ÐÔ¡£
£¨2£©ÈÜÒºµÄµ¼µçÄÜÁ¦´óСÖ÷ÒªÊÇÓÉÀë×ÓŨ¶È´óС¾ö¶¨µÄ£¬ÀûÓá°ÒºÌåµ¼µçÐÔµÄʵÑé×°Öá±£¨ÈçÓÒͼËùʾ£©×÷ÏÂÁÐʵÑ飬µÆÅÝÄܳöÏÖÓÉ
ÁÁ¡ª°µ¡ªÏ¨Ãð¡ªÁÁÏÖÏóµÄÊÇ A£® ÏòBaCl2ÈÜÒºÖеÎÈëÏ¡H2SO4 B£® ÏòBa(OH)2ÈÜÒºÖеÎÈëÏ¡H2SO4 C£®ÏòNaOHÈÜÒºÖеÎÈëÑÎËá
д³öËùÑ¡ÏîµÄÀë×Ó·´Ó¦·½³Ìʽ_________________________________________ £¨ÈôÉÏÌâÑ¡´í£¬¸ÃÏî²»µÃ·Ö£©¡£
£¨3£©ÒÑÖªÂÈÆøÓëÈÈÉÕ¼îÈÜÒº·´Ó¦ÈçÏ£º3Cl2+6NaOH= NaClO3+ 5NaCl+3 H2O£¬
¢ÙÓÃË«ÏßÇÅ·¨±ê³öµç×Ó×ªÒÆµÄ·½ÏòºÍÊýÄ¿¡£
¢Úµ±ÓÐ0.1molµç×Ó×ªÒÆÊ±£¬²Î¼Ó·´Ó¦µÄCl2ÔÚ±ê×¼×´¿öϵÄÌå»ý_________mL¡£
24£®£¨12·Ö£©ÒÑÖªA¡¢B¡¢C¡¢IÊǽðÊôµ¥ÖÊ£¬¼×¡¢ÒÒ¡¢±ûΪ³£¼ûÆøÌ壬ÆäÖÐBÊǵؿÇÖк¬Á¿×î¶àµÄ½ðÊôÔªËØ¡£ËüÃÇÖ®¼äÄÜ·¢ÉúÈçÏ·´Ó¦£¨Í¼ÖÐÓÐЩ·´Ó¦µÄ²úÎïºÍ·´Ó¦µÄÌõ¼þûÓÐÈ«²¿±ê³ö£©¡£
Çë¸ù¾ÝÒÔÉÏÐÅÏ¢»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Ð´³öÏÂÁÐÎïÖʵĻ¯Ñ§Ê½£º C ¡¢H_______________¡£ £¨2£©Ð´³ö·´Ó¦¢ÛµÄ»¯Ñ§·½³Ìʽ£º__________________________________£¬
д³ö·´Ó¦¢ÝµÄÀë×Ó·½³Ìʽ£º__________________________________£¬ д³ö·´Ó¦¢ßµÄÀë×Ó·½³Ìʽ£º__________________________________¡£ £¨3£©ÊµÑéÊÒ¼ìÑéÎïÖÊGÖÐÑôÀë×ӵij£ÓÃÊÔ¼ÁÊÇ£º___________________¡£
25£®£¨10·Ö£©º£ÑóÔ̲Ø×ŷḻµÄ×ÊÔ´£¬ÒÔº£Ë®ÎªÔÁÏ¿ÉÒÔÖÆÈ¡Na¡¢NaOH ¡¢Mg¡¢Cl2 ¡¢Br2µÈ¡£ÏÂͼÊÇij»¯¹¤³§¶Ôº£Ë®×ÊÔ´×ÛºÏÀûÓõÄʾÒâͼ¡£
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Á÷³ÌͼÖвÙ×÷aµÄÃû³ÆÎª_______________¡£
£¨2£©ÔÁÏ´ÖÑÎÖг£º¬ÓÐÄàɳºÍCa2¡¢Mg2¡¢Fe3¡¢SO24µÈÔÓÖÊ£¬±ØÐë¾«ÖÆºó²ÅÄܹ©µç½âʹÓ᣾«ÖÆÊ±£¬
£«
£«
£«
£
