{"id":"chemia-2020-marzec-politechnika-lodzka-probna-rozszerzona/zad/21","paper_id":"chemia-2020-marzec-politechnika-lodzka-probna-rozszerzona","number":"21","points":null,"ptype":"open","subject":"chemia","category":"probna","year":2020,"month":"marzec","level":"rozszerzona","text":"Zadanie 21\nC\nmol/dm\nC\nmol/dm\n#22\nMATURY PRÓBNE 2020, chemia 3\n#23\nMATURY PRÓBNE 2020, chemia 3\nmol\n3\nmol\n3\nMATURA PRÓBNA Z CHEMII\nMatura III 2019/2020 - chemia\nWydział Elektrotechniki, Elektroniki, Informatyki i Automatyki\n90-924 Łódź, ul. Stefanowskiego 18/22, budynek A10\ntel. 42 631 25 00, fax 42 636 47 02, e-mail: deanelec@adm.p.lodz.pl, www.weeia.p.lodz.pl\n16 / 32","answer":null,"answer_text":null,"solution":"Kwas mrówkowy HCOOH, Ka ≈ 1,8·10⁻⁴ (20°C). pH = 2,89 → [H⁺] = 10⁻²·⁸⁹ = 1,288·10⁻³ mol/dm³. Z równania Ka = x²/(c−x), gdzie x = [H⁺]: c = x + x²/Ka = 0,001288 + (0,001288)²/(1,8·10⁻⁴) = 0,001288 + 0,009217 = 0,010505 mol/dm³. Wynik dwukrotnie: 0,01050 mol/dm³.","image":"img/chemia-2020-marzec-politechnika-lodzka-probna-rozszerzona/zad-21.webp","solution_image":null,"topics":null,"page_from":16,"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 21<br>C<br>mol/dm<br>C<br>mol/dm<br>#22<br>MATURY PRÓBNE 2020, chemia 3<br>#23<br>MATURY PRÓBNE 2020, chemia 3<br>mol<br>3<br>mol<br>3<br>MATURA PRÓBNA Z CHEMII<br>Matura III 2019/2020 - chemia<br>Wydział Elektrotechniki, Elektroniki, Informatyki i Automatyki<br>90-924 Łódź, ul. Stefanowskiego 18/22, budynek A10<br>tel. 42 631 25 00, fax 42 636 47 02, e-mail: deanelec@adm.p.lodz.pl, www.weeia.p.lodz.pl<br>16 / 32</p>","solutions":[{"source":"ai","label":"AI","kind":"text","html":"<p>Kwas mrówkowy HCOOH, Ka ≈ 1,8·10⁻⁴ (20°C). pH = 2,89 → [H⁺] = 10⁻²·⁸⁹ = 1,288·10⁻³ mol/dm³. Z równania Ka = x²/(c−x), gdzie x = [H⁺]: c = x + x²/Ka = 0,001288 + (0,001288)²/(1,8·10⁻⁴) = 0,001288 + 0,009217 = 0,010505 mol/dm³. Wynik dwukrotnie: 0,01050 mol/dm³.</p>"}]}