Tiêu đề Chemical Change MCQ & CQ Practice Sheet Solution (HSC 26).pdf ✅

ivmvqwbK cwieZ©b  MCQ & CQ Practice Sheet Solution (HSC 26) 1 Topicwise Board Analysis eûwbe©vPwb cÖkœ UwcK 2016 2017 2018 2019 2021 2022 2023 †gvU wMÖb †Kwgw÷a I ivmvqwbK wewμqv 4 1 Ñ 2 5 3 5 20 ivmvqwbK mvg ̈ve ̄’v I jv-kv‡Zwjqv‡ii bxwZ 5 4 Ñ 2 6 5 10 32 fiwμqvi m~Î I mvg ̈aaæeK (Kp Ges Kc) 11 8 2 9 15 15 12 72 pH I pOH 8 7 1 11 15 12 12 66 evdvi `aeY I evdvi `ae‡Yi wμqv‡KŠkj 3 3 Ñ 5 5 5 6 27 GwmW ÿv‡ii kw3gvÎv I we‡qvRb 7 4 1 4 13 9 5 43 m„Rbkxj cÖkœ UwcK 2016 2017 2018 2019 2021 2022 2023 †gvU ivmvqwbK mvg ̈ve ̄’v I jv-kv‡Zwjqv‡ii bxwZ 1 8 2 5 9 4 7 36 fiwμqvi m~Î I mvg ̈aaæeK (Kp Ges Kc) 2 5 4 6 9 8 9 43 pH I pOH m¤úwK©Z mgm ̈vewj 3 9 8 1 8 7 7 43 evdvi `aeY I evdvi `ae‡Yi wμqv‡KŠkj 4 2 8 6 8 5 9 42 *we.`a.: 2020 mv‡j GBPGmwm cixÿv AbywôZ nqwb| ACS Chemistry Department Gi g‡bvbxZ eûwbe©vPwb cÖkœmg~n wMÖb †Kwgw÷a I ivmvqwbK wewμqv 1. DfgyLx wewμqvi •ewkó ̈ n‡jvÑ [Xv. †ev. 23] wewμqvwU †kl nq Dfqw`‡Ki wewμqvi nvi mgvb nq bv mvg ̈ve ̄’vq Avmvi cÖeYZv cÖfve‡Ki f~wgKv Av‡Q DËi: mvg ̈ve ̄’vq Avmvi cÖeYZv 2. wMÖb †Kwgw÷a‡Z AwaK Zvrch©c~Y© bxwZ †KvbwU? [Xv. †ev. 23] cÖfve‡Ki e ̈envi wbivc` `aveK e ̈envi m‡e©vËg GUg B‡Kvbwg `yN©Ubv cÖwZ‡iva DËi: m‡e©vËg GUg B‡Kvbwg 3. CH3COOH (aq) + NH4OH (aq)  Drcv` (X) + H2O(l); wewμqvwUi Drcv‡`i (X) cÖK...wZ Kx? [Kz. †ev. 23] A¤øagx© ÿviagx© Dfagx© wbi‡cÿ DËi: wbi‡cÿ e ̈vL ̈v: CH3COOH (aq) + NH4OH (aq)  CH3COONH4 (X) + H2O(l) `ye©j GwmW I `ye©j ÿvi‡Ki wewμqvq Drcbœ jeY wbi‡cÿ nq| ZvB (X) wbi‡cÿ| 4. wMÖb †K‡gw÷ai m~Pbv KZ mv‡j n‡qwQj? [g. †ev. 23] 1991 1990 1891 1890 DËi: 1991 5. Gwm‡Wi kw3i †Kvb μgwU mwVK? [g. †ev. 23] H2SO3 > HClO > HNO3 > H3PO4 HNO3 > H3PO4 > H2SO3 > HClO HClO > HNO3 > H3PO4 > H2SO3 H3PO4 > H2SO3 > HNO3 > HClO DËi: HNO3 > H3PO4 > H2SO3 > HClO e ̈vL ̈v: HNO3 +5 , H3PO4 +5 , H2SO3 +4 , HClO +1 N I P Gi g‡a ̈ N Gi AvKvi †QvU ZvB HNO3 kw3kvjx 6. wewμqvwUi •ewkó ̈ n‡jvÑ [iv. †ev. 23] (i) m¤§yL wewμqvi AvqZ‡bi ms‡KvPb N‡U (ii) AwaK cwigvY O2 †hv‡M wewμqvi mvg ̈ve ̄’v ev‡g m‡i hv‡e| (iii) cðvrgyLx wewμqvwU Zvcnvix n‡e wb‡Pi †KvbwU mwVK? ii ii I iii i I iii i, ii I iii DËi: i I iii e ̈vL ̈v: wewμqvwU‡Z SO2 Ges O2 Gi ms‡hv‡M †gvj msL ̈v K‡g| A_©vr m¤§yL wewμqvq AvqZ‡bi ms‡KvPb N‡U| wewμqvwU‡Z wewμq‡Ki NbgvÎv (SO2 ev O2) e„w× Ki‡j mvg ̈e ̄’v Wvbw`‡K m‡i hvq A_©vr m¤§yLgyLx wewμqvi MwZ‡eM e„w× cvq| †h‡nZz wewμqvwU Zv‡cvrcv`x ZvB wecixZgyLx A_©vr cðvrgyLx wewμqvwU Zvcnvix n‡e|
2  Chemistry 1 st Paper Chapter-4 7. meyR imvq‡bi g~jbxwZ KqwU? [iv. †ev. 22; h. †ev. 22] 10 12 14 16 DËi: 12 8. meyR imvq‡b- [e. †ev. 22] i. `aveK wnmv‡e CCl4 e ̈eüZ nq ii. eR© ̈ Drcv`b me©wb¤œ ivLv nq iii. welwμqvgy3 `ae ̈vw` e ̈eüZ nq wb‡Pi †KvbwU mwVK? i, ii ii, iii i, iii i, ii, iii DËi: ii, iii e ̈vL ̈v: CCl4 DØvqx e‡j Gi ev®ú evqy‡Z Qwo‡q c‡o| wbtk¦v‡mi m‡1⁄2 †mB evqy MÖnY Ki‡j wKWwb ÿwZMÖ ̄Í nq ZvB CCl4 `aveK wn‡m‡e e ̈envi Kiv hv‡e bv| 9. meyR imvq‡bi AšÍfy©3- [w`. †ev. 22] i. Kÿ ZvcgvÎv I Pv‡c wewμqv msNU‡bi †Póv Kiv ii. bevqb‡hvM ̈ KuvPvgvj Kg e ̈envi Kiv iii. gva ̈wgK †MŠY Drcv` n«vm Kiv wb‡Pi †KvbwU mwVK? i, ii i, iii ii, iii i, ii, iii DËi: ii, iii 10. CH3 – CH2 – OH + CH3 – COOH  CH3 – COO – CH2 – CH3 + H2O †hŠMwUi GUg BKbwg KZ? [P. †ev. 19] 65% 78% 83% 100% DËi: 83% e ̈vL ̈v: GLv‡b, CH3 – CH2 – OH Gi fi = 12 + 3 + 12 + 2 + 16 + 1 = 46 CH3 – COOH Gi fi = 12 + 3 + 12 + (16 × 2) + 1 = 60 CH3 – COO – CH2 – CH3 Gi fi = 12 + 3 + 12 + 16 + 16 + 12 + 2 + 12 + 3 = 88 Kvw•ÿZ Drcv` CH3 – COO – CH2 – CH3|  GUg BKbwg = CH3 – COO – CH2 – CH3 Gi fi