CH=CH_2` 2. Propene reacts with HBR to form bromopropane `CH_3-CH_2-CH_2Br` as (B).  3. 1- bromopropanereact with aqueouspotassium HYDROXIDE to give 1- propanaol`CH_3-CH_2-CH_2OH` as (C). 4. 2-bromo propanereacts with aqueous KOH to give 2-propanol as (C) 
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| 50054. |
A hydrocarbon C_(5)H_(10) does not react with chlorine in dark but gives C_(5)H_(9)Cl in bright sunlight. Identify the hydrocarbon. |
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Answer» Solution :(i) `C_(5)H_(10)` can be either alkene or cyclo alkane. (ii) Since the hydrocarbon does ot react with CL in dark, hence it is not an alkene, it MAY be cyclopentane. (iii) Since, it forms only single monochloro derivative in BRIGHT sunlight, all the H-atoms should be identical. hence it is cyclopentane 
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| 50055. |
A hydrocarbon, C_(4)H_(8), neither decolourised bromine in carbontetrachloride nor reacted with hydrogen bromide. When heated to 200^(@)C with hydrogen in presence of nickel catalyst, a new hydrocarbon, C_(4)H_(10) was formed. What was the orginal hydrocarbon? |
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Answer» Solution :`C_(4)H_(8)` can have the following isomers: `underset("1-Butene")(CH_(3)CH_(2))CH=CH_(2)""underset("2-Butene")(CH_(3)CH=CHCH_(3))""{:(CH_(2)-CH_(2)),("||"),(CH_(2)-CH_(2)),("Cyclobutane"):}` 1-Butene and 2-butene are ruled out as these decolourise BROMINE in carbontetrachloride and ALSO react with hydrogen bromide. The hydrocarbon is cyclobutane. Cyclo-butane on reduction with hydrogenin presece of NICKEL opens up the ring and forms n-butane. `underset(CH_(2))underset(|)CH_(2)-underset(CH_(2))underset(|)CH_(2)+H_(2)underset(200^(@)C)overset(Ni)toCH_(3)CH_(2)CH_(2)CH_(3)` |
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| 50056. |
A hydrocarbon C_(3)H_(6)(A) reacts with HBr to form compound (B). Compound (B) reacts with aqueous potassium hydroxide to give (C) of molecular formula C_(3)H_(8)O. What are (A) (B) and (C). Explain the reactions. |
Answer» SOLUTION :
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| 50057. |
A hydrocarbon C_3H_6 (A) reacts with HBr to form compound (B). Compound (B) reacts with aqueous potassium hydroxide to give (C) of molecular formula C_3H_8O. What are (A) (B) and (C ). Explain the reactions. |
Answer» SOLUTION : (i)The hydrocarbon with molecular formula `C_3H_6` (A) is identified as propene. `CH_3 - CH = CH_2`. (ii) Propene reacts with HBR to FORM bromopropane `CH_3 - CH_2 - CH_2Br` as (B).  (iii) 1-bromopropane react with aqueous POTASSIUM hydroxide to give l-propanol `CH_3-CH_2 - CH_2OH` as (C ). (iv) 2-bromo propane reacts with aqueous KOH to give 2-propanol as (C) 
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| 50058. |
A hydrocarbon A(V.D=36)forms only one monochloro substitution product.A will be: |
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Answer» iso-pentane |
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| 50059. |
A hydrocarbon (A) of the formula , C_7H_12 on ozonolysis gives a compound (B) which undergoes aldol condensation giving 1-acetylcyclopentene. Identify (A) and (B). |
Answer» Solution :Since, the ozonolysis of hydrocarbon, (A), `C_7H_12` gives only ONE compound (B), it is clear that the compound (A), is cyclo ALKENE, The ozonolysis product , (b) (with no LESS of carbon atom) must have undergone intra- ALDOL CONDENSATION and gives 1-acetyl cyclopentene, i.e.,  .
