ijÐËȤС×éµÄͬѧÃǹ²Í¬Éè¼ÆÁËÏÂͼËùʾµÄʵÑé×°Ö㬴Ë×°ÖüȿÉÓÃÓÚÖÆÈ¡ÆøÌ壬ÓÖ¿ÉÓÃÓÚÑéÖ¤ÎïÖʵÄijЩÐÔÖÊ£®

(1)ÈôÀûÓøÃ×°ÖÃÖÆÈ¡²¢ÊÕ¼¯H2»òNH3£¬ÔÚIÖмÓÈëÒ©Æ·ºóÓ¦¸Ã½øÐеIJÙ×÷ÊÇ________£»ÈôÒªÖÆÈ¡²¢ÊÕ¼¯O2»òNO£¬¸ü»»¢ñÖÐÒ©Æ·ºó£¬²»¸ü»»ÒÇÆ÷£¬Ö»ÐèÒª×÷¼òµ¥¸Ä½ø£¬¼´¿ÉÍê³ÉʵÑ飬Æä¸Ä½øµÄ·½·¨ÊÇ________£®

(2)´ò¿ªK2£¬¹Ø±ÕK1£®

¢ÙÀûÓøÃ×°ÖýøÐÐʵÑ飬¿ÉÒÔÖ¤Ã÷ÒÔÏÂÎïÖʵÄËáÐÔÇ¿Èõ˳ÐòΪ£ºHCl£¾H2CO3£¾H2SiO3£®

ÓÐͬѧÈÏΪ£ºÔÚAÖмÓ________£¬BÖмÓCaCO3£¬CÖмÓ________(¾ùÌîдÎïÖʵĻ¯Ñ§Ê½)£®¹Û²ìµ½________µÄÏÖÏ󣬼´¿ÉÖ¤Ã÷£®µ«ÓеÄͬѧÈÏΪ´ËʵÑéÖ¤Ã÷ËáÐÔH2CO3£¾H2SiO3ʱÓÐȱÏÝ£¬ÄãÈÏΪȱÏÝÊÇ________£®

¢ÚÀûÓøÃ×°ÖÿÉÒÔÖÆÈ¡Cl2£¬²¢ÔÚ70¡æʱÓëNaOHÈÜÒº·´Ó¦Éú³ÉNaClOºÍNaClO3£®

ʵÑé·½°¸ÊÇ£ºÔÚAÖмÓŨÑÎËᣬBÖмӸßÃÌËá¼Ø£¬CÖмÓÇâÑõ»¯ÄÆÈÜÒº£¬ÉÕ±­ÖмÓ________£®

´ýCÖÐÈÜҺǡºÃ·´Ó¦ºó£¬ÏòÆäÖмÓÈë¹ýÁ¿KIÈÜÒº£¬¼ÓÈë´×Ëáµ÷½ÚÈÜÒºµÄËáÐÔ£¬´ËʱֻÓÐNaClO±»»¹Ô­£¬È»ºóÓÃÒ»¶¨Å¨¶ÈµÄNa2S2O3ÈÜÒºµÎ¶¨£»¼ÌÐøÏòÆäÖмÓÈëÑÎËᣬµ÷½ÚÈÜÒºµÄËáÐÔ£¬´ËʱNaClO3±»»¹Ô­£¬ÔÙÓÃͬŨ¶ÈµÄNa2S2O3ÈÜÒºµÎ¶¨£®()ʵÑé½á¹û¼Ç¼ÈçÏ£º

ÊÔͨ¹ý±íÖÐÊý¾Ý¼ÆËãCÈÜÒºÖÐÉú³ÉµÄClO£­ºÍµÄÎïÖʵÄÁ¿Ö®±È________£®

´ð°¸£º
½âÎö£º

¡¡¡¡(1)¹Ø±Õ£¬´ò¿ª¡¢E(2·Ö)¡¡ÔÚ¢ò×°ÖÃÖÐ×°ÂúË®(2·Ö)

¡¡¡¡(2)¢Ù¡¡(¸÷1·Ö)¡¡BÖйÌÌåÖð½¥Èܽ⣬ÓÐÆøÌå²úÉú£¬CÖÐÓа×É«½º×´³ÁµíÉú³É(2·Ö)¡¡IÖвúÉúµÄÖлìÓУ¬Í¨ÈëCÖÐÒ²¿ÉÓë·´Ó¦Éú³É³Áµí(2·Ö)

