Time : 3 Hrs. Max. Marks : 720 For NEET (UG) - 2019
Test Booklet CodeDATE : 05/05/2019Q1SURYAARegd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005. Ph.: 011-47623456Time : 3 hrs.Answers & SolutionsMax. Marks : 720forNEET (UG) - 2019Important Instructions :1.The test is of 3 hours duration and Test Booklet contains 180 questions. Each question carries4 marks. For each correct response, the candidate will get 4 marks. For each incorrect response, onemark will be deducted from the total scores. The maximum marks are 720.2.Use Blue / Black Ball point Pen only for writing particulars on this page/marking responses.3.Rough work is to be done on the space provided for this purpose in the Test Booklet only.4.On completion of the test, the candidate must handover the Answer Sheet to the Invigilator before leavingthe Room / Hall. The candidates are allowed to take away this Test Booklet with them.5.The CODE for this Booklet is Q1.6.The candidates should ensure that the Answer Sheet is not folded. Do not make any stray marks on theAnswer Sheet. Do not write your Roll No. anywhere else except in the specified space in the TestBooklet/Answer Sheet.7.Each candidate must show on demand his/her Admission Card to the Invigilator.8.No candidate, without special permission of the Superintendent or Invigilator, would leave his/her seat.9.Use of Electronic/Manual Calculator is prohibited.10.The candidates are governed by all Rules and Regulations of the examination with regard to their conductin the Examination Hall. All cases of unfair means will be dealt with as per Rules and Regulations of thisexamination.11.No part of the Test Booklet and Answer Sheet shall be detached under any circumstances.12.The candidates will write the Correct Test Booklet Code as given in the Test Booklet / Answer Sheet inthe Attendance Sheet.1
NEET (UG)-2019 (Code-Q 1)1.Answer ( 4 )For the chemical reaction 2NH3 (g)N2 (g) 3H2 (g) S o l . Manganate (MnO42–) :The correct option is:d[NH3 ]d[N2 ] 2(1) dtdtMn–OOO– -bonds are of d –p typed[N2 ] 1 d[NH3 ] (2) dt2 dtO Permanganate (MnO4–) :d[NH3 ]d[H2 ](3) 3 2dtdt(4) OMnOOO– -bonds are of d –p type1 d[H2 ]1 d[NH3 ] 3 dt2 dt5.Which will make basic buffer?Answer ( 2 )(1) 100 mL of 0.1 M CH3COOH 100 mL of 0.1 MNaOH 2NH3S o l . N2 3H2 (2) 100 mL of 0.1 M HCl 200 mL of 0.1 MNH4OHRate of reaction is given as2.(3) 100 mL of 0.1 M HCl 100 mL of 0.1 MNaOHd[N2 ]1 d[H2 ]1 d[NH3 ] dt3 dt2 dtThe non-essential amino acid among thefollowing is:(1) Leucine(2) Alanine(3) Lysine(4) Valine(4) 50 mL of 0.1 M NaOH 25 mL of 0.1 MCH3COOHAnswer ( 2 )Sol.(1)Answer ( 2 )S o l . Alanine3.For an ideal solution, the correct option is :(1) mix V 0 at constant T and PAfter0010 mmol(2) mix H 0 at constant T and PHydrolysis of salt takes place.(3) mix G 0 at constant T and PThis is not basic buffer.