ÌâÄ¿ÄÚÈÝ
ÏÂÁÐʵÑé²Ù×÷ÕýÈ·µÄÊÇ
A.
µãȼ¾Æ¾«µÆ B.
Á¬½ÓÒÇÆ÷
C.
ʹÓÃµÎ¹Ü D.
Ï¡ÊÍŨÁòËá
ÂÁÊÇÒ»ÖÖÓ¦Óù㷺µÄ½ðÊô¡£Ä³ÐËȤС×é¶Ô½ðÊôÂÁÕ¹¿ªÁËϵÁÐÑо¿¡£
I Ñо¿ÂÁµÄ»¯Ñ§ÐÔÖÊ
£¨1£©ÂÁÔÚ³£ÎÂÏ»áÓëÑõÆø·´Ó¦£¬±íÃæÐγÉÒ»²ãÖÂÃܵÄÑõ»¯Ä¤£¬»¯Ñ§·½³ÌʽΪ_________ ¡£ ʵÑéǰ£¬ÐèÒª¶ÔÂÁµÄ±íÃæÏȽøÐÐ______________£¨Ìî²Ù×÷£©´¦Àí¡£ÐËȤС×éͬѧ½«ÂÁ´¦ÀíºÃºó¼ô ³ÉÈô¸É´óСÏàͬµÄСƬ£¬ÓÃÓÚºóÐøÊµÑé¡£
£¨2£©ÂÁºÍËá¡¢¼î¡¢ÑÎÈÜÒºµÄ·´Ó¦
ʵÑé | ²Ù×÷ | ÏÖÏó | ½áÂÛ |
Ò» | ½«ÂÁƬ·ÅÈëÏ¡ ÑÎËáÖÐ | ______________£¬ÊԹܱäÌÌ | ÂÁÄÜÓëÑÎËá·¢Éú·´Ó¦£¬·´Ó¦·ÅÈÈ |
¶þ | ½«ÂÁƬ·ÅÈëÇâ Ñõ»¯ÄÆÈÜÒºÖÐ | ÂÁ±íÃæÓÐÆøÅݲúÉú£¬ÊԹܱä ÌÌ | ÂÁÄÜÓëÇâÑõ»¯ÄÆÈÜÒº·´Ó¦£¬·´Ó¦ ·ÅÈÈ |
Èý | ½«ÂÁ·ÅÈëÁòËá ÍÈÜÒºÖÐ | ÂÁ±íÃæÓкìÉ«ÎïÖʲúÉú£¬ÈÜ ÒºÖð½¥±ä³ÉÎÞÉ« | ½ðÊô»î¶¯ÐÔ£ºÂÁ_____Í |
ʵÑéÒ»µÄÏÖÏóΪ______________________¡£ÊµÑé¶þ²éÔÄ×ÊÁÏ£ºÂÁºÍÇâÑõ»¯ÄÆ¡¢Ë®·´Ó¦Éú³ÉÆ«ÂÁËáÄÆ£¨NaAlO2£©ºÍÇâÆø£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ______________________¡£ÊµÑéÈý±íÃ÷£¬ÂÁµÄ½ðÊô»î¶¯ÐÔ_____________£¨Ìî¡°´óÓÚ¡°»ò¡±Ð¡ÓÚ¡°£©Í¡£
£¨3£©Ð¡×éͬѧ½«ÊµÑéÒ»ÖеÄÏ¡ÑÎËá»»³ÉµÈÖÊÁ¿¡¢ÇâÀë×ÓŨ¶ÈÏàͬµÄÏ¡ÁòËᣬÏàͬʱ¼äÄÚ ·¢ÏÖÂÁƬ±íÃæ²úÉúµÄÆøÅݽÏÉÙ£¬·´Ó¦½ÏÂý¡£
