9£®ÔÚÈçͼÓÃʯī×÷µç¼«µÄµç½â³ØÖУ¬·ÅÈë500mLº¬Ò»ÖÖÈÜÖʵÄijÀ¶É«Ï¡ÈÜÒº½øÐеç½â£¬¹Û²ìµ½Aµç¼«±íÃæÓкìÉ«µÄ¹Ì̬ÎïÖÊÉú³É£¬Bµç¼«ÓÐÎÞÉ«ÆøÌåÉú³É£»µ±ÈÜÒºÖеÄÔ­ÈÜÖÊÍêÈ«µç½âºó£¬Í£Ö¹µç½â£¬È¡³öAµç¼«£¬Ï´µÓ¡¢¸ÉÔï¡¢³ÆÁ¿£¬µç¼«ÔöÖØ1.6g£®Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Aµç¼«ÊǽӵçÔ´µÄ¸º¼«£»Bµç¼«µÄ·´Ó¦Ê½4OH--4e-=2H2O+O2¡ü£®
£¨2£©Ô­ÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈÊÇ0.05mol/L£»µç½âºóÈÜÒºµÄpHΪ1£®£¨¼ÙÉèµç½âÇ°ºóÈÜÒºµÄÌå»ý²»±ä£©
£¨3£©ÇëÄãÉè¼ÆʵÑéÈ·¶¨Ô­ÈÜÒºÖпÉÄÜËùº¬µÄËá¸ùÀë×Ó£¬ÒªÇó£ºÌá³öÁ½ÖÖ¿ÉÄܵļÙÉ裬·Ö±ðд³öÂÛÖ¤ÕâÁ½ÖÖ¼ÙÉèµÄ²Ù×÷²½Ö衢ʵÑéÏÖÏóºÍʵÑé½áÂÛ£®
¼ÙÉè¢Ù¼ÙÉèÔ­ÈÜÒºÖеÄËá¸ùÀë×ÓÊÇÁòËá¸ùÀë×Ó£ºÈ¡µç½âºóµÄÈÜÒº£¬ÏòÆäÖмÓÈëBaCl2ÈÜÒº£¬ÈôÓа×É«³ÁµíÉú³É£¬ÔòÔ­ÈÜÒºÖÐËùº¬µÄÒõÀë×ÓÊÇSO42-
¼ÙÉè¢Ú¼ÙÉèÔ­ÈÜÒºÖеÄËá¸ùÀë×ÓÊÇÏõËá¸ùÀë×Ó£ºÈ¡µç½âºóµÄÈÜÒº£¬ÏòÆäÖмÓÈëÍ­£¬Î¢ÈÈ£¬ÈôÍ­Èܽ⣬²¢ÓÐÎÞÉ«ÆøÌåÉú³É£¬ÇÒÔÚ¿ÕÆøÖбäΪºì×ØÉ«£¬ÔòÔ­ÈÜÒºÖÐËùº¬µÄÒõÀë×ÓÊÇNO3-£®

·ÖÎö £¨1£©ºÍµçÔ´µÄ¸º¼«ÏàÁ¬µÄÊǵç½â³ØµÄÒõ¼«·¢Éú»¹Ô­·´Ó¦£¬ºÍµçÔ´µÄÕý¼«ÏàÁ¬µÄÊǵç½â³ØµÄÑô¼«·¢ÉúÑõ»¯·´Ó¦£»
£¨2£©Òõ¼«Îö³ö½ðÊôÍ­µÄÖÊÁ¿ÊÇ1.6g£¬¸ù¾Ýµç¼«·´Ó¦¼ÆËãÁòËáÍ­µÄÎïÖʵÄÁ¿ÒÔ¼°ÈÜÒºÖÐÇâÀë×ÓµÄÎïÖʵÄÁ¿£¬½ø¶ø¸ù¾ÝŨ¶ÈµÈÓÚÎïÖʵÄÁ¿ºÍÌå»ýÖ®±È¼ÆËãÎïÖʵÄÁ¿Å¨¶ÈÒÔ¼°pH£»
£¨3£©¸ÃÈÜÒºÖÐÒ»¶¨ÊǺ¬Í­Àë×ӵĿÉÈÜÐÔµÄÑÎÈÜÒº£¬ËùÒÔ¿ÉÒÔÊÇÏõËáÍ­£¬»òÊÇÁòËáÍ­£¬¸ù¾ÝÀë×ÓµÄÌØÕ÷Àë×Ó·´Ó¦¼ìÑéÁòËá¸ùÀë×ÓÒÔ¼°ÏõËá¸ùÀë×ӵĴæÔÚ¼´¿É£®

