°´ÒªÇóÌî¿Õ£º

  £¨1£©ÈôijҩƷÖÊÁ¿Ô¼Îª32g£¬ÓÃÍÐÅÌÌìƽ׼ȷ³ÆÆäÖÊÁ¿£¬ÈôÓáý±íʾÏòÓÒÅÌ·ÅÉÏíÀÂ룬Óáü±íʾ½«íÀÂëÈ¡Ï£¬ÔÚÏÂÁбí¸ñµÄ¿Õ¸ñÄÚ£¬ÓáýºÍ¡ü±íʾÏàÓ¦íÀÂëµÄ·ÅÉÏ»òÈ¡Ï¡£

50g

20g

20g

10g

5g

¡ý¡ü

¡ý

 £¨2£©ÅäÖÆ500mL 0.1mol¡¤L -1 Na2CO3ÈÜÒº£¬ÏÂͼ²Ù×÷¢ÚÖÐÓ¦¸ÃÌîдµÄÊý¾ÝΪ        £¬ÊµÑéʱÏÂͼËùʾ²Ù×÷µÄÏȺó˳ÐòΪ                               £¨Ìî±àºÅ£©¡£

                  

                       

 £¨3£©ÔÚÅäÖÆÒ»¶¨Á¿µÄÎïÖʵÄÁ¿Å¨¶ÈÈÜҺʱ£¬Óá°Æ«¸ß¡¢Æ«µÍ¡¢ÎÞÓ°Ï족±íʾÏÂÁвÙ×÷¶ÔËùÅäÈÜҺŨ¶ÈµÄÓ°Ïì¡£

   ¢Ù¶¨ÈÝʱ¸©ÊÓ£¬ËùÅäÖÆÈÜÒºµÄŨ¶È          ¡£

   ¢ÚÈÝÁ¿Æ¿Ï´µÓºóδ¸ÉÔËùÅäÖÆÈÜÒºµÄŨ¶È      ¡£

   ¢Û¶¨ÈÝÒ¡ÔȺó£¬ÓÐÉÙÁ¿ÈÜÒºÍâÁ÷£¬ËùÅäÖÆÈÜÒºµÄŨ¶È           ¡£

 £¨4£©Èô½«Ì¼ËáÄƹÌÌ壬·ÅÈëÂÈË®ÖУ¬ÏÖÏ󣺠                          £¬·´Ó¦µÄ·½³ÌʽΪ                              ¡£

£¨1£©

50g

20g

20g

10g

5g

¡ý¡ü

¡ý

¡ý¡ü

¡ý

¡ý¡ü

£¨2£©5.3 g£»¢Ú¢Ü¢Û¢Ý¢Ù¢Þ£»£¨3£©¢Ù Æ«¸ß£»¢Ú ÎÞÓ°Ï죻¢Û ÎÞÓ°Ï죻£¨4£©ÏÖÏó£ºÈÜÒºÖÐÓÐÆøÅÝð³ö£¨»òÉú³ÉÎÞÉ«ÎÞζµÄÆøÌ壩£»·½³Ìʽ£ºNa2CO3+2HCl =2NaCl+CO2¡ü+H2O¡£


½âÎö:

