ÌâÄ¿ÄÚÈÝ

Ìá´¿º¬ÉÙÁ¿ÄàɳµÄ´ÖÑΣ¬Ò»°ã¾­¹ýϲÙ×÷Á÷³Ì£º
£¨1£©²Ù×÷¢ÜÖбØÐëÓõ½µÄÒ»ÖÖÒÇÆ÷ÊÇ
C
C
£¨ÌîÐòºÅ£©£®
A£®¾Æ¾«µÆ         B£®Á¿Í²         C£®ÉÕ±­         D£®ÊÔ¹Ü
£¨2£©²Ù×÷¢ÜÖÐÐèÒª½«Ô²ÐÎÂËÖ½ÕÛµþ´¦Àí£¬ÏÂÁÐͼʾÖв»¸Ã³öÏÖµÄÇéÐÎÊÇ
D
D
£¨ÌîÐòºÅ£©£®
£¨3£©²Ù×÷¢ÝÖгýÓõ½Èý½Å¼Ü¡¢¾Æ¾«µÆ¡¢²£Á§°ô¡¢ÛáÛöǯÍ⣬»¹ÐèÒªÓõ½
Õô·¢Ãó
Õô·¢Ãó
µÈÒÇÆ÷£®¸Ã²Ù×÷ÖÐÈÝÒ×Ôì³ÉʳÑιÌÌå·É½¦£¬Îª¾¡Á¿¼õÉٷɽ¦£¬¿É²ÉÈ¡ÓÃ
ABC
ABC
µÈ´ëÊ©£®
A  ²£Á§°ô²»¶Ï½Á°è    B À´»ØÒƶ¯¾Æ¾«µÆ¼ÓÈÈ    C À´»ØÒƶ¯Õô·¢Ãó¼ÓÈÈ
£¨4£©ÊµÑé½áÊøºó³ÆÁ¿»ñµÃµÄ¾«ÑΣ¬²¢¼ÆË㾫ÑεÄÖƵÃÂÊ£¬·¢ÏÖÖƵÃÂʽϵͣ¬Æä¿ÉÄÜÔ­ÒòÊÇ
ABD
ABD
£¨ÌîÐòºÅ£©£®
A£®Ê³ÑÎûÓÐÈ«²¿Èܽ⼴¹ýÂË    B£®Õô·¢Ê±Ê³Ñηɽ¦¾çÁÒ
C£®Õô·¢ºó£¬ËùµÃ¾«Ñκܳ±Êª    D£®Æ÷ÃóÉÏÕ´Óеľ«ÑÎûȫ²¿×ªÒƵ½³ÆÁ¿Ö½ÉÏ
£¨5£©ÓÃÖؽᾧ·¨½«´ÖÑÎÌá´¿»ù±¾²½Ö裻
½«ËùµÃ¾§ÌåÈܽ⣬Ȼºó¹ýÂË£¬½«ËùµÃÂËÒº½øÐÐÕô·¢µÄ¾§Ìå
½«ËùµÃ¾§ÌåÈܽ⣬Ȼºó¹ýÂË£¬½«ËùµÃÂËÒº½øÐÐÕô·¢µÄ¾§Ìå
£®
·ÖÎö£º£¨1£©¸ù¾ÝÌâÖÐËù¸øµÄ×°ÖõÄ×÷ÓýøÐзÖÎö£¬
£¨2£©¸ù¾ÝͼʾÖеÄÂËÖ½µÄÕÛµþ´¦Àí½øÐзÖÎö£¬
£¨3£©¸ù¾ÝÂÈ»¯ÄÆÈÜÒºµÄÕô·¢²Ù×÷ÐèÒªµÄÒÇÆ÷½øÐзÖÎö£¬
¸ù¾Ý Õô·¢ÃóµÄÊÜÈȾùÔÈÇé¿ö½øÐзÖÎö£¬
£¨4£©¿ÉÒÔ¸ù¾ÝÖÊÁ¿·ÖÊýµÄ±ÈÀýʽ½øÐзÖÎö£¬¾«ÑεÄÖÊÁ¿ÉÙÁË»òÕß´ÖÑεÄÖÊÁ¿´óÁË£®