½â´ð ½â£º£¨1£©µç½âÒºÊÇÀ¶É«ÈÜÒº£¬Aµç¼«±íÃæÓкìÉ«µÄ¹Ì̬ÎïÖÊÉú³É£¬ËùÒÔA¼«ÊÇÍ­Àë×ӵõç×Ó£¬ËùÒÔA¼«ÊÇÒõ¼«£¬A½ÓµÄÊǵçÔ´µÄ¸º¼«£¬Bµç¼«ÓÐÎÞÉ«ÆøÌåÉú³É£¬ÔòÒ»¶¨ÊÇÇâÑõ¸ùʧµç×Ó²úÉúµÄÑõÆø£¬·´Ó¦£º4OH--4e-=2H2O+O2¡ü£¬¹Ê´ð°¸Îª£º¸º£»4OH--4e-=2H2O+O2¡ü£»
£¨2£©Òõ¼«·´Ó¦£º2Cu2++4e-¡ú2Cu£¬Ñô¼«·´Ó¦£º4OH--4e-=2H2O+O2¡ü£¬µç½âʱ·´Ó¦µÄ×ÜÀë×Ó·½³ÌʽΪ£º2Cu2++2H2O$\frac{\underline{\;ͨµç\;}}{\;}$2Cu+4H++O2¡ü£¬È¡³öAµç¼«£¬Ï´µÓ¡¢¸ÉÔï¡¢³ÆÁ¿¡¢µç¼«ÔöÖØ1.6g£¬ËùÒÔÒõ¼«Îö³ö½ðÊôÍ­µÄÖÊÁ¿ÊÇ1.6g£¬¼´0.025mol£¬ËùÒÔÍ­Àë×ÓµÄŨ¶È=$\frac{0.025mol}{0.5L}$=0.05mol/L£¬Ô­ÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈÊÇ0.05mol/L£¬Éú³ÉÇâÀë×ÓµÄÎïÖʵÄÁ¿Îª0.05mol£¬ËùÒÔc£¨H+£©=$\frac{0.05mol}{0.5L}$=0.1mol/L£¬¼´pH=1£¬¹Ê´ð°¸Îª£º0.05mol/L£»1£»
£¨3£©¸ÃÈÜÒºÖÐÒ»¶¨ÊǺ¬Í­Àë×ӵĿÉÈÜÐÔµÄÑÎÈÜÒº£¬ËùÒÔ¿ÉÒÔÊÇÏõËáÍ­£¬»òÊÇÁòËáÍ­£¬²»ÄÜÊÇÂÈ»¯Í­£¬ÒòΪ²»»á³öÏÖÂÈÆø£¬¼ìÑéÁòËá¸ùÀë×Ó¿ÉÒÔ²ÉÓüÓÈëÂÈ»¯±µÈÜÒºµÄ·½·¨£¬ÈôÓа×É«³ÁµíÉú³É£¬ÔòÔ­ÈÜÒºÖÐËùº¬µÄÒõÀë×ÓÊÇSO42-£»¼ìÑéËá»·¾³ÏµÄÏõËá¸ùÀë×Ó£¬¿ÉÒÔ¼ÓÈë½ðÊôÍ­²¢ÓÐÎÞÉ«ÆøÌåÉú³É£¬ÇÒÔÚ¿ÕÆøÖбäΪºì×ØÉ«£¬ÔòÔ­ÈÜÒºÖÐËùº¬µÄÒõÀë×ÓÊÇNO3-£¬
¹Ê´ð°¸Îª£º¢Ù¼ÙÉèÔ­ÈÜÒºÖеÄËá¸ùÀë×ÓÊÇÁòËá¸ùÀë×Ó£ºÈ¡µç½âºóµÄÈÜÒº£¬ÏòÆäÖмÓÈëBaCl2ÈÜÒº£¬ÈôÓа×É«³ÁµíÉú³É£¬ÔòÔ­ÈÜÒºÖÐËùº¬µÄÒõÀë×ÓÊÇSO42-£»¢Ú¼ÙÉèÔ­ÈÜÒºÖеÄËá¸ùÀë×ÓÊÇÏõËá¸ùÀë×Ó£ºÈ¡µç½âºóµÄÈÜÒº£¬ÏòÆäÖмÓÈëÍ­£¬Î¢ÈÈ£¬ÈôÍ­Èܽ⣬²¢ÓÐÎÞÉ«ÆøÌåÉú³É£¬ÇÒÔÚ¿ÕÆøÖбäΪºì×ØÉ«£¬ÔòÔ­ÈÜÒºÖÐËùº¬µÄÒõÀë×ÓÊÇNO3-£®

