5<\/sub>. \n <\/p>\nQuestion 29: \na. 1-alkynes are weakly acidic in nature. \nGive any two reactions to showthe acidic character of ethyne. \nb. From the following, select the one in which Markownikoffs rule is best applicable. \n \n <\/p>\n
Answer any three from question numbers 30 to 33. Each carries four scores. \n <\/p>\n
Question 30: \nThe enthalpy change in a process is the same, whether the process is carried out in a Single step or in several steps. \na. Identify the law stated here. 1 \nb. Calculate the enthalpy of formation of CH4 from the following data: \n <\/p>\n
Question 31: \na. When nitrogen and hydrogen combines to form ammonia, the ratio between the volumes of gaseous reactants and products is 1:3:2. Name the law of chemical combination illustrated here. \nA compound is made up of two elements A and B, has A= 70%, B= 30%. The relative number of moles of A andB in the compound are 1.25 and 1.88 respectively. If the molecular mass of the compound is 160, find the molecular of the compound. \n <\/p>\n
Question 32: \nPhenol exhibit resonance. \ni. Draw the resonance structures of phenol. \nii. Predict the directive influence of -OH group in Benzene ring. \n <\/p>\n
Question 33: \na. Hydrogen peroxide restore the colour of lead paintings. Give a reason. \nb. How does the atomic hydrogen torch function for cutting and welding purposes? \n <\/p>\n
Answers<\/p>\n
Answer 1: \n\\(_{ 17 }^{ 35 }{ Cl }\\)<\/p>\n
Answer 2: \nBond order<\/p>\n
Answer 3: \nSurface tension.<\/p>\n
Answer 4: \nSolvay process.<\/p>\n
Answer 5: \nSilver halides.<\/p>\n
Answer 6: \nCaCl2<\/sub> or MgSO4<\/sub><\/p>\nAnswer 7: \n <\/p>\n
Answer 8: \nExtensive properties: Internal energy, Heat capacity. \nIntensive properties: Density, Temperature.<\/p>\n
Answer 9: \ni. Reduction species – CuO \nOxidation species – H2<\/sub> \nii. Reductant- H2<\/sub>, Oxidant – CuO<\/p>\nAnswer 10: \n <\/p>\n
Answer 11: \n11. i. 3 – chloropropanal \nii. 3 – methylpentanenitrite.<\/p>\n
Answer 12: \nThere is deviation from ideal behaviour at high pressure and low temperaure. This is due to two faulty assumptions in kinetic theory of gases. It is corrected by van der Waal\u2019s equation. \n \nwhere n is the number of moles of the gas a and b are van der Waal’s constants.<\/p>\n
Answer 13: \na. Diamond, graphite, fullerene \nb. This is due to the absence of vacant d- orbitals in carbon.<\/p>\n
Answer 14: \ni. Acid is a substance which dissociate to give H+<\/sup> ion in aqueous solution. Base is a substance which dissociate to give OH–<\/sup> ion in aqueous solution. \nii. HCl, H2<\/sub>SO4<\/sub>, CH3<\/sub>COOH etc are acids which give free H+<\/sup> ions in aqueous solution.<\/p>\nAnswer 15: \nBlue colour H2<\/sub> gas is evolved during the process. If M is alkali metal. \n \nBlue colour due to the ammoniated electron which absorbs energy. \n <\/p>\nAnswer 16: \n <\/p>\n
Answer 17: \nExtensive – Depends on the quantity of matter, eg., internal energy, heat capacity enthalpy volume, mass etc. \nIntensive – Does not depend on the quantity of matter, eg., Density, S.T.P, T, etc.<\/p>\n
Answer 18: \nOxidation number of fluorine decreases in HF while the oxidation no. of fluorine in-creases in HOF. So it is a redox reaction.<\/p>\n
Answer 19: \na. H2<\/sub>O2<\/sub> decomposes to release nascent oxygen that oxidises the colouring matter to colourless. \nb. Zn(s)+2NaOH(aq) \u2192 Na2<\/sub>ZnO2<\/sub>(aq) +H2<\/sub>(q) \n <\/p>\nAnswer 20: \n <\/p>\n
Answer 21: \na.Oxidation is a process which involves an increase in oxidation number of an element. \nReduction is a process which involves a decrease in oxidation number of an element. \nIMG \nThe reaction (1) is a redox reaction because there is a change in oxidation state. But (2) is not a redox reaction because there is no change in oxidation state.<\/p>\n
