CdSO₄ va Na₂SO₄ dan iborat 98,4 g aralashma 401,6 g suvda eritildi. Kadmiyni batamom ajratib olish uchun eritma orqali 5,36 soat davomida 2 A tok o‘tkazildi. Dastlabki aralashmadagi tuzlarning miqdorlarini (mol) toping.

CdSO₄ − 0,2; Na₂SO₄ − 0,4
CdSO₄ − 0,3; Na₂SO₄ − 0,3
CdSO₄ − 0,4; Na₂SO₄ − 0,4
CdSO₄ − 0,2; Na₂SO₄ − 0,5

📝 Detailed explanation

q = I·t = 2·5,36·3600 = 38 592 C. n(e⁻) = q/F = 38 592/96 500 ≈ 0,4 mol. Cd²⁺ + 2e⁻ → Cd ⇒ n(CdSO₄) = 0,2 mol. m(CdSO₄)=0,2·208=41,6 g. m(Na₂SO₄)=98,4−41,6=56,8 g → n=56,8/142=0,4 mol.

12 / 30 (40%)

⬅ Previous

Question 12 of 30

Next ➡

← Previous → Next Space — Show answer

🎯 Take a test on the topic «Aprobatsiya KIMYO 2026»

This topic contains 30 questions. Test your knowledge!

🔐 Log in to take the test

📋 Related questions

❓ Na₂O va K₂O lardan iborat 12,5 g aralashma 100 g 16,8% li KOH eritmasida eritildi. Hosil bo‘lgan eritmani to‘liq neyt...
❓ A va B moddalardan C va D moddalar olingan. Grafikdagi ma’lumotlar asosida D moddaning massasini (g) toping.
❓ Al₄C₃ + 12H₂O → 4A + 3B reaksiyasida hosil bo‘lgan A moddani ko‘rsating.
❓ Sog‘lom odamning qon plazmasi tarkibidagi Na⁺, K⁺, Ca²⁺ ionlarining mol nisbati doimiy ravishda 1 : 1,4 : 9,6 bo‘ladi...
❓ Ma’lum bir reaksiya +4℃ haroratda 4 soat davom etadi. Agar harorat koeffitsiyenti γ=4 bo‘lsa, ushbu reaksiya +24℃ har...

📚 All tests by subject «Kimyo»

View all collections and topics in Kimyo

Go to subject →