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| 50060. |
A hydrocarbon (A) of molecular weight 54 reacts with an excess of Br_2 in CCl_4 to give a compound (B) whose molecular weight is 593% more than that of (A). However, on catalytic hydrogenation with excess of hyrogen (A) forms (C) whose molecular weight is noly 7.4% more that that of (A). (A) reacts with CH_3CH_2Br in the presence of NaNH_2 to give another hydrocarbon (D) which on ozonolyisis yields kiketone (E). (E) on oxidation gives propionic acid. Give the structure of (A) to (E) with reason. |
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Answer» Solution :step 1. to determine the molecular weights of COMPOUNDS (B) and (C) (i). The molecular weight of compound (A) is 54 while that of compound (B), which it GIVES on treatment with an excess of `Br_2` in `C Cl_4`, is `593%` more than that of (A). Molecular weight of `(B)=((100+593))/(100)xx54=374.22` Thus the increase in weight due to the adiiton of Br atoms is `374.22-54.0=320.22` Since atomic weight of Br is 80, the number of Br atoms added `=(320.22)/(80)=4`. As such the HYDROCARBON (A) must be an alkyne. (ii). Further since the molecular weight of compound (C) which hydrocarbon (A) gives on catalytic HYDROGENATION, is only `7.4%` more than that of (A). The molecular weight of (C) is `((100+7.4)xx54)/(100)=57.994=58`(approx) Thus, the increase in weight due to the addition of H atoms `=58-54=4`. Since the atomic weight of `H` is 1 the number of H atoms added during catalytic hydrogenation is `(4)/(1)=4`. Therefore, hyrocarbon (A) must be an alkyne.  
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| 50061. |
A hydrocarbon A, of C_8H_10, on ozonolysis gives compound (B), C_4H_6O_2 only. The compound (B) can also be obtained from alkyl bromide (C), C_3H_5Br, upon treatment with magnesium in dry ehter, followed by CO_2 and acidification. Identify (A), (B) and (C) and aslo give equations for the reactions. |
Answer» Solution :The compound (B) is formed via the formation of GRIGNARD reagent followed by reaction with `CO_2` and ACIDIFICATION. It shows that `C_3H_5Br` behaves like SATURATED compound. It is only POSSIBLE if `C_3H_5Br` is bromocyclopropane (Cyclopropylbromide).  (A) on ozonolysis GIVES only (B), hence it should be symmetrical unsaturated hydrocarbon. 
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| 50062. |
A hydrocarbon (A) having molecular formula C_(7)H_(14) is capable to exhibit both geometrical and optical isomerism on oxidation with hot conc. MnO_(4) followed by heating with sodalime yields two alkanes (B) and (C). Mixture of B and C can also be formed by oxidatio with hot concentration alkaline MnO_(4)^(-) of hydrocarbon (D) having molecular formula C_(7)H_(12), followed by heating with sodalime. Q. The hydrocarbon (D) is |
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Answer» 5 METHYL 2 hexyne |
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| 50063. |
A hydrocarbon (A) having molecular formula C_(7)H_(14) is capable to exhibit both geometrical and optical isomerism on oxidation with hot conc. MnO_(4) followed by heating with sodalime yields two alkanes (B) and (C). Mixture of B and C can also be formed by oxidatio with hot concentration alkaline MnO_(4)^(-) of hydrocarbon (D) having molecular formula C_(7)H_(12), followed by heating with sodalime. Q. Two compounds (B) and (C) |
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Answer» ETHANE and propane |
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| 50064. |
A hydrocarbon (A) having molecular formula C_(7)H_(14) is capable to exhibit both geometrical and optical isomerism on oxidation with hot conc. MnO_(4) followed by heating with sodalime yields two alkanes (B) and (C). Mixture of B and C can also be formed by oxidatio with hot concentration alkaline MnO_(4)^(-) of hydrocarbon (D) having molecular formula C_(7)H_(12), followed by heating with sodalime. Q. The hydrocarbon (A) is |
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Answer» 2-methyl-2-hexene |
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| 50065. |
