{"id":"chemia-2020-marzec-politechnika-lodzka-probna-rozszerzona/zad/13","paper_id":"chemia-2020-marzec-politechnika-lodzka-probna-rozszerzona","number":"13","points":null,"ptype":"open","subject":"chemia","category":"probna","year":2020,"month":"marzec","level":"rozszerzona","text":"Zadanie 13\nC\nmol/dm\nC\nmol/dm\n#12\nMATURY PRÓBNE 2020, chemia 3\nmol\n3\nmol\n3\n#13\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\n8 / 32","answer":null,"answer_text":null,"solution":"Zadanie 13. Ułamek molowy H2SO4 = 0,0433. Przyjmijmy 1 mol roztworu: n(H2SO4) = 0,0433 mol, n(H2O) = 0,9567 mol. m(H2SO4) = 0,0433*98,08 = 4,247 g; m(H2O) = 0,9567*18,02 = 17,240 g; m(razem) = 21,487 g. V = m/d = 21,487/1,141 = 18,83 cm3 = 0,01883 dm3. C = n/V = 0,0433/0,01883 = 2,30 mol/dm3. Odpowiedź: 2,30 mol/dm3.","image":"img/chemia-2020-marzec-politechnika-lodzka-probna-rozszerzona/zad-13.webp","solution_image":null,"topics":null,"page_from":8,"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 13<br>C<br>mol/dm<br>C<br>mol/dm<br>#12<br>MATURY PRÓBNE 2020, chemia 3<br>mol<br>3<br>mol<br>3<br>#13<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>8 / 32</p>","solutions":[{"source":"ai","label":"AI","kind":"text","html":"<p>Zadanie 13. Ułamek molowy H2SO4 = 0,0433. Przyjmijmy 1 mol roztworu: n(H2SO4) = 0,0433 mol, n(H2O) = 0,9567 mol. m(H2SO4) = 0,0433<em>98,08 = 4,247 g; m(H2O) = 0,9567</em>18,02 = 17,240 g; m(razem) = 21,487 g. V = m/d = 21,487/1,141 = 18,83 cm3 = 0,01883 dm3. C = n/V = 0,0433/0,01883 = 2,30 mol/dm3. Odpowiedź: 2,30 mol/dm3.</p>"}]}