Ñõ»¯ÑÇÍ­ÊÇ´óÐÍË®Ãæ½¢´¬·À»¤Í¿²ãµÄÖØÒªÔ­ÁÏ¡£Ä³Ð¡×é½øÐÐÈçÏÂÑо¿£¬ÇëÌîдÏÂÁпհס£
ʵÑé1£ºÑõ»¯ÑÇÍ­µÄÖÆÈ¡Ñõ»¯ÑÇÍ­¿ÉÓÃÆÏÌÑÌǺÍÐÂÖÆÇâÑõ»¯Í­Ðü×ÇÒº·´Ó¦ÖÆÈ¡¡£ÎÄÏ×±íÃ÷£¬Ìõ¼þ¿ØÖƲ»µ±Ê±»áÓÐÉÙÁ¿CuOÉú³É¡£
£¨1£©ÊµÑéÊÒÖÆÈ¡ÇâÑõ»¯Í­Ðü×ÇÒºµÄÀë×Ó·½³ÌʽΪ____________¡£
£¨2£©ÊµÑéÊÒÓô˷½·¨ÖÆÈ¡²¢»ñµÃÉÙÁ¿Ñõ»¯ÑÇÍ­¹ÌÌ壬ÐèÒªµÄ²£Á§ÒÇÆ÷ÓÐÊԹܡ¢¾Æ¾«µÆ¡¢ÉÕ±­____________ºÍ____________¡£
£¨3£©ÈôҪ̽¾¿¸Ã·´Ó¦·¢ÉúµÄ×îµÍζȣ¬Ó¦Ñ¡ÓõļÓÈÈ·½Ê½Îª____________¡£
ʵÑé2£º²â¶¨Ñõ»¯ÑÇÍ­µÄ´¿¶È
·½°¸1£º³ÆȡʵÑé1ËùµÃ¹ÌÌåm g£¬²ÉÓÃÈçÏÂ×°ÖýøÐÐʵÑé¡£

£¨4£©×°ÖÃaÖÐËù¼ÓµÄËáÊÇ____________£¨Ìѧʽ£©¡£
£¨5£©Í¨¹ý²â³öÏÂÁÐÎïÀíÁ¿£¬ÄܴﵽʵÑéÄ¿µÄµÄÊÇ____________¡£

A£®·´Ó¦Ç°ºó×°ÖÃaµÄÖÊÁ¿
B£®×°ÖÃc³ä·Ö·´Ó¦ºóËùµÃ¹ÌÌåµÄÖÊÁ¿
C£®·´Ó¦Ç°ºó×°ÖÃdµÄÖÊÁ¿
D£®·´Ó¦Ç°ºó×°ÖÃeµÄÖÊÁ¿
£¨6£©ÔÚÇâÆøÑé´¿ºó£¬µãȼװÖÃcÖоƾ«µÆ֮ǰÐèÒª¶ÔK1¡¢K2½øÐеIJÙ×÷ÊÇ  ____________
·½°¸2£º½«ÊµÑélËùµÃ¹ÌÌåmgÈÜÓÚ×ãÁ¿Ï¡ÁòËᣬ¾­¹ýÂË¡¢Ï´µÓ¡¢¸ÉÔïºó³Æ³ö²»ÈÜÎïµÄÖÊÁ¿£¨×ÊÁÏ£ºCu2O+2H+=Cu2++Cu+H2O£©
£¨7£©ÅжϾ­¸ÉÔïÆ÷¸ÉÔïºóµÄ²»ÈÜÎïÊÇ·ñËÈÍêÈ«¸ÉÔïµÄ²Ù×÷·½·¨ÊÇ__________________________________¡£
£¨8£©ÈôʵÑéËùµÃ²»ÈÜÎïΪng£¬Ôò¸ÃÑùÆ·ÖÐÑõ»¯ÑÇÍ­µÄÖÊÁ¿·ÖÊýΪ________________¡£

