UNIVERSAL KIMYO
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Keling!!! Kimyoni birgalikda Γ³rganamizπŸ”₯
Π”Π°Π²Π°ΠΉΡ‚Π΅!!! Π˜Π·ΡƒΡ‡Π°Π΅ΠΌ вмСстС Π₯имиюπŸ”₯
ustoz: Bunyod Raxmanov
Bog'lanish πŸ”— @ChemOfficer_bot
Bog'lanish πŸ”— @ProChem
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🚨 DIQQAT! Sertifikat natijalarini birinchilardan boβ€˜lib biling πŸ™‚

πŸ’₯ Hoziroq kanallarga obuna boβ€˜ling va natijalarni oβ€˜tkazib yubormang ⬇️
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πŸ–₯ Bugun tahlil qiladigan yozma ish topshiriqlarimizdan parcha. Kursimda oddiy test yoki masalalar tahlil qilinmaydi, ayni eng murakkab&mujmal yozma ishlar qismi (41, 42, 43) tahlil qilinadi. Maqsad, murakkab&mujmal topshiriqlarni ham qΓ²rqmasdan yechish kΓΆnikmasini shakllantirish. Unutmang☝🏻 ayni yozma ishlar qismi uchun eng yuqori ballar beriladiπŸ“Ά
⚠️mualliflik huquqi sabab savollar to'liq xolda tashlanmadi.


⚜️MILLIY SERTIFIKAT⚜️ maxsus tayyorlov kursida tahlil qilinayotgan ma'lumotlardan
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❀1
🧠miyani charxlaymizβš”οΈ
βœ”οΈjavobingizni tahlil guruhida bildiring

Yuqoridagi izotoplar asosida vodorod peroksidning:
2️⃣❓necha xil molekulasini
3️⃣❓necha xil massali molekulasini
...olish mumkin
❓

⚜️MILLIY SERTIFIKAT⚜️ maxsus tayyorlov kursida tahlil qilinayotgan ma'lumotlardan
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_________
ustozπŸ‘¨πŸ»β€πŸ« Bunyod Raxmanov
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πŸ” πŸ”€πŸ” πŸ” πŸŸ’πŸ” πŸ” πŸ”€πŸ” 
πŸ” πŸ” πŸ” πŸŸ’πŸ” βž–πŸ” β—οΈ

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ustozπŸ‘¨πŸ»β€πŸ« Bunyod Raxmanov
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πŸ”₯5
Forwarded from Post Bot
❀️‍πŸ”₯❀️‍πŸ”₯❀️‍πŸ”₯πŸ’œπŸ’œπŸ’œπŸ’™πŸ’™πŸ’™

πŸ”΄ Sentyabr oyidagi sertifikat natijalari qanday boΚ»ldi

πŸ”΄ Noyabr sertifikatiga tayyorlanayotganlar,sizga kerak boβ€˜ladigan ENG FOYDALI KANALLAR quyida ⬇️
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Forwarded from Asilbek Yarashev
πŸ”΄πŸ”΄πŸ”΄πŸ”΄πŸ”΄πŸ”΄πŸ”΄πŸ”΄πŸ”΄πŸ”΄πŸ”΄

Sertifikatga tayyorlanyapsizmi unda bu kanallar sizga kerak
⬇️

https://t.me/addlist/M70m8ctFTuxjODUy
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πŸ”΄SAVOL
Assalomu aleykum ustoz yaxshimisiz! Menda bir savol bor edi. Arenlardan p elektronlarning umumiy formulasini 4n+2 ligini qanday isbotlaymiz (repetitor savoli
πŸ’¬)

🟒JAVOB
Voleykum assalom. Avvalo aromatiklik nima degan savolga javob beraylik. Aromatiklik-tutash Ο€ bog'larga (p-orbitallarga) ega yassi halqali (siklik) barqaror birikmalarga aytiladi. Ya'ni, halqa uchi atomlaridagi gibridlanmasdan qolgan p-orbitallari o'zaro delokallashgan (o'zaro tutashgan) rezonans holatda bo'ladi. Endi Xyukkel aromatiklikni tushuntirishda nega aynan 4n+2 formulasini taklif qildi (n-halqa sonini bildiradi). Kvant kimyo niqtai nazardan barqaror tutash Ο€-bog'li birikmalar hosil bo'lishi uchun 2, 6, 10, 14... ya'ni, arifmetik 4 taga ortib borish p-orbitallar zarurligi aniqlandi va shu sababli 4n+2 formulasi vujudga keldi. Jumladan, n=1 bo'lsa (ya'ni, benzol) kimyoviy passiv barqaror birikmani hosil qiladi va bu aromatiklik qoidasiga to'liq amal qiladi. Bundan tashqari, keyinchalik ayrim geterosiklik birikmalar (furan, tiofen, pirrol, imidazol...) ham aromatiklik hususiyatini namoyon qilishi aniqlandi. Ya'ni, geteroatomdagi taqsimlanmagan e-juft delokallashishda ishtirok etib barqaror birikmalar hosil qiladi. Ular ham 4n+2 formulasiga bo'y sungani uchun barqaror birikmalardir.