A hydrocarbon A, adds one mole of hydrogen in presence of platinum catalyst to form n-hexane when A is oxidised vigorously with KMnO_4 , a single carboxylic acid containing three carbon atoms isolated . Give the structure of A and explain. |
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Answer» SOLUTION :(i)SINCE the hydrocarbon A adds one mole of `H_2` in presence of Pt to form n-hexane, THEREFORE, A must be an hexene. (ii)Since A on vigorous oxidation with `KMnO_4` gives a single carboxylic acid containing three carbon atoms, therefore A must be a symmetrical hexene, i.e., hex-3-ene. `UNDERSET"Hex-3-ene (A)"(CH_3CH_2CH=CHCH_2CH_3) underset(KMnO_4)overset"[O]"to underset"Propionic acid"(2CH_3CH_2COOH)` Thus, the GIVEN hydrocarbon A is hex-3-ene. |
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| 50066. |
(A) Hydrazine contains nitrogen but does nto give Lassaigne's test for nitrogen. (R) Hydrazine reacts with fused sodium to give H_(2) gas. |
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Answer» If both ASSERTION and REASON are correct and reason is the correct explanation of the assertion |
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| 50067. |
A hydrated metal chloride contains 27.8%metal and 48.5% halogen by weight.Specific heat ofmetal (M) is 0.67jg^(-1) . What is the valency of the metal? |
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Answer» Solution :SPECIFIC HEAT of M =0.67 `jg^(-1)` =0.16cal `g^(-1)` ATOMIC MASS of M = 6.4/0.16=40 Relative number of M atoms `=(27.8)/(40)=0.695` Relative number of Cl atoms `=(48.5)/(35.5)=1.37` Ratio of number of atoms, of M and C =0.695 :1.37=1:2 FORMULA of metal chloride is `MCl_(2)` Valency of metal is 2 |
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| 50068. |
A hydocarbon having molecular formula C_(8)H_(16) absorbed no hydrogen during hydrogenation. Thehydrocarbon is |
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Answer» CYCLOALKANE |
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| 50069. |
A human adult breathes in apporximately 0.50 L of atm with each breath.If an air tank holds 10 L of air at 200 atm, how many breaths the tank will supply ? |
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Answer» SOLUTION :`P_(1)V_(1)=P_(2) V_(2)` `200xx10=1xxv_(2) "" V_(2)=2000L` NUMBER of Breaths `=(2000l)/(0.5L)=4000` |
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| 50070. |
(a) How many sub-shells are associated with n = 4? (b) How many electrons will be present in the sub-shells having m_(s) value of -1//2 for n = 4? |
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Answer» Solution :(a) `n = 4, l = 0, 1, 2,3` (4 subshells, VIZ , s, p, d and F) (b) No. of orbitals in 4th shell `= n^(2) = 4^(2) =1 6` Each orbital has one ELECTRON with `m_(s) = -1//2`. Hence, there will be 16 electrons with `m_(s) = -1//2` |
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| 50071. |
(a) Howmanysub - shellsare associated withm=4 (b ) Howmany electronwill bepresentin thesub - shellshavingmvalueof (1)/(2)forn= 4 ? |
Answer» SOLUTION :(a) n=4so noof subshell=4for n=4 the VALUEOF l andsubshellis asfollows (b )`m_(1)= (1)/(2) ` andn=4so noof ELECTRON=16 n=4 no oforbitalthereare twoelectroninwhichone has`m_(s) + (1)/(2)` andother `m_(s) = (1)/(2)` So `m_(s ) - (1)/(2) ` andn=4such16 electron . |
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| 50072. |
(a) How many orbitals are possible for a d-subshell ? (b) Draw the shapes of d_(xy) and d_(x^2 - y^2) orbitals? What is common between these and what is difference between these orbitals ? What is the angle between the lobes of these orbitals ? (c ) Name a 3d orbitals which has electron density along all the three axes. |
| Answer» SOLUTION :`1s^2 2s^2 2p^6 3s^2 3p^5 4s^1.` | |
| 50073. |
(a). How many 1,2-shifts are invovled (b). number of alpha-H in major product. |
Answer» 
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| 50074. |
(a) How is H_(2)O_(2) prepared?(b) Explain about the structure of H_(2)O_(2). |