£¨1£©Cu2++2OH¡ª=Cu(OH)2¡ý£¨2·Ö£©
£¨2£©Â©¶·¡¢²£Á§°ô£¨¸÷1·Ö£¬¹²2·Ö£©
£¨3£©Ë®Ô¡¼ÓÈÈ£¨2·Ö£©
£¨4£©H2SO4£¨1·Ö£©
£¨5£©BC£¨Ñ¡¶Ô1¸ö»òÈ«Ñ¡¶¼¸ø2·Ö£©
£¨6£©´ò¿ªK2£¬¹Ø±ÕK1£¨2·Ö£©
£¨7£©½«²»ÈÜÎïÔٴθÉÔïºó³ÆÁ¿£¬Ö±ÖÁ×îºóÁ½´ÎÖÊÁ¿»ù±¾Ïàͬ£¨»òÆäËûºÏÀí´ð°¸£©£¨2·Ö£©  
£¨8£©¡Á100% £¨»òÆäËûºÏÀí´ð°¸£©£¨2·Ö£©

½âÎöÊÔÌâ·ÖÎö£º£¨1£©ÊµÑéÊÒ³£ÓÿÉÈÜÐÔÍ­ÑÎÈÜÒºÓëÇâÑõ»¯ÄÆ·´Ó¦ÖÆÈ¡ÇâÑõ»¯Í­£¬·´Ó¦Ê½ÎªCu2++2OH¡ª=Cu(OH)2¡ý£»£¨2£©ÆÏÌÑÌÇÓëÐÂÖÆÇâÑõ»¯Í­»ìºÏÖ±½Ó¼ÓÈÈÐèÒªÊԹܡ¢¾Æ¾«µÆ£¬´ÓÒºÌåÖзÖÀë³öCu2O³ÁµíÐèÒªÉÕ±­¡¢Â©¶·¡¢²£Á§°ô£»£¨3£©Ì½¾¿¸Ã·´Ó¦·¢ÉúµÄ×îµÍζȣ¬¿ÉÒÔÑ¡Ôñˮԡ¼ÓÈÈ£¬ÐèҪζȼƲâÁ¿Ë®Ô¡µÄζȣ»£¨4£©Ð¿Óë·ÇÑõ»¯ÐÔËá¡¢Äѻӷ¢ÐÔËá·´Ó¦¿ÉÒÔÖÆÈ¡ÇâÆø£¬Ôòa×°ÖÃÓ¦¼ÓÈëÏ¡H2SO4£»£¨5£©×°ÖÃdÔö¼ÓµÄÖÊÁ¿Ô´×ÔcÖз´Ó¦ËùÉú³ÉË®£¬ÓÉ´Ë¿ÉÒÔ¼ÆËã³öÑùÆ·ÖÐÑõÔªËصÄÖÊÁ¿£¬c·´Ó¦ºóËùµÃ¹ÌÌåµÄÖÊÁ¿µÈÓÚÍ­µÄÖÊÁ¿£¬¸ù¾ÝÉÏÊöÁ½ÖÖÔªËصÄÖÊÁ¿¿ÉÒÔ²â³öÑõ»¯ÑÇÍ­µÄ´¿¶È£¬¹ÊBCÕýÈ·£»£¨6£©´ò¿ªK2£¬¹Ø±ÕK1£¬Í¨ÇâÆøÒ»»á¶ùºóÔÙ¼ÓÈÈc×°Ö㻣¨7£©½«²»ÈÜÎïÔٴθÉÔïºó³ÆÁ¿£¬Ö±ÖÁ×îºóÁ½´ÎÖÊÁ¿»ù±¾Ïàͬ£¨»òÆäËûºÏÀí´ð°¸£©£¬ËµÃ÷²»ÈÜÎïÊÇ·ñËÈÍêÈ«¸ÉÔ£¨8£©ÓÉm/M¿ÉÖªn(Cu)=n/64mol£¬ÓÉCu2O+2H+=Cu2++Cu+H2O¿ÉÖªn(Cu2O)=n/64mol£¬ÓÉn?M¿ÉÖªm(Cu2O)=144n/64g=9n/4g£¬ÔòÑùÆ·ÖÐCu2OµÄ´¿¶ÈΪ9n/4m¡Á100%¡£
¿¼µã£º¿¼²éÎïÖÊÖƱ¸ÊµÑé¡¢¼ìÑéÑùÆ·µÄ´¿¶È¡¢»ìºÏÎï·ÖÀëºÍÌá´¿µÄ·½·¨¡¢ÎïÖʵļÓÈÈ¡¢»¯Ñ§¼ÆËãµÈÏà¹Ø֪ʶ¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÊµÑéÌâ

¿ÎÌâʽ¿ÎÌýÌѧÊÇÑо¿ÐÔѧϰµÄÒ»ÖÖ·½Ê½£¬Æä»ù±¾½ÌѧģʽΪ£º

Èçͼ¼×ÊǹØÓÚ¡°Ò»Ñõ»¯Ì¼µÄ»¯Ñ§ÐÔÖÊ¡±µÄ¿ÎÌâʽ¿ÎÌýÌѧÖнâ¾öÎÊÌâ½×¶Î£¬¼×ͬѧÉè¼ÆµÄÖ¤Ã÷CO¾ßÓл¹Ô­ÐÔµÄʵÑé×°Öá£

¼×
(1)ʵÑéʱӦÏȵãȼ_____________(Ìî¡°A¡±»ò¡°B¡±)´¦µÄ¾Æ¾«µÆ¡£
(2)Ó²Öʲ£Á§¹ÜÖз´Ó¦µÄ»¯Ñ§·½³ÌʽΪ__________________________________¡£
(3)ÒÒͬѧÈÏΪ¼××°ÖÃÓдýÓÅ»¯£¬ÈçβÆø¿ÉÏÈ´¢´æÔÚÆ¿ÄÚ£¬È»ºóÔÙ´¦Àí¡£ÈçͼÒÒÊÇËûÉè¼ÆµÄÖüÆøÆ¿£¬Î²ÆøÓ¦´Ó__________(Ìî¡°a¡±»ò¡°b¡±)¿ÚͨÈë(ÆäËû×°ÖÃÂÔ)¡£

ÒÒ
(4)±ûͬѧÖÊÒÉ£ºCOÄÜ·ñʹ³ÎÇåʯ»ÒË®±ä»ë×Ç£¿Òò´Ë£¬ËûÉè¼ÆÔÚCOͨÈëCuO֮ǰ£¬Ó¦ÏÈͨÈë³ÎÇåʯ»ÒË®£¬ÒԱȽÏÅųýCOÓë³ÎÇåʯ»ÒË®·´Ó¦£¬ÊÔ¶Ô´Ë×÷³öÆÀ¼Û¡£ÄãÈÏΪ±ûµÄÉè¼ÆÊÇ·ñÓбØÒª£¿____________£¬ÀíÓÉÊÇ________________¡£
(5)Èç¹û¿ªÊ¼Ê±Í¨ÈëµÄÊÇCOºÍCO2µÄ»ìºÏÆøÌ壬ÔòÓ¦ÈçºÎÉè¼Æ²ÅÄܴﵽʵÑéÄ¿µÄ£¿_________________________________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÊµÑéÌâ

