Tiêu đề 3. Che. Engg. 2nd Paper-Practice Sheet-24_With Solve.pdf ✅

cwigvYMZ imvqb  Engineering Practice Sheet................................................................................................................. 1 WRITTEN weMZ mv‡j BUET-G Avmv cÖkœvejx 1. 0.2 g Al I Cu Gi wgkÖY Av‡Q| 30C ZvcgvÎvq jNy H2SO4 mn‡hv‡M wewμqvq 730 torr Pv‡c 230 mL H2 Drcbœ nq| Rjxq ev®úPvc 30 torr n‡j, Al Gi kZKiv cwigvY KZ? [BUET 22-23] mgvavb: 2Al + 3H2SO4  Al2(SO4)3 + 3H2 Cu + H2SO4  wewμqv K‡i bv| Kcvi nvB‡Wav‡Rb‡K cÖwZ ̄’vcb Ki‡Z cv‡i bv KviY nvB‡Wav‡Rb mwμqZv wmwi‡R Kcv‡ii Dc‡i Aew ̄’Z| PT = Pair + PVapor  Pair = 730 – 30 = 700 torr = 700 mm Hg PV = nRT  700 760 × 0.23 = n × 0.082 × 303  n = 8.526 × 10–3 mol 3 mol H2  2 mol Al  8.526 × 10–3 mol H2  5.68 × 10–3 mol Al = 0.153 g Al  Al Gi kZKiv cwigvY = 0.153 0.2 × 100% = 76.5% 2. wewfbœ Drm †_‡K cÖvß cÖvK...wZK M ̈v‡m H2S GKwU mvaviY Ac`ae ̈| wb‡Pi wewμqvi gva ̈‡g cÖvK...wZK M ̈vm †_‡K H2S AcmviY Kiv nq: 2H2S(g) + SO2(g)  3S(s) + 2H2O(l) GKwU mvaviY cÖwμqvq hw` 290 K ZvcgvÎvq, 0.95 atm Pv‡c 10 wjUvi H2S AwZwi3 SO2 Gi mv‡_ wewμqv K‡i, Z‡e Drcbœ DcRvZ S Gi fi wbY©q Ki| [BUET 21-22] mgvavb: H2S Gi AvqZb, V = 10 L Pvc, P = 0.95 atm; ZvcgvÎv, T = 290 K  H2S Gi †gvj msL ̈v, nH2 S = 0.95 × 10 0.0821 × 290 = 0.399 mol 2 mol H2S †_‡K Drcbœ nq = 3 × 32 = 96 g S  0.399 mol H2S †_‡K Drcbœ nq = 96 × 0.399 2 = 19.152 g S 3. cÖvBgvwi ÷ ̈vÛvW© c`v_© Ges †m‡KÛvwi ÷ ̈vÛvW© c`v_© kbv3 KiÑ H2C2O4 Na2CO3 [BUET 20-21] mgvavb: H2C2O4 †m‡KÛvwi [†KjvwmZ H2C2O4(H2C2O4.2H2O) cÖvBgvwi ÷ ̈vÛvW© c`v_©]; Na2CO3 cÖvBgvwi 4. cvZjv mvjwdDwiK Gwm‡W Aï× 0.14 g †jvnv Mjv‡bv n‡jv| D3 `aeY 20 cm3 0.02 mol dm–3 K2Cr2O7 Gi mv‡_ wewμqv K‡i| (a) Fe2+ Avqb Ges Cr2O 2– 7 Avq‡bi mgZvc~Y© AvqwbK mgxKiY wjL| (b) Aweï× †jvnvi bgybvq f‡ii mv‡c‡ÿ †jvnvi kZKiv nvi wbY©q Ki| [†jvnvi cvigvYweK fi Fe = 56] [BUET 20-21] mgvavb: (a) 6Fe2+ + Cr2O 2– 7 + 14H+  6Fe3+ + 2Cr3+ + 7H2O (b) Cr2O 2– 7 + 6Fe2+ + 14H+  2Cr3+ + 6Fe3+ + 7H2O  1 mol K2Cr2O7  6 mol Fe2+  nK2Cr2O7 = 0.02 × 20 × 10–3 mol = 4 × 10–4 mol  nFe2+ = 6 × 4 × 10–4 mol = 2.4 × 10–3 mol  wFe2+ = 2.4  10–3  56 g = 0.13404 g  %weï×Zv = 0.13404 0.14 × 100% = 95.74% 