(4) mix S 0 at constant T and P(2)Answer ( 2 )S o l . For ideal solution, mixH 0 mixS 0HCl NH4OH NH4Cl H2OBefore 100 mL200 mL0 0.1 M 0.1 M 10 mmol 20 mmolAfter010 mmol10 mmolThis is basic buffer mixG 0(3) mixV 04.CH3COOH NaOH CH3COONa H2OBefore 100 mL100 mL0 0.1 M 0.1 M 10 mmol 10mmolHCl NaOHBefore 100 mL 0.1 MThe manganate and permanganate ions aretetrahedral, due to :(1) There is no -bondingAfter(2) The -bonding involves overlap ofp-orbitals of oxygen with p-orbitals ofmanganese 10 mmol0100 mL 0.1 M 10 mmol0 NaCl H2O010 mmol Neutral solution(4)Before(3) The -bonding involves overlap ofd-orbitals of oxygen with d-orbitals ofmanganeseAfter(4) The -bonding involves overlap ofp-orbitals of oxygen with d-orbitals ofmanganeseCH3COOH NaOH CH3COONa H2O25 mL50 mL0 0.1M 0.1 M 2.5 mmol 5 mmol02.5 mmol2.5 mmolThis is basic solution due to NaOH.This is not basic buffer.2
NEET (UG)-2019 (Code-Q 1)6.Answer ( 4 )The number of moles of hydrogen moleculesrequired to produce 20 moles of ammoniathrough Haber's process is :(1) 20(2) 30(3) 40(4) 10 1S o l . (a) 2Cu 1 2Answer ( 2 ) 6 )( ) (b) 3MnO2(4 4HS o l . Haber's processN2 (g) 3H2 (g) 2NH3 (g) 72 moles of NH3 3 moles of H2Hence 20 moles of NH3 7 4(3) Be2(4) O2 7 45MnO2 4H }10.S o l . MO configuration C2 is:8. *1s2, 2s2, *2s2, 2(d) 2MnO 4 3Mn2( ) 2H2 O Answer ( 2 ) 1s2,0MnO2 O2 } Not a disproportionationWhich of the following diatomic molecularspecies has only bonds according toMolecular Orbital Theory?(2) C2 6 (c) 2KMnO 42 K2MnO4 3 20 30 moles of H22(1) N2 42MnO 4 MnO2 2H2 O} Disproportionation20 moles need to be produced7.0Cu2( ) Cu0 } DisproportionationThe compound that is most difficult toprotonate is :O 2p2x 2p2y(1) H3CHOThe biodegradable polymer is:(2) H3C(1) Nylon-2-Nylon 6CH3O(2) Nylon-6(3) Ph(3) Buna-SHO(4) Nylon-6,6(4) HHAnswer ( 1 )Answer ( 3 )S o l . Nylon-2-Nylon 6Which of the following reactions aredisproportionation reaction?S o l . Due to involvement of lone pair of electrons inresonance in phenol, it will have positivecharge (partial), hence incoming proton willnot be able to attack easily.(a) 2Cu Cu2 Cu011.9.(1) [GeCl6]2–(b) 3MnO24 4H 2MnO 4 MnO2 2H2 O(2) [Sn(OH)6]2– (c) 2KMnO 4 K2MnO 4 MnO2 O2(d)2MnO 42 3MnWhich of the following species is not stable?(3) [SiCl6]2–(4) [SiF6]2– 2H2 O 5MnO2 4HAnswer ( 3 )S o l . Due to presence of d-orbital in Si, Ge andSn they form species like SiF62–, [GeCl6]2–,[Sn(OH)6]2–Select the correct option from the following(1) (a), (b) and (c)(2) (a), (c) and (d) SiCl 62– does not exist because six largechloride ions cannot be accommodatedaround Si4 due to limitation of its size.(3) (a) and (d) only(4) (a) and (b) only3
NEET (UG)-2019 (Code-Q 1)12.Answer ( 3 )An alkene "A" on reaction with O3 and Zn–H2Ogives propanone and ethanal in equimolarratio. Addition of HCl to alkene "A" gives "B"as the major product. The structure ofproduct "B" is:S o l . (a) Pure nitrogenCH2Cl(b) Haber process: Ammonia(c) Contact process: Sulphuric acid(d) Deacon’s process : Chlorine(1) H3 C–CH2 –CH–CH314.CH3(2) H3C–CH2–C–CH3The most suitable reagent for the followingconversion, is :H 3C C CClCH3H3 CCH3HHcis-2-buteneCH3(1) H2, Pd/C, quinoline(3) H3C–CH–CHCl: Sodium azide orBarium azide(2) Zn/HClCH3(3) Hg2 /H , H2O(4) Na/liquid NH3CH3Answer ( 1 )(4) Cl–CH2–CH2–CHCH3S o l . H3C C C CH3Answer ( 2 )H2, Pd/C,H3CCH3CquinolineCHHcis-2-buteneSol.15.The correct order of the