½â´ð£º½â£º£¨1£©ÓɲÙ×÷¹ý³Ì¿ÉÒÔÖªµÀ²Ù×÷¢ÜΪ¹ýÂ˲Ù×÷£¬ËùÒÔ·ÖÎöËù¸øµÄÒÇÆ÷¿ÉÒÔÖªµÀÉÕ±­ÊDZز»¿ÉÉٵģ¬¹ÊÑ¡C£»
£¨2£©A¡¢B¡¢C ÊÇÂËÖ½µÄÕýÈ·µÄÕÛµþ·½·¨£¬DÖеÄÂËÖ½£¬½Ó¿Ú´¦³öÏÖ·ì϶£¬ËùÒÔ²»¿ÉÄܳöÏÖÕâÖÖ×´¿ö£¬¹ÊÑ¡D£¬
£¨3£©²Ù×÷¢ÝΪ¶ÔʳÑÎÈÜÒº½øÐÐÕô·¢£¬ËùÒÔ³ýÁËËù¸øµÄÒÇÆ÷£¬»¹ÐèҪʢ·ÅÒºÌåµÄ¼ÓÈÈ×°ÖÃÕô·¢Ãó£¬ÎªÁËʹҺÌåÊÜÈȾùÔȶø·ÀÖ¹ÒºµÎµÄ·É½¦£¬ËùÒԼȿÉÒÔ½Á°è£¬Ò²¿ÉÒÔͨ¹ýÒƶ¯¾Æ¾«µÆ»òÊÇÒƶ¯Õô·¢ÃóÀ´Ê¹ÈÜÒºÊÜÈȸü¾ùÔÈ£»
£¨4£©´ÖÑεIJúÂʵÍ˵Ã÷ÔÚÌᴿʱ£¬ËðʧÁËÂÈ»¯ÄÆ£¬
A¡¢Ê³ÑÎûÓÐÍêÈ«Èܽ⣬»áʹ¾«ÑεÄÖÊÁ¿¼õÉÙ£¬¹ÊAÕýÈ·£¬
B¡¢Ê³Ñηɽ¦Á˵¼ÖÂÁ˾«ÑεļõÉÙ£¬¹ÊBÕýÈ·£¬
C¡¢¾«Ñκܳ±Ôö´óÁ˾«ÑεÄÖÊÁ¿£¬»áʹÖÊÁ¿·ÖÊýÔö´ó£¬¹ÊC´íÎó£¬
D¡¢Õô·¢ÃóÖеľ«ÑÎûÓÐÈ«²¿×ªÒƵ½³ÆÁ¿Ö½ÉÏ£¬»áʹ¾«ÑεÄÖÊÁ¿¼õС£¬¹ÊDÕýÈ·£¬
¹ÊÑ¡ABD£®
£¨5£©ÓÃÖؽᾧ·¨½«´ÖÑÎÌá´¿ÈÔÈ»ÐèÒª¾­¹ýÈܽ⣬¹ýÂË¡¢Õô·¢µÄÕâÒ»¹ý³Ì£¬¹ÊÆä²½ÖèΪ£º½«ËùµÃ¾§ÌåÈܽ⣬Ȼºó¹ýÂË£¬½«ËùµÃÂËÒº½øÐÐÕô·¢µÃµ½¾§Ì壮
¹Ê´ð°¸Îª£º£¨1£©C£»  
£¨2£©D£»
£¨3£©A B C£»
£¨4£©A B D£»
£¨5£©½«ËùµÃ¾§ÌåÈܽ⣬Ȼºó¹ýÂË£¬½«ËùµÃÂËÒº½øÐÐÕô·¢µÄ¾§Ì壮
µãÆÀ£º±¾ÌâÖ÷Òª¿¼²éÂÈ»¯ÄÆÓë´ÖÑÎÌá´¿µÄ²½Öè¼°×¢ÒâÊÂÏѧ»á̽¾¿ÊµÑéÊý¾Ý´¦Àí»òÕßÎó²î·ÖÎöµÄ·½·¨¼¼ÇÉ£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
ÂÈ»¯ÄÆÊÇÉú»î±ØÐèÆ·£¬Ò²ÊÇÖØÒªµÄ»¯¹¤Ô­ÁÏ£®Ìá´¿º¬ÉÙÁ¿ÄàɳµÄ´ÖÑΣ¬Ò»°ã¾­¹ýÒÔϲÙ×÷Á÷³Ì£º

