3. dissosiya iii 2019 + 2021+2022

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dissosiyalanish – iii hx ⇄ 108. bir asosli kislota eritmasida kislota dissotsilanishidan hosil bo‘lgan barcha zarrachalar konsentratsiyasi dastlabki kislota konsentratsiyasidan 1,8 marta ko‘p bo‘lsa, kislotaning dissotsilanish darajasini (%) aniqlang. a) 80 b) 85 c) 75 d) 90 𝐻++ 𝑋− x ____ 2x 11mol − x +2x=1,8mol 1 + x = 1,8 x = 1,8 – 1=0,8 1mol _____100% 0,8m ____ x = 80% 2 – usul(farq) hx ⇄𝐻++ 𝑋− ______2 _____ f 1 0,8m__________x x = 0,8m 1mol _____100% 0,8m ____ x = 80% 𝑀𝑒2𝑆𝑂4 ⇄ 𝑀𝑒2𝑆𝑂4 eritmasida tuz dissotsilanishidan hosil bo‘lgan barcha zarrachalar konsentratsiyasi dastlabki kislota konsentratsiyasidan 2,8 marta ko‘p bo‘lsa, kislotaning dissotsilanish darajasini (%) aniqlang. a) 80 b) 85 c) 75 d) 90 2𝑀𝑒++ 𝑆𝑂4 2− x _______3x 11mol − x +3x=2,8mol 1 + 2x = 2,8 x = 2,8 – 1=1,8/2=0,9 1mol _____100% 0,9m ____ x = 90% 2 – usul(farq) 𝑀𝑒2𝑆𝑂4⇄2𝑀𝑒++𝑆𝑂4 2− ______3 _____ f 2 …

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arini topamiz. h2xo3 hxo3 - xo3 2- 2 + 𝑥 + 48 2 + 1 + 𝑥 + 48 + 𝑥 + 48 2 = 154 25 + 0,5x + x +49 + 0,5x +24 = 154 2x = 56 x = 28 (si) 124. ikki asosli kislota (h2a) ning dissotsilanish darajasi birinchi bosqichda 80%, ikkinchi bosqichda 25% bo‘lsa, eritmadagi h + , ha - , a 2- ionlari qanday mol nisbatda bo‘ladi? a) 5:4:1 b) 5:3:1 c) 6:4:1 d) 6:3:1 b) 5:3:1 100ta molekula h 2 a olamiz: h 2 a ⇄ ha - + h + 80 80 1) 80 2) ha - ⇄ a 2- + 𝐻+ 80 ∙ 0,25= 20 𝐻2𝐴 20 20 20 = 100 – 80 = 20 𝐻+ = 80 + 20 = 100 𝐻𝐴− =80 – 20 = 60 𝐴2− = 20 100 : 60 : 20 5:3:1 ikki asosli kislota (h2a) …

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ning konsentratsiyalari yig‘indisi 1,6 mol/l ni tashkil etsa, metallni aniqlang. (gidroliz hisobga olinmasin, a = 1) a) fe v) su s) cr d) zn ionlar miqdorini topamiz. 1000ml --------- 1.6mol 250 ml ---------- x = 0.4mol meso4 ⇄ me2+ + so4 2- 1mol --------- 2mol( jami ion) x------------ 0.4mol x = 0.2mol ( tuz miqdori) 0.2mol --------- 30.4g 1mol----------x x = 152g - 96 = 56g(fe) 48 g ikki valentli metall sulfat tuzi suvda eritilib 240 ml eritma olindi. olingan eritmada me2+ va so4 2- ionlarining konsentratsiyalari yig‘indisi 2 mol/l ni tashkil etsa, metallni aniqlang. (gidroliz hisobga olinmasin, a = 0,8) a) fe v) su s) cr d) zn ionlar miqdorini topamiz. 1000ml --------- 2mol 240 ml ---------- x = 0.48mol meso4 ⇄ me2+ + so4 2- 1mol --------- 2mol( jami ion) x ------------ 0.48mol x = 0.24mol / 0,8 = 0,3mol 0.3mol --------- 48g 1mol----------x x = 160g - …

