Tiêu đề Qualilative Chemistry Engg. Practice Sheet_With Solve-25.pdf ✅

2 ....................................................................................................................................  Chemistry 1st Paper Chapter-2 (b) †Zvgvi Dˇii iwk¥ ̧‡jv‡K Zv‡`i Zi1⁄2‣`‡N ̈©i wb¤œZi μgvbymv‡i mvwR‡q wjL| [BUET 18-19] [Easy] mgvavb: (a) (i) WiFi  gvB‡μvI‡qf (ii) Detecting fake currency  AwZ‡e ̧wb iwk¥ (iii) Optical fiber communication  Ae‡jvwnZ iwk¥ (iv) MRI machine  †iwWI Zi1⁄2 (b) †iwWI I‡qf > gvB‡μvI‡qf >Ae‡jvwnZ iwk¥ > AwZ‡e ̧wb iwk¥| 7. wb¤œwjwLZ mswÿß iƒcmg~‡ni c~Y©iƒc wjL| (i) DPF (ii) FTIR (iii) AAS (iv) MRI (v) BTU [BUET 18-19] [Medium] mgvavb: (i) DPF  Diesel Particulate Filter (ii) FTIR  Fourier Transform Infrared Spectroscopy (iii) AAS  Atomic Absorption Spectrometry (iv) MRI  Magnetic Resonance Imaging (v) BTU  British Thermal Unit 8. 200 mL 1.3  10–3 M NbgvÎvi AgNO3 `ae‡Yi mv‡_ 100 mL 4.5  10–5 M NbgvÎvi Na2S `aeY †gkv‡bv nj| G‡Z wK †Kv‡bv Aat‡ÿc co‡e? hyw3 `vI| [Ksp = 1.6  10–49] [BUET 18-19] [Medium] mgvavb: 2AgNO3 + Na2S ⇌ Ag2S + 2NaNO3; Ag2S ⇋ 2Ag+ + S2– Ag+ Gi cÖ_g NbgvÎv = 1.3  10–3 M Ag+ Gi †kl NbgvÎv = V1S1 V2 = 200  1.3 10–3 300 = 8.667  10–4 M S 2– Gi cÖ_g NbgvÎv = 4.5  10–5 M S 2– Gi †kl NbgvÎv = 100 4.5  10–5 300 = 1.5  10–5 M Ag2S Gi Kip = [Ag+ ] 2 × [S2– ] = (8.667  10–4 ) 2  1.5  10–5 = 1.1267  10–11 †h‡nZz Kip > Ksp †m‡nZz `ae‡Y Ag2S Gi Aat‡ÿc co‡e| 9. (a) 4d Dckw3 ͇̄i AiweUvj ̧‡jvi Rb ̈ n, l I m Gi gv‡bi ZvwjKv wjL| (b) cÖavb †Kvqv›Uvg msL ̈v n = 3 n‡j H kw3 ͇̄i me©vwaK KqwU B‡jKUab _vK‡Z cv‡i? [BUET 16-17] [Easy] mgvavb: (a) 4d Dckw3 ͇̄ii Rb ̈, n = 4 l = 2 m = – 2, – 1, 0, + 1, + 2 (b) n = 3 n‡j, H kw3 ͇̄i me©vwaK 2n2 = 2  3 2 = 18wU B‡jKUab _vK‡Z cv‡i| 10. `ae‡Y Cl – Avq‡bi Dcw ̄ wZ Kxfv‡e wbwðZ Ki‡e? [BUET 16-17] [Medium] mgvavb: Cl – Avqb wewkó `ae‡Y K‡qK †dvuUv AgNO3 †hvM Ki‡j AgCl Gi mv`v Aat‡ÿc c‡o| NaCl(aq) + AgNO3(aq)  AgCl(s) + NaNO3(aq) Drcbœ AgCl Gi Aat‡ÿc jNy HNO3 G `aeYxq wKš‘ NH4OH `ae‡Yi mv‡_ [Ag(NH3)2Cl] bvgK RwUj †h.M •Zwi K‡i| AgCl(s) + 2NH4OH(aq)  [Ag(NH3)2]Cl(aq) + 2H2O(l) 11. GKwU B‡jKUab D”P kw3 ̄Íi †_‡K wb¤œ kw3 ͇̄i jvd w`‡j wK N‡U? [BUET 14-15] [Easy] mgvavb: †dvUb wn‡m‡e kw3i wewKiY N‡U| 12. Ag2CO3 (Ksp = 8.2 10–12) Ges CuCO3 (Ksp = 2.5 10–10) Gi `ave ̈Zv wbY©q Ki Ges †Kvb Kve©‡bUwUi `ave ̈Zv Kg D‡jøL Ki| [BUET 16-17] [Easy] mgvavb: awi, Ag2CO3 Gi `ave ̈Zv = S1 Ges CuCO3 Gi `ave ̈Zv = S2 Ag2CO3 ⇌ 2Ag+ + CO2– 3 S1 2S1 S1 S1  (2S1) 2 = Ksp  4S1 3 = 8.2  10–12  S1 = 1.27  10–4 mol L–1 CuCO3 ⇌Cu2+ + CO2– 3 S2 S2 S2 S2  S2 = Ksp = 2.5 10–10  (S2) 2 = 2.5  10–10  S2 = 1.58  10–5 mol L–1  CuCO3 Gi `ave ̈Zv A‡cÿvK...Z Kg| 13. Mg(OH)2 Gi `ave ̈Zv KZ? `ave ̈Zvi ̧Ydj, Ksp = 1.2  10–12 [BUET 15-16] [Easy] mgvavb: Mg(OH)2 ⇌ Mg2+ + 2OH– x x 2x  `ave ̈Zvi ̧Ydj = [Mg2+] × [OH– ] 2 = x  (2x)2 = 4x3 GLb, 4x3 = 1.2  10–12  x = 6.69  10–5 mol L–1