basic strength ofmethyl substituted amines in aqueous solutionis :(1) (CH3)3N CH3NH2 (CH3)2NH(2) (CH3)3N (CH3)2NH CH3NH2(3) CH3NH2 (CH3)2NH (CH3)3N(4) (CH3)2NH CH3NH2 (CH3)3NAnswer ( 4 )13.S o l . In aqueous solution, electron donatinginductive effect, solvation effect (H-bonding)and steric hindrance all together affect basicstrength of substituted aminesMatch the following :(a) Pure nitrogen(i) ChlorineBasic character :(b) Haber process(ii) Sulphuric acid(CH3)2NH CH3NH2 (CH3)3N2 (c) Contact process (iii) Ammonia16.1 3 (a)(b)(c)(d)A gas at 350 K and 15 bar has molar volume20 percent smaller than that for an ideal gasunder the same conditions. The correctoption about the gas and its compressibilityfactor (Z) is :(1) (ii)(iv)(i)(iii)(1) Z 1 and repulsive forces are dominant(2) (iii)(iv)(ii)(i)(2) Z 1 and attractive forces are dominant(3) (iv)(iii)(ii)(i)(3) Z 1 and repulsive forces are dominant(4) (i)(ii)(iii)(iv)(4) Z 1 and attractive forces are dominant(d) Deacon’s process (iv) Sodium azide orBarium azideWhich of the following is the correct option?4
NEET (UG)-2019 (Code-Q 1)Answer ( 2 )Sol. 17.Answer ( 4 )Compressibility factor(Z) VrealVideal Vreal Videal; Hence Z 1 If Z 1, attractive forces are dominantamonggaseousmoleculesandliquefaction of gas will be easy.S o l . Penicillin G20.(1) Clark's method(2) Ion-exchange method(3) Synthetic resins methodThe number of sigma ( ) and pi ( ) bonds inpent-2-en-4-yne is(4) Calgon's methodAnswer ( 1 )(1) 8 bonds and 5 bondsS o l . Clark's method is used to remove temporaryhardness of water, in which bicarbonates ofcalcium and magnesium are reacted withslaked lime Ca(OH)2(2) 11 bonds and 2 bonds(3) 13 bonds and no bonds(4) 10 bonds and 3 bondsCa(HCO3)2 Ca(OH)2 2CaCO3 2H2OAnswer ( 4 )Mg(HCO3)2 2Ca(OH)2 2CaCO3 Mg(OH)2 H 2H2OH S o l . H – C – C C – H C – C 21.H H18.Which of the following series of transitions inthe spectrum of hydrogen atom fall in visibleregion?Number of bonds 10(1) Balmer seriesand number of bonds 3(2) Paschen seriesFor the second period elements the correctincreasing order of first ionisation enthalpy is:(3) Brackett series(4) Lyman series(1) Li B Be C O N F NeAnswer ( 1 )(2) Li B Be C N O F NeS o l . In H-spectrum, Balmer series transitions fallin visible region.(3) Li Be B C O N F Ne22.(4) Li Be B C N O F NeAnswer ( 1 )Match the Xenon compounds in Column-Iwith its structure in Column-II and assign thecorrect code:Column-IS o l . ‘Be’ and ‘N’ have comparatively more stablevalence sub-shell than ‘B’ and ‘O’. Correct order of first ionisation enthalpyis:Li B Be C O N F Ne19.The method used to remove temporaryhardness of water is :Among the following, the narrow spectrumantibiotic is :Column-II(a) XeF4(i) Pyramidal(b) XeF6(ii) Square planar(c) XeOF4(iii) Distorted octahedral(d) XeO3(iv) Square pyramidalCode:(1) Ampicillin(2) Amoxycillin(a)(b)(c)(d)(1) (ii)(iii)(iv)(i)(2) (ii)(iii)(i)(iv)(3) Chloramphenicol(3) (iii)(iv)(i)(ii)(4) Penicillin G(4) (i)(ii)(iii)(iv)5
