UNIVERSAL KIMYO
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Keling!!! Kimyoni birgalikda Γ³rganamizπŸ”₯
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2️⃣7οΈβƒ£πŸ—“sentyabr kuni kimyo fani bo'yicha ⚜️Milliy Sertifikat⚜️ imtihoni bo'lib o'tadi. Shu munosabat bilan talabgorlarga ozmi-ko'pmi o'z xissamni qo'shish maqsadida to imtihongacha yangi nashrdan chiqqan "Milliy Sertifikat β€” yozma ish topshiriqlari" to'plamini (S. Boboyev, Sh. Usmonov) tahlil qilishni boshlaymiz. Nima uchun deganda ayni ushbu qismi, ya'ni 41-, 42-, 43-yozma ishlar ko'pchilikni qiynamoqda va yechmasdan tashlab ketishmoqda. Lekin ayni shu 3 ta topshiriq eng yuqori ballarni to'plashga yordam beradi. Tahlil qilish uchun sizdan albatta kuchli nazariy fundamental ilm talab qilinadi. Buning uchun sizga albatta yangi nashrdan chiqqan UNIVERSAL KIMYO (B. Raxmanov) kerak bo'ladi. Shunday ekan, tahlillarni chuqur o'zlashtirish uchun ushbu kitoblarni tezda oling va maqsadingizga tezroq yetingπŸ™‚

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πŸ’™πŸ’™πŸ’™β–ΆοΈ 🌐NLAYN DARS ✈️ @ChemOfficer kanalida πŸ”— yaqinlarimizga ulashamiz 2️⃣7οΈβƒ£πŸ—“sentyabr kuni kimyo fani bo'yicha ⚜️Milliy Sertifikat⚜️ imtihoni bo'lib o'tadi. Shu munosabat bilan talabgorlarga ozmi-ko'pmi o'z xissamni qo'shish maqsadida to imtihongacha yangi…
πŸ—“bugun 31-32β€”variantlar tahlil qilinadiπŸ“
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πŸ“πŸ” πŸ”€πŸ” πŸ” πŸ” πŸ” 

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Andijon viloyati, Xonabod shaxridagi ixtisoslashtirilgan maxsus maktab o'quvchilari πŸ”₯UNIVERSAL KIMYOπŸ”₯ bilanπŸ™‚
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βš™οΈmiyani charxlaymiz
βœ”οΈjavobingizni tahlil guruhida bildiring

🟒 Reaksiyada dastlabki va mahsulot atomlari o'zaro matematik tenglashsa bo'ldi desak, biroz xato bo'ladi!? Nazariy qonunga asoslangan tenglash esa to'g'ri bo'ladi. Qani aytingchi, testdagi javoblardan qaysi bir(lar)i to'g'ri tenglangan va asosli ilmiy izohingizπŸ’¬

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βš™οΈmiyani charxlaymiz βœ”οΈjavobingizni tahlil guruhida bildiring 🟒 Reaksiyada dastlabki va mahsulot atomlari o'zaro matematik tenglashsa bo'ldi desak, biroz xato bo'ladi!? Nazariy qonunga asoslangan tenglash esa to'g'ri bo'ladi. Qani aytingchi, testdagi javoblardan…
πŸ…°οΈπŸ…°οΈπŸ…°οΈπŸ…°οΈπŸ…°οΈ
Anonymous Quiz
33%
πŸ”€
30%
πŸ”€
25%
πŸ” 
12%
πŸ” 
❀1
84 voters603 viewsπ‚π‘πžπ¦πŽπŸπŸπ’πœπž, 
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πŸ” πŸ”€πŸ” πŸ” πŸ” πŸ” 
Javob variantlarida keltirilgan 3 ta holat matematik nuqtai nazardan barchasi to'g'ri, lekin e-balans tenglash nuqtai nazardan emas. Bu yerda muammoni yechimi termodinamik energetik avfzallikda, ya'ni sistema o'zgarishi (kimyoviy jarayon) eng minimum energiya holatida ro'y berishida. Bilamizki, O3 ning oksidlovchilik hususiyati O2 ga nisbatan yuqori, sababi nosimmetrik O3 da atomar (faol) O holatiga o'tish uchun juda minimum energiya talab qilinadi, ya'ni asosan bitta O ishtirok etib qolgan 2ta O atomi minimum energiyali O2 molekulasi holida saqlashga harakat qiladi (yuqori temperaturada O3 ning barcha atomlari ham ishtirok etishi mumkin). Demak, O3 ning 1ta O atomi reaksion faol ishtirok etadi:
FeS + 4O β€”> FeSO4
demak jami: 4O atomi 4O3 (ya'ni 4O2O) tarkibidan sarflandi, qolgani 4O2, umumiy tenglama:
FeS + 4O3 = FeSO4 + 4O2

✏️ e-balans tenglash orqali isbotlash:
4O^(0) β€”β€”>O4^(-2) |8|1
S^(-2) β€”β€”>S^(+6) |8|1
4O atomi 4O3 (ya'ni 4O2O) tarkibidan sarflandi, qolgani 4O2, umumiy tenglama:
FeS + 4O3 = FeSO4 + 4O2


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2️⃣7οΈβƒ£πŸ—“sentyabr kuni kimyo fani bo'yicha ⚜️Milliy Sertifikat⚜️ imtihoni bo'lib o'tadi. Shu munosabat bilan talabgorlarga ozmi-ko'pmi o'z xissamni qo'shishni maqsad qildim. 1️⃣5οΈβƒ£πŸ”£sentyabrdan…
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