ÈçÏÂͼ£¬AÊÇÖÆÈ¡äå±½µÄʵÑé×°Öã¬B¡¢CÊǸĽøºóµÄ×°Öá£Çë×Ðϸ·ÖÎö£¬¶Ô±ÈÈý¸ö×°Ö㬻شðÒÔÏÂÎÊÌ⣺
¡¡
(1)д³öÈý¸ö×°ÖÃÖÐËù¹²Í¬·¢ÉúµÄÁ½¸ö·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º
________________________________________________________________________¡¢
________________________________________________________________________£»
д³öBµÄÊÔ¹ÜÖÐËù·¢Éú·´Ó¦µÄ»¯Ñ§·½³Ìʽ
________________________________________________________________________
________________________________________________________________________¡£
(2)×°ÖÃAºÍC¾ù²ÉÓÃÁ˳¤²£Á§µ¼¹Ü£¬Æä×÷ÓÃÊÇ
________________________________________________________________________
________________________________________________________________________¡£
(3)ÔÚ°´×°ÖÃB¡¢C×°ºÃÒÇÆ÷¼°Ò©Æ·ºóҪʹ·´Ó¦¿ªÊ¼£¬Ó¦¶Ô×°ÖÃB½øÐеIJÙ×÷ÊÇ________________________________________________________________________
________________________________________________________________________£»
Ó¦¶Ô×°ÖÃC½øÐеIJÙ×÷ÊÇ________________________________________
________________________________________________________________________¡£
(4)BÖвÉÓÃÁËË«ÇòÏ´Æø¹ÜÎüÊÕ×°Öã¬Æä×÷ÓÃÊÇ____________________________
________________________________________________________________________£¬
·´Ó¦ºóË«ÇòÏ´Æø¹ÜÖпÉÄܳöÏÖµÄÏÖÏóÊÇ__________________________¡£
(5)B×°ÖôæÔÚÁ½¸öÃ÷ÏÔµÄȱµã£¬Ê¹ÊµÑéµÄЧ¹û²»ºÃ»ò²»ÄÜÕý³£½øÐС£ÕâÁ½¸öȱµãÊÇ________________________________________________________________________
________________________________________________________________________¡£
(1)2Fe£«3Br2===2FeBr3
(2)µ¼Æø¡¢ÀäÄý
(3)Ðýת·ÖҺ©¶·µÄ»îÈû£¬Ê¹äåºÍ±½µÄ»ìºÏÒºµÎµ½Ìú·ÛÉÏ¡¡ÍÐÆðÈíÏ𽺴üʹÌú·ÛÂäÈëäåºÍ±½×é³ÉµÄ»ìºÏÒºÖÐ
(4)ÎüÊÕËæHBrÒݳöµÄ±½ÕôÆø¡¢Br2¡¡CCl4ÓÉÎÞÉ«±ä³É³ÈÉ«
(5)ËæHBrÒݳöµÄäåÕôÆøºÍ±½ÕôÆø²»ÄÜ»ØÁ÷µ½·´Ó¦Æ÷ÖУ¬ÔÁÏÀûÓÃÂʵͣ»ÓÉÓÚµ¼¹Ü²åÈëAgNO3ÈÜÒºÖжøÒײúÉúµ¹Îü