NEET (UG)-2019 (Code-Q 1)Answer ( 1 )Answer ( 1 )S o l . (a) XeF4 :FFXeFCH Square planarF(b) XeF624.XeFF23.:XeO SquarepyramidalO PyramidalAnswer ( 4 )S o l . The correct structure isO2AHO H2 OOOOO Br – Br – Br OOHCH3The correct structure of tribromooctaoxide is–OCH3OOOOOOO(3) O Br – Br – Br – O – (4) O Br – Br – Br O–OOOOOOThe structure of intermediate A in thefollowing reaction, isCH(A)Cumenehydroperoxide–OOOOOO––(1) O Br – Br – Br – O (2) O – Br – Br – Br O––––OOOOOOO(d) XeO3 H3C – C –CH3FFHOH DistortedoctahedralFFO2H 2OFFH3C – C – O – O – HCumeneXe:(c) XeOF4 :CH3Sol.FFCH3CH3 H3CO O OTribromooctaoxideCH325.CH3Which is the correct thermal stability orderfor H2E (E O, S, Se, Te and Po)?(1) H2O H2S H2Se H2Te H2PoH3C – C – O – O – H(2) H2Po H2Te H2Se H2S H2O(1)(3) H2Se H2Te H2Po H2O H2SCH3(4) H2S H2O H2Se H2Te H2PoO – O – CHAnswer ( 2 )CH3(2)S o l . On going down the group thermal stabilityorder for H2E decreases because H–E bondenergy decreases Order of stability would be:-CH2 – O – O – HHCH2Po H2Te H2Se H2S H2OCH326.(3)(1) Expansion oftemperatureCH3O–In which case change in entropy is negative?CHCH3agas(2) Sublimation of solid to gas(3) 2H(g) H2(g)(4)(4) Evaporation of water6atconstant
NEET (UG)-2019 (Code-Q 1)Answer ( 3 )Sol. 27.29. H2 O v , S 0H2 O Expansionofgastemperature, S 0at Sublimation of solid to gas, S 0 2H(g) H2(g), S 0 ( ng 0)Identify the incorrect statement related toPCl5 from the following:(1) Two axial P–Cl bonds make an angle of180 with each otherconstant(2) Axial P–Cl bonds are longer thanequatorial P–Cl bonds(3) PCl5 molecule is non-reactive(4) Three equatorial P–Cl bonds make anangle of 120 with each otherA compound is formed by cation C andanion A. The anions form hexagonal closepacked (hcp) lattice and the cations occupy75% of octahedral voids. The formula of thecompound is :Answer ( 3 )Axial bondCl240 pmClSol.120 Cl(1) C3A290 PClCl(2) C3A4Cl(3) C4A3(1) True(4) C2A3Answer ( 2 )Sol. Axial bond : 240 pmEquatorial bond : 202 pm(3) FalseDue to longer and hence weaker axialbonds, PCl5 is a reactive molecule.34(4) True18 4 30.Which of the following is an amphoterichydroxide?(1) Ca(OH)292(2) Mg(OH)2So formula of compound will be(3) Be(OH)2C 9 A6 C9 A12(4) Sr(OH)2Answer ( 3 )2C9A12 C3A428.180 (2) TrueCations(C) are in 75% O.V., so number ofcations (C) PClAnions(A) are in hcp, so number of anions(A) 6 6 Equatorial bond202 pmS o l . Be(OH)2 amphoteric in nature, since it canreact both with acid and baseAmong the following, the one that is not agreen house gas isBe(OH)2 2HCl BeCl2 2H2OBe(OH)2 2NaOH Na2 [Be(OH)4](1) Methane31.(3) Sulphur dioxide4d, 5p, 5f and 6p orbitals are arranged in theorder of decreasing energy. The correctoption is(4) Nitrous oxide(1) 6p 5f 5p 4d(2) OzoneAnswer ( 3 )(2) 6p 5f 4d 5pS o l . Fact(3) 5f 6p 4d 5pSO2 (g) is not a greenhouse gas.(4) 5f 6p 5p 4d7