µãÆÀ ±¾Ì⿼²éѧÉúµç½â³ØµÄ¹¤×÷Ô­Àí£¬×¢Òâµç¼«·´Ó¦Ê½µÄÊéдÒÔ¼°µç¼«·´Ó¦¡¢µç×ÓÊغãµÄ¼ÆËãÊǹؼü£¬ÄѶȲ»´ó£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

19£®ÓÃNA±íʾ°¢·ü¼ÓµÂÂÞ³£ÊýµÄÖµ£¬ÏÂÁÐÐðÊöÖÐÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®±ê×¼×´¿öÏ£¬2.24LúÓÍ£¨¼Ù¶¨»¯Ñ§Ê½ÎªC8H18£©Öк¬ÓÐ0.8NA¸ö̼ԭ×Ó
B£®³£Î³£Ñ¹Ï£¬O2ºÍO3µÄ»ìºÏÎï16gÖк¬ÓÐNA¸öÑõÔ­×Ó
C£®25¡æʱ£¬1L 0.1mol•L-1µÄÇâÑõ»¯ÄÆÈÜÒºÖк¬ÓÐNA¸öOH-
D£®0.5mol CH4Öк¬ÓÐ0.5NA¸öµç×Ó

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

20£®ÔÚÒ»¶¨Ìõ¼þÏ£¬xA+yB?zC£¬´ïµ½Æ½ºâ£¬ÊÔÌîдÏÂÁпհףº
£¨1£©ÒÑÖªCÊÇÆøÌ壬ÇÒx+y=z£¬¼ÓѹʱƽºâÈç¹û·¢ÉúÒƶ¯£¬Ôòƽºâ±ØÏòÄæ·´Ó¦·½ÏòÒƶ¯£®
£¨2£©ÈôB¡¢CÊÇÆøÌ壬ÆäËûÌõ¼þ²»±äʱÔö¼ÓAµÄÓÃÁ¿£¬Æ½ºâ²»Òƶ¯£¬ÔòAµÄ״̬Ϊ¹ÌÌå»ò´¿ÒºÌ壮

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

17£®Ìå»ýΪ1LµÄijÈÜÒºÖпÉÄܺ¬ÓÐCl-¡¢NO${\;}_{3}^{-}$¡¢SO${\;}_{4}^{2-}$¡¢CO${\;}_{3}^{2-}$¡¢NH${\;}_{4}^{+}$¡¢Fe3+¡¢Al2+¡¢Ba2+ºÍK+£®È¡¸ÃÈÜÒº100mL£¬¼ÓÈë¹ýÁ¿NaOHÈÜÒº£¬¼ÓÈÈ£¬µÃµ½0.02molÆøÌ壬ͬʱ²úÉúµÄ°×É«³ÁµíѸËÙ±äΪ»ÒÂÌÉ«£»¹ýÂË¡¢Ï´µÓ¡¢×ÆÉÕ£¬µÃµ½1.6g¹ÌÌ壻ÏòÉÏÊöÂËÒºÖмÓÈë×ãÁ¿BaCl2ÈÜÒº£¬µÃµ½4.66g²»ÈÜÓÚÑÎËáµÄ³Áµí£®ÏÂÁÐÐðÊöÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®ÈÜÒºÖеÄÖ÷ÒªÀë×Ó½öΪSO${\;}_{3}^{2-}$¡¢Fe2+¡¢NH${\;}_{4}^{+}$
B£®Cl-Ò»¶¨´æÔÚ£¬ÇÒc£¨Cl-£©=0.2 mol£®L-l
C£®ÏòÔ­ÈÜÒºÖмÓÈëÁòËᣬ¿ÉÄÜÓÐÆøÌåÉú³É
D£®CO${\;}_{3}^{2-}$¡¢Al2+Ò»¶¨²»´æÔÚ£¬K+¿ÉÄÜ´æÔÚ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

