mavzu 17 (qiyin eritma)

PDF 8 pages 359.4 KB Free download

8 pages

FaylHub.uz

Download via Telegram

Size 359.4 KB

File type PDF

Pages 8

Updated May 2026

Page preview (5 pages)

Scroll down 👇

1 / 8

ishlanishlarini ko'rish uchun quydagi telegram kanalimizga kiring: kimyo tv 17-mavzu 1. 20o c da 200 g suvda 148 g cacl2 eritilganda to’yingan eritma hosil bo’lsa, shu tuzning 20oc dagi eruvchanlik koeffitsentini aniqlang. a) 74 b) 148 c) 926 d) 84 2. 20o c da 250 g suvda 220 g nano3 eritilganda to’yingan eritma hosil bo’lsa, shu tuzning 20oc dagi eruvchanlik koeffitsentini aniqlang. a) 88 b) 77 c) 66 d) 55 3. 30o c da 150 g suvda 55,5 g kcl eritilganda to’yingan eritma hosil bo’lsa, shu tuzning 20oc dagi eruvchanlik koeffitsentini aniqlang. a) 47 b) 37 c) 57 d) 67 4. nano3 ning 25 oc dagi eruvchanlik koeffitsenti 91,6 ga teng. shu temperaturada 500 g suvga necha gramm nano3 qo’shilsa, to’yingan eritma hosil bo’ladi? a) 458 b) 183 18,32 d) 500 5. kcl ning 20 oc dagi eruvchanlik koeffitsenti 34 ga teng. shu temperaturada 600 g suvga necha gramm kcl …

2 / 8

irish uchun yana necha gramm nh4cl qo’shish kerak ? a) 32,5 b) 67,4 c) 76,2 d) 42,5 12. nano3ning 20 oc dagi eruvchanlik koeffitsenti 88 ga teng. 300g suvda 200g nano3eritildi. shu eritmani to’yintirish uchun yana necha gramm nano3qo’shish kerak ? a) 64 b) 45 c) 76 d) 23 13. 100oc dagi ba(no3)2 ning 672 g to’yingan eritmasi 25oc gacha sovutilsa, necha gramm tuz kristallanadi ? (s25c=10,5; s100c=34,4) a) 220 b) 320 c) 119,5 d) 200 14. 100oc dagi kcl ning 785 g to’yingan eritmasi 25oc gacha sovutilsa, necha gramm tuz kristallanadi ? (s25c=35; s100c=57) a) 220 b) 320 c) 110 d) 200 15. 100oc dagi kbr ning 408 g to’yingan eritmasi 25oc gacha sovutilsa, necha gramm tuz kristallanadi ? (s25c=66; s100c=104) a) 102 b) 302 c) 252 d) 76 16. 30% li kbr eritmasidan 500 g tayyorlash uchun necha gramm tuz va necha gramm suv kerak bo’ladi ? a) …

3 / 8

r idishga solib aralashtirilganda, hosil bo’lgan eritmaning foiz konsentratsiyasini (%) aniqlang. a) 4,5 b) 3 c) 1,5 d) 2,5 24. 425 g 20% li agno3 eritmasi va 195 g 15% li nacl eritmalari bir idishga solib aralashtirilganda, hosil bo’lgan eritmaning foiz konsentratsiyasini (%) aniqlang. a) 2,25 b) 7,75 c) 5,5 d) 2,75 25. 960 g suvga 82,08 g ba(oh)2 qo’shilishidan hosil bo’lgan eritmaning (ρ=1,2g/ml) normal konsentratsiyasini (n) aniqlang. a) 0,6 b) 1,1 c) 3 c) 0,48 26. 500 g suvga 85,5 g ba(oh)2 qo’shilishidan hosil bo’lgan eritmaning (ρ=1,171g/ml) normal konsentratsiyasini (n) aniqlang. a) 2 b) 4 c) 3 c) 6 27. 200 g suvga 98 g h2so4 qo’shilishidan hosil bo’lgan eritmaning (ρ=1,192g/ml) normal konsentratsiyasini (n) aniqlang. a) 5 b) 8 c) 6 d) 9 28. 9,8 g h2so4 ning 0,2 n li eritmasining hajmini ( l ) aniqlang. a) 1 b) 0,2 c) 0,4 d) 0,04 29. 8,2 g h2so3 …

