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8. 0,2 litr eritmada 94,5 g \( \mathrm{HNO}_{3} \) bor. Shu eritmaning normal konsentratsiyasini toping. A) 2,5 B) 3 C) 2 D) 7,5 9. 0,8 litr critmada \( 100,8 \mathrm{~g} \mathrm{HNO}_{3} \) bor. Shu eritmaning normal konsentratsiyasini toping. A) 2,5 B) 3 C) 2 D) 7,5 10. 0,35 litt eritmada \( 88,2 \mathrm{~g} \mathrm{HNO}_{3} \) bor. Shu eritmaning normal konsentratsiyasini toping. A) 2.5 B) 4 C) 2 D) 7.5

Ask by Schmidt Evans. in Uzbekistan
Jan 11,2025

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### 8. Savol: 0,2 litr eritmada 94,5 g \( \mathrm{HNO}_{3} \) bor. Shu eritmaning normal konsentratsiyasini toping. **Javob: D) 7,5** --- ### 9. Savol: 0,8 litr eritmada 100,8 g \( \mathrm{HNO}_{3} \) bor. Shu eritmaning normal konsentratsiyasini toping. **Javob: C) 2** --- ### 10. Savol: 0,35 litr eritmada 88,2 g \( \mathrm{HNO}_{3} \) bor. Shu eritmaning normal konsentratsiyasini toping. **Javob: B) 4**

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Sizga berilgan masalalarda normal konsentratsiyani hisoblash uchun, avvalo, \( \mathrm{HNO}_{3} \) ning molyar massasi (63 g/mol) dan foydalaning. Keyin, eritmadagi \( \mathrm{HNO}_{3} \) ni molekulyar soniga o'girish uchun, avval berilgan massani molyaqar massaga bo'lish kerak. Masalan, birinchi vaziyatda 94,5 g ni 63 g/mol ga bo'lish orqali mollarni toping, keyin esa mollarni tellash uchun hajmni litrda ajrating. O'zgaruvchan o'lchovlar va hisob-kitoblar jarayonida ko'plab xatolar yuzaga kelishi mumkin. Odatda, o'lchovning aniqligiga e'tibor bermaslik yoki xatoni hisoblashda noaniq qiymatlar ishlatish sababi. Shuning uchun, har doim aniq ma'lumotlardan foydalanishga harakat qiling va natijalarni yana bir bor tekshirib ko'ring!

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