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      <title>Nguyễn Thị Hồng Hoa - GV Kim Giang by Nguyễn Hồng Hoa</title>
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      <description>Nơi chia sẻ các bài làm thú vị của học sinh</description>
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      <pubDate>2021-10-14 02:25:33 UTC</pubDate>
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         <title>Ngo Hong Minh-9a0</title>
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         <title>Cung Thục Ánh-9A0</title>
         <author>cungthucanh</author>
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         <title>Lê Bích Ngọc-9a0</title>
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         <title>An Quốc Hưng</title>
         <author>anquochungp4m8</author>
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         <title></title>
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         <description><![CDATA[<div>Thảo Anh&nbsp;<br><br></div>]]></description>
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         <title>Tạ Ngọc Thủy Tiên-9a0</title>
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         <title>Bài 8: Giải bài toán theo PTHH:Tóm tắt đề bài:   Al     +     CuSO4      Al2(SO4)3   +    Cu                                  M=?g      Vdd=200ml                          m=?g                                           CM=?M  Sau phản ứng thanh kim loại tăng 2,07g.Lời giảia)	Gọi số mol của của kim loại Nhôm phản ứng là xPTHH: 2Al+ 3CuSO4  Al2(SO4)3  + 3Cu         Theo pt: x                                                1,5xKhối lượng kim loại nhôm phản ứng là : mAl =27x(g)    Khối lượng của kim loại đồng sinh ra là: mCu=96x(g)Ta có:  Khối lượng kim loại tăng =    mCu  -  mAl	                               &lt;=&gt;     2,07    =    96x   -  27x                                      2,07     =    69x                                      0.03    =      xPTHH;     2Al     +    3CuSO4     Al2(SO4)3    +    3CuTheo pt:   2mol           3mol                                        3molTheo đb:  0,03mol    0,045mol                                0,045molVậy khối lượng của kim loại nhôm phản ứng có là:           mAl= M . n  = 27 . 0,03 =0,81(g)Vậy khối lượng của kim loại Cu sinh ra là:           mCu= M. n  = 64 . 0,045 =2,88(g)b)	Đổi 200ml=0,2lNồng độ mol của dung dịch CuSO4 phản ứng là :         CM= n/V=0,045/0,2=0,225(M)</title>
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         <description><![CDATA[<div>Bài làm của 29-Nguyễn Phương LinhB (7/8)</div>]]></description>
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         <pubDate>2021-12-22 06:40:01 UTC</pubDate>
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         <title>30- Nguyễn Phương Linh A</title>
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         <title>03-Đỗ Phương Anh-9A0 </title>
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         <title>Nguyễn Yến Nhi-9A0</title>
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