{"id":"chemia-2020-marzec-politechnika-lodzka-probna-rozszerzona/zad/40","paper_id":"chemia-2020-marzec-politechnika-lodzka-probna-rozszerzona","number":"40","points":null,"ptype":"open","subject":"chemia","category":"probna","year":2020,"month":"marzec","level":"rozszerzona","text":"Zadanie 40\nx =","answer":null,"answer_text":null,"solution":"Równanie spalania: C₆H₆Oₓ + nO₂ → 6CO₂ + 3H₂O. Równanie balansu O: x + 2n = 12 + 3 = 15, stąd n = (15−x)/2. Stosunek O₂:CO₂ = n/6 = (15−x)/12 = 1, więc x = 3.","image":"img/chemia-2020-marzec-politechnika-lodzka-probna-rozszerzona/zad-40.webp","solution_image":null,"topics":null,"page_from":28,"source":"ai","answer_source":null,"answer_text_source":null,"solution_source":"ai","text_source":"ocr","source_label":"Chemia · Matura próbna · marzec 2020 (rozszerzona)","subject_label":"Chemia","category_label":"Matura próbna","text_html":"<p>Zadanie 40<br>x =</p>","solutions":[{"source":"ai","label":"AI","kind":"text","html":"<p>Równanie spalania: C₆H₆Oₓ + nO₂ → 6CO₂ + 3H₂O. Równanie balansu O: x + 2n = 12 + 3 = 15, stąd n = (15−x)/2. Stosunek O₂:CO₂ = n/6 = (15−x)/12 = 1, więc x = 3.</p>"}]}