5. GKwU †cwÝj w`‡q ÒBangladesh University of Engineering and Technology (BUET)” wjL‡Z 0.55 mg MÖvdvBU cÖ‡qvRb| Aw·‡R‡b `nb Ki‡j STP †Z D3 Kve©b cigvYymg~n Øviv Drcbœ M ̈v‡mi AvqZb wjUv‡i †ei Ki| [BUET 18-19] mgvavb: C + O2  CO2 1 mol 22.4 L †`Iqv Av‡Q, MÖvdvBU, C = 0.55 mg = 0.55  10–3 12 mol = 4.5833  10–5 mol  Drcbœ M ̈v‡mi AvqZb = (22.4  4.5833  10–5 ) L = 1.0267  10–3 L 6. wKWwb cv_‡ii cÖavb Dcv`vb K ̈vjwmqvg A·v‡jU| K ̈vjwmqvg A·v‡jU‡K i3 †_‡K AvniY K‡i Gwm‡W `aexf~Z K‡i 9.56 × 10–4 M KMnO4 Øviv UvB‡Uakb Kiv nj| 10 mL bgybv i‡3i Rb ̈ 24.2 mL KMnO4 `iKvi nq| cÖwZ wgwjwjUvi i‡3 Kx cwigvY K ̈vjwmqvg Av‡Q wbY©q Ki Ges wewμqvmg~n †`LvI| [Ca = 40] [BUET 18-19] mgvavb: 5CaC2O4 + 5H2SO4  5H2C2O4 + 5CaSO4 2KMnO4 + 5H2C2O4 + 3H2SO4  K2SO4 + 2MnSO4 + 10CO2 + 8H2O GLb, nCa 5 = nKMnO4 2  1 5 × wCa MCa = SV 2  1 5 × wCa 40 = 9.56 × 10–4 × 24.2 × 10–3 2  wCa = 2.31 × 10–3 g [10 mL i‡3]  1 mL i‡3 Ca Av‡Q = 2.31 × 10–3 10 g = 2.31 × 10–4 g
2 ....................................................................................................................................  Chemistry 2nd Paper Chapter-3 7. wgé Ae g ̈vM‡bwkqv nj g ̈vM‡bwkqvg nvB‡Wav·vB‡Wi Rjxq mvm‡cbkb hv mvaviYZ cvK ̄’wji GwmW cÖkwgZ Ki‡Z e ̈eüZ nq| cvK ̄’wji GwmW g~jZ nvB‡Wav‡K¬vwiK GwmW hv wgé Ae g ̈vM‡bwkqvi mv‡_ wewμqv K‡i g ̈vM‡bwkqvg †K¬vivBW I cvwb •Zwi K‡i| cvK ̄’wji Gwm‡Wi NbgvÎv 0.13 M n‡j wgé Ae g ̈vM‡bwkqvi GKwU †meb gvÎvi 100 mg g ̈vM‡bwkqvg nvB‡Wav·vBW KZ wjUvi cvK ̄’wji GwmW cÖkwgZ Ki‡e? [g ̈vM‡bwkqv‡gi cvigvYweK fi 24.3] [BUET 18-19] mgvavb: Mg(OH)2 + 2HCl  MgCl2 + 2H2O nMg(OH)2 1 = nHCl 2  wMg(OH)2 MMg(OH)2 = SHCl × VHCl 2  100 × 10–3 58.3 = 0.13 × VHCl 2  VHCl = 0.02638 L 8. GKwU †cwÝj w`‡q ÒBangladesh University of Engineering and Technology (BUET)Ó wjL‡Z 0.55 mg MÖvg MÖvdvBU cÖ‡qvRbÑ H wjLvi g‡a ̈ Kve©b cigvYyi msL ̈v †ei Ki| [BUET 18-19] mgvavb: awi, wjLvi g‡a ̈ Kve©b cigvYy msL ̈v N wU Avgiv Rvwb, †gvjmsL ̈v n = w M = N NA myZivs, 0.55  10–3 12 = N 6.023  1023  N = 2.76 × 1019 wU 9. GK wjUvi 2.1 Av‡cwÿK ̧iæZ¡ wewkó H2SO4 G Kx cwigvY cvwZZ cvwb †hvM Ki‡j Dnv 1.4 Av‡cwÿK ̧iæZ¡ wewkó H2SO4 G cwiYZ n‡e? [BUET 17-18] mgvavb:   1 V    ∵   = w V  1V1 = 2V2  2.1  1 = 1.4  (1 + V)  V = 0.5 L = 500 mL 10. wb¤œ wjwLZfv‡e GKwU wgkÖY •Zwi Kiv nj: 100 mL 0.05 M Na2CO3 `aeY + 100 mL 0.10 N H2SO4 `aeY + 50 mL 0.1 N NaOH `aeY + 50 mL 0.2 N HCl `aeY | wgkÖYwUi kw3gvÎv wbY©q Ki| [BUET 17-18] mgvavb: awi, wgkÖ‡Yi NbgvÎv St GLb, (ne)Acid – (ne)Base = StVt  (100  0.1 + 50  0.2) – (100  0.05  2 + 50  0.1) = (100 + 100 + 50 + 50)  St St = 1 60 M  pH = – log     1 60 = 1.778 11. wg_vBj A‡iÄ wb‡`©kK e ̈envi K‡i 50 cm3 †mvwWqvg nvB‡Wav·vBW‡K cÖkwgZ Ki‡Z 0.10 mol/dm3 NbgvÎvi 22.3 cm3 mvjwdDwiK GwmW cÖ‡qvRb nq| [BUET 16-17] (a) cÖkgb wewμqvwU wjL| (b) KZ †gvj mvjwdDwiK GwmW cÖ‡qvRb? (c) 50 cm3 †mvwWqvg nvB‡Wav·vBW `ae‡Y KZ †gvj †mvwWqvg nvB‡Wav·vBW Av‡Q? (d) †mvwWqvg nvB‡Wav·vBW `ae‡Yi NbgvÎv †gvjvwiwU‡Z wbY©q Ki| mgvavb: (a) 2NaOH + H2SO4  Na2SO4 + 2H2O (b) mvjwdDwiK Gwm‡Wi cÖ‡qvRbxq †gvj msL ̈v, n = VS = 22.3  10–3  0.1 mol = 2.23  10–3 mol (c) 1 mol H2SO4  2 mol NaOH  2.23  10–3 mol H2SO4  (2  2.23  10–3 ) mol NaOH = 4.46  10–3 mol NaOH (d) NaOH Gi NbgvÎv = 4.46  10–3 50  10–3 M = 0.0892 M 12. Cl – Avq‡bi NbgvÎv 0.25 M n‡j Gi `ae‡Yi ppm KZ? [BUET 15-16] mgvavb: 0.25 M NbgvÎvi Cl – Gi fi = 35.5  0.25 g = 8.875 g = 8875 mg  Cl – Gi gvÎv = 8875 mg L–1 = 8875 ppm 13. 20 mL 0.05 M KMnO4 A¤øxq `aeY Øviv 15 mL AvqZ‡bi H2O2 Gi †Kv‡bv bgybv‡K RvwiZ Kiv hvq| H2O2 Gi fi wbY©q Ki| [BUET 15-16] mgvavb: 2KMnO4 + 3H2SO4 + 5H2O2  K2SO4 + 2MnSO4 + 8H2O + 5O2 (eVS)KMnO4 = (eVS)H2O2  nO4 Gi e = 5; H2O2 Gi e = 2] S  S  1 6  w  1000 M  V = 1 6  w  1000 34  15 = 1 6  w = 0.085 g 14. GKwU we`y ̈r kw3 †K‡›`a evwl©K 2.4% iw¤^K mvjdvi m¤^wjZ 3.1  107 kg Kqjv †cvov‡bv nq| STP †Z Drcbœ SO2 M ̈v‡mi AvqZb wbY©q Ki| [BUET 14-15; CUET 15-16] mgvavb: S8 + 8O2  8SO2(g) 32  8  22.4 =g = 179.2 L = 0.256 kg 3.1  107 kg Gi 2.4% = 7.44  105 kg 0.256 kg mvjdvi †_‡K Drcbœ nq = 179.2 L SO2 7.44  105 kg mvjdvi †_‡K Drcbœ nq = 179.2  7.44  105 0.256 L = 5.21  108 L SO2