4 / 8

sa, shu eritmaning zichligini aniqlang. a) 2 b) 0.6 c) 1,2 0,8 40. 16 % li (ρ=1,4g/ml) noma’lum modda eritmasining molyar konsentratsiyasi 4 m ga teng bo’lsa, shu eritmada erigan noma’lum moddani aniqlang ? a) naoh b) ca(oh)2 c) koh d) ba(oh)2 41. 25 % li (ρ=0,7844g/ml) noma’lum modda eritmasining molyar konsentratsiyasi 2 m ga teng bo’lsa, shu eritmada erigan noma’lum modda(lar)ni aniqlang ? a) naoh ; koh b) h2so4 ; hno3 c) cuso4 ; na2so4 d) h2so4 ; h3po4 42. 16 % li (ρ=1g/ml) noma’lum modda eritmasining molyar konsentratsiyasi 4 m ga teng bo’lsa, shu eritmada erigan noma’lum moddani aniqlang ? a) koh b) naoh c) hno3 d) h2so4 https://t.me/joinchat/aaaaafpdk1rknpqbrl4mca ishlanishlarini ko'rish uchun quydagi telegram kanalimizga kiring: kimyo tv 43. 1,5 n li (ρ=0,98g/ml) h3po4 eritmasining foiz konsentratsiyasini aniqlang. a) 5% b) 10 c) 15 d) 20 44. 2,5 n li (ρ=0,98g/ml) h2so4 eritmasining foiz konsentratsiyasini aniqlang. a) 12,5 …

5 / 8

cha z eritmaning foiz konsentratsiyasini aniqlang. 200 g 20% nacl +50 g h2o x y z +50 g nacl +50 g x eritma a) 56 b) 28 c) 36 d) 18 56. ushbu sxema bo’yicha z eritmaning foiz konsentratsiyasini aniqlang. 500 g 10% nacl -100 g h2o x y z +400 g h2o 200 g 25%li nacl eritmasi a) 40 b) 20 c) 30 d) 10 57. ushbu sxema bo’yicha z eritmaning foiz konsentratsiyasini aniqlang. 500 g 20% nacl +100 g h2o x y z +50 g nacl +100 g x eritma a) 44,4 b) 33,6 c) 22,2 d) 11,1 58. ushbu sxema bo’yicha z eritmaning foiz konsentratsiyasini aniqlang. 400 g 7,5% nacl -100 g h2o x y z +70 g nacl +200 g x eritma a) 42 b) 28 c) 31 d) 21 59. 90 g osh tuzi 410 g suvda eritilishidan hosil bo’lgan eritmaning yarmi bug’latildi, olingan eritmaning …

Want to read more?

Download all 8 pages for free via Telegram.

Download full file

About "mavzu 17 (qiyin eritma)"

ishlanishlarini ko'rish uchun quydagi telegram kanalimizga kiring: kimyo tv 17-mavzu 1. 20o c da 200 g suvda 148 g cacl2 eritilganda to’yingan eritma hosil bo’lsa, shu tuzning 20oc dagi eruvchanlik koeffitsentini aniqlang. a) 74 b) 148 c) 926 d) 84 2. 20o c da 250 g suvda 220 g nano3 eritilganda to’yingan eritma hosil bo’lsa, shu tuzning 20oc dagi eruvchanlik koeffitsentini aniqlang. a) 88 b) 77 c) 66 d) 55 3. 30o c da 150 g suvda 55,5 g kcl eritilganda to’yingan eritma hosil bo’lsa, shu tuzning 20oc dagi eruvchanlik koeffitsentini aniqlang. a) 47 b) 37 c) 57 d) 67 4. nano3 ning 25 oc dagi eruvchanlik koeffitsenti 91,6 ga teng. shu temperaturada 500 g suvga necha gramm nano3 qo’shilsa, …

This file contains 8 pages in PDF format (359.4 KB). To download "mavzu 17 (qiyin eritma)", click the Telegram button on the left.

Tags: mavzu 17 (qiyin eritma) PDF 8 pages Free download Telegram