(ustoz✍️ Bunyod Raxmanov)


manbaπŸ“–UNIVERSAL KIMYO
βœ…TANISHUV
βœ…MUNDARIJA
πŸ’» @universal_kimyo
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#ilmiy_faktπŸ“š

❀️❀️❀️
Nega eritmada muhit har xil β€” ya'ni NaH2PO4 kislotali, Na2HPO4 esa aksincha asosli?

πŸ’šπŸ’šπŸ’š
Eritma muhitida mazkur kuchsiz kislota qoldiqlari ikki yo'nalishda:
1) gidroliz konstantasi
2) dissotsioanish konstantasi
natijasida qayta ionlar taqsimoti vujudga keladi. Km asosida esa muhit (pH) kelib chiqadi. Jumladan:

▢️NaH2PO4
1) H2PO4(-) + H2O = H3PO4 + OH-
Km1=1,4*10(-12)
2) H2PO4(-) = HPO4(2-) + H+
Km2=6,2*10(-8)

Km1<Km2, xulosa: muhit kislotali

▢️Na2HPO4
1) HPO4(2-) + H2O = H2PO4(-) + OH-
Km1=1,6*10(-7)
2) HPO4(2-) = PO4(3-) + H+
Km2=5,0*10(-13)

Km1>Km2, xulosa: muhit asosli
(ustoz
✍️ Bunyod Raxmanov)

manba
πŸ“–UNIVERSAL KIMYO
βœ…TANISHUV
βœ…MUNDARIJA
πŸ’» @universal_kimyo
☎️ 93-585-17-12

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❀3πŸ”₯3
❀️❀️❀️
AsH3 qutblimi yoki qutbsizmi, NEM farqi 0,1 ga teng, iltimos tushuntirib bersangiz, shu kabi savollarda tortishuv bo'lyapti?


πŸ’šπŸ’šπŸ’š
Maktab darsliklari uchun koʻp narsalar kechiriladi, jumladan NEM farqi 0,4 gacha shartli qutbsiz deya qabul qilinishi ham. Ammo, oliy kimyoda unday emas. Masalan, bundanda bahsli molekulalar misolida: CS2 molekulasi qutbsiz, ammo C=S orasi kuchsiz qutbli, NEM farqi esa atigi 2,58-2,55=0,03. Gap shundaki, qutbli va qutbsizlik faqat NEM farqiga borib taqalmaydi, balki molekula tuzilishiga ham borib taqaladi. Yanada ichkariga kiramiz: O2 qutbsiz, O3 esa qutbli, NEM va atom bir xil, bunga nima deysiz!? Xulosa: AsH3 maktab oʻquvchisi uchun shartli qutbsiz, real holat (oliy kimyo)da esa kuchsiz qutbli kovalent boğ tabiatiga ega.
(ustoz✍️ Bunyod Raxmanov)

manbaπŸ“–UNIVERSAL KIMYO
βœ…TANISHUV
βœ…MUNDARIJA
πŸ’» @universal_kimyo
☎️ 93-585-17-12

⚜️MILLIY SERTIFIKAT⚜️
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❀3⚑2πŸ”₯1
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πŸ”‘πŸ”‘β“βŒ¨οΈβŒ¨οΈβŒ¨οΈβŒ¨οΈβŒ¨οΈ

manbaπŸ“–UNIVERSAL KIMYO
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βœ…MUNDARIJA
πŸ’» @universal_kimyo
☎️ 93-585-17-12

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πŸ”ΉKVANT SONLARπŸ”Ή

manbaπŸ“–UNIVERSAL KIMYO
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βœ…MUNDARIJA
πŸ’» @universal_kimyo
☎️ 93-585-17-12

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ustozπŸ‘¨πŸ»β€πŸ« Bunyod Raxmanov
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βŒ¨οΈβŒ¨οΈβŒ¨οΈβŒ¨οΈπŸ”»βŒ¨οΈβŒ¨οΈβŒ¨οΈβŒ¨οΈβŒ¨οΈβŒ¨οΈ

πŸ™‚sodda yodlash metodi: 43, 61, 84β€”118

manbaπŸ“–UNIVERSAL KIMYO
βœ…TANISHUV
βœ…MUNDARIJA
πŸ’» @universal_kimyo
☎️ 93-585-17-12

⚜️MILLIY SERTIFIKAT⚜️
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ustozπŸ‘¨πŸ»β€πŸ« Bunyod Raxmanov
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🧠miyani charxlaymizβš”οΈ
βœ”οΈjavobingizni tahlil guruhida bildiring

manbaπŸ“–UNIVERSAL KIMYO
βœ…TANISHUV
βœ…MUNDARIJA
πŸ’» @universal_kimyo
☎️ 93-585-17-12

⚜️MILLIY SERTIFIKAT⚜️
✍️qabulga yozilish
@online_dars_qabul
@qabul_online_dars
_____________________________
ustozπŸ‘¨πŸ»β€πŸ« Bunyod Raxmanov
kanalπŸ“£
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πŸ‘€sinus&cosinus kimyoda nima qilib yuribdiπŸ’¬

manbaπŸ“–UNIVERSAL KIMYO
βœ…TANISHUV
βœ…MUNDARIJA
πŸ’» @universal_kimyo
☎️ 93-585-17-12

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_____________________________
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