£¨1£©Ä³¶ÌÖÜÆÚÖ÷×åÔªËصÄÔ­×ÓM²ãÉÏÓÐÒ»¸ö°ë³äÂúµÄÑDz㣬ÕâÖÖÔ­×ÓµÄÖÊ×ÓÊýÊÇ
 
£¬Ð´³öËüµÄÍâΧµç×ÓÅŲ¼Í¼
 
£®
£¨2£©VIA×åµÄÎø£¨Se£©£®ÔÚ»¯ºÏÎïÖг£±íÏÖ³ö¶àÖÖÑõ»¯Ì¬£¬H2SeO4±ÈH2SeO3ËáÐÔ
 
£¨  ÌîÇ¿»òÈõ£©£¬H2SeµÄËáÐÔ±ÈH2S
 
£¨Ìî¡°Ç¿¡±»ò¡°Èõ¡±£©£®
£¨3£©¹ÌÌåAµÄ»¯Ñ§Ê½ÎªNH5£¬ËüµÄËùÓÐÔ­×ÓµÄ×îÍâ²ã¶¼·ûºÏÏàӦϡÓÐÆøÌåÔ­×ÓµÄ×îÍâµç×Ó²ã½á¹¹£¬ÔòÏÂÁÐÓйØ˵·¨ÖÐÕýÈ·µÄÊÇ
 

A£®NH5ÖмÈÓÐÀë×Ó¼üÓÖÓй²¼Û¼ü
B£®NH5µÄÈÛ¡¢·Ðµã¸ßÓÚNH3
C£®NH5¹ÌÌåͶÈëÉÙÁ¿Ë®ÖУ¬¿É²úÉúÁ½ÖÖÆøÌå
D£®0.1mol NH5Öк¬ÓÐ5mol N-H¼ü
£¨4£©ÓÃÇâ¼ü±íʾʽд³öHFÈÜÒºÖдæÔÚµÄËùÓÐÇâ¼ü
 
