Question

15£®ÏÂÁÐ×Öĸ±íʾ¶ÌÖÜÆÚÖм¸ÖÖÔªËصÄλÖã¬Çë»Ø´ðÏÂÁÐÎÊÌ⣮

A
			B	C	D
E	F	G			H	I

£¨1£©D^2-µÄÀë×ӽṹʾÒâͼΪ£¬EÓëIÐγɵĻ¯ºÏÎïµÄµç×ÓʽΪ£®
£¨2£©E£¬G£¬HÈýÔªËصÄ×î¸ß¼ÛÑõ»¯Îï¶ÔӦˮ»¯ÎïµÄÈÜÒº¼äÏ໥·´Ó¦µÄÀë×Ó·½³ÌʽΪAl£¨OH£©₃+OH^-=AlO₂^-+2H₂O£¬Al£¨OH£©₃+3H⁺=Al³⁺+3H₂O£¬OH^-+H⁺=H₂O£®
£¨3£©FºÍIÐγɵĻ¯ºÏÎïÖдæÔڵĻ¯Ñ§¼üÀàÐÍÊÇÀë×Ó¼ü£¬ÔÚA¡«I¾ÅÖÖÔªËØÖУ¬ÍêÈ«ÓɷǽðÊôÔªËØÐγɵÄÒ»¸öÀë×Ó»¯ºÏÎïµÄ»¯Ñ§Ê½ÊÇNH₄NO₃£®
£¨4£©4.8gµ¥ÖÊFÄÜÓë±ê×¼×´¿öÏÂ2.24LµÄBµÄ×î¸ß¼ÛÑõ»¯ÎïÇ¡ºÃ·´Ó¦£®

Answer 1

·ÖÎö ÓÉÔªËØÔÙÖÜÆÚ±íÖÐλÖ㬿ÉÖªAΪÇ⣬BΪ̼£¬CΪN£¬DΪO£¬EΪNa£¬FΪMg£¬GΪAl£¬HΪS£¬IΪCl£®
£¨1£©O^2-µÄÀë×ÓºËÍâµç×ÓÊýΪ10£¬¸÷²ãµç×ÓÊýΪ2¡¢8£»EÓëIÐγɵĻ¯ºÏÎïΪNaCl£¬ÓÉÄÆÀë×ÓÓëÂÈÀë×Ó¹¹³É£»
£¨2£©E¡¢G¡¢HÈýÔªËصÄ×î¸ß¼ÛÑõ»¯Îï¶ÔӦˮ»¯Îï·Ö±ðΪNaOH¡¢Al£¨OH£©₃¡¢H₂SO₄£»
£¨3£©FºÍIÐγɵĻ¯ºÏÎïΪMgCl₂£¬ÓÉþÀë×ÓÓëÂÈÀë×Ó¹¹³É£»ÔÚA¡«I¾ÅÖÖÔªËØÖУ¬ÍêÈ«ÓɷǽðÊôÔªËØÐγɵÄÒ»¸öÀë×Ó»¯ºÏÎïΪÏõËá淋ȣ»
£¨4£©·¢Éú·´Ó¦2Mg+CO₂$\frac{\underline{\;µãȼ\;}}{\;}$2MgO+C£¬¸ù¾Ý·½³Ìʽ¼ÆËãÐèÒª¶þÑõ»¯Ì¼µÄÌå»ý£®

½â´ð ½â£ºÓÉÔªËØÔÚÖÜÆÚ±íÖÐλÖ㬿ÉÖªAΪÇ⣬BΪ̼£¬CΪN£¬DΪO£¬EΪNa£¬FΪMg£¬GΪAl£¬HΪS£¬IΪCl£®
£¨1£©O^2-µÄÀë×ÓºËÍâµç×ÓÊýΪ10£¬¸÷²ãµç×ÓÊýΪ2¡¢8£¬Àë×ӽṹʾÒâͼΪ£»EÓëIÐγɵĻ¯ºÏÎïΪNaCl£¬ÓÉÄÆÀë×ÓÓëÂÈÀë×Ó¹¹³É£¬µç×ÓʽΪ£¬
¹Ê´ð°¸Îª£º£»£»
£¨2£©E¡¢G¡¢HÈýÔªËصÄ×î¸ß¼ÛÑõ»¯Îï¶ÔӦˮ»¯Îï·Ö±ðΪNaOH¡¢Al£¨OH£©₃¡¢H₂SO₄£¬Ï໥·´Ó¦Àë×Ó·½³ÌʽΪ£ºAl£¨OH£©₃+OH^-=AlO₂^-+2H₂O¡¢Al£¨OH£©₃+3H⁺=Al³⁺+3H₂O¡¢OH^-+H⁺=H₂O£¬
¹Ê´ð°¸Îª£ºAl£¨OH£©₃+OH^-=AlO₂^-+2H₂O¡¢Al£¨OH£©₃+3H⁺=Al³⁺+3H₂O¡¢OH^-+H⁺=H₂O£»
£¨3£©FºÍIÐγɵĻ¯ºÏÎïΪMgCl₂£¬ÓÉþÀë×ÓÓëÂÈÀë×Ó¹¹³É£¬º¬ÓÐÀë×Ó¼ü£»ÔÚA¡«I¾ÅÖÖÔªËØÖУ¬ÍêÈ«ÓɷǽðÊôÔªËØÐγɵÄÒ»¸öÀë×Ó»¯ºÏÎïΪNH₄NO₃µÈ£¬
¹Ê´ð°¸Îª£ºÀë×Ó¼ü£»NH₄NO₃£»
£¨4£©·¢Éú·´Ó¦2Mg+CO₂$\frac{\underline{\;µãȼ\;}}{\;}$2MgO+C£¬4.8gMgµÄÎïÖʵÄÁ¿Îª$\frac{4.8g}{24g/mol}$=0.2mol£¬ÓÉ·½³Ìʽ¿ÉÖªÐèÒª¶þÑõ»¯Ì¼Îª0.2mol¡Á$\frac{1}{2}$=0.1mol£¬¹ÊÐèÒª¶þÑõ»¯Ì¼µÄÌå»ýΪ0.1mol¡Á22.4L/mol=2.24L£¬
¹Ê´ð°¸Îª£º2.24£®

µãÆÀ ±¾Ì⿼²éÔªËØÖÜÆÚ±íºÍÔªËØÖÜÆÚÂɵÄ×ÛºÏÓ¦Óã¬ÊìÁ·ÕÆÎÕÔªËØÖÜÆÚ±íµÄ½á¹¹£¬²àÖضԻ¯Ñ§ÓÃÓïµÄ¿¼²é£¬×¢Òâ¶Ô»ù´¡ÖªÊ¶µÄÀí½âÕÆÎÕ£®