´ÖÑÎÈÜÓÚË®¹ýÂ˺󣬻¹Òª¼ÓÈëµÄÊÔ¼Á·Ö±ðΪ¢ÙNa2CO3¡¢¢ÚHCl(ÑÎËá)¡¢¢ÛBa(OH)2£¬Õâ3ÖÖÊÔ¼ÁÌí¼ÓµÄºÏÀí˳ÐòÊÇ____________(ÌîÐòºÅ)¡£
£¨3£©Ã¾ÊÇÒ»ÖÖÓÃ;ºÜ¹ãµÄ½ðÊô²ÄÁÏ£¬Ä¿Ç°ÊÀ½çÉÏ60%µÄþ´Óº£Ë®ÖÐÌáÈ¡¡£ µç½âËùµÃµÄþÕôÆøÐèÔÚÌØ¶¨µÄ»·¾³ÖÐÀäÈ´£¬²»ÄÜÓÃCO2 ×÷ÀäÈ´¼ÁµÄÔÒòÊÇ____________________________________£¨Óû¯Ñ§·½³Ìʽ±íʾ£©¡£ £¨4£©ÉÏÊöÁ÷³ÌûÓÐÉæ¼°µ½µÄËÄ´ó»ù±¾·´Ó¦ÀàÐÍÊÇ________________ A »¯ºÏ·´Ó¦ B ·Ö½â·´Ó¦ C Öû»·´Ó¦ D ¸´·Ö½â·´Ó¦
£¨5£©Ôڸû¯¹¤³§ÖУ¬º£Ë®ÌáÈ¡ÂÈ»¯ÄƺóµÄĸҺ¾¹ýÌáÈ¡ÂÈ»¯Ã¾ºóÓÖÐγÉÁËеÄĸҺ£¬ÏòÐÂĸҺÖмÓÈëÂÈÆø£¬ÓÖÖÆÈ¡ÁËÖØÒªµÄ»¯¹¤ÔÁÏäåµ¥ÖÊ£¬Éú³Éäåµ¥ÖÊ·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ________________________________ ¡£ 26£®£¨23·Ö£©
£¨1£©ÈôÓÃʵÑé¢ôÀ´Ñé֤̼ËáÄÆºÍ̼ËáÇâÄÆµÄÈÈÎȶ¨ÐÔ£¬ÔòÊÔ¹ÜBÖÐ×°ÈëµÄ¹ÌÌå×îºÃÊÇ __________£¬ÏÖÏóÊÇ______________________________________¡£
£¨2£©ÈôÓÃʵÑéÀ´¼ø±ð̼ËáÄÆºÍ̼ËáÇâÄÆÁ½ÖÖ¹ÌÌ壬ÄܴﵽʵÑéÄ¿µÄµÄÊÇ£¨Ìî×°ÖÃÐòºÅ£©_____________ ¡£ £¨3£©Îª²â¶¨Na2CO3ºÍNaHCO3»ìºÏÎïÖÐNa2CO3ÖÊÁ¿·ÖÊý£¬²ÉÓÃʵÑé¢ó½øÐÐʵÑ飬ȡa¿Ë»ìºÏÎï³ä·Ö¼ÓÈÈ£¬×îºó¼õÖØb¿Ë£¬Ôò̼ËáÄÆµÄÖÊÁ¿·ÖÊýΪ £¨Óú¬ÓÐa¡¢bµÄÊýѧʽ±íʾ£©¡£
II£®£¨14·Ö£©¸ÃÑо¿ÐÔѧϰС×éΪÁË̽¾¿ÔÚʵÑéÊÒÖÆ±¸Cl2µÄ¹ý³ÌÖÐÓÐË®ÕôÆøºÍHCl»Ó·¢³öÀ´£¬Í¬Ê±Ö¤Ã÷ÂÈÆøµÄijЩÐÔÖÊ£¬¼×ͬѧÉè¼ÆÁËÈçͼËùʾµÄʵÑé×°Öã¨Ö§³ÅÓõÄÌú¼Ų̈ʡÂÔ£©¡£ ÒÑÖª£ºCl2Ò×ÈÜÓÚCCl4¶øHCl²»ÈÜÓÚCCl4
°´ÒªÇ󻨴ðÎÊÌ⣺
£¨1£©Ð´³ö×°ÖÃAÖÆÈ¡Cl2µÄ»¯Ñ§·½³Ìʽ_______________________________¡£ £¨2£©×°ÖÃBÖеÄÊÔ¼ÁÊÇ_______________£¬ÏÖÏóÊÇ_____________________¡£ £¨3£©×°ÖÃDºÍEÖгöÏֵIJ»Í¬ÏÖÏó˵Ã÷µÄÎÊÌâÊÇ_______________________¡£ £¨4£©×°ÖÃFµÄ×÷ÓÃÊÇ______________________________________________¡£