£®
¿¼µã£ºÔ­×ÓºËÍâµç×ÓÅŲ¼,ÔªËØÖÜÆÚÂɵÄ×÷ÓÃ,Àë×Ó»¯ºÏÎïµÄ½á¹¹ÌØÕ÷ÓëÐÔÖÊ,¹²¼Û¼üµÄÐγɼ°¹²¼Û¼üµÄÖ÷ÒªÀàÐÍ,º¬ÓÐÇâ¼üµÄÎïÖÊ
רÌ⣺ԭ×Ó×é³ÉÓë½á¹¹×¨Ìâ,»¯Ñ§¼üÓ뾧Ìå½á¹¹
·ÖÎö£º£¨1£©Ä³¶ÌÖÜÆÚÖ÷×åÔªËصÄÔ­×ÓM²ãÉÏÓÐÒ»¸ö°ë³äÂúµÄÑDz㣬ÕâÑùµÄÔ­×Óµç×ÓÅŲ¼Îª3s1»ò3s23p3£»
£¨2£©H2SeO4ºÍ H2SeO3Ïà±È½Ï£¬SeµÄ»¯ºÏ¼ÛÔ½¸ß£¬ËáÐÔԽǿ£¬·Ç½ðÊôÐÔԽǿµÄÔªËØ£¬ÆäÓëÇâÔªËصĽáºÏÄÜÁ¦Ô½Ç¿£¬ÔòÆäÇ⻯ÎïÔÚË®ÈÜÒºÖоÍÔ½Äѵç½â£¬ËáÐÔ¾ÍÔ½Èõ£»
£¨3£©AµÄ»¯Ñ§Ê½ÎªNH5£¬ËüµÄËùÓÐÔ­×ÓµÄ×îÍâ²ã¶¼·ûºÏÏàӦϡÓÐÆøÌåÔ­×ÓµÄ×îÍâµç×Ó²ã½á¹¹£¬ÔòAÖдæÔÚ笠ùÀë×ÓºÍH-£¬ÎªÀë×Ó»¯ºÏÎÒÔ´ËÀ´½â´ð£»
£¨4£©ÄÜÐγÉÇâ¼üµÄÔ­×ÓΪN¡¢O¡¢FºÍÇâÔ­×ÓÖ®¼ä£¬Çâ¼üÊÇ·Ö×ÓÖ®¼äµÄÇ¿Ï໥×÷ÓÃÁ¦£®
½â´ð£º ½â£º£¨1£©Ä³¶ÌÖÜÆÚÖ÷×åÔªËصÄÔ­×ÓM²ãÉÏÓÐÒ»¸ö°ë³äÂúµÄÑDz㣬ÕâÑùµÄÔ­×Óµç×ÓÅŲ¼Îª3s1»ò3s23p3£»
M²ãΪ3s1µÄÔ­×ÓºËÍâµç×ÓÅŲ¼Ê½Îª[Ne]3s1£¬ÖÊ×ÓÊýΪ11£¬ÎªNa£¬Ô­×ÓµÄÍâΧµç×ÓÅŲ¼Í¼Îª£¬
M²ãΪ3s23p3µÄÔ­×ÓºËÍâµç×ÓÅŲ¼Ê½Îª[Ne]3s23p3£¬ÖÊ×ÓÊýΪ15£¬ÎªP£¬Ô­×ÓµÄÍâΧµç×ÓÅŲ¼Í¼£¬
¹Ê´ð°¸Îª£º11»ò15£»»ò£»
£¨2£©Í¬Ò»ÔªËصĺ¬ÑõËáÖл¯ºÏ¼ÛÔ½¸ß£¬ËáÐÔԽǿ£¬Òò´ËH2SeO4±ÈH2SeO3ËáÐÔÇ¿£¬·Ç½ðÊôÐÔԽǿµÄÔªËØ£¬ÆäÓëÇâÔªËصĽáºÏÄÜÁ¦Ô½Ç¿£¬ÔòÆäÇ⻯ÎïÔÚË®ÈÜÒºÖоÍÔ½Äѵç½â£¬ËáÐÔ¾ÍÔ½Èõ£¬·Ç½ðÊôÐÔS£¾Se£¬ËùÒÔH2SeµÄËáÐÔ±ÈH2SÇ¿£¬
¹Ê´ð°¸Îª£ºÇ¿£»Ç¿£»
£¨3£©A£®NH5ÖдæÔÚ笠ùÀë×ÓºÍH-£¬ÎªÀë×Ó»¯ºÏÎ笠ùÀë×ÓÖк¬N-H¹²¼Û¼ü£¬¹ÊAÕýÈ·£»
B£®NH5ΪÀë×Ó¾§Ì壬°±ÆøΪ·Ö×Ó¾§Ì壬ÔòNH5µÄÈ۷еã¸ßÓÚNH3£¬¹ÊBÕýÈ·£»
C£®NH5¹ÌÌåͶÈëÉÙÁ¿Ë®ÖУ¬¿É²úÉú°±Æø¡¢ÇâÆøÁ½ÖÖÆøÌ壬¹ÊCÕýÈ·£»
D£®0.1mol NH5Öк¬ÓÐ0.4mol N-H¼ü£¬¹ÊD´íÎó£»
¹Ê´ð°¸Îª£ºABC£»
£¨4£©Çâ·úËáµÄË®ÈÜÒº£¬º¬ÓÐHF·Ö×ÓÒÔ¼°H2O·Ö×Ó£¬´æÔÚµÄËùÓÐÇâ¼üΪ£ºF-H¡­F¡¢F-H¡­O¡¢O-H¡­F¡¢O-H¡­O£¬
¹Ê´ð°¸Îª£ºF-H¡­F¡¢F-H¡­O¡¢O-H¡­F¡¢O-H¡­O£®
µãÆÀ£º±¾Ì⿼²éÁËÔªËصÄÐÔÖʵÄÍƶϺͻ¯Ñ§¼ü¶ÔÎïÖÊÐÔÖʵÄÓ°ÏìÒÔ¼°Çâ¼üµÄ±íʾ·½·¨£¬×ÛºÏÐÔ½ÌÇ¿£¬Îª¸ßƵ¿¼µã£¬ÄѶÈÖеȣ¬×¢ÒâÕÆÎÕͬһԪËصĺ¬ÑõËá¡¢Ç⻯ÎïËáÐÔÇ¿ÈõµÄ±È½Ï·½·¨£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÏÂÁÐÒÀ¾ÝÏà¹ØʵÑéµÃ³öµÄ½áÂÛÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A¡¢ÏòijÈÜÒºÖмÓÈëÏ¡ÑÎËᣬ²úÉúµÄÆøÌåͨÈë³ÎÇåʯ»ÒË®£¬Ê¯»ÒË®±ä»ë×Ç£¬¸ÃÈÜÒºÒ»¶¨ÊÇ̼ËáÑÎÈÜÒº
B¡¢Óò¬Ë¿ÕºÈ¡ÉÙÁ¿Ä³ÈÜÒº½øÐÐÑÕÉ«·´Ó¦£¬»ðÑæ³Ê»ÆÉ«£¬¸ÃÈÜÒºÒ»¶¨ÊÇÄÆÑÎÈÜÒº
C¡¢Al¡¢Fe¡¢Cu¶ÔÓ¦µÄÑõ»¯Îï¾ùÄÜÓëËá·´Ó¦Éú³ÉÑκÍË®£¬ÈýÖÖ½ðÊôµÄÑõ»¯Îï¾ùΪ¼îÐÔÑõ»¯Îï
D¡¢ÏòijÈÜÒºÖеμÓÂÈË®ºó£¬ÔٵμÓKSCNÈÜÒººóÈÜÒºÏÔѪºìÉ«£¬¸ÃÈÜÒºÖв»Ò»¶¨º¬Fe2+