¡¡¡¡¢Ú70¡æµÄÈÈË®»òÈÈË®(1·Ö)¡¡1¡Ã2(2·Ö)


Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â

ij»¯Ñ§ÐËȤС×éµÄͬѧÃÇ°´ÕÕÏÂÃæµÄʵÑé·½·¨ÖƱ¸ÇâÑõ»¯Ìú½ºÌ壺Ê×ÏÈÈ¡ÉÙÁ¿ÕôÁóË®ÓڽྻµÄÉÕ±­ÖУ¬Óþƾ«µÆ¼ÓÈÈÖÁ·ÐÌÚ£¬ÏòÉÕ±­ÖÐÖðµÎµÎ¼Ó±¥ºÍµÄFeCl3ÈÜÒº¼ÌÐøÖó·Ð£¬ÖÁÒºÌå³Ê͸Ã÷µÄºìºÖÉ«£®FeCl3+3H2O
  ¡÷  
.
 
Fe£¨OH£©3£¨½ºÌ壩+3HCl
£¨1£©ÅжϽºÌåÖƱ¸ÊÇ·ñ³É¹¦£¬¿ÉÀûÓýºÌåµÄ
¶¡´ï¶ûЧӦ
¶¡´ï¶ûЧӦ
£®
£¨2£©ÔÚ×öÖƱ¸ÇâÑõ»¯Ìú½ºÌåµÄʵÑéʱ£¬ÓÐЩͬѧûÓа´ÒªÇó½øÐУ¬½á¹ûûÓй۲쵽½ºÌ壬ÇëÄãÔ¤²âÆäÏÖÏó²¢·ÖÎöÔ­Òò£º
¢Ù¼×ͬѧûÓÐÑ¡Óñ¥ºÍÂÈ»¯ÌúÈÜÒº£¬¶øÊǽ«Ï¡ÂÈ»¯ÌúÈÜÒºµÎÈë·ÐË®ÖУ¬½á¹ûûÓй۲쵽
ºìºÖÉ«ÒºÌå
ºìºÖÉ«ÒºÌå
£¬ÆäÔ­ÒòÊÇ
FeCl3ÈÜҺ̫ϡ£¬Éú³ÉµÄFe£¨OH£©3Ì«ÉÙ
FeCl3ÈÜҺ̫ϡ£¬Éú³ÉµÄFe£¨OH£©3Ì«ÉÙ
£®
¢ÚÒÒͬѧÔÚʵÑéÖÐûÓÐʹÓÃÕôÁóË®£¬¶øÊÇÓÃ×ÔÀ´Ë®£¬½á¹û»á
Éú³ÉºìºÖÉ«³Áµí
Éú³ÉºìºÖÉ«³Áµí
£¬Ô­Òò
×ÔÀ´Ë®Öк¬Óеç½âÖÊ£¬½ºÌå·¢Éú¾Û³Á
×ÔÀ´Ë®Öк¬Óеç½âÖÊ£¬½ºÌå·¢Éú¾Û³Á
£®
¢Û±ûͬѧÏò·ÐË®Öеμӱ¥ºÍÂÈ»¯ÌúÈÜÒººó£¬³¤Ê±¼ä¼ÓÈÈ£¬½á¹û»á
Éú³ÉºìºÖÉ«³Áµí
Éú³ÉºìºÖÉ«³Áµí
£¬Ô­ÒòÊÇ
³¤Ê±¼ä¼ÓÈȽºÌå·¢Éú¾Û³Á
³¤Ê±¼ä¼ÓÈȽºÌå·¢Éú¾Û³Á
£®
£¨3£©¶¡Í¬Ñ§°´ÒªÇóÖƱ¸ÁËFe£¨OH£©3½ºÌ壬µ«ÊÇËûÓÖÏòFe£¨OH£©3½ºÌåÖÐÖðµÎ¼ÓÈëÁËÏ¡H2SO4ÈÜÒº£¬½á¹û³öÏÖÁËһϵÁб仯£®
¢ÙÏȳöÏÖºìºÖÉ«³Áµí£¬Ô­ÒòÊÇ
µç½âÖÊH2SO4ʹFe£¨OH£©3½ºÌå¾Û³Á¶ø²úÉú³Áµí
µç½âÖÊH2SO4ʹFe£¨OH£©3½ºÌå¾Û³Á¶ø²úÉú³Áµí
£®
¢ÚËæºó³ÁµíÈܽ⣬´Ë·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ
Fe£¨OH£©3+3H+¨TFe3++3H2O
Fe£¨OH£©3+3H+¨TFe3++3H2O
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â