cwigvYMZ imvqb  Engineering Practice Sheet................................................................................................................ 3 15. 15.82 g †fRvjhy3 K ̈vjwmqvg Kve©‡bU nvB‡Wav‡K¬vwiK Gwm‡Wi mv‡_ wewμqv K‡i 37C ZvcgvÎv I 750 mm (Hg) Pv‡c 2.53 dm3 Kve©b WvB A·vBW Drcbœ K‡i| H K ̈vjwmqvg Kve©‡b‡U †fRv‡ji cwigvY wbY©q Ki| [BUET 14-15] mgvavb: CaCO3(s) + 2HCl(aq)  CaCl2 + CO2(g) + H2O(aq) (ne)CaCO3 = (ne)CO2  nCaCO3  2 = nCO2  2  nCaCO3 = nCO2  wCaCO3 MCaCO3 = PV RT  wCaCO3 100 = 0.987  2.53 0.0821  310  wCaCO3 = 9.81 g P = 750 mm (Hg) = 750 760 atm = 0.987 atm V = 2.53 dm3 T = 37C = 310 K  †fRv‡ji cwigvY = (15.82 – 9.81) g = 6.01 g 16. 3.78 g bvBwUaK GwmW‡K 400 mL cvwb‡Z `aexf~Z Kiv nj| †gvjvwiwU‡Z GB `ae‡Yi NbgvÎv KZ n‡e? AviI Kx cwigvY cvwb GB `ae‡Y †hvM Ki‡j Gi NbgvÎv 0.1 M n‡e? [BUET 14-15] mgvavb: 3.78 g HNO3 = 3.78 63 = 0.06 mol HNO3  †gvjvwiwU, S1 = 0.06 400  1000 = 0.15 M Avevi, V1S1 = V2S2  V2 = V1S1 S2 = 400  0.15 0.1 mL = 600 mL cvwb †hvM Ki‡Z n‡e = (600 – 400) mL = 200 mL 17. wb‡¤œi wewμqvmg~n c~Y© Ki: [BUET 14-15] [Hard] (a) CuSO4 + KI = (b) KMnO4 + H2C2O4 + H2SO4 = (c) Al2O3 + NaOH = (d) MnO– 4 + Cl – + H + = (e) Cl2 + Na2SO3 + H2O = mgvavb: (a) 2CuSO4 + 4KI = Cu2I2 + 2K2SO4 + I2 (b) 2KMnO4 + 5H2C2O4 + 3H2SO4 = K2SO4 + 2MnSO4 + 10CO2 + 8H2O (c) Al2O3 + 2NaOH = 2NaAlO2 + H2O (d) 2MnO– 4 + 10Cl – + 16H+ = 2Mn2+ + 5Cl2 + 8H2O (e) Cl2 + Na2SO3 + H2O = Na2SO4 + 2HCl 18. GKwU 0.204 g ÷x‡ji bgybv mvjwdDwiK Gwm‡W `aexf~Z Kiv nj| Drcbœ `ae‡Yi mv‡_ m¤ú~Y© wewμqv Ki‡Z 0.0220 mol/dm3 K2Cr2O7 `ae‡Yi 27.4 cm3 cÖ‡qvRb n‡jv| ÷xj bgybvq †jvnvi kZKiv cwigvY wbY©q Ki| wewμqvwUi AvqwbK mgxKiYwU wjL| [BUET 13-14] mgvavb: (ne)Fe2+ = (ne)K2Cr2O7 wFe2+ 55.85  1 = 0.0220  27.4 1000  6 wFe = 0.201998 g  †jvnvi cwigvY = 0.201998 0.204  100% = 99.019% wewμqvwUi AvqwbK mgxKiY: Fe2+ + 14H+ + Cr2O 2– 7  6Fe3+ + 2Cr3+ + 7H2O 19. 90.0 g cvwb‡Z 20.0 g NaOH Ges 81.0 g cvwb‡Z 18.25 g HCl c„_Kfv‡e `aexf~Z K‡i `ywU Rjxq `aeY cÖ ̄‘Z Kiv nj| Zvici `aeY `ywU‡K GK‡Î wgwkÖZ Kiv nj| Zvc cÖ‡qv‡M wgkÖYwU‡K m¤ú~Y©iƒ‡c ïKv‡bv nj| m¤ú~Y© ïKv‡Z 10 N›Uv mgq jvM‡jv| m¤ú~Y©iƒ‡c ïKv‡bvi ci KZ MÖvg jeY cvIqv hv‡e Ges ïKv‡bvi mgq cÖwZ †m‡K‡Û KZwU cvwbi AYy ev®úxf~Z n‡q‡Q Zv wbY©q Ki| [BUET 13-14] mgvavb: NaOH + HCl = NaCl + H2O 40 g  20 g 36.5 g 18.25 g 58.5 g 29.25 g 18 g 9 g  29.25 g salt cvIqv hv‡e| Total H2O = 90 + 81 + 9 = 180 g = 10 mol  cÖwZ sec G ev®úxf~Z AYy = 10  6.023  1023 3600  10 wU = 1.673  1020 wU 20. 