£¨5£©ÒÒͬѧÈÏΪ¼×ͬѧµÄʵÑéÓÐȱÏÝ£¬²»ÄÜÈ·±£×îÖÕͨÈëAgNO3ÈÜÒºÖÐµÄÆøÌåÖ»ÓÐÒ»ÖÖ£¬ÎªÁËÈ·±£ÊµÑé½áÂ۵Ŀɿ¿ÐÔ£¬Ö¤Ã÷×îÖÕͨÈëAgNO3ÈÜÒºµÄÆøÌåÖ»ÓÐÒ»ÖÖ£¬ÒÒͬѧÌá³öÔÚFºÍDÖ®¼äÔÙÁ¬½ÓÒ»¸öÏ´ÆøÆ¿×°Öã¬Ôò×°ÖÃÖÐÓ¦·ÅÈë_______________________£¨ÌîдÊÔ¼Á»òÓÃÆ·Ãû³Æ£©£¬¸ÄÕýºóÖ¤Ã÷ÓÐHCl»Ó·¢³öÀ´µÄÏÖÏóÊÇ____________________________________________________¡£
1 D 12 A 2 C 13 B 3 A 14 B 4 B 15 B 5 D 16 C 6 B 17 D 7 B 18 C 8 C 19 B 9 C 20 C 10 B 21 B 11 D 22 A ¶þ¡¢Ìî¿ÕÌ⣨56·Ö£¬Ã¿¿Õ2·Ö£¬·½³ÌʽδÅ䯽¿Û1·Ö£¬ÌØÊâÁí×¢£© 23¡¢£¨11·Ö£©
£¨1£©Al(SO4)2 £½ +AlO2+2SO4£¬Îü¸½ÐÔ£¨1·Ö£© £¨2£©B£¬Ba+2OH+2 H+SO4£½Ba SO4 ¡ý+ 2 H2O £¨3£©ÂÔ£¬1344mL
2+
£
+
2£
+
£
2£
24¡¢£¨12·Ö£©£¨1£©Fe£¬Fe(OH)3
£¨2£©2Al+2NaOH+2H2O£½2Na AlO2 +3H2 ¡ü £¬
2Fe + Cl2£½2Fe + 2Cl
3+
2+
2+
3+
£
2Fe + Cu £½ Cu + 2Fe £¨3£©SCNÈÜÒº 25¡¢£¨10·Ö£©
£¨1£©Õô·¢½á¾§ £¨2£©¢Û¢Ù¢Ú£¬ £¨3£© µãȼ
2Mg+ CO2£½2MgO + C£¨4£© C
£¨5£©2Br+ Cl2 £½ Br 2 + 2Cl 26¡¢£¨23·Ö£©
I£®£¨¸÷3·Ö£¬¹²9·Ö£©
£¨1£© NaHCO3£¨1·Ö£©£¬×ó±ß³ÎÇåʯ»ÒË®±ä»ë×Ç£¬Óұ߳ÎÇåʯ»ÒË®²»±ä»ë×Ç£¨2·Ö£©¡£
£
£
2+
£¨2£© II¡¢III¡¢IV,£¨Ò»¸ö1·Ö£¬´íÑ¡1¸öµ¹¿Û1·Ö£¬¹²3·Ö£© £¨3£©(84b-53a)/31a£¨3·Ö£© II£®£¨Ã¿¿Õ¸÷2·Ö£¬¹²14·Ö£© £¨1£©MnO2+4HCl£¨Å¨£©
MnCl2 + Cl2¡ü+2H2O
£¨2£©ÎÞË®CuSO4 £¬°×É«¹ÌÌå±äΪÀ¶É«
£¨3£©ÂÈÆøÃ»ÓÐÆ¯°×ÐÔ£¬ÂÈË®ÓÐÆ¯°×ÐÔ £¨4£©³ýÈ¥ÂÈÆø£¬·ÀÖ¹¶ÔHClÆøÌå¼ìÑéµÄ¸ÉÈÅ
£¨5£©µí·ÛIÈÜÒº£¬µí·ÛIÈÜÒº²»±äÀ¶£¬GÖвúÉú°×É«³Áµí¡£