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Home Lớp 8 Khoa học tự nhiên 8 - sách Kết nối tri thức Bài 11: MuốiBài tập trung bình

{"common":{"save":0,"post_id":"8578","level":2,"total":10,"point":10,"point_extra":0},"segment":[{"id":"7999","post_id":"8578","mon_id":"1159539","chapter_id":"1159543","question":"\u0110\u1ec3 có \u0111\u01b0\u1ee3c dung d\u1ecbch NaCl 32%, thì kh\u1ed1i l\u01b0\u1ee3ng NaCl c\u1ea7n l\u1ea5y hoà tan vào 200 gam n\u01b0\u1edbc là:       <\/p>","options":["A. 90g.","B. 94,12 g.","C. 100g.","D. 141,18 g."],"correct":"2","level":"2","hint":"Áp d\u1ee5ng công th\u1ee9c tính n\u1ed3ng \u0111\u1ed9 %.<\/p>","answer":"\u0110áp án \u0111úng: B. 94,12 g.<\/strong><\/p>G\u1ecdi kh\u1ed1i l\u01b0\u1ee3ng NaCl c\u1ea7n dùng là m (gam)<\/p>=> mdung d\u1ecbch = mNaCl + mH2O = m + 200<\/p>C% = mct \/ mdd.100% => m \/ (m+200).100% = 32% => m = 94,12 g<\/p>","type":"choose","extra_type":"classic","time":"0","user_id":"139","test":"0","date":"2025-03-04 22:12:13","option_type":"txt","len":1},{"id":"8000","post_id":"8578","mon_id":"1159539","chapter_id":"1159543","question":"Cho 200 ml dung d\u1ecbch Ba(OH)\u2082 0,5M<\/strong> ph\u1ea3n \u1ee9ng hoàn toàn v\u1edbi dung d\u1ecbch K\u2082CO\u2083 d\u01b0.<\/p>1. Xác \u0111\u1ecbnh kh\u1ed1i l\u01b0\u1ee3ng Ba(OH)\u2082 có trong dung d\u1ecbch, sau \u0111ó tính s\u1ed1 mol Ba(OH)\u2082.<\/p>2. Vi\u1ebft ph\u01b0\u01a1ng trình ph\u1ea3n \u1ee9ng v\u1edbi K\u2082CO\u2083 và tìm s\u1ed1 mol BaCO\u2083 t\u1ea1o thành.<\/p>3. Cu\u1ed1i cùng, tính kh\u1ed1i l\u01b0\u1ee3ng k\u1ebft t\u1ee7a BaCO\u2083 t\u1ea1o thành.<\/p>","options":["A. 4,89 gam","B. 19,7 gam","C. 6,19 gam","D. 5,45 gam"],"correct":"2","level":"2","hint":"- Xác \u0111\u1ecbnh s\u1ed1 mol Ba(OH)2\u200b trong dung d\u1ecbch ban \u0111\u1ea7u.<\/p>- Vi\u1ebft ph\u01b0\u01a1ng trình ph\u1ea3n \u1ee9ng gi\u1eefa Ba(OH)2 và K2\u200bCO3\u200b.<\/p>- Xác \u0111\u1ecbnh s\u1ed1 mol BaCO3 t\u1ea1o thành và tính kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a nó.<\/p>","answer":"<h2 data-end=\"512\" data-start=\"486\">Tính s\u1ed1 mol Ba(OH)\u2082<\/h2>Công th\u1ee9c:<\/p> <\/p>n=C×V<\/p>Trong \u0111ó:<\/p><ul data-end=\"601\" data-start=\"558\"> <li data-end=\"574\" data-start=\"558\">C=0,5M<\/p><\/li> <li data-end=\"601\" data-start=\"575\">V=200ml=0,2L<\/p><\/li><\/ul>\\n <\/p>nBa(OH)\u2082 = 0,5×0,2=0,1 mol<\/p><h2 data-end=\"693\" data-start=\"663\">Tính kh\u1ed1i l\u01b0\u1ee3ng Ba(OH)\u2082<\/h2>Kh\u1ed1i l\u01b0\u1ee3ng mol:<\/p>MBa(OH)2=137+2(16+1)=171 g\/mol<\/p>m=n×M<\/p> mBa(OH)2=0,1×171=17,1 g<\/p><h2 data-end=\"877\" data-start=\"849\">Ph\u01b0\u01a1ng trình ph\u1ea3n \u1ee9ng<\/h2>Ba(OH)2+K2CO3→BaCO3↓+2KOH<\/p>T\u1ec9 l\u1ec7 mol: 1 : 1<\/strong><\/p><h2 data-end=\"1003\" data-start=\"969\">Tính s\u1ed1 mol BaCO\u2083 t\u1ea1o thành<\/h2>Vì t\u1ec9 l\u1ec7 1:1 và K\u2082CO\u2083 d\u01b0 nên:<\/p>nBaCO3 \u200b\u200b= nBa(OH)2 \u200b\u200b= 0,1 mol<\/p><h2 data-end=\"1120\" data-start=\"1092\">Tính kh\u1ed1i l\u01b0\u1ee3ng BaCO\u2083<\/h2>Kh\u1ed1i l\u01b0\u1ee3ng mol:<\/p>MBaCO3 = 137+12+3×16 = 197 g\/mol<\/p>mBaCO3\u200b\u200b = 0,1×197 = 19,7 g<\/p> <\/p>\u0110áp án \u0111úng: B. 19,7 <\/strong>gam<\/strong><\/p>","type":"choose","extra_type":"classic","time":"0","user_id":"139","test":"0","date":"2025-03-04 22:15:41","option_type":"txt","len":1},{"id":"8001","post_id":"8578","mon_id":"1159539","chapter_id":"1159543","question":"Cho dung d\u1ecbch NaOH vào \u1ed1ng nghi\u1ec7m \u0111\u1ef1ng dung d\u1ecbch FeCl3, ta quan sát \u0111\u01b0\u1ee3c hi\u1ec7n t\u01b0\u1ee3ng là:<\/p>","options":["A. Có khí thoát ra","B. Xu\u1ea5t hi\u1ec7n k\u1ebft t\u1ee7a màu tr\u1eafng","C. Xu\u1ea5t hi\u1ec7n k\u1ebft t\u1ee7a xanh lam","D. Xu\u1ea5t hi\u1ec7n k\u1ebft t\u1ee7a màu \u0111\u1ecf nâu"],"correct":"4","level":"2","hint":"Ph\u1ea3n \u1ee9ng gi\u1eefa NaOH và FeCl\u2083 t\u1ea1o ra hi\u0111roxit c\u1ee7a s\u1eaft. Hãy nh\u1edb màu s\u1eafc \u0111\u1eb7c tr\u01b0ng c\u1ee7a Fe(OH)\u2083.<\/p>","answer":"\u0110áp án \u0111úng: D. Xu\u1ea5t hi\u1ec7n k\u1ebft t\u1ee7a màu \u0111\u1ecf nâu<\/strong><\/p>Khi cho NaOH vào dung d\u1ecbch FeCl\u2083, x\u1ea3y ra ph\u1ea3n \u1ee9ng: FeCl3 + 3NaOH → Fe(OH)3↓ + 3NaCl<\/p>Fe(OH)\u2083 là m\u1ed9t hi\u0111roxit không tan, có màu \u0111\u1ecf nâu \u0111\u1eb7c tr\u01b0ng.<\/p>","type":"choose","extra_type":"classic","time":"0","user_id":"139","test":"0","date":"2025-03-04 22:17:07","option_type":"txt","len":2}]}