ÌâÄ¿ÄÚÈÝ
ÈçͼÊÇÔªËØÖÜÆÚ±íµÄÒ»²¿·Ö£¬±íÖеĢ١«¢âÖÐÔªËØ£¬ÓÃÔªËØ·ûºÅ»ò»¯Ñ§Ê½Ìî¿Õ×å ÖÜÆÚ | ¢ñA | ¢òA | ¢óA | ¢ôA | ¢õA | ¢öA | ¢÷A | |
| ¶þ | ¢Ù | ¢Ú | ||||||
| Èý | ¢Û | ¢Ü | ¢Ý | ¢Þ | ¢ß | ¢à | ||
| ËÄ | ¢á | ¢â |
£¨2£©ÔÚÕâÐ©ÔªËØÖУ¬µØ¿ÇÖк¬Á¿×î¶àµÄ½ðÊôÔªËØÊÇ
£¨3£©Óõç×Óʽ±íʾ¢ÚÓë¢ÜÐγɻ¯ºÏÎïµÄ¹ý³Ì £®
£¨4£©ÕâÐ©ÔªËØÖеÄ×î¸ß¼ÛÑõ»¯Îï¶ÔÓ¦µÄË®»¯ÎïÖУ¬ËáÐÔ×îÇ¿µÄÊÇ £¬¼îÐÔ×îÇ¿µÄÊÇ £¬³ÊÁ½ÐÔµÄÇâÑõ»¯ÎïÊÇ
£¨5£©Ð´³ö¢ÝÓëÇâÑõ»¯ÄÆÈÜÒº·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º £®
д³ö¢ÝµÄÇâÑõ»¯ÎïÓëÇâÑõ»¯ÄÆÈÜÒº·´Ó¦µÄÀë×Ó·½³Ìʽ£º £®
£¨6£©Ä³ÔªËØRµÄÆøÌ¬Ç⻯ÎïΪHXR£¬ÇÒRÔÚ¸ÃÇ⻯ÎïÖеÄÖÊÁ¿·ÖÊýΪ94%£¬8.5gµÄHXRÆøÌåÔÚ±ê׼״̬ϵÄÌå»ýÊÇ5.6L£®ÔòHXRµÄÏà¶Ô·Ö×ÓÁ¿Îª £»HXRµÄ»¯Ñ§Ê½Îª £®
¡¾´ð°¸¡¿·ÖÎö£º£¨1£©¸ù¾Ý»¯Ñ§ÐÔÖÊ×î²»»îÆÃÊÇÏ¡ÓÐÆøÌåÕÒ³öÔªËØÐ´³öÆäÔ×ӽṹʾÒâͼ£»
£¨2£©¸ù¾ÝµØ¿ÇÖÐÔªËØµÄº¬Á¿¸ßµÍ£»
£¨3£©ÏÈÅжϢÚÓë¢ÜÊÇÊ²Ã´ÔªËØ£¬È»ºóµç×Óʽд³öÐγɻ¯ºÏÎïµÄ¹ý³Ì£»
£¨4£©¸ù¾ÝÔªËØµÄ·Ç½ðÊôÐÔԽǿ£¬×î¸ß¼ÛÑõ»¯Îï¶ÔÓ¦µÄË®»¯ÎïËáÐÔԽǿ£¬ÔªËصĽðÊôÐÔԽǿ£¬×î¸ß¼ÛÑõ»¯Îï¶ÔÓ¦µÄË®»¯Îï¼îÐÔԽǿ£¬Á½ÐÔµÄÇâÑõ»¯ÎïΪAl£¨OH£©3£»
£¨5£©ÏÈÅжϢÝÊÇÊ²Ã´ÔªËØ£¬È»ºó¸ù¾ÝÒªÇóд³ö¶ÔÓ¦µÄ»¯Ñ§·½³Ìʽ£»
£¨6£©Ïȸù¾ÝHXRÆøÌåµÄÖÊÁ¿ºÍÌå»ýÇó³öÆäĦ¶ûÖÊÁ¿£¬È»ºó¸ù¾ÝÏà¶Ô·Ö×ÓÖÊÁ¿ºÍRÔÚ¸ÃÇ⻯ÎïÖеÄÖÊÁ¿·ÖÊýÇó³öRµÄÔ×ÓÁ¿£¬µÃ³öÔªËØRÊǺÎÖÖÔªËØ£»
½â´ð£º½â£º£¨1£©Òò¢àÊÇArÔªËØ£¬Îª¶èÐÔÔªËØ£¬»¯Ñ§ÐÔÖÊÎȶ¨£¬ÆäÔ×ӽṹʾÒâͼΪ
£»
£¨2£©µØ¿ÇÖÐÔªËØº¬Á¿´Ó¸ßµ½µÍµÄ˳ÐòΪ£ºÑõ¡¢¹è¡¢ÂÁ¡¢Ìú¡¢¸Æ¡¢ÄÆ¡¢¼Ø¡¢Ã¾µÈ£»