4£®Ä³Î¶ÈÏ H2£¨g£©+I2£¨g£©?2HI£¨g£©µÄƽºâ³£ÊýK=O.25£¬¸ÃζÈÏÂÔڼס¢ÒÒ¡¢±ûÈý¸öºãÈÝÃܱÕÈÝÆ÷ÖгäÈëH2£¨g£©¡¢I2£¨g£©£¬ÆðʼŨ¶ÈÈçϱíËùʾ£º
ÆðʼŨ¶È¼×ÒÒ±û
c£¨H2/mol•L-10.0100.0200.020
 c£¨I2£©/mol•L-10.0100.010 0.020
ÏÂÁÐÅжϲ»ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®Æ½ºâʱ¼×ÖкͱûÖÐH2µÄת»¯ÂÊÏàͬ
B£®Æ½ºâʱÒÒÖÐI2µÄת»¯ÂÊСÓÚ40%
C£®Æ½ºâʱÒÒÖÐc£¨HI£©±È¼×ÖеÄ2±¶´ó
D£®·´Ó¦Æ½ºâʱ£¬±ûÖÐÆøÌåÑÕÉ«×îÉÒÒÖÐÆøÌåÑÕÉ«×îdz

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

14£®ÊµÑéÊÒÖо­³£ÒòΪÊÔ¼ÁµÄ±£´æ·½·¨²»µ±µ¼Ö´óÁ¿Ò©Æ·±äÖÊ£¬ÏÂÁÐÊÔ¼Á±£´æ·½·¨ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®½ðÊôÄƱ£´æÔÚË®ÖÐ
B£®äåË®·ÅÔÚ´øÏð½ºÈûµÄÊÔ¼ÁÆ¿ÖÐ
C£®ÏõËáÒø±£´æÔÚ×ØÉ«¹ã¿ÚÆ¿ÖУ¬²¢ÖÃÓÚÒõÁ¹´¦
D£®ÇâÑõ»¯¼ØÈÜÒº±£´æÔÚÄ¥¿Ú²£Á§ÈûµÄÊÔ¼ÁÆ¿ÖÐ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

1£®ÄãÈÏΪÏÂÁÐ˵·¨²»ºÏÀíµÄÊÇ£¨¡¡¡¡£©
A£®ËÜÁÏÖÆÆ·³¤ÆÚ²»Ê¹ÓÃÒ²»áÀÏ»¯
B£®¿ÉÒÔͨ¹ýÓëÑÎËáµÄ»¥µÎʵÑé¼ø±ð̼ËáÄƺÍ̼ËáÇâÄÆ
C£®¹ýÑõ»¯ÄÆ¿ÉÒÔ×÷¹©Ñõ¼Á£¬¿ÉÒԺͶþÑõ»¯Áò·´Ó¦Éú³ÉÑÇÁòËáÄƺÍÑõÆø
D£®´¿¼îÊÇ Na2C03£¬Ð¡ËÕ´òÊÇ NaHC03£¬ÆäË®ÈÜÒº¾ù³Ê¼îÐÔ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

18£®ÏÂÁз´Ó¦ÖУ¬Ë®×öÑõ»¯¼ÁµÄÊÇ£¨¡¡¡¡£©
A£®SO3+H2O¨TH2SO4B£®2Na+2H2O¨T2NaOH+H2¡ü
C£®2H2+O2$\frac{\underline{\;µãȼ\;}}{\;}$2H2OD£®2F2+2H2O¨T4HF+O2

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

19£®ÏÂÁÐÅäÖƵÄÈÜҺŨ¶ÈÆ«¸ßµÄÊÇ£¨¡¡¡¡£©
A£®ÅäÖÆÏ¡ÑÎËáÓÃÁ¿Í²Á¿È¡Å¨ÑÎËáʱ£¬¸©Êӿ̶ÈÏß
B£®ÓÃÁ¿Í²Á¿È¡ËùÐèµÄŨÑÎËáµ¹ÈëÉÕ±­ºó£¬ÔÙÓÃˮϴÁ¿Í² 2¡«3 ´Î£¬Ï´Òºµ¹ÈëÉÕ±­ÖÐ
C£®³ÆÁ¿ 11.7 g NaCl ÅäÖÆ 0.2 mol/L NaCl ÈÜҺʱ£¬íÀÂë´í·ÅÔÚ×óÅÌ
D£®¶¨ÈÝʱÑöÊӿ̶ÈÏß

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