ÕÅÃ÷ͬѧÔÚѧϰ¡°ÁòËá¼°ÆäÑεÄijЩÐÔÖÊÓëÓÃ;¡°ÖУ¬½øÐÐÈçÏÂʵÑé̽¾¿£®
[ʵÑéÒ»]̽¾¿Å¨ÁòËáµÄÑõ»¯ÐÔ
½«ÊµÑéÊÒ³£ÓõÄÒ©Æ··ÅÈëÈçͼËùʾµÄʵÑé×°Öú󣬼ÓÈÈ×°Öüף®£¨¼Ð³ÖºÍ¼ÓÈÈ×°ÖÃÊ¡ÂÔ£©
£¨1£©¸Ã×°ÖÃÉè¼ÆÉÏ´æÔÚÃ÷ÏÔȱÏÝ£¬ÇëÖ¸³ö£º______£®
£¨2£©Ð´³ö×°Öü×Öз´Ó¦µÄ»¯Ñ§·½³Ìʽ£º______£®
£¨3£©×°ÖÃÒÒÖеÄÊÔ¼ÁÊÇ______£®
[ʵÑé¶þ]̽¾¿Ä³ÁòËáÑÇÌúÑιÌÌåÊÇ·ñ±äÖÊ
£¨4£©ÇëÄã°ïÖúÕÅÃ÷ͬѧÍê³ÉÈçÏÂʵÑé·½°¸£º
ʵÑé²Ù×÷ Ô¤ÆÚÏÖÏóºÍ½áÂÛ
______ ______
[ʵÑéÈý]ÓÃKHSO4ÖÆÈ¡H2O2²¢²âÆäÖÊÁ¿·ÖÊý
²éÔÄ×ÊÁϵÃÖª£º¹¤ÒµÉÏÓõç½âKHSO4±¥ºÍÈÜÒºÖÆÈ¡H2O2£¬Ê¾ÒâͼÈçÏ£º

¾«Ó¢¼Ò½ÌÍø

ÕÅÃ÷Óô˷¨ÖÆÈ¡Ò»¶¨Å¨¶ÈµÄH2O2£¬²¢½øÐÐÏÂÁÐʵÑé²â¶¨H2O2µÄÖÊÁ¿·ÖÊý£º£¨Àë×Ó·½³Ìʽ£º2MnO4-+5H2O2+6H+=2Mn2++8H2O+5O2¡ü£©
¢ÙÈ¡5.00mL H2O2ÈÜÒº£¨ÃܶÈΪ1.00g/mL£©ÖÃÓÚ׶ÐÎÆ¿ÖмÓˮϡÊÍ£¬ÔÙ¼ÓÏ¡ÁòËáËữ£»
¢ÚÓÃ0.1000mol/L KMnO4ÈÜÒºµÎ¶¨£»
¢ÛÓÃͬÑù·½·¨µÎ¶¨£¬Èý´ÎÏûºÄKMnO4ÈÜÒºµÄÌå»ý·Ö±ðΪ20.00mL¡¢19.98mL¡¢20.02mL£®
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨5£©µç½â±¥ºÍKHSO4ÈÜҺʱ£¬Ñô¼«µÄµç¼«·´Ó¦Ê½Îª______£®
£¨6£©²Ù×÷¢ÚÖУ¬µÎÈëµÚÒ»µÎKMnO4ÈÜÒº£¬ÈÜÒº×ϺìÉ«ÏûʧºÜÂý£¬Ëæ×ŵζ¨¹ý³ÌÖÐMn2+µÄÔö¶à£¬ÈÜÒº×ϺìÉ«ÏûʧËÙÂʼӿ죮Mn2+µÄ×÷ÓÃÊÇ______£®
£¨7£©Ô­H2O2ÈÜÒºÖÐÈÜÖʵÄÖÊÁ¿·ÖÊýΪ______£®
¾«Ó¢¼Ò½ÌÍø
£¨1£©×°ÖÃÒÒÊÇÓÃÓÚÎüÊÕ¶þÑõ»¯ÁòÆøÌ壬ËùÒÔûÓбØÒªÓÐÆ¿Èû£»¹Ê´ð°¸Îª£º×°ÖÃÒÒ²»Ó¦ÓÐÆ¿Èû£»
£¨2£©Í­ºÍŨÁòËá·´Ó¦Éú³ÉÁòËáÍ­¡¢¶þÑõ»¯ÁòºÍË®£¬·´Ó¦µÄ»¯Ñ§·½³Ìʽ£ºCu+2H2SO4£¨Å¨£©
  ¡÷  
.
 
