{"id":"chemia-2020-marzec-politechnika-lodzka-probna-rozszerzona/zad/10","paper_id":"chemia-2020-marzec-politechnika-lodzka-probna-rozszerzona","number":"10","points":null,"ptype":"open","subject":"chemia","category":"probna","year":2020,"month":"marzec","level":"rozszerzona","text":"Zadanie 10\n%\n%","answer":null,"answer_text":null,"solution":"Zadanie 10. Masa CO₂ = (20,0 + 200,0 + 300,0) − 511,0 = 9,0 g. n(CO₂) = 9,0/44,01 = 0,20450 mol. Niech x = masa Na₂CO₃, y = masa NaHCO₃. x + y = 20,0 g oraz x/105,99 + y/84,01 = 0,20450. Rozwiązanie: x = 13,60 g (Na₂CO₃), y = 6,40 g (NaHCO₃). %Na₂CO₃ = 13,60/20,0 × 100% = 68,0%; %NaHCO₃ = 32,0%.","image":"img/chemia-2020-marzec-politechnika-lodzka-probna-rozszerzona/zad-10.webp","solution_image":null,"topics":null,"page_from":7,"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 10<br>%<br>%</p>","solutions":[{"source":"ai","label":"AI","kind":"text","html":"<p>Zadanie 10. Masa CO₂ = (20,0 + 200,0 + 300,0) − 511,0 = 9,0 g. n(CO₂) = 9,0/44,01 = 0,20450 mol. Niech x = masa Na₂CO₃, y = masa NaHCO₃. x + y = 20,0 g oraz x/105,99 + y/84,01 = 0,20450. Rozwiązanie: x = 13,60 g (Na₂CO₃), y = 6,40 g (NaHCO₃). %Na₂CO₃ = 13,60/20,0 × 100% = 68,0%; %NaHCO₃ = 32,0%.</p>"}]}