ijÐËȤС×éµÄͬѧÔÚ̽¾¿½ðÊô»î¶¯ÐÔ˳Ðòʱ£¬¾ö¶¨°ÑÂÁƬ·ÅÈëÁòËáÍ­µÄË®ÈÜÒºÖнøÐÐʵÑ飬ÒÔÑéÖ¤ÂÁºÍÍ­µÄ»î¶¯ÐÔ˳Ðò£®Ð¡×éµÄͬѧ½øÐÐÁË·Ö¹¤£¬¼×ͬѧȡһ¶¨Á¿µÄÎÞË®ÁòËáÍ­·ÛÄ©¼Óµ½Ò»¶¨Á¿µÄÕôÁóË®ÖÐÈܽ⣬·¢ÏÖÈÜÒº±ä»ë×Ç£¬¼ÓˮϡÊÍÈÔÈ»»ë×Ç£®
£¨1£©ÄãÈÏΪÊÇʲôԭÒòÔì³ÉµÄ
ÁòËáÍ­Ë®½âÉú³ÉÇâÑõ»¯Í­³Áµí
ÁòËáÍ­Ë®½âÉú³ÉÇâÑõ»¯Í­³Áµí
£¬Ð´³ö·´Ó¦µÄ»¯Ñ§·½³Ìʽ
CuSO4+2H2O=Cu£¨OH£©2¡ý+H2SO4
CuSO4+2H2O=Cu£¨OH£©2¡ý+H2SO4
£®
£¨2£©ÇëÄã¸ø³öÅäÖÆ´ËÈÜÒºµÄ½¨Òé
È¡ÁòËáÍ­·ÛÄ©¼ÓÈëÉÙÁ¿ÁòËáÔÙ¼ÓÊÊÁ¿ÕôÁóˮϡÊÍ
È¡ÁòËáÍ­·ÛÄ©¼ÓÈëÉÙÁ¿ÁòËáÔÙ¼ÓÊÊÁ¿ÕôÁóˮϡÊÍ
£®
£¨3£©¾­¹ýС×éÌÖÂÛºó£¬ÖÕÓÚÅäÖƳɳÎÇåÈÜÒº£®¼×ͬѧ½ØÈ¡Ò»¶ÎÂÁƬ·ÅÈëÊ¢ÓиÃÁòËáÍ­ÈÜÒºµÄÊÔ¹ÜÖУ¬¹Û²ì£¬Ò»¿ªÊ¼ÎÞÃ÷ÏÔÏÖÏó£¬Ò»¶Îʱ¼äºóÂÁƬ±íÃæÏȱä
°µ
°µ
£¬ËæºóÈÜÒºÑÕÉ«±ädz²¢ÔÚÂÁƬ±íÃæ³Á»ýÒ»²ãÊèËɵĺìÉ«ÎïÖÊ£®
£¨4£©ÒÒͬѧÈÏΪ½«ÂÁƬ·ÅÈëÇ°Ó¦
ÏÈÓÃÉ°Ö½´òÄ¥¹âÁÁ
ÏÈÓÃÉ°Ö½´òÄ¥¹âÁÁ
£¬²¢ÖØ×ö´ËʵÑ飬³ý·¢ÏÖ¸ú¼×ͬѧһÑùµÄÏÖÏóÍ⣬»¹¿´µ½ÓÐÆøÅݲúÉú£¬ÓÚÊǼ¤·¢ÁËËûÃǵÄ̽¾¿ÓûÍû£®
[Ìá³öÎÊÌâ]¸ÃÆøÌåÊÇʲôÄØ£¿
[²ÂÏë]¢Ù¼×ͬѧÈÏΪ¿ÉÄÜÊÇÇâÆø   ¢ÚÒÒͬѧÈÏΪ¿ÉÄÜÊǶþÑõ»¯Áò£®
[²é×ÊÁÏ]Í­ÑÎÈÜÒºÖУ¬ÈÜÒºµÄpH¾ö¶¨ÈÜÒºÖÐCu2+Ũ¶È´óС£¬Í¨¹ý²é×ÊÁϵõ½ÈçÏÂÊý¾Ý£¨Cu2+Ũ¶ÈСÓÚ1.0¡Á10-5molL-1ʱºöÂÔ²»¼Æ£©£º
Cu2+/molL-1 0.1 0.01 0.001 0.0001 0.00001
ÈÜÒºpH 4.7 5.2 5.7 6.2 6.7
[·½°¸Éè¼Æ]ͬѧÃÇÉè¼Æ·½°¸ÖðÒ»¼ÓÒÔÑéÖ¤£º
¢Ù
½«²úÉúµÄÆøÌåÊÕ¼¯µ½Ð¡ÊÔ¹ÜÖУ¬×ö±¬ÃùÆøʵÑé
½«²úÉúµÄÆøÌåÊÕ¼¯µ½Ð¡ÊÔ¹ÜÖУ¬×ö±¬ÃùÆøʵÑé
£»
¢Ú
½«²úÉúµÄÆøÌåͨÈëÆ·ºìÈÜÒº»ã×Ü£¬¹Û²ìÊÇ·ñÍÊÉ«
½«²úÉúµÄÆøÌåͨÈëÆ·ºìÈÜÒº»ã×Ü£¬¹Û²ìÊÇ·ñÍÊÉ«
£®
[ÌÖÂÛ½»Á÷]Äã¾õµÃ»¹¿ÉÒÔͨ¹ýʲô²Ù×÷ÑéÖ¤AlºÍCuµÄ»î¶¯ÐÔ
·Ö±ðȡͭƬºÍÂÁƬ£¬¼ÓÈëÏ¡ÑÎËáÖй۲ì
·Ö±ðȡͭƬºÍÂÁƬ£¬¼ÓÈëÏ¡ÑÎËáÖй۲ì
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