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m 100g _____ x x = 0,5mol/l 1m-----1m----3m 0,5mol 151. ma’lum hajmdagi suvda 1 mol aluminiy sulfat va 2 mol aluminiy xlorid tuzlari eritildi. hosil bo‘lgan 2 litr eritmadagi ionlar konsentratsiyalari c(al3+) = 1,5 mol/1 va c(so4 2-) = 1,2 mol/1 bo‘lsa, tuzlaming dissotsilanish darajalarini (%) mos ravishda aniqlang. (gidrolizlanish jarayoni inobatga olinmasin). a) 70; 80 b) 80; 60 c) 80; 70 d) 60; 80 al2(so4)3 ↔ 2al3+ + 3so4 2- 1m--------2m-----3m 1,2m x----------y---- x=0,4m y=0,8 al3+ = 1,5m – 0,8 = 0,7mol/l alcl3 ↔ al3+ +3cl- 0,7m x----- x=0,7mol →dis.langan dastlabki tuzlar konsentratsyalarini topamiz 1𝑚 2𝑙 = 0,5𝑚𝑜𝑙/𝑙 2𝑚 2𝑙 = 1𝑚𝑜𝑙/𝑙 -------100% 0,4m -------x x= 80% 1mol -------100% 0,7m -------x x= 70% al2(so4)3 alcl3 ma’lum hajmdagi suvda 1,5 mol aluminiy sulfat va 2,4 mol aluminiy xlorid tuzlari eritildi. hosil bo‘lgan 3 litr eritmadagi ionlar konsentratsiyalari c(al3+) = 1,2 mol/1 va c(so4 2-) = 0,9 mol/1 bo‘lsa, tuzlaming dissotsilanish …

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n 75% bo‘lsa, ikkinchi bosqich uchun dissotsilanish darajasini (%) aniqlang. (suvning dissotsilanishi inobatga olinmasin.) a) 50 b) 40 c) 25 d) 20 𝐾𝐻𝑆𝑂4 x + →𝐾++𝐻𝑆𝑂4 − 75 ___75___75 𝐻𝑆𝑂4 −→𝐻++𝑆𝑂4 2− ___ x ___ kation x 75+ x_______ 7 = anion 75− x x__________ 5 375 + 5x = 525 5x = 150 x = 30_____ 75 = 0,4=40% 157. 0,25 mol x2uo4 tuzi 100 ml suvda eritildi. tuz erishidan hosil bo‘lgan kation tarkibida 4∙na, anion tarkibida 10∙na ta elektron bo‘lsa, tuzni aniqlang. (a = 80%) a) na2so4 b) k2seo4 c) na2seo4 d) k2so4 0.2mol = 0.25mol -------- 100% x ------- 80% x2yo4 ⇄ 2x1+ + yo4 2- 0.2mol --- 0.4mol — 0.2mol x+ ioni: 1m -------- 2mol —-- 1mol x=10 + 1 (zaryad) = 11e (na) 0.4mol ------- 4•na 1mol ----- x yo4 2- ioni 10•na 0.2mol -------- 1mol ------ x x = 50 – 2 = 48 …

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dissosiyalanish – iii hx ⇄ 108. bir asosli kislota eritmasida kislota dissotsilanishidan hosil bo‘lgan barcha zarrachalar konsentratsiyasi dastlabki kislota konsentratsiyasidan 1,8 marta ko‘p bo‘lsa, kislotaning dissotsilanish darajasini (%) aniqlang. a) 80 b) 85 c) 75 d) 90 𝐻++ 𝑋− x ____ 2x 11mol − x +2x=1,8mol 1 + x = 1,8 x = 1,8 – 1=0,8 1mol _____100% 0,8m ____ x = 80% 2 – usul(farq) hx ⇄𝐻++ 𝑋− ______2 _____ f 1 0,8m__________x x = 0,8m 1mol _____100% 0,8m ____ x = 80% 𝑀𝑒2𝑆𝑂4 ⇄ 𝑀𝑒2𝑆𝑂4 eritmasida tuz dissotsilanishidan hosil bo‘lgan barcha zarrachalar konsentratsiyasi dastlabki kislota konsentratsiyasidan 2,8 marta ko‘p bo‘lsa, kislotaning dissotsilanish darajasini (%) aniqlang. a) 80 b) 85 c) 75 d) 90 2𝑀𝑒++ 𝑆𝑂4 2− x _______3x …

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