£¨1£©ÓÉÌâÒâÖª£¬30gΪíÀÂëµÄÖÊÁ¿£¬ÒѼÓ20gíÀÂë¡£ÔòÔÙ¼Ó10gíÀÂë¼´¿É£¬¶ø2gÒƶ¯ÓÎÂë¡££¨2£©m(Na2CO3)=0.5L ¡Á0.1mol¡¤L -1¡Á106g¡¤mol -1=5.3g£»£¨3£©¢Ù¸©ÊÓÔòÈÜÒºÌå»ý±äС£¬Å¨¶ÈÆ«¸ß£»¢Û¶¨ÈÝÒ¡ÔȺó£¬ÈÜÒºÒÑÅäÖÆÍ꣬ÓÐÉÙÁ¿ÈÜÒºÍâÁ÷¶ÔŨ¶ÈÎÞÓ°Ïì¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÔÚÏÂÁм¸ÖÖ¼òµ¥µÄÔ­µç³ØÖУ¬°´ÒªÇóÌî¿Õ£º
£¨1£©½«Ð¿¡¢Í­Óõ¼ÏßÁ¬½Óºó¹²Í¬½þÈë2mol/LµÄÏ¡ÁòËáÖУ¬Õý¼«µç¼«·´Ó¦Ê½Îª
2H++2e-=H2¡ü
2H++2e-=H2¡ü
£¬·¢Éú
»¹Ô­
»¹Ô­
·´Ó¦£¨ÌîÑõ»¯»ò»¹Ô­£©£»µç×Ó´Ó
п
п
¾­Íâµç·µ½
Í­
Í­
£¨Ìîп»òÍ­£©£¬ÈÜÒºÖÐH+Ïò
Õý¼«
Õý¼«
Òƶ¯£¨ÌîÕý¼«»ò¸º¼«£©£¬·ÅµçÒ»¶Îʱ¼äºó£¬Õý¼«ÇøpHÖµ
±ä´ó
±ä´ó
£¨Ìî±ä´ó¡¢±äС»ò²»±ä£©£®
£¨2£©½«Í­Æ¬ºÍÌúƬÓõ¼ÏßÁ¬½Óºó²åÈëÈýÂÈ»¯ÌúÈÜÒºÖУ¬Ò²ÓеçÁ÷ͨ¹ýµçÁ÷±í£®»­³ö¸Ãµç³Ø½á¹¹µÄʾÒâͼ£¬ÔÚͼÉϱê³öÕý¡¢¸º¼«£¬µç½âÖʺ͵ç×ÓÁ÷Ïò
д³ö¸º¼«µç¼«·´Ó¦Ê½Îª
Cu-2e-=Cu2+
Cu-2e-=Cu2+
£¬
×Ü·´Ó¦µÄÀë×Ó·½³ÌʽΪ
Cu+2Fe3+=Cu2++2Fe2+
Cu+2Fe3+=Cu2++2Fe2+
£®
£¨3£©½«ÌúƬºÍʯīµç¼«Óõ¼ÏßÁ¬½Óºó²åÈëÂÈ»¯ÄÆÈÜÒºÖУ¬Ò²ÓеçÁ÷ͨ¹ýµçÁ÷±í£¬Çëд³ö¸º¼«µç¼«·´Ó¦Ê½
2Fe-4e-=2Fe2+
2Fe-4e-=2Fe2+
£¬Õý¼«µç¼«·´Ó¦Ê½Îª
O2+4e-+2H2O=4OH-
O2+4e-+2H2O=4OH-
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

Çë°´ÒªÇóÌî¿Õ£º
£¨1£©Óû³ýÈ¥ÏõËá¼Ø¹ÌÌåÖлìÓеÄÉÙÁ¿ÂÈ»¯ÄƹÌÌ壬Ëù½øÐеÄʵÑé²Ù×÷ÒÀ´ÎÊÇ
Èܽâ
Èܽâ
¡¢Õô·¢Å¨Ëõ¡¢½µÎ½ᾧ¡¢
¹ýÂË
¹ýÂË
£®
£¨2£©Óû³ýÈ¥ÂÈ»¯ÄÆÈÜÒºÖлìÓеÄÁòËáÄÆ£¬ÒÀ´Î¼ÓÈëµÄÈÜҺΪ£¨ÌîÈÜÖʵĻ¯Ñ§Ê½£©
BaCl2
BaCl2
£®
£¨3£©³ýÈ¥»ìÈëNaClÈÜÒºÖÐÉÙÁ¿CuSO4ÔÓÖÊÑ¡ÓõÄÒ»ÖÖÊÔ¼ÁÊÇ
Ba£¨OH£©2
Ba£¨OH£©2
£¬ÓйØÀë×Ó·½³ÌʽΪ
Ba2++2OH-+SO42-+Cu2+=BaSO4¡ý+Cu£¨OH£©2¡ý
Ba2++2OH-+SO42-+Cu2+=BaSO4¡ý+Cu£¨OH£©2¡ý
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

