11. oleum

PDF 94 стр. 1,3 МБ Бесплатная загрузка

94 стр.

FaylHub.uz

Скачать через Telegram

Размер 1,3 МБ

Тип файла PDF

Страницы 94

Обновлено Май 2026

Предварительный просмотр (5 стр.)

Прокрутите вниз 👇

1 / 94

oleum erituvchi → sardor voxidov oleum → sulfat angidridning ( 𝑆𝑂3) sulfat kislotadagi eritmasidir. 𝐻2𝑆𝑂4 erigan modda → 𝑆𝑂3 oleum → sulfat angidridning ( 𝑆𝑂3) sulfat kislotadagi eritmasidir. oleum tarkibi 2 xil ifodalanishi mumkin. 1. formula bilan. 𝐻2𝑆𝑂4 ∙ 𝑥𝑆𝑂3 → 𝑥𝐻2 𝑆𝑂4 ∙ 𝑦𝑆𝑂3 x butun son bo’lishi shart emas! 2. foizda ifodalash oleum eritmasi tarkibida 𝑆𝑂3 𝑠𝑢𝑙𝑓𝑎𝑡 𝑎𝑛𝑔𝑖𝑑𝑟𝑖𝑑 miqdori uning foizi hisoblanadi. 20% li oleum→ 𝐻2𝑆𝑂4 + 𝑆𝑂3 80g ____20g____100g(eritma) formula va foiz tarkibni bog’lash. a) formuladan foizga o’tish. 𝐻2𝑆𝑂4 ∙ 1,4𝑆𝑂3 oleumning foiz tarkibini aniqlang. 𝐻2𝑆𝑂4 ∙ 1,4𝑆𝑂3 98g + 112 =210g _____100% 112g ____ x = 53,33% 3𝐻2𝑆𝑂4 ∙ 2𝑆𝑂3 oleumning foiz tarkibini aniqlang. 3𝐻2𝑆𝑂4 ∙ 2𝑆𝑂3 294g + 160 =454g _____100% 160g ____ x = 35,24% sardor voxidov ______ b) foizda berilsa formulani topish. 1. 28,98% li oleumning formulasini aniqlang. oleum 100g 1 – usul (proporsiya) 𝐻2𝑆𝑂4___71,02g 𝑆𝑂3____28,98g 98g____x x = 40g 80 =0,5 2 …

2 / 94

oleumdan olish mumkin bo’lgan toza kislota massasi. 𝑆𝑂3 +𝐻2𝑂→𝐻2𝑆𝑂4 80g ________98g 100% _______ x = 122,5% → 𝑆𝑂3 𝑠𝑢𝑙𝑓𝑎𝑡 𝑓𝑜𝑖𝑧𝑖. 𝐻2𝑆𝑂4 ∙ 0,5𝑆𝑂3 +0,5𝐻2𝑂→1,5 138g _____________147g 100 _____________ x 106,52 sardor voxidov 𝐻2𝑆𝑂4 = 𝐻2𝑆𝑂4 ∙ 1,4𝑆𝑂3 +1,4𝐻2𝑂→2,4 210g _____________235,2g 100 _____________ x 112% 𝐻2𝑆𝑂4 = 𝐻2𝑆𝑂4 ∙ 2𝑆𝑂3 +2𝐻2𝑂 → 3 258g _____________294g 100 _____________ x 113,95% 𝐻2𝑆𝑂4 = 𝐻2𝑆𝑂4 ∙ 1,25𝑆𝑂3+1,25𝐻2𝑂→ 2,25 198g _____________220,5g 100 _____________ x 111,36% sardor voxidov sulfat foiz = + + 40% li oleumning sulfat foizini aniqlang. 100g____ 𝐻2𝑆𝑂4 60 ____ 𝑆𝑂3 40g 𝑆𝑂3 + 𝐻2𝑂 → 𝐻2𝑆𝑂4 80g __________ 98g 40g __________x = 49g 60g = 49g=109g 100 𝐶%(𝑜𝑙𝑒𝑢𝑚)∙0,225 sulfat foiz = +100 40 ∙ 0,225 109 sardor voxidov + 30% li oleumning sulfat foizini aniqlang. 100g____ 𝐻2𝑆𝑂4 70 ____ 𝑆𝑂3 30g 𝑆𝑂3 + 𝐻2𝑂 → 𝐻2𝑆𝑂4 80g __________ 98g 30g __________x = 36,75g 70g 36,75g= 106,75g sardor voxidov 3. yuduzcha ( oleumning …

3 / 94

eum tarkibida s va o massa ulshlari farqi 25.69% bo’lgan oleumning 43.6g miqdorini to’liq neytralash uchun qanday massada 32% li naoh eritmasi zarur? javob:125g + 𝐻2𝑆𝑂4 ∙ 𝑥𝑆𝑂3 __________32 + 16x 98 + 80x =0,2569 32 + 16x =25,18 + 20,55x 4,55x = 6,82 x = 1,5 𝐻2𝑆𝑂4 ∙ 1,5𝑆𝑂3 naoh → 𝑁𝑎2𝑆𝑂4 + 𝐻2𝑂2,55 3,5 218g ______ 200g 43,6g ____ x = 40g_____ 0,32 =125g sardor voxidov oleum tarkibida h va o massa ulushlari yig’indisi s massa ulushidan 31,08% ko’p bo’lgan oleumning 39g miqdorini to’liq sulfat kislotaga aylantirish uchun qanday masada suv qo’shish zarur? javob:2,16g + 𝐻2𝑆𝑂4 ∙ 𝑥𝑆𝑂3 __________34 + 16x 98 + 80x =0,3108 34 + 16x =30,46 + 24,86x 8,86x =3,54 x = 0,4 𝐻2𝑆𝑂4 ∙ 0,4𝑆𝑂3 𝐻2𝑂 → 𝐻2𝑆𝑂41,40,4 130g ______ 7,2g 39g ____ x = 2,16g sardor voxidov m(o) 13. oleumda s massa ulushi kislorod massa ulushidan 1,727 marta kichik bo’lsa, shu oleumning foizi …