NEET (UG)-2019 (Code-Q 1)Answer ( 4 )Answer ( 3 )S o l . (n l) values for, 4d 4 2 6S o l . Malachite : CuCO3.Cu(OH)2 (Green colour)5p 5 1 65f 5 3 8pH of a saturated solution of Ca(OH)2 is 9. Thesolubility product (Ksp) of Ca(OH)2 is:6p 6 1 7(1) 0.25 10–10(2) 0.125 10–15(3) 0.5 10–10(4) 0.5 10–1535. Correct order of energy would beAnswer ( 4 )5f 6p 5p 4d32.For the cell reaction2 S o l . Ca(OH)2 Ca 2OH2Fe3 (aq) 2I (aq) 2Fe2 (aq) I2 (aq)EΘcellpH 9 0.24 V at 298 K. The standard GibbsHence [Ca2 ] [Given that Faraday constant F 96500 C mol–1](1) – 23.16 kJ mol–1(2) 46.32 kJ mol–1(3) 23.16 kJ mol–1(4) – 46.32 kJ mol–1 10 5 5 2 (10 ) 2 ΘS o l . GΘ nF Ecell 0.5 10–15 – 2 96500 0.24 Jmol–136. – 46320 J mol–1 – 46.32 kJ(2) GeX4 (X F, Cl, Br, I) is more stable thanGeX2Conjugate base for Br onsted acids H2O andHF are :(3) SnF4 is ionic in natureH3O and F–, respectivelyOH– and F–, respectively(4) PbF4 is covalent in natureAnswer ( 4 )(3) H3O and H2F , respectivelyS o l . PbF4 and SnF4 are ionic in nature.(4) OH– and H2F , respectively37.Answer ( 2 )OHS o l . H2O–Conjugate base(2) 50 mL of 2 M AgNO3 50 mL of 1.5 M KIHF on loss of H ion becomes F – is theconjugate base of HF(3) 50 mL of 0.1 M AgNO3 50 mL of 0.1 M KI(4) 50 mL of 1 M AgNO3 50 mL of 1.5 M KIExample :34.Which mixture of the solutions will lead to theformation of negatively charged colloidal[Agl]l– sol ?(1) 50 mL of 1 M AgNO3 50 mL of 2 M KI H3O Conjugate acidHF H2OAcidBaseWhich of the following is incorrect statement?(1) SiCl4 is easily hydrolysedmol–1ii(2)10 52Thus Ksp [Ca2 ][OH–]2Answer ( 4 )(1)pOH 14 – 9 5[OH–] 10–5 Menergy ( r GΘ ) of the cell reaction is :33.HenceAnswer ( 1 )–F Conjugatebase H3OConjugateacidS o l . Generally charge present on the colloid is dueto adsorption of common ion from dispersionmedium. Millimole of KI is maximum inoption (2) (50 2 100) so act as solvent andanion I– is adsorbed by the colloid AgI formedWhich one is malachite from the following?(1) Cu(OH)2(2) Fe3O4(3) CuCO3.Cu(OH)2(4) CuFeS28
NEET (UG)-2019 (Code-Q 1)38.The mixture that forms maximum boilingazeotrope is:heatCH 2OH HCl(3)(1) Ethanol WaterCH 2Cl H 2O(2) Acetone Carbon disulphide(3) Heptane Octane(4) Water Nitric acidCu 2Cl2 –N 2CI(4)Answer ( 4 )S o l . Solutions showing negative deviation fromRaoult's law form maximum boiling azeotropeAnswer ( 1 )ClWater and Nitric acid forms maximumboiling azeotrope39.CI N 2 Cl2Sol.For a cell involving one electron E cell 0.59 Vat 298 K, the equilibrium constant for the cellreaction is :AlCl 3 HClGeneration of electrophile:Cl.Cl. AlCl3 Cl.2.303 RT 0.059 V at T 298 K Given thatF (1) 1.0 105–ClAlCl3. –Cl. AlCl4.(2) 1.0 1010Electrophile(3) 1.0 1030Cl(4) 1.0 102Answer ( 2 )0.059log Qnlog Keq Cl0.059log Keq (from equation (i))1 E cell0.59 100.059 0.05941.The major product of the following reactionis:COOH NH3Among the following, the reaction thatproceedsthroughanelectrophilicsubstitution, is:(1) Cl2AlCl 3ClH(ii)Keq 1010 1 101040.etc.(i)(At equilibrium, Q Keq and Ecell 0)0 E cell – Cl (i)S o l . Ecell E cell –HCOOHOCI HClCOOHNH(1)CIstrong heating(2)NH2CIO(2) Cl2UV lightCINH2CI(3)CINH2CI9COOH(4)CONH 2
