Tiêu đề 4. Che. Engg. Practice Sheet-(2nd Paper) With Solve.pdf ✅

Zwor imvqb  Engineering Practice Sheet ....................................................................................................................... 3 15. GKwU Zwor ivmvqwbK †Kv‡l GKwU †jŠn `ЇK wb‡Kj w`‡q Zwor cÖ‡jcb Ki‡Z n‡e| G m¤úwK©Z wb‡Pi cÖkœ ̧‡jvi DËi `vI| (i) †Kvb avZzwU K ̈v‡_vW wnmv‡e e ̈envi Ki‡Z n‡e? (ii) †Kvb avZzwU A ̈v‡bvW wnmv‡e e ̈envi Ki‡Z n‡e? (iii) †Kvb B‡jK‡UavW †_‡K ewnt ̄’ mvwK©‡U B‡jKUab cÖevwnZ n‡e? (iv) A ̈v‡bvW B‡jK‡UavW Gi wewμqvwU wjL| (v) K ̈v‡_vW B‡jK‡UavW Gi wewμqvwU wjL| [BUET 11-12] mgvavb: (i) †jŠn `Ð (ii) wb‡Kj (iii) A ̈v‡bvW (wb‡Kj `Ð) †_‡K (iv) Ni – 2e–  Ni2+ (v) Ni2+ + 2e–  Ni 16. Kcvi mvj‡d‡Ui Rjxq `ae‡Y ` ̄Ívi P~Y© †hvM Kiv n‡j Kx N‡U? [BUET 10-11] mgvavb: Zn Øviv cÖwZ ̄’vcb wewμqv N‡U: CuSO4 + Zn  ZnSO4 + Cu 17. 50 wgwbU a‡i 0.2 A ̈vw¤úqvi we`y ̈r cÖevwnZ nIqvi 0.2 MÖvg Kcvi Rgv nq| Kcv‡ii ivmvqwbK Zzj ̈ IRb wbY©q Ki| [BUET 09-10] mgvavb: w = ZIt  w It = 0.20 0.20  50  60  Z = 3.33  10–4 g C–1 = 3.33  10–7 kg C–1  ivmvqwbK Zzj ̈ IRb, Z = 3.33  10–7 kg C–1 18. h_v_©Zv wePvi K‡i ev‡K ̈ ÒmZ ̈Ó ev Òwg_ ̈vÓ emvI| M ̈vjfvbvBwRs †jvnvi cvBc‡K cathodic protection cÖ`vb K‡i| [BUET 04-05] mgvavb: wg_ ̈v 19. jNy NaCl `ae‡Yi Zwor we‡køl‡Y Kx cwigvY we`y ̈r cÖev‡ni d‡j 1 NÈv mg‡q Pt K ̈v‡_v‡W cÖgvY Pvc I ZvcgvÎvq 0.112 L H2 Drcbœ n‡e| [BUET 03-04] mgvavb: It = neF  I = PV RT  e  F t = 1  0.112 0.0821  273   96500 3600 = 0.268 A 20. 5 mM Fe2+ m¤^wjZ GKwU `aeY Ce4+ w`‡q UvB‡UaU Kiv n‡jv Ges UvB‡Uak‡bi GK ch©v‡q 1 mM Fe3+ •Zix nj| 25C ZvcgvÎvq Aa©‡KvlwUi B.Gg.Gd EFe2+/Fe3+ wnmve Ki| †`Iqv Av‡Q, E 0 Fe2+/Fe3+ = 0.770 V| [BUET 02-03] mgvavb: E 0 Fe2+/Fe3+ = E0 cell – 0.0591 e log [Fe3+] [Fe2+] = 0.77 – 0.0591 1 log     1 5 = 0.8113 Volt 21. GKwU †jvnvi PvgP‡K K ̈v‡_vW wn‡m‡e e ̈envi K‡i Zwor we‡køl ̈ Ni(NO3)2 je‡Yi ga ̈ w`‡q 5.0 amp we`y ̈r 30 wgwbU a‡i Pvjbv Kiv n‡jv| †jvnvi Pvg‡Pi DcwiZ‡ji †ÿÎdj 7.0 cm2 | †jvnvi Pvg‡Pi Dci me©Î mgvbfv‡e m„ó wb‡Kj cÖ‡j‡ci cyiæZ¡ MYbv Ki| [Ni Gi cvigvYweK fi = 58.7 Ges wb‡Kj cÖ‡j‡ci Av‡cwÿK ̧iæZ¡ = 1.4] [BUET 02-03] mgvavb: It = neF t = weF M  w = MIt eF = 2.74 gm w = V × s = A × d × s s = Av‡cwÿK ̧iæZ¡ = 58.7  5  30  60 2  96500 g = 2.74 g w = Ads  d = w As = 2.74 7  1.4 = 0.28 cm 22. hLb GKwU †jvnvi †c‡iK‡K GKwU bxj Kcvi mvj‡dU `ae‡Yi g‡a ̈ Wzev‡bv nq ZLb †h wewμqv N‡U Zvi bvg wjL Ges wewμqvwU †`LvI| [BUET 01-02] mgvavb: Kcvi mvj‡d‡Ui Kcvi cigvYy †jvnvi cigvYy Øviv cÖwZ ̄’vwcZ nq I Kcv‡ii Zvgv‡U Aat‡ÿc c‡i, GwU GKwU cÖwZ ̄’vcb wewμqv| CuSO4(aq) + Fe(s) = Cu(s) + FeSO4(aq) 23. Rjxq CuSO4 `ae‡Yi ga ̈ w`‡q 0.40 amp we`y ̈r GK NÈv Pwjøk wgwbU a‡i cÖevwnZ Ki‡j 0.80 gm Kcvi Rgv nq| Kcv‡ii Zwor ivmvqwbK Zzj ̈v1⁄4 wbY©q Ki| [BUET 01-02] mgvavb: w = ZIt  Z = w It = 0.8 0.4  {(1  60) + 40}  60 = 3.33  10–4 g/C 24. wb‡¤œi †KvlwUi ZworØvi wewμqv I †Kvl wewμqv ̧wj †`LvI| Hg/Hg2Cl2(s), KCl(aq)||Cu++(aq) /Cu [BUET 00-01] mgvavb: GLv‡b, ZworØviwU n‡jv K ̈v‡jv‡gj ZworØvi| A ̈v‡bv‡Wi wewμqv: 2Hg(l) ⇌ Hg2+ 2 + 2e– Hg2+ 2 + 2Cl – (KCl †_‡K) ⇌ Hg2Cl2 2Hg(l) + 2Cl –  Hg2Cl2(s) + 2e– K ̈v‡_v‡Wi wewμqv: Cu2+(aq) + 2e–  Cu(s) mgMÖ wewμqv: 2Hg(l) + 2Cl – (aq) + Cu2+(aq)  Hg2Cl2(s) + Cu(s)