£¨3£©Òò¢ÚÊÇFÔªËØ£¬¢ÜÊÇMgÔªËØ£¬ËùÒÔMgF2µÄÐγɹý³ÌÊÇ£º
£¨4£©Òò·Ç½ðÊôÐÔÔªËØ×îÇ¿µÄÊÇClÔªËØ£¬½ðÊôÐÔ×îÇ¿µÄÔªËØÊÇKÔªËØ£¬ËùÒÔËáÐÔ×îÇ¿µÄÊÇHClO4£¬¼îÐÔ×îÇ¿µÄÊÇKOH£¬Á½ÐÔµÄÇâÑõ»¯ÎïÖ»ÓÐAl£¨OH£©3£»
£¨5£©Òò¢ÝÊÇÂÁ£¬ËùÒÔÂÁÓëÇâÑõ»¯ÄÆÈÜÒº·´Ó¦ÒÔ¼°ÇâÑõ»¯ÂÁÓëÇâÑõ»¯ÄÆÈÜÒº·´Ó¦µÄ·½³Ìʽ·Ö±ðΪ£º2Al+2NaOH+2H2O=2NaAlO2+3H2¡ü£¬Al£¨OH£©3+OH-=AlO2-+2H2O£»
£¨6£©HxRÆøÌåÔÚ±ê×¼×´¿öϵÄÌå»ýÊÇ5.6L£¬ËµÃ÷ÎïÖʵÄÁ¿Îª5.6/22.4=0.25mol£¬ÓÖÒòÎªÆøÌåÖÊÁ¿Îª8.5g£¬ËùÒÔĦ¶ûÖÊÁ¿Îª8.5/0.25=34Òò´ËHxRµÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª34£¬RÔÚ¸ÃÇ⻯ÎïÖеÄÖÊÁ¿·ÖÊýΪ94%£¬ËùÒÔRµÄÔ×ÓÁ¿Îª34*0.94=31.96=32£¬ËùÒÔRÊÇÁò£¬Òò´ËHxRµÄ»¯Ñ§Ê½ÎªH2S£®
¹Ê´ð°¸Îª£º£¨1£©
£»
£¨2£©Al£»
£¨3£©
£¨4£©HClO4£»KOH£»Al£¨OH£©3£»
£¨5£©2Al+2NaOH+2H2O=2NaAlO2+3H2¡ü£¬Al£¨OH£©3+OH-=AlO2-+2H2O£»
£¨6£©34£¬H2S£®
µãÆÀ£º±¾Ì⿼²éѧÉú¶ÔÔªËØÖÜÆÚ±íÖеÄһЩ±ä»¯¹æÂɵĹéÄɺÍ×ܽᣬ²¢Ñ§ÒÔÖÂÓã¬Í¬Ê±¿¼²éÁËÔªËØ¼°Æä»¯ºÏÎïµÄÓйØÖªÊ¶£®
£¨2£©¸ù¾ÝµØ¿ÇÖÐÔªËØµÄº¬Á¿¸ßµÍ£»
£¨3£©ÏÈÅжϢÚÓë¢ÜÊÇÊ²Ã´ÔªËØ£¬È»ºóµç×Óʽд³öÐγɻ¯ºÏÎïµÄ¹ý³Ì£»
£¨4£©¸ù¾ÝÔªËØµÄ·Ç½ðÊôÐÔԽǿ£¬×î¸ß¼ÛÑõ»¯Îï¶ÔÓ¦µÄË®»¯ÎïËáÐÔԽǿ£¬ÔªËصĽðÊôÐÔԽǿ£¬×î¸ß¼ÛÑõ»¯Îï¶ÔÓ¦µÄË®»¯Îï¼îÐÔԽǿ£¬Á½ÐÔµÄÇâÑõ»¯ÎïΪAl£¨OH£©3£»
£¨5£©ÏÈÅжϢÝÊÇÊ²Ã´ÔªËØ£¬È»ºó¸ù¾ÝÒªÇóд³ö¶ÔÓ¦µÄ»¯Ñ§·½³Ìʽ£»
£¨6£©Ïȸù¾ÝHXRÆøÌåµÄÖÊÁ¿ºÍÌå»ýÇó³öÆäĦ¶ûÖÊÁ¿£¬È»ºó¸ù¾ÝÏà¶Ô·Ö×ÓÖÊÁ¿ºÍRÔÚ¸ÃÇ⻯ÎïÖеÄÖÊÁ¿·ÖÊýÇó³öRµÄÔ×ÓÁ¿£¬µÃ³öÔªËØRÊǺÎÖÖÔªËØ£»
