ÌâÄ¿ÄÚÈÝ
2£®Çë¸ù¾ÝÈçͼװÖÃͼ£¬»Ø´ðÓйØÎÊÌ⣺£¨1£©Ð´³öA¡«F×°ÖÃÖеÄÒ»ÖÖ²£Á§ÒÇÆ÷µÄÃû³Æ£ºÊԹܣ»
£¨2£©ÊµÑéÊÒÓùýÑõ»¯ÇâÖÆÈ¡ÑõÆøµÄ»¯Ñ§·½³ÌʽΪ2H2O2$\frac{\underline{\;MnO_2\;}}{\;}$2H2O+O2¡ü£¬¿ÉÑ¡Óõķ¢Éú×°ÖÃÊÇB £¨Ìî×Öĸ£¬ÏÂͬ£©£»¸Ã×°Öû¹¿ÉÓÃÓÚʵÑéÊÒÖÆÈ¡¶þÑõ»¯Ì¼£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪCaCO3+2HCl¨TCaCl2+H2O+CO2¡ü£»ÊÕ¼¯ÑõÆø¿ÉÓÃE×°ÖõÄÔÒòÊÇÑõÆø²»Ò×ÈÜÓÚË®
£¨3£©ÈôÓÃF×°ÖøÉÔïÉú³ÉµÄÑõÆø£¬FÖÐÊ¢·ÅµÄÒºÌåÊÔ¼ÁÊÇŨÁòËá£¬ÆøÌåÓ¦´Óa£¨Ñ¡Ìî¡°a¡±»ò¡°b¡±£©¶Ëµ¼È룮ÔÚÒ½ÔºÀï¸ø²¡ÈËÊäÑõʱ£¬ÔÚÑõÆø¸ÖÆ¿ºÍ²¡ÈËÎüÑõÆ÷Ö®¼äͬÑù»áÁ¬½ÓÒ»¸öÀàËÆFµÄ×°Ö㬸Ã×°ÖõÄ×÷ÓÃÓУ¨ÌîдһÌõ£©ÊªÈóÑõÆø£®
·ÖÎö £¨1£©ÊìϤ³£¼ûÒÇÆ÷£¬Á˽âÃû³Æ£»
£¨2£©¸ù¾ÝÑõÆø¡¢¶þÑõ»¯Ì¼µÄÖÆÈ¡ÔÀí¡¢·¢Éú×°ÖÃÒÔ¼°ÊÕ¼¯×°ÖõÈ֪ʶ·ÖÎö£»
£¨3£©¸ù¾ÝŨÁòËáµÄÐÔÖÊÒÔ¼°Í¼ÖÐ×°ÖõÄ×÷ÓýøÐзÖÎö£®
½â´ð ½â£º£¨1£©A¡«F×°ÖÃÖеIJ£Á§ÒÇÆ÷ÓÐÊԹܡ¢³¤¾±Â©¶·¡¢×¶ÐÎÆ¿¡¢¼¯ÆøÆ¿¡¢Ë®²ÛµÈ£¬¹ÊÌÊԹܣ»
£¨2£©ÊµÑéÊÒÓùýÑõ»¯ÇâÖÆÈ¡ÑõÆøµÄ»¯Ñ§·½³ÌʽΪ£º2H2O2$\frac{\underline{\;MnO_2\;}}{\;}$2H2O+O2¡ü£»ÊôÓÚ¹ÌÒºÔÚ³£ÎÂÏ·´Ó¦£¬·¢Éú×°ÖÃÑ¡B£»ÓôóÀíʯºÍÏ¡ÑÎËáÖÆÈ¡¶þÑõ»¯Ì¼Ê±£¬ÊǹÌÒº²»¼ÓÈÈ·´Ó¦£¬¿ÉÓÃ×°ÖÃB×÷Ϊ·¢Éú×°Ö㬻¯Ñ§·´Ó¦Ê½ÊÇ£ºCaCO3+2HCl¨TCaCl2+H2O+CO2¡ü£»ÊÕ¼¯ÑõÆø¿ÉÓÃE×°ÖõÄÔÒòÊÇÑõÆø²»Ò×ÈÜÓÚË®£¬¹ÊÌ2H2O2$\frac{\underline{\;MnO_2\;}}{\;}$2H2O+O2¡ü£»B£»¶þÑõ»¯Ì¼£»CaCO3+2HCl¨TCaCl2+H2O+CO2¡ü£»ÑõÆø²»Ò×ÈÜÓÚË®£®
£¨3£©ÈôÓÃF×°ÖøÉÔïÉú³ÉµÄÑõÆø£¬FÖÐÊ¢·ÅµÄÒºÌåÊÔ¼ÁÊÇŨÁòËá£¬ÆøÌåÓ¦´Óa¶Ëµ¼È룮ÔÚÒ½ÔºÀï¸ø²¡ÈËÊäÑõʱ£¬ÔÚÑõÆø¸ÖÆ¿ºÍ²¡ÈËÎüÑõÆ÷Ö®¼äͬÑù»áÁ¬½ÓÒ»¸öÀàËÆFµÄ×°Ö㬸Ã×°ÖõÄ×÷ÓÃÊÇʪÈóÑõÆø£¬¹Û²ì¹©¸øÑõÆøµÄËÙÂʵȣ®¹Ê´ð°¸Îª£ºÅ¨ÁòË᣻a£»ÊªÈóÑõÆø£®