CuSO4+SO2¡ü+2H2O£¨»òC+2H2SO4£¨Å¨£©
  ¡÷  
.
 
CO2¡ü+2SO2¡ü+2H2O£©£»
¹Ê´ð°¸Îª£ºCu+2H2SO4£¨Å¨£©
  ¡÷  
.
 
CuSO4+SO2¡ü+2H2O£¨»òC+2H2SO4£¨Å¨£©
  ¡÷  
.
 
CO2¡ü+2SO2¡ü+2H2O£©£»
£¨3£©¶þÑõ»¯ÁòÊôÓÚËáÐÔÑõ»¯Î¿ÉÓüîÒºÎüÊÕ£»¹Ê´ð°¸Îª£ºÇâÑõ»¯ÄÆÈÜÒº£¨»òÇâÑõ»¯¼ØÈÜÒºµÈ£©£»
£¨4£©ÑÇÌúÀë×ÓÒ׺ÜÈÝÒ×±»Ñõ»¯ÎªÈý¼ÛÌúÀë×Ó£¬¿ÉÒÔÓÃÁòÇèËá¼ØÀ´¼ìÑ飬¼´Èý¼ÛÌúÀë×ÓÍùÒ»Ö§ÊÔ¹ÜÖмÓÈëÉÙÁ¿¹ÌÌ壬¼ÓË®Èܽ⣬ÔٵμӼ¸µÎKSCNÈÜÒº£»ÈôÈÜÒº±äΪºìÉ«£¬ËµÃ÷¸Ã¹ÌÌåÒѱäÖÊ£¬ÈôÈÜҺδ±äºìÉ«£¬ËµÃ÷¸Ã¹ÌÌåûÓбäÖÊ£¬
¹Ê´ð°¸Îª£ºÈý¼ÛÌúÀë×ÓÍùÒ»Ö§ÊÔ¹ÜÖмÓÈëÉÙÁ¿¹ÌÌ壬¼ÓË®Èܽ⣬ÔٵμӼ¸µÎKSCNÈÜÒº£»ÈôÈÜÒº±äΪºìÉ«£¬ËµÃ÷¸Ã¹ÌÌåÒѱäÖÊ£¬ÈôÈÜҺδ±äºìÉ«£¬ËµÃ÷¸Ã¹ÌÌåûÓбäÖÊ£»
£¨5£©µç½â±¥ºÍKHSO4ÈÜҺʱ£¬Ñô¼«ÊÇÒõÀë×Ó·¢Éúʧµç×ÓµÄÑõ»¯·´Ó¦£¬µç¼«·´Ó¦Ê½Îª2HSO4--2e-=S2O82-+2H+
£¨»ò2SO42--2e-=S2O82-£©£¬¹Ê´ð°¸Îª£º2HSO4--2e-=S2O82-+2H+£¨»ò2SO42--2e-=S2O82-£©£»
£¨6£©ÔÚË«ÑõË®µÄ·Ö½â¹ý³ÌÖУ¬¸ßÃÌËá¼Ø¾ßÓÐÑõ»¯ÐÔ£¬¶ÔÓ¦µÄ»¹Ô­²úÎïÊÇÃÌÀë×Ó£¬µ±¼ÓÈëµÚÒ»µÎKMnO4ÈÜÒº£¬ÈÜÒº×ϺìÉ«ÏûʧºÜÂý£¬Ëæ×ŵζ¨¹ý³ÌÖÐMn2+µÄÔö¶à£¬ÈÜÒº×ϺìÉ«ÏûʧËÙÂʼӿ죬¿É¼ûÊÇÃÌÀë×ÓÆðµ½´ß»¯¼ÁµÄ×÷Óã¬
¹Ê´ð°¸Îª£º´ß»¯¼Á£¨»ò¼Ó¿ì·´Ó¦ËÙÂÊ£©£»
£¨7£©Èý´ÎÏûºÄKMnO4ÈÜÒºµÄÌå»ý·Ö±ðΪ20.00mL¡¢19.98mL¡¢20.02mL£¬ÔòÌå»ýƽ¾ùֵΪ£º20.00mL£¬ÔòÏûºÄ¸ßÃÌËá¸ùµÄÁ¿£º0.1mol/L¡Á0.02L=0.002mol£¬ÉèË«ÑõË®µÄÎïÖʵÄÁ¿Îªn£¬Ôò
2MnO4-+5H2O2+6H+=2Mn2++8H2O+5O2¡ü
2        5
0.002mol n
½âµÃn=0.005mol£¬ËùÒÔË«ÑõË®µÄÖÊÁ¿Îª£º0.005mol¡Á34g/mol=0.17g£¬Ë«ÑõË®µÄÖÊÁ¿·ÖÊý=
0.17g
5.00mL¡Á1.00g/mL
¡Á100%=3.4%£¬¹Ê´ð°¸Îª£º3.4%£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â