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Answer» Solution :(a) Hydrogen PEROXIDE can be made by adding a metal peroxide to dilute acid. `BaO_(2(s))+H_(2)SO_(4(aq))toBaSO_(4(aq))+H_(2)O_(2(aq))` (b) Structure of `H_(2)O_(2)`.  (i) `H_(2)O_(2)`has a non-polar structure. The molecular dimensions in the gas phase and solid phase differ as shown in the figure. (ii) Both in gas phase and solid phase, the `H_(2)O_(2)`, MOLECULE adopt a skew configuration due to repulsive interaction of the -OH bonds with lone pairs of electrons on each oxygen atom. (iii) Indeed, it is the smallest molecule known to show hindrance rotation about a single bond. In solid phase, the dihedral angle is SENSITIVE and hydrogen bonding decreasing from `115.5^(@)` in the gas phase to 90.2o, in the solid phase. (iv) Structurally, `H_2O_2` is represented by the dihydroxyl formula in which the TWO O-H groups do not lie in the same plane. In the solid phase of molecule, the dihedral angle reduces to 90.2" due to hydrogen bonding and the `O - O - H` angle expands from `94.8^(@)` to `101.9^(@)`. (v) One way of explaining the shape of hydrogen peroxide is that the hydrogen atoms would Tie on the pages of a partly opened book, and the oxygen atoms along the spin. |
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| 50075. |
(a) How bond energy veries fromN_(2) ^(-)toN_(2)^(+)and why ? (b) On the basis of molecular orbital theory what is similartiy between (i)F_(2), O_(2) ^(- )(ii)CO, N_(2) , NO^(+) ? |
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Answer» Solution :(a) Bond energy of ` N_(2)^(+)` = Bond energy of ` N_(2)^(-)` . This is because they have the same bond order. (Strictly SPEAKING , ` N_(2)^(-)`is SLIGHTLY less stable and hence has less bond energy then `1 N_(2)^(+)`due to presnece of greater NUMBER of electronsin the antibonding molecular orbitals ) (b) (i) Same bond order and bond lenght . (II) Same bond order and bond length . |
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| 50076. |
A homogeneous mixture contains 90g of water, 18.4g of glycerol and 6.4g of methanol. What is the mole fraction of methanol in the givenmixture? |
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| 50077. |
(A): Holes in ozone layer are observed at the north and the south poles by scientists. (R): UV radiation damages eyes causing cataract of eyes. |
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Answer» Both (A) and (R) are TRUE and (R) is the CORRECT explanation of (A) |
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| 50078. |
A highly viscous liquid was heated from 10^(@)C to 14^(@)C. The per cent decrease in viscosity will be about |
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| 50079. |
A highly viscous compound (X) on reacting with conc. HNO_(3) gives an inorganic ester (Z) which on detonation gives CO_(2), N_(2), O_(2) as gaseous products. With KMnO_(4) it gives a compound (Y) of the molecular formula C_(2)O_(4)H_(2) which can also be obtained from glycol using same reagent. The compound (X) is |
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Answer» Phenol The reactions are `{:(CH_(2)_OH),("|"),(CH-OH overset(Conc. HNO_(3))RARR underset((Z))underset("Glycerltrinitrate")(C_(3)H_(5)(ONO_(2))_(3))overset("Detonating")rarrCO_(2)+N_(2)+O_(2)+H_(2)O),("|"),(underset((X))(CH_(2)-OH)):}` |
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| 50080. |
A highly concentrated solution (about 5 M) of an alkali metal in ammonia (pick the odd one out) |
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Answer» is called EXPANDED metal |
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| 50081. |
A hexagonal close-packed lattic can be represented by figures (a) and (b) below. If c=asqrt((8)/(3))=1.633a , there is an atom at each corner of the unit cell and another atom which can be located by moving one-third the distance along the diagonal of the rhombus base, starting at the lower left hand corner and moving perpendicularly upward by c/2 . Mg crystallizes in this lattice and has a density of 1.74 g cm^(-3). What is the distance between nearest heighbours? |
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Answer» 1.6 |
