ÌâÄ¿ÄÚÈÝ

ÒÑÖª¹¤ÒµÊ³Ñκ¬Ca2+£¬Mg2+£¬µÈÔÓÖÊÀë×Ó£¬ÒªÏ뾫Á¶³É¾«ÑΣ®Çë»Ø´ðÏÂÁÐÎÊÌ⣺

(1)ΪÁËÓÐЧ³ýÈ¥Ca2+£¬Mg2+£¬£¬¼ÓÈëÊÔ¼ÁµÄºÏÀí˳ÐòΪ£º________£®

A£®ÏȼÓNaOH£¬ºó¼ÓNa2CO3£¬ÔÙ¼Óº¬Ba2+µÄÊÔ¼Á

B£®ÏȼÓNaOH£¬ºó¼Óº¬Ba2+µÄÊÔ¼Á£¬ÔÙ¼ÓNa2CO3

C£®ÏȼӺ¬Ba2+µÄÊÔ¼Á£¬ºó¼ÓNaOH£¬ÔÙ¼ÓNa2CO3

(2)³ýÈ¥£¬±ØÐëÌí¼Óº¬Ba2+µÄÊÔ¼Á£¬¸ÃÊÔ¼Á¿ÉÒÔÊÇ

[¡¡¡¡]

A£®Ba(OH)2
B£®Ba(NO3)2
C£®BaCl2

(3)¾«Öƹý³ÌÖз¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ£º________£¬________£¬________£®

´ð°¸£º
½âÎö£º

¡¡¡¡(1)BC

¡¡¡¡(2)AC

¡¡¡¡(3)Ba2+£«£½BaSO4¡ý,Ca2+£«£½CaCO3¡ý,Mg2+£«2OH£­£½Mg(OH)2¡ý

¡¡¡¡µ¼½â£ºÒª³ýNaClÖеÄCa2+£¬Mg2+£¬£¬¾ÍÊÇÒª½«ÆäÒÔ³ÁµíµÄÐÎʽ³ýÈ¥£¬ÇÒ²»Ó°ÏìÆäËûÀë×Ó(¼´²»ÒýÈëÔÓÖÊ)£®Ca2+Óóý£¬Mg2+ÓÃOH£­³ý£¬ÓÃBa2+³ý£®


Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
£¨2013?½­Î÷¶þÄ££©ÒÔʳÑÎΪԭÁϽøÐÐÉú²ú²¢×ÛºÏÀûÓõÄijЩ¹ý³ÌÈçͼËùʾ£®

£¨1£©³ýÈ¥´ÖÑÎÖеÄCa2+¡¢Mg2+ºÍSO4-2Àë×Ó£¬¼ÓÈëÏÂÁгÁµí¼ÁµÄ˳ÐòÊÇ£¨ÌîÐòºÅ£©
cab»òcba»òbca
cab»òcba»òbca
£®
a£®Na2CO3         b£®NaOH        c£®BaCl2
£¨2£©½«ÂËÒºµÄpHµ÷ÖÁËáÐÔ³ýÈ¥µÄÀë×ÓÊÇ
CO32-ºÍOH-
CO32-ºÍOH-
£®
£¨3£©µç½âÂÈ»¯ÄÆÏ¡ÈÜÒº¿ÉÖÆ±¸¡°84Ïû¶¾Òº¡±£¬Í¨µçʱÂÈÆø±»ÈÜÒºÍêÈ«ÎüÊÕ£¬ÈôËùµÃÏû¶¾Òº½öº¬Ò»ÖÖÈÜÖÊ£¬Ð´³öÏàÓ¦µÄ»¯Ñ§·½³Ìʽ£º
NaCl+H2O
 Í¨µç 
.
 
NaClO+H2¡ü
NaCl+H2O
 Í¨µç 
.
 