þ¡¢ÂÁ¼°Æ仯ºÏÎï¹ã·ºÓ¦ÓÃÓÚ²ÄÁϹ¤Òµ¡¢Ò©ÎïºÏ³ÉµÈ¹¤ÒµÉú²úÖУ®Ä³»¯Ñ§ÐËȤС×éµÄͬѧÃǶÔÈýÖÖ²»Í¬µÄþÂÁºÏ½ðÑùÆ·½øÐбàºÅ£¬²¢Õ¹¿ªÁËʵÑé̽¾¿£®
£¨1£©È¡ÑùÆ·¢Ùm1 g£¬¼ÓÈë¹ýÁ¿NaOHÈÜÒº£¬³ä·Ö·´Ó¦ºó¹ýÂË£»ÍùÂËÒºÖÐͨÈë¹ýÁ¿CO2ÆøÌ壻½«ËùµÃ³Áµí¹ýÂË¡¢Ï´µÓ¡¢ºæ¸É¡¢×ÆÉÕ£¬µÃµ½¹ÌÌåµÄÖÊÁ¿ÈÔȻΪm1 g£®ºÏ½ðÖÐþµÄÖÊÁ¿·ÖÊýΪ
 
£¨±£Áô2λСÊý£¬ÏÂͬ£©£®
£¨2£©È¡ÑùÆ·¢Úm2 g£¬Óë×ãÁ¿Ï¡ÁòËá·´Ó¦£¬¹ÌÌåÍêÈ«ÈܽâʱµÃµ½ÆøÌåµÄÌå»ýΪV L£¨±ê×¼×´¿ö£©£®m2µÄÈ¡Öµ·¶Î§ÊÇ
 
£®
£¨3£©È¡ÑùÆ·¢Û0.918g£¬Óë30.00mL¡¢2.00mol/LÑÎËá³ä·Ö·´Ó¦ºó£¬µÃµ½672mLÆøÌ壨±ê×¼×´¿ö£©£¬Ê£ÓàºÏ½ð0.306g£®È»ºó¸Ä¼Ó1.00mol/LµÄNaOHÈÜÒº£¬Ê¹Ê£ÓàºÏ½ðÖеÄÂÁÇ¡ºÃÍêÈ«Èܽ⣮ÐèÒª¼ÓÈë1.00mol/LµÄNaOHÈÜÒº
 