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

°´ÒªÇóÍê³ÉÏÂÁÐÎÊÌ⣺
£¨1£©Ð´³ö»ù̬ÌúÔ­×ӵĵç×ÓÅŲ¼Ê½
 
£¬Mg2+µÄµç×ÓÅŲ¼Í¼
 
£®
£¨2£©ÏÂÁзÖ×ÓÈôÊÇÊÖÐÔ·Ö×Ó£¬ÇëÓá°*¡±±ê³öÆäÊÖÐÔ̼ԭ×Ó£®

£¨3£©Ö¸³öÅäºÏÎïK3[Co£¨CN£©6]ÖеÄÖÐÐÄÀë×Ó¡¢ÅäÌå¼°ÆäÅäλÊý£º
 
¡¢
 
¡¢
 
£®
£¨4£©ÅжÏBCl3·Ö×ӵĿռ乹ÐÍ¡¢ÖÐÐÄÔ­×ӳɼüʱ²ÉÈ¡µÄÔÓ»¯¹ìµÀÀàÐͼ°·Ö×ÓÖй²¼Û¼üµÄ¼ü½Ç
 
¡¢
 
¡¢
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÏÂÁÐÓйرí´ïÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A¡¢ÁòÀë×ӵĵç×ÓÅŲ¼Ê½£º1s22s22p63s23p4
B¡¢H2OµÄµç×Óʽ£º
C¡¢NÔ­×Ó×îÍâ²ãµç×ӵĹìµÀ±íʾʽ£º
D¡¢µÄÃû³Æ£º2-ÒÒ»ù±ûÍé

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÏÂÁм¸ÖÖÔªËصıí´ïʽ´íÎóµÄÊÇ£¨¡¡¡¡£©
A¡¢F-µÄµç×ÓÅŲ¼Í¼£º
B¡¢Na+µÄ½á¹¹Ê¾Òâͼ£º
C¡¢Mg2+µÄµç×ÓÅŲ¼Ê½£º1s22s22p6
D¡¢CrµÄ¼ò»¯µç×ÓÅŲ¼Ê½£º[Ar]3d44s2