¶Ô±ÈÉÏÊöÁ½×éʵÑé²úÉú²ÂÏë¡£ ²ÂÏë¢Ù£ºÏ¡ÑÎËáÖеÄÂÈÀë×Ó¶Ô·´Ó¦¿ÉÄÜÓдٽø×÷Óᣠ²ÂÏë¢Ú£º_____¡£
ΪÑéÖ¤²ÂÏë¢ÙÊÇ·ñºÏÀí£¬Ó¦ÔÚÂÁºÍÏ¡ÁòËáÖмÓÈë___________£¨Ìî×Öĸ£©£¬¹Û²ìÏÖÏó¡£
A£®Na2SO4 B£®Na2CO3 C£®NaCl
II ²â¶¨Ä³ÂÁÑùÆ·ÖнðÊôÂÁµÄÖÊÁ¿·ÖÊý
¡¾×ÊÁÏ1¡¿ÇâÑõ»¯ÂÁÄÜÓëÇâÑõ»¯ÄÆÈÜÒº·´Ó¦£¬µ«²»Ó백ˮ·´Ó¦¡£
¡¾×ÊÁÏ2¡¿AlCl3+3NaOH=Al(OH)3¡ý+3NaCl£»AlCl3+3NH3¡¤H2O=Al(OH)3¡ý+3NH4Cl¡£
£¨4£©Ð¡×éͬѧ³ÆÈ¡4.62gij½ðÊôÂÁÑùÆ·£¨ÑùÆ·ÖÐÔÓÖʽöΪÑõ»¯ÂÁ£©£¬ÖÃÓÚͼһÉÕÆ¿ÖУ¬¼ÓÈë×ãÁ¿Ï¡ÑÎËáÖÁÍêÈ«·´Ó¦¡£½«·´Ó¦ºóµÄÒºÌå·Ö³ÉÈÜÒº1ºÍÈÜÒº2Á½µÈ·Ý£¬Éè¼ÆÁ½ÖÖʵÑé·½°¸£¨ÈçͼÈý£©£¬Í¨¹ý³ÁµíÖÊÁ¿²â¶¨ÑùÆ·ÖнðÊôÂÁµÄÖÊÁ¿·ÖÊý¡£
![]()
¢ÙÑ¡ÔñÕýÈ·µÄÒ»ÖÖʵÑé·½°¸£¬¼ÆËãÑùÆ·ÖнðÊôÂÁµÄÖÊÁ¿·ÖÊý£¨Çëд³ö¼ÆËã¹ý³Ì£©.
¢ÚС×éͬѧ×éºÏͼһºÍͼ¶þ×°ÖòâÇâÆøÌå»ý£¬ÉÕÆ¿Öз´Ó¦Í£Ö¹¼´¶Á³öÁ¿Í²ÄÚË®µÄÌå»ý£¬ ¼ÆËãºó·¢ÏÖ½ðÊôÂÁµÄÖÊÁ¿·ÖÊýÆ«´ó£¬¿ÉÄܵÄÔÒòÊÇ?
4Al+3O2=2Al2O3 ´òÄ¥£¨ºÏÀí½Ô¿É£© ÂÁ±íÃæÓÐÆøÅݲúÉú 2Al+2 NaOH + 2H2O =2NaAlO2 + 3H2¡ü ´óÓÚ Ï¡ÁòËáÖеÄÁòËá¸ùÀë×Ó¶Ô·´Ó¦¿ÉÄÜÓÐÒÖÖÆ×÷Óà C ¢ÙËã³öÿ·ÝÖÐÂÁÔªËØµÄÖÊÁ¿ 2.07g£¨»òËã³öÕû¸öÑùÆ·ÖÐÂÁÔªËØµÄÖÊÁ¿ 4.14g£©£¬ Ëã³öÿ·ÝÖÐÑõ»¯ÂÁÖÊÁ¿ 0.51g£¨»òËã³öÕû¸öÑùÆ·ÖÐÑõ»¯ÂÁµÄÖÊÁ¿ 1.02g£©£¬ Ëã³öÿ·ÝÖÐÂÁµÄÖÊÁ¿ 1.8g£¬ÂÁµÄÖÊÁ¿·ÖÊýΪ 77...