µ¨·¯£¨CuSO4¡¤5H2O£©ÊÇÍ­µÄÖØÒª»¯ºÏÎÓÐ׏㷺µÄÓ¦Óá£ÒÔÏÂÊÇCuSO4¡¤5H2OµÄʵÑéÊÒÖƱ¸Á÷³Ìͼ¡£

Íê³ÉÏÂÁи÷Ì⣺
¢ñ£®£¨1£©Ïòº¬Í­·ÛµÄÏ¡ÁòËáÖеμÓŨÏõËᣬËæ×ÅÍ­·ÛµÄÈܽâ¿ÉÄܹ۲쵽µÄʵÑéÏÖÏó    ¡¢    ¡£
£¨2£©ÖƵõĵ¨·¯¾§Ì壨CuSO4¡¤5H2O£©ÖпÉÄÜ´æÔÚµÄÔÓÖÊÊÇ       £¨Ð´»¯Ñ§Ê½£©¡£
£¨3£©²ÉÓÃÖØÁ¿·¨²â¶¨CuSO4¡¤5H2OµÄº¬Á¿Ê±£¬²½ÖèÈçÏ£º
¢ÙÈ¡Ñù£¬³ÆÁ¿    ¢Ú¼ÓË®Èܽ⠠ ¢Û¼ÓÂÈ»¯±µÈÜÒºÉú³É³Áµí  ¢Ü¹ýÂË£¨ÆäÓಽÖèÊ¡ÂÔ£©
ÔÚ¹ýÂËÇ°£¬ÐèÒª¼ìÑéÊÇ·ñ³ÁµíÍêÈ«£¬Æä²Ù×÷ÊÇ                   ¡£
¢ò. ijÑо¿ÐÔѧϰС×éÓÃZRY-1ÐÍÈÈÖØ·ÖÎöÒǶÔ12.5¿ËÁòËáÍ­¾§Ìå(CuSO4¡¤5H2O)½øÐÐÈÈÖØ·ÖÎö£¬ËæζȵÄÉý¸ß£¬ÁòËáÍ­¾§ÌåÒÀ´Î·¢ÉúÏÂÁз´Ó¦¡£
a.CuSO4¡¤5H2OCuSO4£«5H2O
b.CuSO4CuO£«SO3¡ü£¬2SO32SO2£«O2
c.4CuO2Cu 2 O£«O2¡ü
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÊµÑé¹ý³ÌÖÐÈÈÖØ·ÖÎöÒDzâµÃ²ÐÁô¹ÌÌåÖÊÁ¿Îª3.8 g£¬ÊÔÍƶϸùÌÌåµÄ×é·ÖÊÇ   (д»¯Ñ§Ê½)£¬Æä¶ÔÓ¦µÄÎïÖʵÄÁ¿Ö®±ÈÊÇ     ¡£
£¨2£©ÈÈÖØ·ÖÎöÒǼÓÈȾ§ÌåÖÁºãÖغ󣬽«È«²¿ÆøÌåµ¼³ö£¬Í¨ÈëÇâÑõ»¯±µÈÜÒº³ä·Ö·´Ó¦£¬ËùµÃ³Áµí¾­¹ýÂË¡¢Ï´µÓ¡¢¸ÉÔÖÊÁ¿Îª       g¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÊµÑéÌâ

ÈýÂÈ»¯Á×£¨PCl3£©ÊÇÒ»ÖÖÖØÒªµÄÓлúºÏ³É´ß»¯¼Á¡£ÊµÑéÊÒ³£ÓúìÁ×Óë¸ÉÔïµÄCl2ÖÆÈ¡PCl3£¬×°ÖÃÈçÏÂͼËùʾ¡£

ÒÑÖª£ººìÁ×ÓëÉÙÁ¿Cl2·´Ó¦Éú³ÉPCl3£¬Óë¹ýÁ¿Cl2·´Ó¦Éú³ÉPCl5¡£PCl3ÓöO2»áÉú³ÉPOCl3(ÈýÂÈÑõÁ×)£¬ POCl3ÈÜÓÚPCl3£¬PCl3ÓöË®»áÇ¿ÁÒË®½âÉú³ÉH3PO3ºÍHCl¡£PCl3¡¢POCl3µÄÈ۷еã¼ûÏÂ±í¡£

ÎïÖÊ
ÈÛµã/¡æ
·Ðµã/¡æ
PCl3
-112
75.5
POCl3
2
105.3
Çë´ðÏÂÃæÎÊÌ⣺
£¨1£©BÖÐËù×°ÊÔ¼ÁÊÇ      £¬FÖмîʯ»ÒµÄ×÷ÓÃÊÇ       ¡£
£¨2£©ÊµÑéʱ£¬¼ì²é×°ÖÃÆøÃÜÐÔºó£¬ÏòD×°ÖõÄÇú¾±êµÖмÓÈëºìÁ×£¬´ò¿ªK3ͨÈë¸ÉÔïµÄCO2£¬Ò»¶Îʱ¼äºó£¬¹Ø±ÕK3£¬¼ÓÈÈÇú¾±êµÖÁÉϲ¿ÓлÆÉ«Éý»ªÎï³öÏÖʱͨÈëÂÈÆø£¬·´Ó¦Á¢¼´½øÐС£Í¨¸ÉÔïCO2µÄ×÷ÓÃÊÇ           £¬
£¨3£©ÊµÑéÖƵõĴֲúÆ·Öг£º¬ÓÐPOCl3¡¢PCl5µÈ¡£¼ÓÈëºìÁ×¼ÓÈȳýÈ¥PCl5ºó£¬Í¨¹ý     £¨ÌîʵÑé²Ù×÷Ãû³Æ£©£¬¼´¿ÉµÃµ½½Ï´¿¾»µÄPCl3¡£
£¨4£©C×°ÖÃÖеÄK1¡¢K2µÄÉè¼ÆÒ²³öÓÚÀàËƵÄÄ¿µÄ£¬ÎªÁË´ïµ½ÕâһʵÑéÄ¿µÄ£¬ÊµÑéʱÓëK1¡¢K2ÓйصIJÙ×÷ÊÇ                     ¡£
£¨5£©ÊµÑéºóÆڹرÕK1£¬´ò¿ªK2£¬½«ÆøÌåͨÈëC×°ÖÃÖз¢Éú·´Ó¦£¬·´Ó¦ºóµÄÈÜҺΪX¡£Ä³Í¬Ñ§Éè¼ÆʵÑéÀ´È·¶¨ÈÜÒºXÖк¬ÓеÄijЩÀë×Ó£¬Çë²¹³äÍê³ÉʵÑé²½ÖèºÍÏÖÏó¡£
ʵÑé²½Öè
ʵÑéÏÖÏó
ʵÑé½áÂÛ
¢Ù
 