½â´ð£º½â£º£¨1£©Òò¢àÊÇArÔªËØ£¬Îª¶èÐÔÔªËØ£¬»¯Ñ§ÐÔÖÊÎȶ¨£¬ÆäÔ×ӽṹʾÒâͼΪ
£¨2£©µØ¿ÇÖÐÔªËØº¬Á¿´Ó¸ßµ½µÍµÄ˳ÐòΪ£ºÑõ¡¢¹è¡¢ÂÁ¡¢Ìú¡¢¸Æ¡¢ÄÆ¡¢¼Ø¡¢Ã¾µÈ£»
£¨3£©Òò¢ÚÊÇFÔªËØ£¬¢ÜÊÇMgÔªËØ£¬ËùÒÔMgF2µÄÐγɹý³ÌÊÇ£º
£¨4£©Òò·Ç½ðÊôÐÔÔªËØ×îÇ¿µÄÊÇClÔªËØ£¬½ðÊôÐÔ×îÇ¿µÄÔªËØÊÇKÔªËØ£¬ËùÒÔËáÐÔ×îÇ¿µÄÊÇHClO4£¬¼îÐÔ×îÇ¿µÄÊÇKOH£¬Á½ÐÔµÄÇâÑõ»¯ÎïÖ»ÓÐAl£¨OH£©3£»
£¨5£©Òò¢ÝÊÇÂÁ£¬ËùÒÔÂÁÓëÇâÑõ»¯ÄÆÈÜÒº·´Ó¦ÒÔ¼°ÇâÑõ»¯ÂÁÓëÇâÑõ»¯ÄÆÈÜÒº·´Ó¦µÄ·½³Ìʽ·Ö±ðΪ£º2Al+2NaOH+2H2O=2NaAlO2+3H2¡ü£¬Al£¨OH£©3+OH-=AlO2-+2H2O£»
£¨6£©HxRÆøÌåÔÚ±ê×¼×´¿öϵÄÌå»ýÊÇ5.6L£¬ËµÃ÷ÎïÖʵÄÁ¿Îª5.6/22.4=0.25mol£¬ÓÖÒòÎªÆøÌåÖÊÁ¿Îª8.5g£¬ËùÒÔĦ¶ûÖÊÁ¿Îª8.5/0.25=34Òò´ËHxRµÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª34£¬RÔÚ¸ÃÇ⻯ÎïÖеÄÖÊÁ¿·ÖÊýΪ94%£¬ËùÒÔRµÄÔ×ÓÁ¿Îª34*0.94=31.96=32£¬ËùÒÔRÊÇÁò£¬Òò´ËHxRµÄ»¯Ñ§Ê½ÎªH2S£®
¹Ê´ð°¸Îª£º£¨1£©
£¨2£©Al£»
£¨3£©
£¨4£©HClO4£»KOH£»Al£¨OH£©3£»
£¨5£©2Al+2NaOH+2H2O=2NaAlO2+3H2¡ü£¬Al£¨OH£©3+OH-=AlO2-+2H2O£»
£¨6£©34£¬H2S£®
µãÆÀ£º±¾Ì⿼²éѧÉú¶ÔÔªËØÖÜÆÚ±íÖеÄһЩ±ä»¯¹æÂɵĹéÄɺÍ×ܽᣬ²¢Ñ§ÒÔÖÂÓã¬Í¬Ê±¿¼²éÁËÔªËØ¼°Æä»¯ºÏÎïµÄÓйØÖªÊ¶£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
| A¡¢XÔªËØ×î¶à¿ÉÐγÉÁùÖÖÑõ»¯Îï | B¡¢YÔªËØµÄ×î¸ß¼ÛÑõ»¯ÎïµÄË®»¯ÎïÊǺ¬ÑõËáÖÐËáÐÔ×îÇ¿µÄ | C¡¢XÔªËØµÄ·Ç½ðÊôÐÔ±ÈYÔªËØ·Ç½ðÊôÐÔÇ¿ | D¡¢ZºÍXÄÜÒÔ¹²¼Û¼ü½áºÏÐγÉÒ»ÖÖÎÞ»ú·Ç½ðÊô²ÄÁÏ |