£¨1£©²Ù×÷¢ÙÖбØÐëÓõ½µÄÒ»ÖÖÒÇÆ÷ÊÇ
A
A
£¨ÌîÐòºÅ£©£®
A£®Ñв§       B£®Á¿Í²         C£®ÉÕ±­            D£®ÊÔ¹Ü
£¨2£©²Ù×÷¢ÜÖÐÐèÒª½«Ô²ÐÎÂËÖ½ÕÛµþ´¦Àí£¬ÏÂÁÐͼʾÖв»¸Ã³öÏÖµÄÇéÐÎÊÇ
D
D
£¨ÌîÐòºÅ£©£®

£¨3£©²Ù×÷¢ÝÖгýÓõ½Èý½Å¼Ü¡¢¾Æ¾«µÆ¡¢²£Á§°ô¡¢ÛáÛöǯÍ⣬»¹ÐèÒªÓõ½
Õô·¢Ãó
Õô·¢Ãó
µÈÒÇÆ÷£®¸Ã²Ù×÷ÖÐÈÝÒ×Ôì³ÉʳÑιÌÌå·É½¦£¬Îª¾¡Á¿¼õÉٷɽ¦£¬³ýÁ¬Ðø½Á°èÍ⻹¿É²ÉÈ¡
À´»ØÒƶ¯¾Æ¾«µÆ
À´»ØÒƶ¯¾Æ¾«µÆ

µæʯÃÞÍø
µæʯÃÞÍø
µÈ´ëÊ©£®
£¨4£©Èç¹û¹ýÂ˺óµÃµ½µÄÂËÒºÈÔÈ»»ë×Ç£¬Çëд³öÔì³ÉÂËÒº»ë×ǵĿÉÄÜÔ­Òò£º
ÒºÃæ¸ßÓÚÂËÖ½±ßÔµ
ÒºÃæ¸ßÓÚÂËÖ½±ßÔµ

»òÂËÖ½ÆÆËð
»òÂËÖ½ÆÆËð
 £¨Ð´Ò»Ìõ£©£®
£¨5£©ÊµÑé½áÊøºó³ÆÁ¿»ñµÃµÄ¾«ÑΣ¬²¢¼ÆË㾫ÑεÄÖƵÃÂÊ£¬·¢ÏÖÖƵÃÂʽϵͣ¬Æä¿ÉÄÜÔ­ÒòÊÇ
ABD
ABD
£¨ÌîÐòºÅ£©£®
A£®Ê³ÑÎûÓÐÈ«²¿Èܽ⼴¹ýÂË          B£®Õô·¢Ê±Ê³Ñηɽ¦¾çÁÒ
C£®Õô·¢ºó£¬ËùµÃ¾«Ñκܳ±Êª          D£®Æ÷ÃóÉÏÕ´Óеľ«ÑÎûȫ²¿×ªÒƵ½³ÆÁ¿Ö½ÉÏ£®
£¨2012?¹ÄÂ¥Çøһģ£©ÂÈ»¯ÄÆÊÇÉú»î±ØÐèÆ·£¬Ò²ÊÇÖØÒªµÄ»¯¹¤Ô­ÁÏ£®Ìá´¿º¬ÉÙÁ¿ÄàɳµÄ´ÖÑΣ¬Ò»°ã¾­¹ýÒÔϲÙ×÷Á÷³Ì£º

