7£®ÅäÖÆÈÜÖÊÖÊÁ¿·ÖÊýÒ»¶¨µÄÂÈ»¯ÄÆÈÜÒº³£°´ÒÔϲÙ×÷˳Ðò½øÐУ®

Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÅäÖÆ80gÈÜÖÊÖÊÁ¿·ÖÊýΪ15%µÄNaClÈÜÒº£¬ÐèNaClµÄÖÊÁ¿Îª12g£¬ÐèË®µÄÖÊÁ¿Îª68g
£¨2£©ÓÃÉÏͼËùʾµÄÐòºÅ±íʾÕýÈ·ÅäÖƸÃÈÜÒºµÄ²Ù×÷˳ÐòΪCBDEA£®
£¨3£©ÓÃÍÐÅÌÌìƽ³ÆÁ¿ËùÐèµÄÂÈ»¯ÄÆʱ£¬·¢ÏÖÍÐÅÌÌìƽµÄÖ¸ÕëÆ«Ïò×óÅÌ£¬Ó¦B£®
A£®Ôö¼ÓÊÊÁ¿ÂÈ»¯ÄƹÌÌå                     B£®¼õÉÙÊÊÁ¿ÂÈ»¯ÄƹÌÌå
C£®µ÷½ÚÓÎÂë                               D£®Ìí¼ÓíÀÂë
£¨4£©Ëù³ÆÁ¿µÄ¹ÌÌåNaClº¬ÓÐÉÙÐíÄàɳ£¬ÔòËùÅäÖƵÄÈÜÒºÈÜÖÊÖÊÁ¿·ÖÊý£¼15%£¨Ì»ò£¼»ò=£©

·ÖÎö £¨1£©¸ù¾ÝÈÜÖÊÖÊÁ¿·ÖÊýµÄ¹«Ê½½øÐмÆË㣻
£¨2£©¸ù¾ÝÅäÖƸÃÈÜÒºµÄ²½ÖèÓУº¼ÆËã¡¢³ÆÁ¿¡¢Èܽ⡢װƿµÈ½øÐзÖÎö£»
£¨3£©¸ù¾ÝÍÐÅÌÌìƽ³ÆÁ¿Ò©Æ·µÄ·½·¨½øÐзÖÎö£»
£¨4£©¸ù¾ÝËù³ÆÁ¿µÄ¹ÌÌåNaClº¬ÓÐÉÙÐíÄàɳ£¬½«Ê¹ÂÈ»¯ÄÆÖÊÁ¿¼õÉÙ½øÐзÖÎö£®

½â´ð ½â£º£¨1£©ÅäÖÆ80gÈÜÖÊÖÊÁ¿·ÖÊýΪ15%µÄÂÈ»¯ÄÆÈÜÒºËùÐ裺ÂÈ»¯ÄƵÄÖÊÁ¿Îª£º80g¡Á15%=12g£¬Ë®µÄÖÊÁ¿ÊÇ£º80g-12g=68g£»
£¨2£©ÅäÖƸÃÈÜÒºµÄ²½ÖèÓУº¼ÆËã¡¢³ÆÁ¿¡¢Èܽ⡢װƿ£¬ËùÒÔÕýÈ·ÅäÖƸÃÈÜÒºµÄ²Ù×÷˳ÐòΪ£ºCBDEA£»
£¨3£©ÓÃÍÐÅÌÌìƽ³ÆÁ¿ËùÐèµÄÂÈ»¯ÄÆʱ£¬·¢ÏÖÍÐÅÌÌìƽµÄÖ¸ÕëÆ«Ïò×óÅÌ£¬ËµÃ÷¼ÓÈëµÄʳÑεÄÖÊÁ¿¹ý¶à£¬Ó¦¼õÉÙÊÊÁ¿ÂÈ»¯ÄƹÌÌ壬¹ÊÑ¡£ºB£»
£¨4£©Ëù³ÆÁ¿µÄ¹ÌÌåNaClº¬ÓÐÉÙÐíÄàɳ£¬½«Ê¹ÂÈ»¯ÄÆÖÊÁ¿¼õÉÙ£¬ÈܼÁÖÊÁ¿²»±ä£¬ËùÒÔËùÅäÖƵÄÈÜÒºÈÜÖÊÖÊÁ¿·ÖÊý£¼15%£®
¹Ê´ð°¸Îª£º£¨1£©12g£¬68g£»
£¨2£©CBDEA£»
£¨3£©B£»
£¨4£©£¼£®

µãÆÀ ±¾Ìâ½ÏΪ¼òµ¥£¬ÊìϤÈÜÖÊÖÊÁ¿·ÖÊýµÄ¹«Ê½ºÍÕýȷʹÓÃÍÐÅÌÌìƽÊǽâ´ð±¾ÌâµÄ»ù´¡£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

5£®Ä³»¯¹¤³§ÓûÒÔ´ÖÑÎΪԭÁÏ̽¾¿¹ØÓÚµç½â·¨ÖƱ¸ÇâÑõ»¯ÄƵĻ¯¹¤¹¤ÒÕ£¬ÊµÏÖ×ÊÔ´µÄ×ÛºÏÀûÓã¬ÏÖÄ£Ä⹤ÒÕÖÆ·¨£¬Éè¼ÆÁËÈçÏÂ΢Ð͹¤ÒÕÁ÷³Ì£®

²éÔÄ×ÊÁÏ£º
£¨1£©ÇâÆøºÍÂÈÆø»ìºÏÓöÃ÷»ð¼«Ò×·¢Éú±¬Õ¨£»
£¨2£©µç½â±¥ºÍÂÈ»¯ÄÆÈÜÒº¿ÉÒԵõ½ÂÈÆø£¨Cl2£©¡¢ÇâÆø¡¢ÇâÑõ»¯ÄÆÈýÖÖ²úÆ·£®
Çë»Ø´ðÏà¹ØÎÊÌ⣺
£¨1£©¸Ã¹¤ÒÕÁ÷³ÌÖвÙ×÷1ӦΪ¹ýÂË£»
£¨2£©¹¤ÒÕÁ÷³ÌÖвÙ×÷2µÄӦΪÕô·¢½á¾§£¨Ìî¡°Õô·¢½á¾§¡±»ò¡°½µÎ½ᾧ¡±£©£»
£¨3£©¸Ã¹¤ÒÕÁ÷³ÌÖпÉÒÔÑ­»·ÀûÓõÄÎïÖÊΪÇâÆø£»
£¨4£©µç½â±¥ºÍÂÈ»¯ÄÆÈÜҺӦֱͨÁ÷µç£¨Ìî¡°Ö±Á÷µç¡±»ò¡°½»Á÷µç¡±£©£»
£¨5£©2NaCl+2H2O$\frac{\underline{\;ͨµç\;}}{\;}$2NaOH+H2¡ü+Cl2¡ü
£¨6£©Çë½áºÏÄãÒÑѧ¹ýµÄ֪ʶ£¬ÇâÑõ»¯ÄƵÄÒ»ÖÖÓÃ;ÔìÖ½£¨´ð°¸ºÏÀí¼´¿É£©£»
£¨7£©ÓÐͬѧÈÏΪ²Ù×÷2¡¢²Ù×÷3²»ÐèÒª·Ö²½½øÐÐÒ²¿ÉÒԴﵽĿµÄ£¬ÄãÈÏΪËûµÄÀíÓÉÊÇʲô£¿¶¼ÊÇͨ¹ýÕô·¢ÈܼÁʵÏÖ£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

6£®ÎïÖÊÃû³Æ£¨»òË×Ãû£©Ó뻯ѧʽһÖµÄÊÇ£¨¡¡¡¡£©
A£®¿ÁÐÔÄÆ-Na2SO4B£®Ïûʯ»Ò-CaOC£®Ì¼ËáÄÆ-Na2CO3D£®¸É±ù--H2O

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

15£®ÏÖÓÐÈܽâ¶ÈÇúÏßÈçͼ£¬ÏÂÁÐÓйØ˵·¨ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®t2¡æʱa¡¢c Á½ÎïÖÊÈܽâ¶ÈÏàµÈ
B£®½«cµÄ²»±¥ºÍÈÜÒº½µÎ¿ÉÒÔת»¯Îª±¥ºÍÈÜÒº
C£®t1¡æʱa¡¢c±¥ºÍÈÜÒºµÄÈÜÖÊÖÊÁ¿·ÖÊýÏàµÈ
D£®½«t3¡æµÄa¡¢b¡¢c ÈýÖÖÎïÖʵı¥ºÍÈÜÒºÀäÈ´µ½t1 Ê±£¬aÎïÖÊÎö³ö×î¶à£¬ÎÞcÎïÖÊÎö³ö

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

2£®¸ù¾ÝÔªËØÖÜÆÚ±í»Ø´ðÒÔÏÂÎÊÌâ

£¨1£©10ºÅÔªËØ·ÖÀàÊÇÏ¡ÓÐÆøÌ壬16ºÅÔªËصÄÀë×Ó·ûºÅÊÇS2-£®
£¨2£©Ð´³ö12ºÅÔªËØÓë17ºÅÔªËØÐγɻ¯ºÏÎïµÄ»¯Ñ§Ê½MgCl2
£¨3£©´Ó±íÖÐÕÒ³öÒ»Ìõ¹æÂÉÔªËØÖÜÆÚ±íÖУ¬Ã¿¸öÖÜÆÚ¶¼ÊÇ°´Ô­×ÓÐòÊýµÝÔöµÄ˳Ðò½øÐÐÅÅÁеģ®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