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| 50082. |
A heated copper block at 130^(@)C loses 340 J ofheat to the surroundings which are at room temperature of 32^(@)C . Calculate (i) the entropy change of the system ( copper block )(ii) the entropy change in the surrounding (iii) the total entropy changein the universe due to this process Assume that the temperature of the block and the surroundings remains constant. |
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Answer» Solution :`T_("system") 130^(@)C = 130 + 273 K = 403 K, T _(surr) = 32^(@)C = 32 + 273 K = 305K` `q_("system") = -340 J , Q _(surr) =+ 340 J` (i) `DeltaS_("system") = (q _("system"))/(T_("system")) = ( -340J)/(403K) = - 0.84JK ^(-1) `(ii) `DeltaS_(surr) = (q_(surr))/(T_(surr))= (+340J)/(350K) = + 1.11 JK^(-1)` (III) `DeltaS_("total")`or `DeltaS_("system") + DeltaS_(surr) = - 0.84 + ( +1.11) JK^(-1)= 0.27 JK^(_1)` |
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| 50083. |
Assertion (A) : Heavy water is used as moderator in nuclear reactors.Reason (R) : Heavy water undergoes exchange reaction. |
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Answer» Both A and R are FALSE |
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| 50084. |
A heat engine absorbs heat q_(1) from a source at temperature T_(1) and heat q_(2) from a source at temperature T_(2). Work done is found to be J (q_(1) + q_(2)). This is in accordance with |
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Answer» VIOLATES 1st LAW of thermodynamics |
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| 50085. |
(A) : Heat flows always from a hotter body to a colder body by itself (R ): without an external aid heat cannot flow from colder body to a hotter body |
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Answer» Both A and R are CORRECT and R is the correct EXPLANATION of A |
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| 50086. |
A hArr B,K=8---(I) BhArr C,K, =10 ---(II) C hArr D,K=0.01 ---(III) The correct order of Delta G values of processes at the same temperature is |
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Answer» `III GT I gt II` |
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| 50087. |
A hard water sample contains magnesium bicarbonate only. 100 ml of the hard water sample requires 10 ml. of standard soap solution (1 ml soap solution = 1 mg CaCO_3 ) to get good lather. Find out of the weight of Ca(OH_2)to be added to 10 Kg water sample to make it soft water. |
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| 50088. |
A hard water sample has 131 ppm CaSO_(4). What fraction of the water must be evporated in a container before solid CaSO_(4) begins to deposit. K_(sp) of CaSO_(4) = 9.0 xx 10^(-6). |
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Answer» Solution :Maximum solubility of `CaSO_(4)` in water `S = sqrt(K_(sp)) = 3 xx 10^(-3)mol L^(-1)` Let `V` litre of sample is taken, then `CaSO_(4)` present `= (131 xx V xx 10^(3))/(10^(6)) G` [`because` ppm = g of `CaSO_(4)` in `10^(6) g` of sample] `= 131 xx 10^(-3) V g` `= (131 xx 10^(-3)xxV)/(136)` mole in V L If water is evaporated on HEATING so that just precipitation of `CaSO_(4)` occurs. Let `V_(1)L` of water is left, then `(131 xx 10^(-3)xxV)/(136)` mol is present in `V_(1)L` solution is equal to `3 xx 10^(-3) xx V_(1)` mol `:. (131 xx 10^(-3) xx V)/(136) = 3 xx 10^(-3) xx V_(1) :. V_(1) = 0.32V` Thus, volume evaporated =` V - 0.32 V = 0.68V` or `68%` of water should be evaporated. |
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| 50089. |
A hard water sample contains 162 ppm of calcium bicorbonate only. Find out the volume of standard soap solution (1 ml =10^(-3) g CaCO_(3)) to be added to the 100mlf of hard water sample. |
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| 50090. |
A hard water sample contains dissolved Mg(HCO_3)_2only. If degree of hardness of that sample is 150ppm, calculate the weight of Mg(HCO_3)_2present in 1kg of water sample. |