mL£®
£¨4£©ÎªÁËÈ·¶¨¿ÉÄܺ¬ÓÐþ¡¢ÂÁÔªËصÄijδ֪ÁòËáÑξ§ÌåµÄ»¯Ñ§Ê½£¬³ÆÈ¡67.80g¾§ÌåÈÜÓÚË®£¬ÏòÆäÖÐÖðµÎ¼ÓÈë×ãÁ¿Ba£¨OH£©2ÈÜÒººó¹ýÂË£¬µÃµ½99.00g°×É«³Áµí£¬ÓÃÏ¡ÏõËá´¦Àí³ÁµíÎ¾­Ï´µÓºÍ¸ÉÔ×îÖյõ½°×É«¹ÌÌå93.20g£®ÊÔͨ¹ý¼ÆËãÈ·¶¨¸Ã¾§ÌåµÄ»¯Ñ§Ê½£®£¨Ð´³öÍÆËã¹ý³Ì£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ijÐËȤС×éµÄͬѧÃǹ²Í¬Éè¼ÆÁËÏÂͼËùʾµÄʵÑé×°Ö㬴Ë×°Öü´¿ÉÓÃÓÚÖÆÈ¡ÆøÌ壬ÓÖ¿ÉÓÃÓÚÑéÖ¤ÎïÖʵÄijЩÐÔÖÊ¡£

   

£¨1£©ÈôÀûÓøÃ×°ÖÃÖÆÈ¡²¢ÊÕ¼¯H2»òNH3£¬ÔÚIÖмÓÈëÒ©Æ·ºóÓ¦¸Ã½øÐеIJÙ×÷ÊÇ              

                                 £»ÈôÒªÖÆÈ¡²¢ÊÕ¼¯O2»òNO£¬¸ü»»IÖÐÒ©Æ·ºó£¬²»¸ü»»ÒÇÆ÷£¬Ö»ÐèÒª×÷¼òµ¥¸Ä½ø£¬¼´¿ÉÍê³ÉʵÑ飬Æä¸Ä½øµÄ·½·¨ÊÇ                                              

               ¡£

£¨2£©´ò¿ªK2£¬¹Ø±ÕK1¡£

¢ÙÀûÓøÃ×°ÖýøÐÐʵÑ飬¿ÉÒÔÖ¤Ã÷ÒÔÏÂÎïÖʵÄËáÐÔÇ¿Èõ˳ÐòΪ£ºHC1>H2CO3>HSiO3¡£ÓÐͬѧÈÏΪ£ºÔÚAÖмӠ                        £¬BÖмÓCaCO3£¬CÖмӠ             £¨¾ùÌîдÎïÖʵĻ¯Ñ§Ê½£©¡£¹Û²ìµ½                         µÄÏÖÏ󣬼´¿ÉÖ¤Ã÷¡£µ«ÓеÄͬѧÈÏΪ´ËʵÑéÔÚÖ¤Ã÷ËáÐÔH2CO3>H2SiO3ʱÓÐȱÏÝ£¬ÄãÈÏΪȱÏÝÊÇ                          ¡£

¢ÚÀûÓøÃ×°ÖÿÉÒÔÖÆÈ¡C12£¬²¢ÔÚ70¡æʱÓëNaOHÈÜÒº·´Ó¦Éú³ÉNaC1OºÍNaC1O3¡£ÊµÑé·½°¸ÊÇ£ºÔÚAÖмÓŨÑÎËᣬBÖмӸßÃÌËá¼Ø£¬CÖмÓÇâÑõ»¯ÄÆÈÜÒº£¬ÉÕ±­ÖмӠ            ¡£´ýCÖÐÈÜҺǡºÃ·´Ó¦ºó£¬ÏòÆäÖмÓÈë¹ýÁ¿KIÈÜÒº£¬¼ÓÈë´×Ëáµ÷½ÚÈÜÒºµÄËáÐÔ£¬´ËʱֻÓÐNaC1O±»»¹Ô­£¬È»ºóÓÃÒ»¶¨Å¨¶ÈµÄNa2S2O3ÈÜÒºµÎ¶¨£»¼ÌÐøÏòÆäÖмÓÈëÑÎËᣬµ÷½ÚÈÜÒºµÄËáÐÔ£¬´ËʱNaC1O3±»»¹Ô­£¬ÔÙÓÃͬŨ¶ÈµÄNa2S2O3ÈÜÒºµÎ¶¨¡££¨I2+2S2O32¨D=S4O62¨D+2I¨D£©ÊµÑé½á¹û¼Ç¼ÈçÏ£º

½«KIת»¯ÎªI2

µÎ¶¨I2£¬ÏûºÄNa2S2O3ÈÜÒºµÄÌå»ý

KII2

5.00mL

KII2

30.00mL

ͨ¹ý±íÖÐÊý¾Ý¼ÆËãCÈÜÒºÖÐÉú³ÉC1O¨DºÍC1O3¨DµÄÎïÖʵÄÁ¿Ö®±È              ¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