CH3 – CH2 – OH Gi fi + CH3 – COOH Gi fi  100% = 88 46 + 60  100% = 83% 11. meyR imvq‡bi AšÍf©~3Ñ [Xv. †ev. 19] i. Kg ÿwZKi ivmvqwbK ms‡kølY ii. bevqb‡hvM ̈ KvuPvgvj e ̈envi iii. eR© ̈ c`v_© cÖwZ‡iva wb‡Pi †KvbwU mwVK? i I ii i I iii ii I iii i, ii I ii DËi: i, ii I ii e ̈vL ̈v: meyR imvq‡bi g~jbxwZ 12wU: 1. eR© ̈ c`v_© wbeviY 2. cvigvYweK wgZe ̈wqZv 3. Kg SzuwKc~Y© c`v_© ms‡kølY 4. wbivc` Ges ̄^v ̄’ ̈m¤§Z ivmvqwbK c`v_© DTM¢veb 5. wbivc` `aveK I weKvi‡Ki e ̈envi 6. Aí kw3 Øviv AwaK DTM¢veb 7. bevqb‡hvM ̈ KuvPvgvj e ̈envi 8. cvk¦© Drcv` n«vmKiY 9. cÖfveK e ̈envi 10. eR© ̈ e ̈e ̄’vcbvi wWRvBb •Zwi 11. mwVK mg‡q eR© ̈ we‡kølY 12. `yN©Ubv cÖwZ‡iv‡ai Rb ̈ mnRmva ̈ I wbivc` imvqb| 12. CH3CH2Br + NaOH(aq)  CH3CH2OH + NaBr wewμqvwUi GUg BKbwg KZ? [iv. †ev. 16] 29.11% 30.89% 36.12% 39.12% DËi: 30.89% e ̈vL ̈v: CH3CH2Br Gi fi = 12 + 3 + 12 + 2 + 79.9 = 108.9 NaOH Gi fi = 23 + 16 + 1 = 40 CH3CH2OH Gi fi = 12 + 3 + 12 + 2 + 16 + 1 = 46 Kvw•ÿZ Drcv`, B_vbj (CH3CH2OH)|  GUg BKbwg = Kw•ÿZ Drcv‡`i fi †gvU wewμq‡Ki fi × 100% = 46 (108.9 + 40) × 100% = 30.89% (cÖvq)  DÏxcKwUi Av‡jv‡K wb‡¤œi 13 I 14 bs cÖ‡kœi DËi `vI: e×cv‡Î CaCO3 wb¤œiƒ‡c we‡qvwRZ nqÑ CaCO3(s)  ⇌ CaO(s) + CO2(s) 13. wewμqvwU‡K GKgyLx Ki‡Z n‡jÑ [w`. †ev. 16] i. cv‡Îi XvKbv Ly‡j w`‡Z n‡e ii. Drcv`‡K Kw÷K †mvWv `ae‡Y Pvjbv Ki‡Z n‡e iii. cÖfveK e ̈envi Ki‡Z n‡e wb‡Pi †KvbwU mwVK? i I ii ii I iii i I iii i, ii I iii DËi: i I ii e ̈vL ̈v: wewμqvwU n‡jv: CaCO3(s)  ⇌ CaO(s) + CO2(g) wewμqvwU GKgyLx Ki‡Z n‡j cv‡Îi XvKbv Ly‡j w`‡Z n‡e Zvn‡j CO2 D‡o hv‡e d‡j CaCO3 cybivq Drcbœ n‡Z cvi‡e bv| ZvB wewμqvwU GKgyLx wewμqvq iƒcvšÍwiZ n‡e| Drcv`‡K KwóK †mvWv `ae‡Y Pvjbv Ki‡j KwóK †mvWv CO2 Gi mv‡_ wewμqv K‡i †mvwWqvg evBKve©‡bU Drcbœ Ki‡e hv A`aeYxq Aat‡ÿc •Zwi K‡i| d‡j wewμqvwU GKgyLx nq|