NEET (UG)-2019 (Code-Q 1)Answer ( 1 )Answer ( 2 )COOH NH3Sol.– – 4S o l . First order rate constant is given as,COO NH4COOHCOO NH– 2H2O– NH3[A ]2.303log 0t[A]tk 99% completed reaction,CONH 2k 2.303100logt1CONH 2 2.303log102tStrong heatingOk 2.303 2log10tt 2.3034.606 2 kkt 4.606kCNHCOPhthalimide42.What is the correct electronic configurationof the central atom in K4[Fe(CN)6] based oncrystal field theory?44.60(1) t 2g e gUnder isothermal condition, a gas at 300 Kexpands from 0.1 L to 0.25 L against aconstant external pressure of 2 bar. The workdone by the gas is(Given that 1 L bar 100 J)(2) e3 t 32(3) e 4 t 224e2g(4) t 2g(1) 5 kJ(2) 25 J(3) 30 J(4) –30 JAnswer ( 4 )S o l . Wirr – Pext VAnswer ( 1 ) – 2 bar (0.25 – 0.1) LS o l . K4[Fe(CN)6]Fe ground state: [Ar]3d64s2 – 2 0.15 L-barFe2 : 3d64s0 – 0.30 L-bareg – 0.30 100 J – 30 JInSphericalfieldEnergy2 Fe43.t 62g e0g45.t2gIn the presence of–6CN strong fieldIf the rate constant for a first order reactionis k, the time (t) required for the completionof 99% of the reaction is given by:(1) t 6.909/k(2) t 4.606/k(3) t 2.303/k(4) t 0.693/kEnzymes that utilize ATP in phosphate transferrequire an alkaline earth metal (M) as thecofactor. M is :(1) Mg(2) Ca(3) Sr(4) BeAnswer ( 1 )S o l . All enzymes that utilize ATP in phosphatetransfer require magnesium(Mg) as theco-factor.10
NEET (UG)-2019 (Code-Q 1)46.Select the correct option.Cells in G0 phase :(1) enter the cell cycle(2) suspend the cell cycle(3) terminate the cell cycle(4) exit the cell cycleAnswer (4)(d)(1) (iii)(ii)(i)(iv)(2) (iii)(iv)(i)(ii)(3) (iv)(iii)(i)(ii)(4) (iii)(i)(iv)(ii)Homo habilis—650-800 ccPhloem in gymnosperms lacks :Homo neanderthalensis —1400 cc(1) Sieve tubes onlyHomo erectus—900 cc(2) Companion cells onlyHomo sapiens—1350 cc50.(4) Albuminous cells and sieve cellsAnswer (3)S o l . Phloem in Gymnosperms lacks both sievetube and companion cells.Which of the following factors is responsiblefor the formation of concentrated urine?(2) SecretionoferythropoietinJuxtaglomerular complex(a) i gene(i) -galactosidase(b) z gene(ii) Permease(c) a gene(iii) Repressor(d) y gene(iv) Transacetylase(a)by(3) Hydrostatic pressure during glomerularfiltration(b)(c)(d)(1) (iii)(i)(ii)(iv)(2) (iii)(i)(iv)(ii)(3) (iii)(iv)(i)(ii)(4) (i)(iii)(ii)(iv)Answer (2)S o l . In lac operon(4) Low levels of antidiuretic hormonei gene RepressorAnswer (1)z gene -galactosidaseS o l . The proximity between loop of henle and vasarecta as well as counter current in them helpin maintaining an increasing osmolalritytowards the inner medullary interstitium. Thismechanism help to maintain a concentrationgradient in medullary interstitium so humanurine is nearly four times concentrated thaninitial filtrate formed.y gene Permeasea gene Transacetylase51.(2) Syngamy(3) Parthenogenesis(4) Autogamy(i) 900 ccAnswer (3)S o l . The phenomenon in which female gametedevelops into embryo without getting fusedwith male gamete (fertilisation) is calledparthenogenesis.(b) Homo neanderthalensis (ii) 1350 cc(c) Homo erectus(iii) 650-800 cc(d) Homo sapiens(iv) 1400 ccIn some plants, the female gamete developsinto embryo without fertilization. Thisphenomenon is known as(1) ParthenocarpyMatch the hominids with their correct brainsize :(a) Homo habilisMatch the following genes of the Lac operonwith their respective products :Select the correct option.(1) Maintaining hyperosmolarity towardsinner medullary interstitium in the kidneys.49.(c)S o l . The correct match of hominids and their brainsizes are :(3) Both sieve tubes and companion cells48.(b)Answer (2)S o l . Cells in G0 phase are said to exit cell cycle.These are at quiescent stage and do notproliferate unless called upon to do so.47.(a)11