2FeCl3 + SnCl2 = 2FeCl2 + SnCl4 GKwU RviY-weRviY wewμqv| RviY-weRviY cwieZ©‡bi Ask `yBwU Avjv`vfv‡e †`LvI| AvaywbK gZev` Abymv‡i cwieZ©b `yBwUi h_v_©Zv e ̈vL ̈v Ki| [BUET 13-14] mgvavb: +3 +2 +2 +4 2FeCl3 + SnCl2 = 2FeCl2 + SnCl4 RviY Aa©wewμqv: Sn2+  Sn4+ + 2e– weRviY Aa©wewμqv: 2Fe3+  2Fe2+ – 2e– GLv‡b, Sn2+ 2wU e – †Q‡o w`‡q RvwiZ n‡q Sn4+ MVb K‡i Ges `ywU Fe3+ GKwU K‡i e – MÖnY K‡i Fe2+ MVb K‡i weRvwiZ nq| 21. 25 mL 0.1 M HCl `aeY‡K 24.99 mL 0.1 M NaOH `aeY Øviv UvB‡Uak‡bi mgq pH Gi gvb †ei Ki| [BUET 13-14] mgvavb: V1S1e1 – V2S2e2 = VS  25  0.1  1 – 24.99  0.1  1 = (25 + 24.99) S  S = 2.0004  10–5 M  pH = – log (2.004  10–5 ) = 4.699 22. 10 g Aweï× GKwU cvwbi bgybv we‡kølY K‡i 10 mg Ac`ae ̈ cvIqv †Mj| bgybvwU‡Z KZ AYy weï× cvwb Av‡Q? [BUET 13-14] mgvavb: weï× cvwbi cwigvY = (10 – 10  10–3 ) g = 9.99 g  weï× cvwbi AYy =     9.99 18  6.022  1023 wU = 3.3428  1023 wU 23. CaC2 + H2O  CaO + HC  CH  H2C = CH2 n(H2C = CH2)  – CH2 – CH2 –n Dc‡ii wewμqv ̧‡jvi gva ̈‡g K ̈vjwmqvg Kve©vBW †_‡K cwjw_‡bi cwigvY wbY©q Kiv hvq| 80 kg K ̈vjwmqvg Kve©vBW †_‡K cÖ ̄‘Z cwjw_‡bi cwigvY wnmve Ki| [BUET 12-13] mgvavb: CaC2 + H2O  CaO + HC  CH H2C = CH2 n(H2C = CH2)  – CH2 – CH2 – n  64  10–3 kg CaC2 †_‡K Bw_wjb cvIqv hvq = 28  10–3 kg  80 kg CaC2 †_‡K Bw_wjb cvIqv hvq = 28  10–3 64  10–3  80 kg = 35 kg  Bw_wjb †_‡K cwjw_b mgvb cwigv‡Y cvIqv hvq  Drcbœ cwjw_b = 35 kg
4 ....................................................................................................................................  Chemistry 2nd Paper Chapter-3 24. 1.0 g GKwU avZe Kve©‡bU jeY 25.0 mL 1.0 M HCl `ae‡Y `aexf~Z Kiv nj| cÖvß jeY‡K m¤ú~Y©iƒ‡c cÖkwgZ Ki‡Z 50.0 mL †Wwm‡gvjvi Kw÷K †mvWvi `aeY jv‡M| jeYwUi AvYweK fi I ms‡KZ †ei Ki| [BUET 12-13] mgvavb: Kve©‡bU jeYmg~‡ni m¤¢ve ̈ ms‡KZ nj m2(CO3)x| GLv‡b x = m avZzi †hvR ̈Zv| Kve©‡bU je‡Yi Zzj ̈ msL ̈v = 2 w = je‡Yi fi I M = je‡Yi AvYweK fi, w = 1 g  w M  2 + 50 1000  0.1 = 25 1000  1  1 M  = 25 1000 – 5 1000 = 20 1000 = 1 50  M = 100 g GLb, CO2– 3 g~j‡Ki fi = 60 g  evwK avZzi fi = 40 g  avZzwU Ca  jeYwU CaCO3 25. CuSO4 `ae‡Yi mv‡_ AwaK cwigvY KI †hvM