£¨2009?ÈýÃ÷һģ£©ÕÅÃ÷ͬѧÔÚѧϰ¡°ÁòËá¼°ÆäÑεÄijЩÐÔÖÊÓëÓÃ;¡°ÖУ¬½øÐÐÈçÏÂʵÑé̽¾¿£®
[ʵÑéÒ»]̽¾¿Å¨ÁòËáµÄÑõ»¯ÐÔ
½«ÊµÑéÊÒ³£ÓõÄÒ©Æ··ÅÈëÈçͼËùʾµÄʵÑé×°Öú󣬼ÓÈÈ×°Öüף®£¨¼Ð³ÖºÍ¼ÓÈÈ×°ÖÃÊ¡ÂÔ£©
£¨1£©¸Ã×°ÖÃÉè¼ÆÉÏ´æÔÚÃ÷ÏÔȱÏÝ£¬ÇëÖ¸³ö£º
×°ÖÃÒÒ²»Ó¦ÓÐÆ¿Èû
×°ÖÃÒÒ²»Ó¦ÓÐÆ¿Èû
£®
£¨2£©Ð´³ö×°Öü×Öз´Ó¦µÄ»¯Ñ§·½³Ìʽ£º
Cu+2H2SO4£¨Å¨£©
  ¡÷  
.
 
CuSO4+SO2¡ü+2H2O£¨»òC+2H2SO4£¨Å¨£©
  ¡÷  
.
 
CO2¡ü+2SO2¡ü+2H2O£©
Cu+2H2SO4£¨Å¨£©
  ¡÷  
.
 
CuSO4+SO2¡ü+2H2O£¨»òC+2H2SO4£¨Å¨£©
  ¡÷  
.
 