NaClO+H2¡ü
£®
£¨4£©ÈôÏò·ÖÀë³öNaHCO3¾§ÌåºóµÄĸҺÖмÓÈë¹ýÁ¿Éúʯ»Ò£¬Ôò¿É»ñµÃÒ»ÖÖ¿ÉÒÔÑ­»·Ê¹ÓõÄÎïÖÊ£¬Æä»¯Ñ§Ê½ÊÇ
NH3
NH3
£®
£¨5£©Ä³Ì½¾¿»î¶¯Ð¡×齫¶þÑõ»¯Ì¼ÆøÌåͨÈ뺬°±µÄ±¥ºÍʳÑÎË®ÖÐÖÆ±¸Ì¼ËáÇâÄÆ£¬ÊµÑé×°ÖÃÈçͼËùʾ£¨Í¼Öмг֡¢¹Ì¶¨ÓõÄÒÇÆ÷δ»­³ö£©£®

ÊԻشðÏÂÁÐÓйØÎÊÌ⣺
¢ÙÒÒ×°ÖÃÖеÄÊÔ¼ÁÊÇ
±¥ºÍ̼ËáÇâÄÆÈÜÒº
±¥ºÍ̼ËáÇâÄÆÈÜÒº
£»
¢Ú¶¡×°ÖÃÖÐÏ¡ÁòËáµÄ×÷ÓÃÊÇ
ÎüÊÕδ·´Ó¦µÄ°±Æø
ÎüÊÕδ·´Ó¦µÄ°±Æø
£»
¢ÛʵÑé½áÊøºó£¬·ÖÀë³öNaHCO3 ¾§ÌåµÄ²Ù×÷ÊÇ
¹ýÂË
¹ýÂË
£¨Ìî·ÖÀë²Ù×÷µÄÃû³Æ£©£®
£¨6£©´¿¼îÔÚÉú²úÉú»îÖÐÓй㷺µÄÓ¦Óã®
¢Ù´¿¼î¿ÉÓÃÓÚ³ýÔį̂ÓÍÎÛ£®ÆäÔ­ÒòÊÇ£¨½áºÏÀë×Ó·½³Ìʽ±íÊö£©
̼Ëá¸ùÀë×ÓË®½âÏÔ¼îÐÔ£¬CO32-+H2O?HCO3-+OH-£¬ÓÍÎÛÔÚ¼îÐÔÌõ¼þÏÂË®½â¶ø±»³ýÈ¥
̼Ëá¸ùÀë×ÓË®½âÏÔ¼îÐÔ£¬CO32-+H2O?HCO3-+OH-£¬ÓÍÎÛÔÚ¼îÐÔÌõ¼þÏÂË®½â¶ø±»³ýÈ¥
£®
¢Ú¹¤ÒµÉÏ£¬¿ÉÒÔÓô¿¼î´úÌæÉÕ¼îÉú²úijЩ»¯¹¤²úÆ·£®ÈçÓñ¥ºÍ´¿¼îÈÜÒºÓëCl2·´Ó¦ÖÆÈ¡ÓÐЧ³É·ÖΪNaClOµÄÏû¶¾Òº£¬Æä·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ
2CO32-+Cl2+H2O=2HCO3-+Cl-+ClO-
2CO32-+Cl2+H2O=2HCO3-+Cl-+ClO-
£®£¨ÒÑ֪̼ËáµÄËáÐÔÇ¿ÓÚ´ÎÂÈËᣩ£®

¶þÑõ»¯ÂÈ£¨ClO2£©ÊÇÒ»ÖÖ¸ßЧ¡¢¹ãÆ×¡¢°²È«µÄɱ¾ú¡¢Ïû¶¾¼Á¡£
£¨1£©ÂÈ»¯ÄƵç½â·¨ÊÇÒ»ÖÖ¿É¿¿µÄ¹¤ÒµÉú²úClO2·½·¨¡£
¢ÙÓÃÓÚµç½âµÄʳÑÎË®ÐèÏȳýÈ¥ÆäÖеÄCa2+¡¢Mg2+¡¢SO42-µÈÔÓÖÊ¡£Æä´Î³ýÔÓ²Ù×÷ʱ£¬Íù´ÖÑÎË®ÖÐÏȼÓÈë¹ýÁ¿µÄ________£¨Ìѧʽ£©£¬ÖÁ³Áµí²»ÔÙ²úÉúºó£¬ÔÙ¼ÓÈë¹ýÁ¿µÄNa2CO3ºÍNaOH£¬³ä·Ö·´Ó¦ºó½«³ÁµíÒ»²¢ÂËÈ¥¡£¾­¼ì²â·¢ÏÖÂËÒºÖÐÈÔº¬ÓÐÒ»¶¨Á¿µÄSO42-£¬ÆäÔ­ÒòÊÇ___________¡¾ÒÑÖª£ºKsp(BaSO4)= 1.1 ¡Á10-10 Ksp(BaCO3)= 5.1 ¡Á10-9¡¿