K‡i Na2S2O3 `aeY w`‡q UvB‡UaU Ki‡j Kx Drcbœ nq? [BUET 12-13] mgvavb: CuSO4 I KI Gi wewμqvq Av‡qvwWb Drcbœ nq| 2CuSO4 + 4KI  Cu2I2 + I2 + 2K2SO4 I2, Na2S2O3 Gi mv‡_ wewμqv K‡i Na2S4O6 I 2NaI Drcbœ K‡i| G‡Z I2 weRvwiZ nq| 2Na2S2O3 + I2  Na2S4O6 + 2NaI 26. Kx N‡U ivmvqwbK wewμqvi mvnv‡h ̈ wjL: (a) CaCO3 †K HCl Gwm‡W `aexf~Z K‡i Kw÷K †mvWv `aeY wgkv‡j/wgwkÖZ Ki‡ZÑ (b) CuSO4 `ae‡Yi mv‡_ AwaK cwigvY KI †hvM K‡i Na2S2O3 `aeY w`‡q UvB‡UaU Ki‡jÑ (c) NaHCO3 †K 180C ZvcgvÎvq DËß Ki‡jÑ (d) wmwjKv I Na2CO3 †_‡K cvwb KuvP •Zwi Ki‡ZÑ [BUET 12-13] mgvavb: (a) CaCO3 + HCl  CaCl2 + H2O + CO2  CaCl2,H2O  cÖkg 2NaOH + CO2  Na2CO3 + H2O CO2  A¤øagx© (b) 2CuSO4 + 4KI  2K2SO4 + Cu2I2 + I2 I2 + 2Na2S2O3  Na2S4O6 + 2NaI (c) 2NaHCO3 180C  Na2CO3 + H2O + CO2 (d) Na2CO3 + SiO2  Na2SiO3 + CO2 27. NaCl Ges AgNO3 Gi Rjxq `aeY †gkv‡j †h wewμqv msNwUZ nq, Zvi mgxKiYwU wjL| [BUET 10-11] mgvavb: NaCl(aq) + AgNO3(aq)  AgCl(s)  + NaNO3(aq) mv`v Aat‡ÿc 28. GK UzKiv mv`v dmdivm (cv. fi = 31) evZv‡m wb‡ÿc Kiv nj Ges cÖ3⁄4¡wjZ nj| GB NUbvi Dci wfwË K‡i wb‡Pi cÖkœ ̧‡jvi DËi `vI| (a) evZv‡m NUgvb ivmvqwbK wewμqvwU wewμqK Ges Drcv‡`i †fŠZ Ae ̄’v mn wjL| (b) Drcv‡`i cwigvY wQj 2.84 g| evZv‡m Kx cwigvY (MÖv‡g) dmdivm wb‡ÿc Kiv n‡qwQj? KZ †gvj dmdivm wb‡ÿc Kiv n‡qwQj? [BUET 10-11] mgvavb: (a) P4(s) + 5O2(g)  2P2O5(g) 124 g 284 g (b) 284 g Drcv` Av‡m = 124 g dmdivm n‡Z  2.84 g Drcv` Av‡m = 1.24 g dmdivm n‡Z = 1.24 31  4 mol dmdivm n‡Z = 0.01 mol dmdivm n‡Z = 1.24 g dmdivm n‡Z 29. 25 g wRsK cvDWvi‡K ZnCl2 G cwiYZ Kivi Rb ̈ b~ ̈bZg Kx cwigvY (cm3 G) 0.05 mol/dm3 HCl Gi cÖ‡qvRb n‡e Zv wbY©q Ki| [BUET 10-11] mgvavb: Zn + 2HCl  ZnCl2 + H2 nZn = nHCl 2  m M = S  V 2  25 65.4  2 = 0.05  V  1  V = 15.29052 dm3 = 15290.52 cm3 30. wb‡¤œi wewμqv ̧‡jv c~Y© Ki: [BUET 10-11] A. CH2(COOH)2 + P4O10  B. MnO– 4 + COO– | COO– + H+  C. Cr2O 2– 7 + H+ + I–  D. NaHSO3 + H2S  E. Na2CO3 + NO2  mgvavb: A. CH2(COOH)2 + P4O10  CO CO CH2 O + 4H3PO4 (wbiæ`b wewμqv) B. MnO– 4 + COO– | COO– + H+  Mn2+ + CO2 + H2O (RviY-weRviY) C. Cr2O 2– 7 + H+ + I–  2Cr3+ + I2 + H2O (RviY-weRviY) D. NaHSO3 + H2S  3S + 2H2O + NaOH (mvgÄm ̈Zv wewμqv) E. Na2CO3 + NO2  NaNO3 + NaNO2 + CO2 (AmvgÄm ̈Zv wewμqv)