CO2¡ü+2SO2¡ü+2H2O£©
£®
£¨3£©×°ÖÃÒÒÖеÄÊÔ¼ÁÊÇ
ÇâÑõ»¯ÄÆÈÜÒº£¨»òÇâÑõ»¯¼ØÈÜÒºµÈ£©
ÇâÑõ»¯ÄÆÈÜÒº£¨»òÇâÑõ»¯¼ØÈÜÒºµÈ£©
£®
[ʵÑé¶þ]̽¾¿Ä³ÁòËáÑÇÌúÑιÌÌåÊÇ·ñ±äÖÊ
£¨4£©ÇëÄã°ïÖúÕÅÃ÷ͬѧÍê³ÉÈçÏÂʵÑé·½°¸£º
ʵÑé²Ù×÷ Ô¤ÆÚÏÖÏóºÍ½áÂÛ
ÍùÒ»Ö§ÊÔ¹ÜÖмÓÈëÉÙÁ¿¹ÌÌ壬¼ÓË®Èܽ⣬ÔٵμӼ¸µÎKSCNÈÜÒº
ÍùÒ»Ö§ÊÔ¹ÜÖмÓÈëÉÙÁ¿¹ÌÌ壬¼ÓË®Èܽ⣬ÔٵμӼ¸µÎKSCNÈÜÒº
ÈôÈÜÒº±äΪºìÉ«£¬ËµÃ÷¸Ã¹ÌÌåÒѱäÖÊ£»ÈôÈÜҺδ±äºìÉ«£¬ËµÃ÷¸Ã¹ÌÌåûÓбäÖÊ
ÈôÈÜÒº±äΪºìÉ«£¬ËµÃ÷¸Ã¹ÌÌåÒѱäÖÊ£»ÈôÈÜҺδ±äºìÉ«£¬ËµÃ÷¸Ã¹ÌÌåûÓбäÖÊ
[ʵÑéÈý]ÓÃKHSO4ÖÆÈ¡H2O2²¢²âÆäÖÊÁ¿·ÖÊý
²éÔÄ×ÊÁϵÃÖª£º¹¤ÒµÉÏÓõç½âKHSO4±¥ºÍÈÜÒºÖÆÈ¡H2O2£¬Ê¾ÒâͼÈçÏ£º

ÕÅÃ÷Óô˷¨ÖÆÈ¡Ò»¶¨Å¨¶ÈµÄH2O2£¬²¢½øÐÐÏÂÁÐʵÑé²â¶¨H2O2µÄÖÊÁ¿·ÖÊý£º£¨Àë×Ó·½³Ìʽ£º2MnO4-+5H2O2+6H+=2Mn2++8H2O+5O2¡ü£©
¢ÙÈ¡5.00mL H2O2ÈÜÒº£¨ÃܶÈΪ1.00g/mL£©ÖÃÓÚ׶ÐÎÆ¿ÖмÓˮϡÊÍ£¬ÔÙ¼ÓÏ¡ÁòËáËữ£»
¢ÚÓÃ0.1000mol/L KMnO4ÈÜÒºµÎ¶¨£»
¢ÛÓÃͬÑù·½·¨µÎ¶¨£¬Èý´ÎÏûºÄKMnO4ÈÜÒºµÄÌå»ý·Ö±ðΪ20.00mL¡¢19.98mL¡¢20.02mL£®
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨5£©µç½â±¥ºÍKHSO4ÈÜҺʱ£¬Ñô¼«µÄµç¼«·´Ó¦Ê½Îª
2HSO4--2e-=S2O82-+2H+£¨»ò2SO42--2e-=S2O82-£©
2HSO4--2e-=S2O82-+2H+£¨»ò2SO42--2e-=S2O82-£©
£®
£¨6£©²Ù×÷¢ÚÖУ¬µÎÈëµÚÒ»µÎKMnO4ÈÜÒº£¬ÈÜÒº×ϺìÉ«ÏûʧºÜÂý£¬Ëæ×ŵζ¨¹ý³ÌÖÐMn2+µÄÔö¶à£¬ÈÜÒº×ϺìÉ«ÏûʧËÙÂʼӿ죮Mn2+µÄ×÷ÓÃÊÇ
´ß»¯¼Á£¨»ò¼Ó¿ì·´Ó¦ËÙÂÊ£©
´ß»¯¼Á£¨»ò¼Ó¿ì·´Ó¦ËÙÂÊ£©
£®
£¨7£©Ô­H2O2ÈÜÒºÖÐÈÜÖʵÄÖÊÁ¿·ÖÊýΪ
3.4%
3.4%
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