¢Ú¸Ã·¨¹¤ÒÕÔ­ÀíÈçÓÒ¡£Æä¹ý³ÌÊǽ«Ê³ÑÎË®ÔÚÌØ¶¨Ìõ¼þϵç½âµÃµ½µÄÂÈËáÄÆ£¨NaClO3£©ÓëÑÎËá·´Ó¦Éú³ÉClO2¡£
¹¤ÒÕÖпÉÒÔÀûÓõĵ¥ÖÊÓÐ____________£¨Ìѧʽ£©£¬·¢ÉúÆ÷ÖÐÉú³ÉClO2µÄ»¯Ñ§·½³ÌʽΪ___________¡£
£¨2£©ÏËÎ¬ËØ»¹Ô­·¨ÖÆClO2ÊÇÒ»ÖÖз½·¨£¬ÆäÔ­ÀíÊÇ£ºÏËÎ¬ËØË®½âµÃµ½µÄ×îÖÕ²úÎïDÓëNaClO3·´Ó¦Éú³ÉClO2¡£Íê³É·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º
¡õ £¨D£© +24NaClO3+12H2SO4=¡õClO2¡ü+¡õCO2¡ü+18H2O+¡õ_________
£¨3£©ClO2ºÍCl2¾ùÄܽ«µç¶Æ·ÏË®ÖеÄCN-Ñõ»¯ÎªÎÞ¶¾µÄÎïÖÊ£¬×ÔÉí±»»¹Ô­ÎªCl-¡£´¦Àíº¬CN-ÏàͬÁ¿µÃµç¶Æ·ÏË®£¬ËùÐèCl2µÄÎïÖʵÄÁ¿ÊÇClO2µÄ_______±¶

ÂȼҵÖÐÓÃÀë×Ó½»»»Ä¤·¨µç½âÖÆ¼îµÄÖ÷ÒªÉú²úÁ÷³ÌʾÒâͼÈçÏ£º

ÒÀ¾ÝÉÏͼÍê³ÉÏÂÁÐÌî¿Õ£º

(1)ÓëµçÔ´Õý¼«ÏàÁ¬µÄµç¼«ÉÏËù·¢Éú·´Ó¦µÄµç¼«·´Ó¦Ê½Îª____________________£»ÓëµçÔ´¸º¼«ÏàÁ¬µÄµç¼«¸½½ü£¬ÈÜÒºµÄpH________(Ìî¡°²»±ä¡±¡¢¡°Éý¸ß¡±»ò¡°½µµÍ¡±)¡£

(2)¹¤ÒµÊ³Ñκ¬Ca2£«¡¢Mg2£«¡¢Fe3£«µÈÔÓÖÊ£¬¾«Öƹý³ÌÖз¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ_________________________________________________________________________¡£

(3)Èç¹û´ÖÑÎÖк¬Á¿½Ï¸ß£¬±ØÐëÌí¼Ó±µÊÔ¼Á³ýÈ¥£¬¸Ã±µÊÔ¼Á¿ÉÒÔÊÇ________(Ìî×Öĸ´úºÅ)¡£

a£®Ba(OH)2¡¡¡¡ b£®Ba(NO3)2¡¡¡¡ c£®BaCl2

(4)ΪÓÐЧ³ýÈ¥Ca2£«¡¢Mg2£«¡¢£¬¼ÓÈëÊÔ¼ÁµÄºÏÀí˳ÐòΪ________(Ìî×Öĸ´úºÅ)¡£

a£®ÏȼÓNaOH£¬ºó¼ÓNa2CO3£¬ÔÙ¼Ó±µÊÔ¼Á

b£®ÏȼÓNaOH£¬ºó¼Ó±µÊÔ¼Á£¬ÔÙ¼ÓNa2CO3

c£®ÏȼӱµÊÔ¼Á£¬ºó¼ÓNaOH£¬ÔÙ¼ÓNa2CO3

(5)ÍÑÑι¤ÐòÖÐÀûÓÃNaOHºÍNaClÔÚÈܽâ¶ÈÉϵIJîÒ죬ͨ¹ý______¡¢ÀäÈ´¡¢______(Ìîд²Ù×÷Ãû³Æ)³ýÈ¥NaCl¡£