½âÎö¡¡(1)±½ÔÚFeBr3×÷ÓÃÏ£¬¿ÉÓëÒºäå·¢ÉúÈ¡´ú·´Ó¦Éú³É£¬Í¬Ê±Éú³ÉHBr£¬HBrÓëAgNO3ÈÜÒºÉú³ÉAgBrdz»ÆÉ«³Áµí£¬2Fe£«3Br2===2FeBr3£¬
(2)Òò·´Ó¦·ÅÈÈ£¬±½¡¢äåÒ×»Ó·¢£¬Èô½øÈë¿ÕÆøÖлáÔì³ÉÎÛȾ£¬Òò´ËA¡¢C×°ÖÃÖоù²ÉÓÃÁ˳¤²£Á§µ¼¹Ü£¬Æðµ¼³öHBr£¬¼æÆðÀäÄýÆ÷µÄ×÷Óá£
(3)BÖÐÐýת·ÖҺ©¶·µÄ»îÈû£¬Ê¹äåºÍ±½µÄ»ìºÏÒºµÎµ½Ìú·ÛÉÏ£¬CÖÐÍÐÆðÈíÏ𽺴üʹÌú·ÛÂäÈëäåºÍ±½×é³ÉµÄ»ìºÏÒºÖС£
(4)ÎüÊÕ·´Ó¦ÖÐËæHBrÒݳöµÄBr2ºÍ±½ÕôÆø£¬ÓÉÓÚCCl4ÖÐÈܽâÁËä壬CCl4ÓÉÎÞÉ«±ä³É³ÈÉ«¡£
(5)ËæHBrÒݳöµÄäåÕôÆøºÍ±½ÕôÆø²»ÄÜ»ØÁ÷µ½·´Ó¦Æ÷ÖУ¬ÔÁÏÀûÓÃÂʵͣ»ÓÉÓÚµ¼¹Ü²åÈëAgNO3ÈÜÒºÖжøÒײúÉúµ¹Îü¡£
Ä꼶 | ¸ßÖÐ¿Î³Ì | Ä꼶 | ³õÖÐ¿Î³Ì |
¸ßÒ» | ¸ßÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ | ³õÒ» | ³õÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ |
¸ß¶þ | ¸ß¶þÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õ¶þ | ³õ¶þÃâ·Ñ¿Î³ÌÍƼö£¡ |
¸ßÈý | ¸ßÈýÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õÈý | ³õÈýÃâ·Ñ¿Î³ÌÍƼö£¡ |
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
ÏÂÁÐʵÑéµÄÇ°ºó£¬ÍµÄÖÊÁ¿Ò»¶¨·¢Éú±ä»¯µÄÊÇ £¨ £©
A£®ÍË¿ÔÚ¿ÕÆøÖÐ×ÆÉÕºóÁ¢¼´²åÈëÒÒ´¼ B£®ÍƬ·ÅÈËËữµÄÏõËáÄÆÈÜÒº
C£®ÍºÍÌúµÄ»ìºÏÎï·ÅÈËÏ¡ÏõËá D£®Í¡¢Ð¿¡¢Ï¡ÁòËá¹¹³ÉµÄÔµç³Ø·Åµç
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¼×±½ÊÇúµÄ×ÛºÏÀûÓõõ½µÄ²úÎïÖ®Ò»£¬Æä½á¹¹¼òʽΪ¡£ÊԻشðÏÂÁÐÎÊÌ⣺
(1)±½Óë¼×±½µÄÏ໥¹ØϵΪ________¡£
A£®Í¬·ÖÒì¹¹Ìå B£®Í¬Î»ËØ
C£®Í¬ËØÒìÐÎÌå D£®Í¬ÏµÎï
(2)¼×±½È¼ÉÕʱµÄÏÖÏóΪ____________________________________________£¬
1 mol ¼×±½ÍêȫȼÉÕÏûºÄÑõÆøµÄÎïÖʵÄÁ¿Îª______________________¡£
(3)¼×±½±½»·ÉϵÄÒ»ÂÈ´úÎïÓÐ________ÖÖ¡£
(4)ÒÑÖª¾ßÓнṹµÄÎïÖʿɱ»ËáÐÔ¸ßÃÌËá¼ØÈÜÒºÑõ»¯¡£Çø·Ö±½ºÍ¼×±½µÄ·½·¨ÊÇ_____________________________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
ÏÂÁйØÓÚ±½µÄÐÔÖʵÄÐðÊöÖУ¬²»ÕýÈ·µÄÊÇ(¡¡¡¡)
A£®±½ÊÇÎÞÉ«´øÓÐÌØÊâÆøζµÄÒºÌå
B£®³£ÎÂϱ½ÊÇÒ»ÖÖ²»ÈÜÓÚË®ÇÒÃܶÈСÓÚË®µÄÒºÌå