µãÆÀ ¶Ô³£¼û»¯Ñ§·´Ó¦Ê½µÄÊéд¡¢³£¼ûʵÑé×°ÖõÄÑ¡Ôñ¡¢³£¼ûÆøÌåµÄÊÕ¼¯·½·¨Ò»¶¨ÒªÁËÈ»ÓÚÐÄ£®
| A£® | Ò»¶¨Êǵ¥ÖÊ | B£® | Ò»¶¨ÊÇ´¿¾»Îï | C£® | Ò»¶¨ÊÇ»ìºÏÎï | D£® | Ò»¶¨²»ÊÇ»¯ºÏÎï |
| ÐòºÅ | »¯Ñ§Öеġ°²»Ò»¶¨¡± | Ëù¾ÙÀýÖ¤ |
| A | º¬ÓÐÑõÔªËØµÄ»¯ºÏÎï²»Ò»¶¨ÊÇÑõ»¯Îï | H2O |
| B | ȼÉÕ²»Ò»¶¨ÊǸúÑõÆø | 2Mg+CO2 $\frac{\underline{\;µãȼ\;}}{\;}$2MgO+C |
| C | Éú³ÉÑκÍË®µÄ·´Ó¦²»Ò»¶¨ÊÇÖкͷ´Ó¦ | NaOH+HCl¨TNaCl+H2O |
| D | »¯ºÏ·´Ó¦²»Ò»¶¨ÊÇÑõ»¯·´Ó¦ | CO2+H2O=H2CO3 |
| Æ·Ãû | X X X |
| ÓªÑø³É·Ö | µ°°×ÖÊ¡¢Ö¬·¾¡¢ÌÇÀà¡¢¸Æ¡¢Ã¾¡¢¼Ø¡¢ÄÆ¡¢Ð¿¡¢Î¬ÉúËØA¡¢Î¬ÉúËØD¡¢Î¬ÉúËØE¡¢Ë® |
| Öü´æ·½·¨ | ÖÃÓÚÒõÁ¹¸ÉÔï´¦¡¢±ÜÃâÑô¹âÖ±½ÓÕÕÉ䣮ÔÊÐíÓÐÉÙÁ¿³Áµí£¬ÒûÓÃǰÇëÒ¡ÔÈ£® |
| ÉúÖ÷ÈÕÆÚ£º2009.1.17 ±£ÖÊÆÚ£ºÁù¸öÔ | |
£¨2£©ÄÜÌṩÄÜÁ¿µÄÓлúÎïÊÇÌÇÀֻࣨдһÖÖ£©£»
£¨3£©¸ÃÒûÁϵİü×°µÄ²ÄÁÏÊÇËÜÁÏ£¬Èçͼͼ±êÖбíʾËÜÁϰü×°ÖÆÆ·»ØÊÕ±êÖ¾µÄÊÇC£®
£¨4£©¸ÃʳƷÄÜ·ñʳÓã¿ÎªÊ²Ã´£¿ÄÜ£»Ê³Æ·ÔÚ±£ÖÊÆÚÄÚ£®
£¨1£©ÇëÄã°ïÖúÍê³ÉÈç±í̽¾¿£º
| ·½°¸ | ËùÑ¡ÊÔ¼Á | ÏÖÏó | »¯Ñ§·½³Ìʽ |
| ·½°¸Ò»£º²â¶¨·´Ó¦ºóÉú³ÉCO2µÄÖÊÁ¿£¬¼ÆËã³öÑùÆ·ÖÐNa2CO3µÄº¬Á¿ | Ï¡ÑÎËá | ²úÉúÆøÅÝ | Na2CO3+2HCl¨T 2NaCl+H2O+CO2¡ü |
| ·½°¸¶þ£º²â¶¨·´Ó¦ºóÉú³É³ÁµíµÄÖÊÁ¿£¬¼ÆËã³öÑùÆ·ÖÐNa2CO3µÄº¬Á¿ | Ë®¡¢ ÂÈ»¯¸Æ | ²úÉú°× É«³Áµí |
¢ÙÉú³ÉCO2 µÄÖÊÁ¿£º15g+100g+89.4g-200g=4.4g£¨ÁÐʽ²¢Ð´³ö½á¹û£©£»
¢ÚÑùÆ·ÖÐNa2CO3 µÄÖÊÁ¿·ÖÊý£»
¢ÛÏ¡ÑÎËáÖÐÈÜÖʵÄÖÊÁ¿·ÖÊý£®
| A£® | µØ¿ÇÖк¬Á¿×î¸ßµÄ½ðÊôÔªËØÊÇÑõÔªËØ | |
| B£® | ´¿¾»µÄ¿ÕÆøÖУ¬º¬ÑõÆø78%£¬º¤Æø21% | |
| C£® | Å©×÷ÎïÒ»°ãÊÊÒËÔÚÖÐÐÔ»ò½Ó½üÖÐÐÔµÄÍÁÈÀÖÐÉú³¤£¬Îª´Ë³£ÓÃŨÁòËá¸ÄÁ¼¼îÐÔÍÁÈÀ | |
| D£® | ÏõËá¼Ø£¨KNO3£©ÊÇÒ»ÖÖ¸´ºÏ·ÊÁÏ£¬¿ÉÒÔÌṩֲÎïÉú³¤±ØÐèµÄ¼ØºÍµªÁ½ÖÖÓªÑøÔªËØ |