(6)ÓÉͼʾ¿ÉÖªÓÃÀë×Ó½»»»Ä¤·¨µç½âÖÆ¼î¹¤ÒÕÖÐ________²úÆ·¿ÉÑ­»·Ê¹Óá£

(7)ÒÑÖªNaClÔÚ60 ¡æµÄÈܽâ¶ÈΪ37.1 g£¬ÏÖµç½â60 ¡æ¾«ÖƱ¥ºÍʳÑÎË®1371 g£¬¾­·ÖÎö£¬µç½âºóÈÜÒºÃܶÈΪ1.37 g¡¤cm£­3£¬ÆäÖк¬ÓÐ20 g NaCl£¬Ôòµç½âºóNaOHµÄÎïÖʵÄÁ¿Å¨¶ÈΪ________mol¡¤L£­1¡£

 

ÒÔʳÑÎΪԭÁϽøÐÐÉú²ú²¢×ÛºÏÀûÓõÄijЩ¹ý³ÌÈçÏÂͼËùʾ¡£

£¨1£©³ýÈ¥´ÖÑÎÖеÄCa2+¡¢Mg2+ºÍSOÀë×Ó£¬¼ÓÈëÏÂÁгÁµí¼ÁµÄ˳ÐòÊÇ£¨ÌîÐòºÅ£©      ¡£

a£®Na2CO3         b£®NaOH        c£®BaCl2  

£¨2£©½«ÂËÒºµÄpHµ÷ÖÁËáÐÔ³ýÈ¥µÄÀë×ÓÊÇ            ¡£

£¨3£©µç½âÂÈ»¯ÄÆÏ¡ÈÜÒº¿ÉÖÆ±¸¡°84Ïû¶¾Òº¡±£¬Í¨µçʱÂÈÆø±»ÈÜÒºÍêÈ«ÎüÊÕ£¬ÈôËùµÃÏû¶¾Òº½öº¬Ò»ÖÖÈÜÖÊ£¬Ð´³öÏàÓ¦µÄ»¯Ñ§·½³Ìʽ£º___________________________¡£

£¨4£©ÈôÏò·ÖÀë³öNaHCO3¾§ÌåºóµÄĸҺÖмÓÈë¹ýÁ¿Éúʯ»Ò£¬Ôò¿É»ñµÃÒ»ÖÖ¿ÉÒÔÑ­»·Ê¹ÓõÄÎïÖÊ£¬Æä»¯Ñ§Ê½ÊÇ            ¡£

£¨5£©Ä³Ì½¾¿»î¶¯Ð¡×齫¶þÑõ»¯Ì¼ÆøÌåͨÈ뺬°±µÄ±¥ºÍʳÑÎË®ÖÐÖÆ±¸Ì¼ËáÇâÄÆ£¬ÊµÑé×°ÖÃÈçÏÂͼËùʾ£¨Í¼Öмг֡¢¹Ì¶¨ÓõÄÒÇÆ÷δ»­³ö£©¡£

ÊԻشðÏÂÁÐÓйØÎÊÌ⣺

¢ÙÒÒ×°ÖÃÖеÄÊÔ¼ÁÊÇ             £»

¢Ú¶¡×°ÖÃÖÐÏ¡ÁòËáµÄ×÷ÓÃÊÇ                                   £»

¢ÛʵÑé½áÊøºó£¬·ÖÀë³öNaHCO3 ¾§ÌåµÄ²Ù×÷ÊÇ            £¨Ìî·ÖÀë²Ù×÷µÄÃû³Æ£©¡£

£¨6£©´¿¼îÔÚÉú²úÉú»îÖÐÓй㷺µÄÓ¦Óá£

¢Ù ´¿¼î¿ÉÓÃÓÚ³ýÔį̂ÓÍÎÛ¡£ÆäÔ­ÒòÊÇ£¨½áºÏÀë×Ó·½³Ìʽ±íÊö£©            ¡£

¢Ú ¹¤ÒµÉÏ£¬¿ÉÒÔÓô¿¼î´úÌæÉÕ¼îÉú²úijЩ»¯¹¤²úÆ·¡£ÈçÓñ¥ºÍ´¿¼îÈÜÒºÓëCl2·´Ó¦ÖÆÈ¡ÓÐЧ³É·ÖΪNaClOµÄÏû¶¾Òº£¬Æä·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ            ¡££¨ÒÑ֪̼ËáµÄËáÐÔÇ¿ÓÚ´ÎÂÈËᣩ¡£

 

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

¾«Ó¢¼Ò½ÌÍø