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Answer» Solution :Degree of hardness is 150ppm. It means that `10^6`G of WATER sample contains 150 g of `CaCO_3`(which is equivalent to `Mg(HCO_3)_2` . Weight of `CaCO_3`in 1KG `(10^3 g)` of water `= (10^3 xx 150)/(10^6) =0.15 g ` 0.15 g of `CaCO_3 -= 0.219 g ` of `Mg(HCO_3)_2` 0.15 g of `CaCO_3 -= (146 xx 0.15)/(100)` `= 0.219g Mg (HCO_3)_2` Weight of `Mg(HCO_3)_2`present in 1kg of water = 0.219 gram |
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| 50091. |
A halide with formula C_6H_13I is found to give two isomeric alkenes 2-methyl-2-pentene and 4-methyl-2-pentene on dehydrohalogenation with alcoholic KOH.Suggest its structure. |
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| 50092. |
A halide of Be(X) sublimes on heating and is a bad conductor of electricity in molten state. From its aqueous solution it is diffucult to obtain anhydrous salt. This halide of berylium (X) is obtained by heating berylium oxide with carbon tetrachloride at 800^(@)C. This halide of berylium forms a complex of the M_(2)[BeX_(4)] What is the compound formed when BeCI_(2) disolves in water in cold conditions ? |
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Answer» `Be(OH)_(2)` |
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| 50093. |
A halide of Be(X) sublimes on heating and is a bad conductor of electricity in molten state. From its aqueous solution it is diffucult to obtain anhydrous salt. This halide of berylium (X) is obtained by heating berylium oxide with carbon tetrachloride at 800^(@)C. This halide of berylium forms a complex of the M_(2)[BeX_(4)] What is the compound (X) |
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Answer» `BeBr_(2)` `2BeO + C CI_(4) overset"(800^(@)C)rarr 2BeCI_(2) + CO_(2)` |
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| 50094. |
A half cell rectio A^(-)-E^(-)rarr a hasa large negative reduction potential.If follows that |
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Answer» A is easily REDUCED |
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| 50095. |
A+H_(2)OrarrNaOH, A+O_(2)overset(400^(@)C)rarrBunderset(at 25^(@)C)overset(H_(2)O)rarr NaOH + O_(2) Which of the following statement is false regarding B. |
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Answer» B turns GREEN CHROMIC salt SOLUTION to yellow |
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| 50096. |
(a) H_(2)O_(2) +O_(3) to H_(2)O+2O_(2) (b) H_(2)O_(2) +Ag_(2)O to 2Ag+H_(2)O +O_(2) Role of hydrogen peroxide in the above reactions is respectively : |
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Answer» OXIDIZING in (a) and reducing in (b) |
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| 50097. |
A + H_2O_2 to B + O_2.In this reaction H_2O_2 is (B is the compound of chlorine) |
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Answer» OXIDANT |
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| 50098. |
A H-like species is in some excited state "A" and on absorbing a photon of energy 10.2eV gets promotedto new stateB. When the electrons from state "B" return back to ground statedirectly or indirectly, photons of a fifteen different wavelengths are observed in which only mine photon have energy greater than 10.2.eV. [Given : hc=1240 eV-nm] Determine orbit number of state "A" of hydrogen like speci. |
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Answer» 2 `n_(2)(n_(2)-1)=30rArrn_(2)=6` `10.2=13.6""Z^(2)[(1)/(n_(2))-(1)/(36)]` `Z& n_(1)` is an INTEGRE `thereforeZ=3,"" n_(1)=3=A` |
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| 50099. |
A gas can be liquefied |
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Answer» above its CRITICAL temperature |
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| 50100. |
A group which deactivites the benzene ring towards eletrophilic substitution but which directs the incoming group principally to the o- and p- positons is |
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Answer» `-NH_(2)` Ortho`-` and para`-` directing groups are as follow `-CH_(3),C_(2)H_(5)(-R),-NH_(2),-OH, `halogens, `(Cl,Br,I)` |
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