4 / 94

avob: 3h 2 so 4 ∙2so 3 𝐻2𝑆𝑂4 ∙ 𝑥𝑆𝑂3 s o ______ = ______ 1 3,6 1 + x 4 + 3x 3,6 + 3,6x = 4 + 3x 0,6x = 0,4 x = 0,667 𝐻2𝑆𝑂4 ∙ 0,667𝑆𝑂3 3𝐻2𝑆𝑂4 ∙ 2𝑆𝑂3 80x = 64 + 48x oleum tarkibida o massa ulushi so 3 massa ulushiga teng bo’lsa. shu oleum formulasi topilsin. javob: h 2 so 4 ∙2so 3𝐻2𝑆𝑂4 ∙ 𝑥𝑆𝑂3 ______ = ______ 1 80x 64 + 48x 1 32x = 64 x = 2 𝐻2𝑆𝑂4 ∙ 2𝑆𝑂3 sardor voxidov 27. oleumda σ va π bog’lar nisbati 2:1 bo’lsa , oleum formulasi topilsin. javob: 3h2so4∙2so3 𝐻2𝑆𝑂4 ∙ 𝑥𝑆𝑂3 σ π ______ = ______ 2 1 6 + 3x 2 + 3x 6 + 3x = 4 + 6x 2 = 3x x = 0,667 𝐻2𝑆𝑂4 ∙ 0,667𝑆𝑂3 3𝐻2𝑆𝑂4 ∙ 2𝑆𝑂3 31. 0.8mol oleum 73g bo’lsa, formulasi topilsin. javob: h2so4∙0.6so3 …

5 / 94

topilsin. a) 31,36 b) 58,24 c) 28,6 d) 30,2 146g +h 2 so 4 ∙ 0.6so 3 0,6𝐻2𝑂→ 𝐻2𝑆𝑂41,6 ________________ 156,8g 102,2g ___________ x x = 109,76g → hosil bo’lgan kislota 102,2g + 247,8g = 350g _______ 100% 109,76g ____ x x = 31,36% sardor voxidov h2so4∙0.8so3 tarkibli 121,5g oleum, 128,5g suvda eritilganda hosil bo’lgan eritma konsentratsiyasi topilsin. javob: 52,92% 162g +h 2 so 4 ∙ 0.8so 3 0,8𝐻2𝑂→ 𝐻2𝑆𝑂41,8 ________________ 176,4g 121,5g ___________ x x = 132,3g → hosil bo’lgan kislota 121,5g + 128,5g = 250g _______ 100% 132,3g ____ x x = 52,92% sardor voxidov 1 – usul tenglama 44. 120g suvda qanday massada (g) h 2 so 4 ∙1,1so 3 eritilganda hosil bo’lgan eritma 26,18% li bo’ladi? a) 27,9 b) 36,36 c) 54 d) 37.2 h 2 so 4 ∙1,1so 3 205,8x + 1,1 𝐻2𝑂 → 2,1 𝐻2𝑆𝑂4 186x _______________ 205,8x 120g + 186x ______ 100% 205,8x …

Хотите читать дальше?

Скачайте все 94 страниц бесплатно через Telegram.

Скачать полный файл

О "11. oleum"

oleum erituvchi → sardor voxidov oleum → sulfat angidridning ( 𝑆𝑂3) sulfat kislotadagi eritmasidir. 𝐻2𝑆𝑂4 erigan modda → 𝑆𝑂3 oleum → sulfat angidridning ( 𝑆𝑂3) sulfat kislotadagi eritmasidir. oleum tarkibi 2 xil ifodalanishi mumkin. 1. formula bilan. 𝐻2𝑆𝑂4 ∙ 𝑥𝑆𝑂3 → 𝑥𝐻2 𝑆𝑂4 ∙ 𝑦𝑆𝑂3 x butun son bo’lishi shart emas! 2. foizda ifodalash oleum eritmasi tarkibida 𝑆𝑂3 𝑠𝑢𝑙𝑓𝑎𝑡 𝑎𝑛𝑔𝑖𝑑𝑟𝑖𝑑 miqdori uning foizi hisoblanadi. 20% li oleum→ 𝐻2𝑆𝑂4 + 𝑆𝑂3 80g ____20g____100g(eritma) formula va foiz tarkibni bog’lash. a) formuladan foizga o’tish. 𝐻2𝑆𝑂4 ∙ 1,4𝑆𝑂3 oleumning foiz tarkibini aniqlang. 𝐻2𝑆𝑂4 ∙ 1,4𝑆𝑂3 98g + 112 =210g _____100% 112g ____ x = 53,33% 3𝐻2𝑆𝑂4 ∙ 2𝑆𝑂3 oleumning foiz tarkibini aniqlang. 3𝐻2𝑆𝑂4 ∙ 2𝑆𝑂3 294g + 160 =454g _____100% 160g ____ x = …

Этот файл содержит 94 стр. в формате PDF (1,3 МБ). Чтобы скачать "11. oleum", нажмите кнопку Telegram слева.

Теги: 11. oleum PDF 94 стр. Бесплатная загрузка Telegram