ÈÜÒºXÖк¬ÓÐNa+
¢Ú
 
ÈÜÒºXÖк¬ÓÐCl-
 
£¨5£©²£Á§¹ÜÖ®¼äµÄÁ¬½ÓÐèÒªÓõ½½ºÆ¤¹Ü£¬Á¬½ÓµÄ·½·¨ÊÇ£ºÏÈ°Ñ        £¬È»ºóÉÔÉÔÓÃÁ¦¼´¿É°Ñ²£Á§¹Ü²åÈëÏðƤ¹Ü¡£¼×ͬѧ½«×°ÖÃAµÄʾÒâͼ»­³ÉÓÒͼ£¬¸ÃʾÒâͼÖÐÃ÷ÏԵĴíÎóÊÇ     ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÊµÑéÌâ

ʵÑéÊÒ³£ÓÃMnO2ÓëŨÑÎËá·´Ó¦ÖƱ¸Cl2(·´Ó¦×°ÖÃÈçͼËùʾ)¡£

£¨1£©ÖƱ¸ÊµÑ鿪ʼʱ£¬Ïȼì²é×°ÖÃÆøÃÜÐÔ£¬½ÓÏÂÀ´µÄ²Ù×÷ÒÀ´ÎÊÇ__________£¨ÌîÐòºÅ£©¡£
A£®ÍùÉÕÆ¿ÖмÓÈëMnO2·ÛÄ©   B£®¼ÓÈÈ    C£®ÍùÉÕÆ¿ÖмÓÈëŨÑÎËá
ÖƱ¸·´Ó¦»áÒòÑÎËáŨ¶ÈϽµ¶øÍ£Ö¹¡£Îª²â¶¨ÒÑ·ÖÀë³ö¹ýÁ¿MnO2ºóµÄ·´Ó¦²ÐÓàÒºÖÐÑÎËáµÄŨ¶È£¬Ì½¾¿Ð¡×éͬѧÌá³öµÄÏÂÁÐʵÑé·½°¸£º
¼×·½°¸£ºÓë×ãÁ¿AgNO3ÈÜÒº·´Ó¦£¬³ÆÁ¿Éú³ÉµÄAgClÖÊÁ¿¡£
ÒÒ·½°¸£ºÓëÒÑÖªÁ¿CaCO3(¹ýÁ¿)·´Ó¦£¬³ÆÁ¿Ê£Óà¹ÌÌåµÄÖÊÁ¿¡£
±û·½°¸£ºÓë×ãÁ¿Zn·´Ó¦£¬²âÁ¿Éú³ÉµÄH2Ìå»ý¡£
¼Ì¶ø½øÐÐÏÂÁÐÅжϺÍʵÑ飺
£¨2£©Åж¨¼×·½°¸²»¿ÉÐУ¬ÀíÓÉÊÇ______________________________________________________¡£
£¨3£©ÒÒ·½°¸µÄʵÑé·¢ÏÖ£¬¹ÌÌåÖк¬ÓÐMnCO3£¬ËµÃ÷̼Ëá¸ÆÔÚË®ÖдæÔÚ__________________,²â¶¨µÄ½á¹û»á£º______________________£¨Ìî¡°Æ«´ó¡±¡¢¡°Æ«Ð¡¡±»ò¡°×¼È·¡±£©¡£
½øÐбû·½°¸ÊµÑ飺װÖÃÈçͼËùʾ£¨¼Ð³ÖÆ÷¾ßÒÑÂÔÈ¥£©¡£