C£®±½ÔÚÒ»¶¨Ìõ¼þÏÂÄÜÓëäå·¢ÉúÈ¡´ú·´Ó¦
D£®±½²»¾ßÓеäÐ͵ÄË«¼üËùÓ¦¾ßÓеķ¢Éú¼Ó³É·´Ó¦µÄÌØÐÔ£¬¹Ê²»¿ÉÄÜ·¢Éú¼Ó³É·´Ó¦
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¿ÉÒÔÓ÷ÖҺ©¶··ÖÀëµÄÒ»×éÒºÌå»ìºÏÎïÊÇ(¡¡¡¡)
A£®äåºÍ±½ B£®±½ºÍäå±½
C£®Ë®ºÍÏõ»ù±½ D£®±½ºÍÆûÓÍ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
50 ¡æʱ£¬ÏÂÁи÷ÈÜÒºÖУ¬Àë×ÓµÄÎïÖʵÄÁ¿Å¨¶È¹ØϵÕýÈ·µÄÊÇ (¡¡¡¡)¡£
A£®pH£½4µÄ´×ËáÖУºc(H£«)£½4.0 mol¡¤L£1
B£®±¥ºÍСËÕ´òÈÜÒºÖУºc(Na£«)£½c(HCO)
C£®±¥ºÍʳÑÎË®ÖУºc(Na£«)£«c(H£«)£½c(Cl£)£«c(OH£)
D£®pH£½12µÄ´¿¼îÈÜÒºÖУºc(OH£)£½1.0¡Á10£2 mol¡¤L£1
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
³£ÎÂÏ£¬ÓÃ0.100 0 mol¡¤L£1 NaOHÈÜÒºµÎ¶¨20.00 mL 0.100 0 mol¡¤L£1 CH3COOHÈÜÒºËùµÃµÎ¶¨ÇúÏßÈçÓÒͼ¡£ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ (¡¡¡¡)¡£
A£®µã¢ÙËùʾÈÜÒºÖУºc(CH3COO£)£«c(OH£)£½c(CH3COOH)£«c(H£«)
B£®µã¢ÚËùʾÈÜÒºÖУºc(Na£«)£½c(CH3COOH)£«c(CH3COO£)
C£®µã¢ÛËùʾÈÜÒºÖУºc(Na£«)>c(OH£)>c(CH3COO£)>c(H£«)
D£®µÎ¶¨¹ý³ÌÖпÉÄܳöÏÖ£ºc(CH3COOH)>c(CH3COO£)>c(H£«)>c(Na£«)>c(OH£)
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
ºÏ³É°±¹¤ÒµÖпØÖƵķ´Ó¦Ìõ¼þÓ¦ (¡¡¡¡)¡£
A£®Î¶ÈÔ½¸ßÔ½ºÃ
B£®Ñ¹Ç¿Ô½´óÔ½ºÃ
C£®»ìºÏÆøÌåÖÐÇâÆøº¬Á¿Ô½¸ßÔ½ºÃ
D£®ËùÑ¡µÄ´ß»¯¼Á»îÐÔÔ½´óÔ½ºÃ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
ÔÚÈÜÒºÖмÓÈë×ãÁ¿Na2O2ºóÈÔÄÜ´óÁ¿¹²´æµÄÀë×Ó×éÊÇ£¨ £©
A£®NH4+¡¢Ba2+¡¢Cl£¡¢NO3£ B£®K+¡¢AlO2£¡¢Cl£¡¢SO42£
C£®Ca2+¡¢Mg2+¡¢NO3£¡¢HCO3£ D£®Na+¡¢Cl£¡¢CO32£¡¢SO32£
²é¿´´ð°¸ºÍ½âÎö>>
°Ù¶ÈÖÂÐÅ - Á·Ï°²áÁбí - ÊÔÌâÁбí
ºþ±±Ê¡»¥ÁªÍøÎ¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨Æ½Ì¨ | ÍøÉÏÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | µçÐÅթƾٱ¨×¨Çø | ÉæÀúÊ·ÐéÎÞÖ÷ÒåÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | ÉæÆóÇÖȨ¾Ù±¨×¨Çø
Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com