{"id":"chemia-2020-marzec-politechnika-lodzka-probna-rozszerzona/zad/11","paper_id":"chemia-2020-marzec-politechnika-lodzka-probna-rozszerzona","number":"11","points":null,"ptype":"open","subject":"chemia","category":"probna","year":2020,"month":"marzec","level":"rozszerzona","text":"Zadanie 11\nx =\nM =\ng/mol\n#10\nMATURY PRÓBNE 2020, chemia 3\n#11\nMATURY PRÓBNE 2020, chemia 3\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\n7 / 32","answer":null,"answer_text":null,"solution":"Zadanie 11. n(NaOH) = 0,045 dm3 * 0,1 mol/dm3 = 0,0045 mol; n(kwasu) = 0,010 dm3 * 0,15 mol/dm3 = 0,0015 mol. Stosunek NaOH:kwas = 3:1, więc kwas jest trójprotonowy, x = 3. Masa soli = 55 g * 0,4473% = 0,2460 g; n(soli) = n(kwasu) = 0,0015 mol; M(soli Na3R) = 0,2460/0,0015 = 164,0 g/mol. M(R) = 164,0 - 3*22,99 = 95,0; M(H3R) = 3*1,008 + 95,0 = 98,1 ≈ 98 g/mol (kwas fosforowy(V) H3PO4). Odpowiedzi: x = 3, M = 98 g/mol.","image":"img/chemia-2020-marzec-politechnika-lodzka-probna-rozszerzona/zad-11.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 11<br>x =<br>M =<br>g/mol<br>#10<br>MATURY PRÓBNE 2020, chemia 3<br>#11<br>MATURY PRÓBNE 2020, chemia 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>7 / 32</p>","solutions":[{"source":"ai","label":"AI","kind":"text","html":"<p>Zadanie 11. n(NaOH) = 0,045 dm3 * 0,1 mol/dm3 = 0,0045 mol; n(kwasu) = 0,010 dm3 * 0,15 mol/dm3 = 0,0015 mol. Stosunek NaOH:kwas = 3:1, więc kwas jest trójprotonowy, x = 3. Masa soli = 55 g * 0,4473% = 0,2460 g; n(soli) = n(kwasu) = 0,0015 mol; M(soli Na3R) = 0,2460/0,0015 = 164,0 g/mol. M(R) = 164,0 - 3<em>22,99 = 95,0; M(H3R) = 3</em>1,008 + 95,0 = 98,1 ≈ 98 g/mol (kwas fosforowy(V) H3PO4). Odpowiedzi: x = 3, M = 98 g/mol.</p>"}]}