£¨4£©Ê¹YÐιÜÖеIJÐÓàÇåÒºÓëпÁ£·´Ó¦µÄÕýÈ·²Ù×÷Êǽ«________תÒƵ½______ÖС£
£¨5£©·´Ó¦Íê±Ï£¬Ã¿¼ä¸ô1·ÖÖÓ¶ÁÈ¡ÆøÌåÌå»ý£¬·¢ÏÖÆøÌåÌå»ýÖð½¥¼õС£¬Ö±ÖÁ²»±ä¡£ ÆøÌåÌå»ýÖð´Î¼õСµÄÔ­ÒòÊÇ____________£¨Åųý×°ÖúÍʵÑé²Ù×÷µÄÓ°ÏìÒòËØ£©¡£
£¨6£©Ð¡×éÄÚÓÖÓÐͬѧÌá³ö»¹¿É²ÉÓÃËá¼îÖк͵ζ¨·¨²â¶¨²ÐÓàÒºÖÐÑÎËáµÄŨ¶È£¬µ«»¹Ðè¾­²éÔÄ×ÊÁÏÖªµÀ£º________________________________________________________ ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÊµÑéÌâ

ÀûÓà Y Ð͹ÜÓëÆäËüÒÇÆ÷×éºÏ¿ÉÒÔ½øÐÐÐí¶àʵÑ飨¹Ì¶¨×°ÖÃÂÔ£©¡£·ÖÎö²¢»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©ÊµÑéÄ¿µÄ£ºÑéÖ¤SO2µÄÑõ»¯ÐÔ¡£½«½ºÍ·µÎ¹ÜÖÐŨÁòËá·Ö±ðµÎÈë YÐ͹ܵÄÁ½¸öÖ§¹ÜÖУ¬Ëù²úÉúµÄÁ½ÖÖÆøÌåÏàÓö·¢Éú·´Ó¦£ºSO2+2H2S=3S+2H2O£¬ÔòÔÚÖ§¹Ü½»²æ´¦ÊµÑéÏÖÏóΪ                                                        £¬Áò»¯ÑÇÌú´¦¼ÓË®µÄÄ¿µÄÊÇ                                                ¡£
£¨2£©ÊµÑéÄ¿µÄ£ºÌ½¾¿ SO2ÓëBaCl2·´Ó¦Éú³É³ÁµíµÄÌõ¼þ¡£SO2ͨÈëBaCl2ÈÜÒº²¢²»²úÉú³Áµí£¬ÔÙͨÈëÁíÒ»ÖÖÆøÌåºó¾Í²úÉúÁË°×É«³Áµí¡£³£ÎÂÏ£¬ÈôÓÉÓÒ²à YÐ͹ܲúÉúÁíÒ»ÖÖÆøÌ壬ÔòÔÚÆä×óÓÒÖ§¹ÜÓ¦·ÅÖõÄÒ©Æ·ÊÇ           ºÍ             £¬µ¼Æø¹ÜAµÄ×÷ÓÃÊÇ                                                      ¡£
£¨3£©ÊµÑéÄ¿µÄ£ºÌúþºÏ½ðÖÐÌúº¬Á¿µÄ²â¶¨¡£¢Ù¶ÁÈ¡Á¿Æø¹ÜÖÐÊý¾Ýʱ£¬Èô·¢ÏÖË®×¼¹ÜÖеÄÒºÃæ¸ßÓÚÁ¿Æø¹ÜÖÐÒºÃ棬Ӧ²ÉÈ¡µÄ´ëÊ©ÊÇ                                       £»¢ÚÈô³ÆµÃÌúþºÏ½ðµÄÖÊÁ¿Îª 0.080g£¬Á¿Æø¹ÜÖгõ¶ÁÊýΪ1.00mL£¬Ä©¶ÁÊýΪ 45.80mL£¨ÒÑÕÛËãΪ±ê×¼×´¿ö£©£¬ÔòºÏ½ðÖÐÌúµÄ°Ù·Öº¬Á¿Îª              £¨¾«È·µ½0.01%£©¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÊµÑéÌâ

ÁòËáÍ­ÊÇÒ»ÖÖÓ¦Óù㷺µÄ»¯¹¤Ô­ÁÏ£¬ÊµÑéÊÒÖпÉͨ¹ý²»Í¬Í¾¾¶ÖÆÈ¡ÁòËáÍ­ÈÜÒººÍµ¨·¯(CuSO4¡¤5H2O)£¬ÆäÖÐÒ»ÖÖÁ÷³ÌÈçÏ£º

(1)²Ù×÷I¾ßÌåΪ__________¡¢___________¡¢¹ýÂË¡¢ºæ¸É¡£
(2)ÔÓÍ­(º¬ÉÙÁ¿ÓлúÎï)×ÆÉÕºóµÄ²úÎï³ýÑõ»¯Í­»¹º¬ÉÙÁ¿Í­£¬Ô­Òò¿ÉÄÜÊÇ___________(Ìî×Öĸ´úºÅ)¡£
a£®×ÆÉÕ¹ý³ÌÖв¿·ÖÑõ»¯Í­±»»¹Ô­
b£®×ÆÉÕ²»³ä·Ö£®Í­Î´±»ÍêÈ«Ñõ»¯
c£®Ñõ»¯Í­ÔÚ¼ÓÈȹý³ÌÖзֽâÉú³ÉÍ­
d£®¸ÃÌõ¼þÏÂÍ­ÎÞ·¨±»ÑõÆøÑõ»¯
(3)Èô½«ÔÓÍ­»»Îª´¿¾»µÄÍ­·Û£¬¿ÉÖ±½ÓÏòÆäÖмÓÈëÏ¡ÁòËáºÍFe2(SO4)3ÈÜÒº£¬²»¶ÏͨÈëÑõÆø¡£·´Ó¦ÍêÈ«ºóÏòÆäÖмÓÈë¹ýÁ¿_______________(Ìѧʽ£¬ÏÂͬ)£¬µ÷½ÚpHÖÁ4£¬Éú³É_________³Áµí£¬¹ýÂ˵ÃÁòËáÍ­ÈÜÒº£ÛÒÑÖªFe(OH)3ºÍCu(OH)2ÍêÈ«³ÁµíʱµÄpH·Ö±ðΪ3£®7¡¢6£®4]¡£
£¨4)ͨ¹ýÏÂÊöͼ1×°ÖÃÒ²¿ÉÖÆÈ¡ÁòËáÍ­ÈÜÒº(ÒÑÖª£º2NaOH+2NO2£½NaNO3+NaNO2+H2O)¡£

ÉÕÆ¿ÄÚ·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ____________________________________________£»
ͼ2ÊÇͼ1µÄ¸Ä½ø×°Öã¬ÆäÓŵãÓТÙ__________________________£¬¢Ú_________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÊµÑéÌâ

ÔÚûÓÐÏֳɵÄCO2ÆøÌå·¢ÉúÆ÷µÄÇé¿öÏ£¬ÇëÄãÑ¡ÓÃÏÂͼËùʾ²¿·ÖÒÇÆ÷£¬×°Åä³ÉÒ»¸ö¼òÒ׵ġ¢ÄÜË濪ËæÓá¢Ëæ¹ØËæÍ£µÄCO2ÆøÌå·¢Éú×°Öá£

£¨1£©Ó¦Ñ¡ÓõÄÒÇÆ÷ÊÇ________________(ÌîÈë±àºÅ)£»
£¨2£©ÈôÓÃÉÏÊö×°ÖÃÖÆÈ¡CO2ÆøÌ壬¶øʵÑéÊÒÖ»ÓÐÏ¡ÁòËᡢŨÏõËᡢˮ¡¢¿é×´´¿¼î¡¢¿é×´´óÀíʯ£¬±È½ÏºÏÀíµÄ·½°¸£¬Ó¦Ñ¡ÓõÄÒ©Æ·ÊÇ________                   _________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÏÂÁÐʵÑé¶ÔÓ¦µÄ½áÂÛ²»ÕýÈ·µÄÊÇ(  )

A£®¢ÙÄÜ×é³ÉZn£­CuÔ­µç³Ø
B£®¢ÚÄÜÖ¤Ã÷·Ç½ðÊôÐÔCl>C>Si
C£®¢Û˵Ã÷·´Ó¦2NO2(g) N2O4(g)£¬¦¤H<0
D£®¢Ü°×É«³ÁµíΪBaSO4

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