ÕÅÃ÷ͬѧÔÚѧϰ¡°ÁòËá¼°ÆäÑεÄijЩÐÔÖÊÓëÓÃ;¡°ÖУ¬½øÐÐÈçÏÂʵÑé̽¾¿£®
[ʵÑéÒ»]̽¾¿Å¨ÁòËáµÄÑõ»¯ÐÔ
½«ÊµÑéÊÒ³£ÓõÄÒ©Æ··ÅÈëÈçͼËùʾµÄʵÑé×°Öú󣬼ÓÈÈ×°Öüף®£¨¼Ð³ÖºÍ¼ÓÈÈ×°ÖÃÊ¡ÂÔ£©
£¨1£©¸Ã×°ÖÃÉè¼ÆÉÏ´æÔÚÃ÷ÏÔȱÏÝ£¬ÇëÖ¸³ö£º______£®
£¨2£©Ð´³ö×°Öü×Öз´Ó¦µÄ»¯Ñ§·½³Ìʽ£º______£®
£¨3£©×°ÖÃÒÒÖеÄÊÔ¼ÁÊÇ______£®
[ʵÑé¶þ]̽¾¿Ä³ÁòËáÑÇÌúÑιÌÌåÊÇ·ñ±äÖÊ
£¨4£©ÇëÄã°ïÖúÕÅÃ÷ͬѧÍê³ÉÈçÏÂʵÑé·½°¸£º
ʵÑé²Ù×÷Ô¤ÆÚÏÖÏóºÍ½áÂÛ
____________
[ʵÑéÈý]ÓÃKHSO4ÖÆÈ¡H2O2²¢²âÆäÖÊÁ¿·ÖÊý
²éÔÄ×ÊÁϵÃÖª£º¹¤ÒµÉÏÓõç½âKHSO4±¥ºÍÈÜÒºÖÆÈ¡H2O2£¬Ê¾ÒâͼÈçÏ£º

ÕÅÃ÷Óô˷¨ÖÆÈ¡Ò»¶¨Å¨¶ÈµÄH2O2£¬²¢½øÐÐÏÂÁÐʵÑé²â¶¨H2O2µÄÖÊÁ¿·ÖÊý£º£¨Àë×Ó·½³Ìʽ£º2MnO4-+5H2O2+6H+=2Mn2++8H2O+5O2¡ü£©
¢ÙÈ¡5.00mL H2O2ÈÜÒº£¨ÃܶÈΪ1.00g/mL£©ÖÃÓÚ׶ÐÎÆ¿ÖмÓˮϡÊÍ£¬ÔÙ¼ÓÏ¡ÁòËáËữ£»
¢ÚÓÃ0.1000mol/L KMnO4ÈÜÒºµÎ¶¨£»
¢ÛÓÃͬÑù·½·¨µÎ¶¨£¬Èý´ÎÏûºÄKMnO4ÈÜÒºµÄÌå»ý·Ö±ðΪ20.00mL¡¢19.98mL¡¢20.02mL£®
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨5£©µç½â±¥ºÍKHSO4ÈÜҺʱ£¬Ñô¼«µÄµç¼«·´Ó¦Ê½Îª______£®
£¨6£©²Ù×÷¢ÚÖУ¬µÎÈëµÚÒ»µÎKMnO4ÈÜÒº£¬ÈÜÒº×ϺìÉ«ÏûʧºÜÂý£¬Ëæ×ŵζ¨¹ý³ÌÖÐMn2+µÄÔö¶à£¬ÈÜÒº×ϺìÉ«ÏûʧËÙÂʼӿ죮Mn2+µÄ×÷ÓÃÊÇ______£®
£¨7£©Ô­H2O2ÈÜÒºÖÐÈÜÖʵÄÖÊÁ¿·ÖÊýΪ______£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£ºÁÉÄþÊ¡Ä£ÄâÌâ ÌâÐÍ£ºÊµÑéÌâ