NEET (UG)-2019 (Code-Q 1)52.Under which of the following conditions willthere be no change in the reading frame offollowing mRNA?Sol. Vertebrosternal ribs are true ribs, dorsallythey are attached to the thoracicvertebrae and ventrally connected to thesternum with the help of hyaline cartilage.First seven pairs of ribs are called trueribs. 8 th , 9 th and 10 th pairs of ribs do notarticulate directly with the sternum b
NEET (UG)-2019 (Code- 1)Q 6. The number of moles of hydrogen molecules required to produce 20 moles of ammonia through Haber's process is : (1) 20 (2) 30 (3) 40 (4) 10 Answer (2) Sol. Haber's process N22 3(g) 3H (g) 2NH (g) 20 moles need to be produced 2 moles of NH 3 3 mol
Family Questionnaire Regarding Autism Spectrum Disorder (ASD) 41. How many hours per week is your family member with ASD engaged in each of the following activities? 0 hr 1 hr 2 hrs 3 hrs 4 hrs 5 hrs 10 hrs 15 hrs 20 hrs 25 hrs 30 hrs 35 hrs 40 hrs College/Post-Secondary School Adult Day Habilitation Employed Social Activities with
portable power systems Recharge from a Renewable Power Source Recharge Through Household AC Power Emergency Backup for Power Outages or Rolling Blackouts . AP Portable Power System 1800 800 Specifications 400 5W 8W 35W 5W 8W 40W 40W 60W 65W 5W 80W 120W 300W 700W 1000W 1200W 95 hrs 60 hrs 15.5 hrs 90 hrs 55 hrs 13 hrs 65 hrs 40 hrs 9.5 hrs 55 hrs
Science 30 19 30 19 30 19 30 19 30 19 30 19 30 19 30 19 30 19 30 19 30 19 30 19 30 19 30 19 30 19 30 19 30 19 30 19 Abbreviated Battery 230 212 210 189 240 203 240 201 220 187 220 187 220 186 220 185 200 160 200 160 200 160 Total Testing Time 3 hrs. 32 min. 3 hrs. 9 min. 3 hrs. 23 min. 3 hrs. 21 min. 3 hrs. 7 min. 3 hrs. 7 min. 3 hrs. 6 min. 3 .
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Toledo, Ohio 109 2 hrs Columbus, Ohio 124 2 hrs Pittsburgh, Pennsylvania 130 2.5 hrs Detroit, Michigan 162 3 hrs Buffalo, New York 213 4 hrs Cincinnati, Ohio 226 3.45 hrs Indianapolis, Indiana 297 5.15 hrs Chi
ENGLISH MATHEMATICS _2022 WEEKLY TEACHING PLAN _ GRADE 9 TERM 1 Week 1 3 days Week 2 5 days Week 3 5 days Week 4 . GRADE 8 WORK WHOLE NUMBERS Properties of numbers . 4.5 hrs. 9 hrs. 9 hrs. 2 hrs. 4.5 hrs. 4.5 hrs. 3 hrs. Topics, concepts and skills
Paper -I (MCQ) Test of General Science and awareness (Level-Post graduate) Time: 2 hrs.* Max. Marks: 300 marks (150 questions) Paper – II Subject specific laboratory based practical questions Time: 3 hrs.* Max. Marks: 150 marks Skill Test Skills pertaining to subject matter of the concerned post would be
The Department of Basic Education (DBE) has set the question papers for Physical Sciences . Geography P1 (3 hrs) THEORY Geography P2 (1½ hrs) MAPWORK Friday 18/09/2015 Mathematics P1 (3 hrs) Mathematical Literacy P1 (3 hrs) Afrikaans HL P2 (2½ hrs) Afrikaans FAL P2 (2 hrs) ANNEXURE B (Page 1 of 2) ASSESSMENT INSTRUCTION 29 OF 2015 .