Çë°´ÒªÇóÌî¿Õ£º
£¨1£©Ä³ÔªËØÔ­×ӵļ۵ç×Ó¹¹ÐÍΪ3d104s2£¬ËüÊôÓÚµÚ
4
4
ÖÜÆÚ£¬µÚ
¢òB
¢òB
×壬
ds
ds
ÇøÔªËØ£¬ÔªËØ·ûºÅÊÇ
Zn
Zn
£¬Ä³ÔªËØÔ­×ӵļ۵ç×Ó¹¹ÐÍΪ3d54s1£¬ËüÊôÓÚµÚËÄÖÜÆÚ£¬dÇøÔªËØ£¬ÔªËØ·ûºÅÊÇ
Cr
Cr
£®
£¨2£©D3+µÄ3dÑDzãΪ°ë³äÂú£¬Æä»ù̬ԭ×ӵĵç×ÓÅŲ¼Ê½Îª
[Ar]3d64s2
[Ar]3d64s2
£¬E+µÄK¡¢L¡¢M²ãÈ«³äÂú£¬EµÄÔªËØ·ûºÅÊÇ
Cu
Cu
£¬Æä»ù̬ԭ×ӵĵç×ÓÅŲ¼Ê½Îª
[Ar]3d104s1
[Ar]3d104s1

£¨3£©ÒÑÖªÏÂÁÐÔªËØÔÚÖÜÆÚ±íÖеÄλÖã¬Ð´³öËüÃÇ×îÍâ²ãµç×ÓµÄÅŲ¼Ê½ºÍÔªËØ·ûºÅ£º
¢ÙµÚ4ÖÜÆÚ¢ôB×å
Ti£¬3d64s2
Ti£¬3d64s2
£»
¢ÚµÚ5ÖÜÆÚ¢÷A×å
I£¬5s25p5
I£¬5s25p5
£®
£¨4£©¢ÙCaC2ÖÐC22-ÓëO22+»¥ÎªµÈµç×ÓÌ壬O22+µÄµç×Óʽ¿É±íʾΪ
 1mol O22+Öк¬ÓеĦмüÊýĿΪ
2
2
£®
¢ÚÒÒȲÓëÇâÇèËá·´Ó¦¿ÉµÃ±ûÏ©ë棨H2C=CH-C¡ÔN£©£®±ûÏ©ëæ·Ö×ÓÖÐ̼ԭ×Ó¹ìµÀÔÓ»¯ÀàÐÍÓÐ
spÔÓ»¯¡¢sp2 ÔÓ»¯
spÔÓ»¯¡¢sp2 ÔÓ»¯
£»·Ö×ÓÖд¦ÓÚͬһֱÏßÉϵÄÔ­×ÓÊýÄ¿×î¶àΪ
3
3
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

°´ÒªÇóÌî¿Õ£º
£¨1£©µÄϵͳÃüÃûÊÇ£º
2£¬6-¶þ¼×»ùÐÁÍé
2£¬6-¶þ¼×»ùÐÁÍé

£¨2£©µÄ¼üÏßʽÊÇ

£¨3£©3-¼×»ù-2-ÎìÏ©µÄ½á¹¹¼òʽÊÇ
CH3CH=C£¨CH3£©CH2CH3
CH3CH=C£¨CH3£©CH2CH3
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

Çë°´ÒªÇóÌî¿Õ£º
£¨1£©Æøֱ̬Á´ÍéÌþµÄ̼ԭ×ÓÊý¡Ü
 
£»
£¨2£©C4H9ClµÄͬ·ÖÒì¹¹Ìå¹²ÓÐ
 
ÖÖ£»
£¨3£©±½·¢ÉúÏõ»¯·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