{"id":"chemia-2020-marzec-politechnika-lodzka-probna-rozszerzona/zad/12","paper_id":"chemia-2020-marzec-politechnika-lodzka-probna-rozszerzona","number":"12","points":null,"ptype":"open","subject":"chemia","category":"probna","year":2020,"month":"marzec","level":"rozszerzona","text":"Zadanie 12\nC\nmol/dm\nC\nmol/dm","answer":null,"answer_text":null,"solution":"Zadanie 12. n(HCl) = 44,8 cm3 / 22400 cm3/mol = 0,002 mol. n(H2SO4) = 0,010 dm3 * 0,05 mol/dm3 = 0,0005 mol; n(H+) z H2SO4 = 2*0,0005 = 0,001 mol (nadmiar OH-). OH- przereagowane z HCl = 0,002 mol. n(KOH) całkowite = 0,002 + 0,001 = 0,003 mol. C(KOH) = 0,003 mol / 0,015 dm3 = 0,2 mol/dm3. Odpowiedź: 0,2 mol/dm3.","image":"img/chemia-2020-marzec-politechnika-lodzka-probna-rozszerzona/zad-12.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 12<br>C<br>mol/dm<br>C<br>mol/dm</p>","solutions":[{"source":"ai","label":"AI","kind":"text","html":"<p>Zadanie 12. n(HCl) = 44,8 cm3 / 22400 cm3/mol = 0,002 mol. n(H2SO4) = 0,010 dm3 * 0,05 mol/dm3 = 0,0005 mol; n(H+) z H2SO4 = 2*0,0005 = 0,001 mol (nadmiar OH-). OH- przereagowane z HCl = 0,002 mol. n(KOH) całkowite = 0,002 + 0,001 = 0,003 mol. C(KOH) = 0,003 mol / 0,015 dm3 = 0,2 mol/dm3. Odpowiedź: 0,2 mol/dm3.</p>"}]}