12£®¸ù¾ÝÏÂÁÐʵÑéʾÒâͼ»Ø´ðÏà¹ØÎÊÌ⣺
 
£¨1£©¼×ʵÑéÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ3Fe+2O2$\frac{\underline{\;µãȼ\;}}{\;}$Fe3O4£®
£¨2£©ÒÒʵÑéÖУ¬ÈÈË®µÄ×÷ÓÃÊǸô¾øÑõÆøºÍ¼ÓÈÈ£®
£¨3£©±û×°ÖõIJ»×ãÖ®´¦ÊÇûÓÐβÆø´¦Àí×°Öã®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

19£®Æû³µ¼ÓÓÍÕ¾ÄÚÓ¦ÕÅÌùµÄͼ±êÊÇ£¨¡¡¡¡£©
A£®B£®C£®D£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

16£®ÏàÐÅͬѧÃÇÒѾ­³õ²½ÕÆÎÕÁËʵÑéÊÒÖÆÈ¡ÆøÌåµÄÓйØ֪ʶ£®Çë½áºÏͼʾ»Ø´ðÎÊÌ⣮

£¨1£©×°ÖÃB¾­³£ÓÃÀ´¸ø¹ÌÌå¼ÓÈÈ£¬Ð´³öͼÖбêʾµÄÒÇÆ÷Ãû³Æ£º¢Ù¾Æ¾«µÆ_£»
£¨2£©×¢ÉäÆ÷A¿ÉÓÃÓÚ¼ì²é×°ÖÃCµÄÆøÃÜÐÔ£¬²½ÖèÈçÏ£º
¢ÙÏò׶ÐÎÆ¿ÖмÓÈëÉÙÁ¿Ë®ÖÁ½þû³¤¾±Â©¶·µÄÏ¿ڴ¦£»
¢Ú½«×¢ÉäÆ÷AÁ¬½Óµ½×°ÖÃCµÄµ¼¹Ü¿Ú´¦£»
¢Û»ºÂýÏòÍâÀ­¶¯×¢ÉäÆ÷AµÄ»îÈû£¬¹Û²ìµ½³¤¾±Â©¶·µÄÏ¿ÚÓÐÆøÅÝð³ö£¬±íʾװÖÃCµÄÆøÃÜÐÔÁ¼ºÃ£®
£¨3£©ÊµÑéÊÒÖƶþÑõ»¯Ì¼µÄÔ­ÀíÊÇ£¨Óû¯Ñ§·½³Ìʽ±íʾ£©CaCO3+2HCl=CaCl2+H2O+CO2¡ü£»ÓÃ×°ÖÃCÖÆÈ¡¶þÑõ»¯Ì¼Ê±£¬ÈçÓÃ×¢ÉäÆ÷AÌæ»»³¤¾±Â©¶·£¬ÓŵãÊÇ¿ÉÒÔ¿ØÖÆ·´Ó¦µÄËÙÂÊ£»
£¨4£©ÊµÑéÊÒ·ñ£¨Ìî¡°ÊÇ¡±»ò¡°·ñ¡±£©ÓÃÅÅË®·¨ÊÕ¼¯¶þÑõ»¯Ì¼£»Èô½«G×°ÖÃÖÐÔ¤ÏÈ×°Âú±¥ºÍ̼ËáÇâÄÆÈÜÒº£¬ÓÃÅű¥ºÍ̼ËáÇâÄÆÈÜÒº¿ÉÒÔÊÕ¼¯CO2£¬ÔòÆøÌå´Ób¶Ë£¨Ìîa»òb£©½øÈ룮
£¨5£©ÈôÓÃ×°ÖÃH¶ÔCO2½øÐиÉÔϴÆøÆ¿ÖÐ×°µÄÒ©Æ·¿ÉÒÔÊÇŨÁòËᣮÇ뽫ͼH²¹»­ÍêÕû£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

17£®µÇɽÔ˶¯Ô±µÇɽʱËù´øµÄÄÜÔ´ÊÇÇ⻯¸Æ£¨CaH2£©£¬ËüÓëË®·¢Éú¾çÁÒ·´Ó¦Éú³ÉÇâÆøºÍÇâÑõ»¯¸Æ£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£ºCaH2+2H2O=Ca£¨OH£©2+2H2¡ü£®µ±µÇɽÔ˶¯Ô±ÐèÒªÈÈÁ¿Ê±£¬½«ÉÏÊö·´Ó¦ÖÐÉú³ÉµÄÇâÆøµãȼ£¬¸Ã¹ý³ÌÖÐʵÏÖÁËÄÜÁ¿×ª»¯£¬¼´»¯Ñ§ÄÜת»¯ÎªÈÈÄÜ£®

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