ÕÅÃ÷ͬѧÔÚѧϰ¡°ÁòËá¼°ÆäÑεÄijЩÐÔÖÊÓëÓÃ;¡±ÖУ¬½øÐÐÈçÏÂʵÑé̽¾¿¡£
[ʵÑéÒ»]̽¾¿Å¨ÁòËáµÄÑõ»¯ÐÔ¡£
½«ÊµÑéÊÒ³£ÓõÄÒ©Æ··ÅÈëÈçͼËùʾµÄʵÑé×°Öú󣬼ÓÈÈ×°Öüס££¨¼Ð³ÖºÍ¼ÓÈÈ×°ÖÃÊ¡ÂÔ£©
(1)¸Ã×°ÖÃÔÚÉè¼ÆÉÏ´æÔÚÃ÷ÏÔµÄȱÏÝ£¬ÇëÖ¸³ö£º_____________________¡£
(2)д³ö×°Öü×Öз´Ó¦µÄ»¯Ñ§·½³Ìʽ£º______________________¡£
(3)×°ÖÃÒÒÖеÄÊÔ¼ÁÊÇ_____________¡£
[ʵÑé¶þ]̽¾¿Ä³ÁòËáÑÇÌúÑιÌÌåÊÇ·ñ±äÖÊ¡£
(4)ÇëÄã°ïÖúÕÅÃ÷ͬѧÍê³ÉÈçÏÂʵÑé·½°¸£º
[ʵÑéÈý]ÓÃKHSO4ÖÆÈ¡H2O2²¢²âÆäÖÊÁ¿·ÖÊý¡£²éÔÄ×ÊÁϵÃÖª£¬¹¤ÒµÉϵç½âKHSO4±¥ºÍÈÜÒºÖÆÈ¡H2O2µÄʾÒâͼÈçÏ£º
ÕÅÃ÷Óô˷¨ÖÆÈ¡Ò»¶¨Å¨¶ÈµÄH2O2£¬²¢½øÐÐÏÂÁÐʵÑé²â¶¨H2O2µÄÖÊÁ¿·ÖÊý£º(Àë×Ó·½³Ìʽ£º2MnO4-+5H2O2+6H+=2Mn2++8H2O+5O2¡ü)
¢ÙÈ¡5. 00 mL H2O2ÈÜÒº£¨ÃܶÈΪ1.00 g/mL£©ÖÃÓÚ׶ÐÎÆ¿ÖУ¬¼ÓˮϡÊÍ£¬ÔÙ¼ÓÏ¡ÁòËáËữ£»
¢ÚÓÃ0. 100 0 mol/L KMnO4ÈÜÒºµÎ¶¨£»
¢ÛÓÃͬÑù·½·¨µÎ¶¨£¬Èý´ÎÏûºÄKMnO4ÈÜÒºµÄÌå»ý·Ö±ðΪ20.00 mL¡¢19. 98 mL¡¢20.02 mL¡£
»Ø´ðÏÂÁÐÎÊÌ⣺
(5)µç½â±¥ºÍKHSO4ÈÜҺʱ£¬Ñô¼«µÄµç¼«·´Ó¦Ê½Îª__________________¡£
(6)²Ù×÷¢ÚÖУ¬µÎÈëµÚÒ»µÎKMnO4ÈÜÒº£¬ÈÜÒº×ϺìÉ«Ïû ʧºÜÂý£¬Ëæ×ŵζ¨¹ý³ÌÖÐMn2+µÄÔö¶à£¬ÈÜÒº×ϺìÉ«ÏûʧËÙÂʼӿ졣Mn2+µÄ×÷ÓÃÊÇ________________¡£
(7)Ô­H2O2ÈÜÒºÖÐÈÜÖʵÄÖÊÁ¿·ÖÊýΪ_________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2009Ä긣½¨Ê¡ÈýÃ÷Êи߿¼»¯Ñ§Ò»Ä£ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÌî¿ÕÌâ