Answer 22: \na. 5, 5 -dimethyl heptan – 2 – ol \nb. 2 – pentanone \nc. 2, 4, 6 – tribromophenol.<\/p>\n
Answer 23: \na. Common ion effect. It the suppression of dissociation of weak electrolyte \n(CH3<\/sub>COOH) by addition of a salt containing common ion (CH3<\/sub>COONa). \nIMG \nc. When HCI gas is passed through AgCl solution. Solubility of AgCI decreases while the solubility product does not change.<\/p>\nAnswer 24: \na. CO2<\/sub>, methane, chlorofluoro carbons etc are the major gases which contribute to- wards global warming. \nb. 1. Minimise the use of automobiles. \n2. Plant more trees. \n3. Avoid burning of dry leaves, wood.<\/p>\nAnswer 25: \n \n <\/p>\n
Answer 26: \ni. because of the half filled electronic configuration of N. (N = 1s2<\/sup> 2s2<\/sup> 2p3<\/sup>) \nii. due to their small size, high electrone gativity and ionisation enthalpy,large charge\/radius ratio and due to the absence of vacant d-orbitals.<\/p>\nAnswer 27: \na. Low ionisation enthalpy of the electropositive atom (metal atom). \nHigh negative electron gain enthalpy of the electronegative atom (non-metal atom) \nHigh lattice enthalpy of the ionic compound formed. \ni. NH4<\/sup>+<\/sup> – Tetrahedral \nii. HgCl2<\/sup> – Linear<\/p>\nAnswer 28: \na. The solution becomes blue colour due to the formation of Cu2+<\/sup> ions. \nb. Ag+<\/sup> (AgNO3<\/sub>) is the oxidising agent and Cu is the reducing agent. \nc. Oxidation number of Cr in K2<\/sub>Cr2<\/sub>O7<\/sub> is +6 and P in H2<\/sub>P2<\/sub>O5<\/sub> is +4.<\/p>\nAnswer 29: \na. Ethyne reacts with sodium ethynide and liberate H2<\/sub>. It reacts with sodamide (NaNH2<\/sub>) to form sodium ethynide and liberate ammonia. \n <\/p>\nAnswer 30: \n \n <\/p>\n
Answer 31: \na. Gay – Lussac\u2019s law of Gaseous volumes \nb. The relative number of moles means % \/ atomic mass<\/p>\n
\n\n\nElement<\/strong><\/td>\n%<\/strong><\/td>\nRelative\u00a0no.of\u00a0moles<\/strong><\/td>\nSimple ratio<\/strong><\/td>\nSimplest\u00a0whole\u00a0no.ratio<\/strong><\/td>\n<\/tr>\n\nA<\/td>\n 70<\/td>\n 1.25<\/td>\n 1.25\/1.25= 1<\/td>\n 2<\/td>\n<\/tr>\n \nB<\/td>\n 30<\/td>\n 1.88<\/td>\n 1.88\/1.25 = 1.5<\/td>\n 3<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\nEmpirical formula is A2<\/sub>B2<\/sub> \nAtomic mass of A = % \/ no.of moles = 70 \/ 1.25 = 56 \nAtomic mass of B = % \/ no.of moles = 30\/ 1.88 = 15.96 \nSo, emp. Formula mass = 56 x 2 + 15.96 x 3 = 159.88 = 160 \nn = mol. mass \/ emp.Formula mass = 160\/160 = 1 \nMolecular formula = (emp.Formula)n = (A2<\/sub>B3<\/sub>)X 1 = A2<\/sub>B3<\/sub><\/p>\nAnswer 32: \n \nii. -OH group is an ortho-para directing group.<\/p>\n
Answer 33: \na. This is due to the oxidising action of H2<\/sub>O2<\/sub>. \n \nb. Here H atoms are allowed to recombine on the surface to be welded to generate the temperature of 4000K.<\/p>\nWe hope the Plus One Chemistry Model Question Papers Paper 3 help you. If you have any query regarding Plus One Chemistry Model Question Papers Paper 3, drop a comment below and we will get back to you at the earliest.<\/p>\n","protected":false},"excerpt":{"rendered":"
Plus One Chemistry Model Question Papers Paper 3 are part of Plus One Chemistry Previous Year Question Papers and Answers. Here we have given Plus One Chemistry Model Question Papers Paper 3. Board SCERT Class Plus One Subject Chemistry Category Plus One Previous Year Question Papers Plus One Chemistry Model Question Papers Paper 3 Time […]<\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_genesis_hide_title":false,"_genesis_hide_breadcrumbs":false,"_genesis_hide_singular_image":false,"_genesis_hide_footer_widgets":false,"_genesis_custom_body_class":"","_genesis_custom_post_class":"","_genesis_layout":"","footnotes":""},"categories":[42728],"tags":[],"yoast_head":"\n
Plus One Chemistry Model Question Papers Paper 3 - A Plus Topper<\/title>\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\t \n\t \n\t \n