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ºÍÄÊÔ­×ÓÓÐÏàͬµÄµç×Ó²ã½á¹¹µÄ΢Á£ÊÇ£¨¡¡¡¡£©
A¡¢He
B¡¢K+
C¡¢Cl-
D¡¢

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

µ¥ÖÊAÓë·ÛÄ©»¯ºÏÎïB×é³ÉµÄ»ìºÏÎïת»¯¹ØϵÈçͼ1Ëùʾ£º

»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©×é³Éµ¥ÖÊAµÄÔªËØÔÚÖÜÆÚ±íÖеÄλÖÃÊÇ
 
£¬¹¤ÒµÉÏÖƱ¸µ¥ÖÊAµÄ»¯Ñ§·½³ÌʽΪ
 
£®
£¨2£©»¯ºÏÎïBµÄµç×ÓʽΪ
 
£®
£¨3£©DÓëGÁ½ÈÜÒº»ìºÏºó·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ
 
£®
£¨4£©ÓÃÀë×Ó·½³Ìʽ±íʾ³£ÎÂÏÂDÈÜÒº³ÊÏÖËá¼îÐÔµÄÔ­Òò
 
£®
£¨5£©ÓÃͼ2ËùʾװÖÃÑо¿Óйص绯ѧµÄÎÊÌ⣨µç¼«¾ùΪ¶èÐԵ缫£¬¼×¡¢ÒÒ¡¢±ûÈý³ØÖÐÈÜÖʾù×ãÁ¿£©£¬±ÕºÏµç¼üʱ£¬¹Û²ìµ½µç¼«b±íÃæÓкìÉ«¹ÌÌåÎö³ö£®
¢Ùд³öEÆøÌå·¢ÉúµÄµç¼«·´Ó¦·½³Ìʽ
 
£®
¢Ú±û³ØÖÐΪ100g 8%µÄCÈÜÒº£¬µ±µç½âµ½ÈÜÖʵÄÖÊÁ¿·ÖÊýΪ12.5%ʱֹͣµç½â£¬Ôòµç½â¹ý³ÌÖУ¬Á½¼«Éú³ÉµÄÆøÌåÔÚ±ê×¼×´¿öϵÄÌå»ý¹²Îª
 
L£®
¢ÛÒ»¶Îʱ¼äºó£¬¶Ï¿ªK£¬¼ÓÈëÏÂÁÐÎïÖÊÄÜʹÒҳػָ´µ½·´Ó¦Ç°Å¨¶ÈµÄÊÇ
 
£¨ÌîÑ¡Ïî×Öĸ£©£®
A£®Cu      B£®CuO     C£®Cu£¨OH£©2    D£®Cu2£¨OH£©2CO3£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÏÂÁзÖ×Ó»òÀë×ÓÖУ¬ÖÐÐÄÔ­×Óº¬Óйµç×Ó¶ÔµÄÊÇ£¨¡¡¡¡£©
A¡¢H3O+
B¡¢SiH4
C¡¢PH3
D¡¢SO42-

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A¡¢Ò»¶¨Î¶ÈÏ£¬Ä³ÈÜÒºµÄpH£¼7£¬Ôò¸ÃÈÜÒº³ÊËáÐÔ
B¡¢ÔÚË®ÖмÓÈëÉÙÁ¿Ì¼ËáÄƹÌÌ彫ÒÖÖÆË®µÄµçÀë
C¡¢0.02mol?L-1CH3COOHÈÜÒººÍ0.01mol?L-1NaOHÈÜÒºµÈÌå»ý»ìºÏ£¬ÔòÈÜÒºÖУº2c£¨H+£©+c£¨CH3COOH£©=2 c£¨OH-£©+c£¨CH3COO-£©
D¡¢Å¨¶È¾ùΪ0.1mol/LµÄNH4ClÈÜÒººÍNH4HSO4ÈÜÒº£¬Ç°ÕßµÄc£¨NH4+£©´óÓÚºóÕß

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