ÕÅÃ÷ͬѧÔÚѧϰ¡°ÁòËá¼°ÆäÑεÄijЩÐÔÖÊÓëÓÃ;¡°ÖУ¬½øÐÐÈçÏÂʵÑé̽¾¿£®
[ʵÑéÒ»]̽¾¿Å¨ÁòËáµÄÑõ»¯ÐÔ
½«ÊµÑéÊÒ³£ÓõÄÒ©Æ··ÅÈëÈçͼËùʾµÄʵÑé×°Öú󣬼ÓÈÈ×°Öüף®£¨¼Ð³ÖºÍ¼ÓÈÈ×°ÖÃÊ¡ÂÔ£©
£¨1£©¸Ã×°ÖÃÉè¼ÆÉÏ´æÔÚÃ÷ÏÔȱÏÝ£¬ÇëÖ¸³ö£º    £®
£¨2£©Ð´³ö×°Öü×Öз´Ó¦µÄ»¯Ñ§·½³Ìʽ£º    £®
£¨3£©×°ÖÃÒÒÖеÄÊÔ¼ÁÊÇ    £®
[ʵÑé¶þ]̽¾¿Ä³ÁòËáÑÇÌúÑιÌÌåÊÇ·ñ±äÖÊ
£¨4£©ÇëÄã°ïÖúÕÅÃ÷ͬѧÍê³ÉÈçÏÂʵÑé·½°¸£º
ʵÑé²Ù×÷Ô¤ÆÚÏÖÏóºÍ½áÂÛ
       
[ʵÑéÈý]ÓÃKHSO4ÖÆÈ¡H2O2²¢²âÆäÖÊÁ¿·ÖÊý
²éÔÄ×ÊÁϵÃÖª£º¹¤ÒµÉÏÓõç½âKHSO4±¥ºÍÈÜÒºÖÆÈ¡H2O2£¬Ê¾ÒâͼÈçÏ£º

ÕÅÃ÷Óô˷¨ÖÆÈ¡Ò»¶¨Å¨¶ÈµÄH2O2£¬²¢½øÐÐÏÂÁÐʵÑé²â¶¨H2O2µÄÖÊÁ¿·ÖÊý£º£¨Àë×Ó·½³Ìʽ£º2MnO4-+5H2O2+6H+=2Mn2++8H2O+5O2¡ü£©
¢ÙÈ¡5.00mL H2O2ÈÜÒº£¨ÃܶÈΪ1.00g/mL£©ÖÃÓÚ׶ÐÎÆ¿ÖмÓˮϡÊÍ£¬ÔÙ¼ÓÏ¡ÁòËáËữ£»
¢ÚÓÃ0.1000mol/L KMnO4ÈÜÒºµÎ¶¨£»
¢ÛÓÃͬÑù·½·¨µÎ¶¨£¬Èý´ÎÏûºÄKMnO4ÈÜÒºµÄÌå»ý·Ö±ðΪ20.00mL¡¢19.98mL¡¢20.02mL£®
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨5£©µç½â±¥ºÍKHSO4ÈÜҺʱ£¬Ñô¼«µÄµç¼«·´Ó¦Ê½Îª    £®
£¨6£©²Ù×÷¢ÚÖУ¬µÎÈëµÚÒ»µÎKMnO4ÈÜÒº£¬ÈÜÒº×ϺìÉ«ÏûʧºÜÂý£¬Ëæ×ŵζ¨¹ý³ÌÖÐMn2+µÄÔö¶à£¬ÈÜÒº×ϺìÉ«ÏûʧËÙÂʼӿ죮Mn2+µÄ×÷ÓÃÊÇ    £®
£¨7£©Ô­H2O2ÈÜÒºÖÐÈÜÖʵÄÖÊÁ¿·ÖÊýΪ    £®

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