£¨1£©²Ù×÷¢ÙÖбØÐëÓõ½µÄÒ»ÖÖÒÇÆ÷ÊÇ
A
A
£¨ÌîÐòºÅ£©£®
A£®Ñв§       B£®Á¿Í²       C£®ÉÕ±­       D£®ÊÔ¹Ü
£¨2£©Ð¡¸ÕÍê³É¸ÃʵÑéµÄ²¿·Ö²Ù×÷¹ý³ÌÈçͼËùʾ£¬ÆäÖÐÓÐÃ÷ÏÔ´íÎóµÄÊÇ£®£¨Ìî×ÖĸÐòºÅ£©

£¨3£©²Ù×÷¢ÝÖгýÓõ½Ìú¼Ų̈£¨´øÌúȦ£©¡¢¾Æ¾«µÆ¡¢²£Á§°ô¡¢ÛáÛöǯÍ⣬»¹ÐèÒªÓõ½
Õô·¢Ãó
Õô·¢Ãó
µÈÒÇÆ÷£®¸Ã²Ù×÷ÖÐÈÝÒ×Ôì³ÉʳÑιÌÌå·É½¦£¬Îª¾¡Á¿¼õÉٷɽ¦£¬³ýÁ¬Ðø½Á°èÍ⻹¿É²ÉÈ¡
À´»ØÒƶ¯¾Æ¾«µÆ
À´»ØÒƶ¯¾Æ¾«µÆ
µÈ´ëÊ©£®
£¨4£©ÊµÑé½áÊøºó³ÆÁ¿»ñµÃµÄ¾«ÑΣ¬²¢¼ÆË㾫ÑεÄÖƵÃÂÊ£¬·¢ÏÖÖƵÃÂʽϵͣ¬Æä¿ÉÄÜÔ­ÒòÊÇ
ABD
ABD
£¨ÌîÐòºÅ£©£®
A£®Ê³ÑÎûÓÐÈ«²¿Èܽ⼴¹ýÂË    B£®Õô·¢Ê±Ê³Ñηɽ¦¾çÁÒ
C£®Õô·¢ºó£¬ËùµÃ¾«Ñκܳ±Êª    D£®Æ÷ÃóÉÏÕ´Óеľ«ÑÎûȫ²¿×ªÒƵ½³ÆÁ¿Ö½ÉÏ£®
£¨5£©Ð¡¸Õ²éÔÄÏà¹Ø×ÊÁϵÃÖª£º´ÖÑÎÖгýº¬ÄàɳµÈ²»ÈÜÐÔÔÓÖÊÍ⣬»¹º¬ÓÐÉÙÁ¿µÄMgCl2¡¢CaCl2µÈ¿ÉÈÜÐÔÔÓÖÊ£»ÎªÁ˵õ½½Ï´¿¾»µÄÂÈ»¯ÄÆ£¬Ð¡¸Õ½«ÉÏÊöÁ÷³ÌͼÖвÙ×÷¢ÝµÃµ½µÄ¡°¾«ÑΡ±ÓÖ×÷ÁËÈçÏ´¦Àí£¨¼Ù¶¨ÔÓÖÊÖ»ÓÐMgCl2¡¢CaCl2Á½ÖÖ£©£º

¢ÙÌṩµÄÊÔ¼Á£ºNa2CO3ÈÜÒº¡¢K2CO3ÈÜÒº¡¢NaOHÈÜÒº¡¢KOHÈÜÒº¡¢±¥ºÍNaClÈÜÒº£®´ÓÌṩµÄÊÔ¼ÁÖÐÑ¡³öaËù´ú±íµÄÊÔ¼ÁÊÇ
NaOHÈÜÒº
NaOHÈÜÒº
¡¢
Na2CO3ÈÜÒº
Na2CO3ÈÜÒº

¢ÚÔÚÂËÒºÖмÓÑÎËáµÄ×÷ÓÃÊÇ
HCl+NaOH=NaCl+H2O
HCl+NaOH=NaCl+H2O
¡¢
2HCl+Na2CO3=2NaCl+CO2¡ü+H2O
2HCl+Na2CO3=2NaCl+CO2¡ü+H2O
£¨Óû¯Ñ§·½³Ìʽ±íʾ£©£®

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø