ÏÂÃæÊÇijÐËȤС×éͬѧ×öµÄ3¸öСʵÑé¡£¸ù¾ÝʵÑéÄÚÈݻشðÏÂÁÐÎÊÌ⣨ÒÇÆ÷µÄ¹Ì¶¨×°ÖÃÒѾ­Ê¡ÂÔ)¡£

£¨1£©Èô¼×ÖÐΪ¶þÑõ»¯ÃÌ£¬Á¬½Ó¼×ÒÒ×°Öã¬______________(Ì¹Ø¡°K1¡±ºÍ¡°K2¡±µÄ²Ù×÷)£¬´Ó·ÖҺ©¶·Ïò¼×ÖмÓÈëÊÊÁ¿Ë«ÑõË®£¬¸ÃʵÑé¿ÉÒÔÖ¤Ã÷¿ÉȼÎïȼÉÕÐèÒªÓëÑõÆø½Ó´¥¡£¼×Öз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ____________________¡£
£¨2£©Èô¼×ÖÐΪˮ£¬±ûÖÐΪºìÉ«ÈÜÒº£¬Á¬½Ó¼×±û£¬´ò¿ªK1£¬¹Ø±ÕK2£¬´Ó·ÖҺ©¶·Ïò¼×ÖмÓÈëijҩƷºó£¬±ûÖÐÖ»¿´µ½ÈÜÒºÓɺìÉ«±äΪÎÞÉ«¡£Ð´³ö±ûÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³Ìʽ______¡£
£¨3£©Èô¼×ÖÐΪ¶þÑõ»¯Ì¼£¨K1¡¢K2¹Ø±Õ£©£¬Á¬½Ó¼×¶¡£¬´Ó·ÖҺ©¶·Ïò¼×ÖмÓÈëÊÊÁ¿µÄ³ÎÇåʯ»ÒË®£¬Ò»¶Îʱ¼äºó£¬´ò¿ªK1£¬Õû¸öʵÑé¹ý³ÌÖм×Öй۲쵽ÏÖÏóÊÇ______________________¡£

MnO2

 
£¨1£©¹Ø±ÕK1,´ò¿ªK2£»

   2H2O2       2H2O + O2¡ü
£¨2£©2NaOH + H2SO4 = Na2SO4 + 2H2O
£¨3£©³ÎÇåʯ»ÒË®±ä»ë×Ç£¬´ò¿ªK1ºó£¬¶¡ÖÐÒºÌåµ¹ÎüÈë¼×£¬²úÉúÎÞÉ«ÆøÅÝ£¬»ë×ǼõÉÙ£¨Ïûʧ£©£¬¼×ÖÐÒºÌåÁ÷È붡¡£

ÊÔÌâ·ÖÎö£º£¨1£©Ë«ÑõË®ÔÚ¶þÑõ»¯Ã̵Ĵ߻¯×÷ÓÃÏ·ֽ⣬Éú³ÉË®ºÍÑõÆø£¬¹Ê·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ2H2O22H2O + O2¡ü£»ÓÉÓÚÔÚË«ÑõË®¼ÓÈëºó£¬Ï¶˵ĵ¼Æø¹Ü¿Ú»á±»Òº·â£¬ÇÒÆøÌåµÄÃܶȶ¼½ÏС£¬»á´ÓÉÏÃæµÄ¹Ü¿ÚÒݳö£¬¹ÊÐè¹Ø±ÕK1,´ò¿ªK2¡£
£¨2£©¸ù¾Ý¡°Ö»¿´µ½±ûÖÐÈÜÒºÓɺìÉ«±äΪÎÞÉ«¡±£¬ËµÃ÷±ûÖÐÈÜҺԭΪµÎÓзÓ̪µÄ¼îÒº£¬Óöµ½ÁËÀ´×ÔÓÚ¼×ÖеÄËᣬËáÖкÍÁ˱ûÖеļ¹Ê·Ó̪±äΪÎÞÉ«£¬ÇÒ¸ÃËá¼îÖкͷ´Ó¦ÎÞÆäËûÏÖÏó£»ÔÙ¸ù¾ÝÌâÒ⣬¡°Á¬½Ó¼×±û£¬´ò¿ªK1£¬¹Ø±ÕK2£¬´Ó·ÖҺ©¶·Ïò¼×ÖмÓÈëijҩƷºó£¬±ûÖÐÖ»¿´µ½ÈÜÒºÓɺìÉ«±äΪÎÞÉ«¡±£¬ËµÃ÷¼×ÖеÄËáÊÇÔÚÆøѹµÄ×÷ÓÃÏ£¬´Ó϶˵ĵ¼Æø¹Ü¿Ú½øÈë±ûµÄ£¬¶ø¼×ÖÐÔ­ÓÐË®£¬ÓÉ´Ë¿ÉÖª£¬´Ó·ÖҺ©¶·Ïò¼×ÖмÓÈëµÄÊÇŨÁòËᣬÓöË®·ÅÈÈʹ¼×ÖÐÆøÌåÅòÕÍ£¬Ñ¹Ç¿Ôö´ó£¬´Ó¶ø½«ÁòËáѹÈë±û£¬ÓëÆäÖеļîÒº·´Ó¦£¬Èç2NaOH + H2SO4 = Na2SO4 + 2H2O¡£
£¨3£©Èô¼×ÖÐΪ¶þÑõ»¯Ì¼£¨K1¡¢K2¹Ø±Õ£©£¬Á¬½Ó¼×¶¡£¬´Ó·ÖҺ©¶·Ïò¼×ÖмÓÈëÊÊÁ¿µÄ³ÎÇåʯ»ÒË®£¬¶þÕß·´Ó¦Éú³É̼Ëá¸Æ³Áµí£¬¹Ê¿É¼û³ÎÇåʯ»ÒË®±ä»ë×Ç£»Ò»¶Îʱ¼äºó£¬´ò¿ªK1£¬ÓÉÓÚ¼×ÖÐÆøÌå¼õÉÙ£¬Ñ¹Ç¿¼õС£¬¹ÊÔÚ´óÆøѹµÄ×÷ÓÃÏ£¬¶¡ÖÐÏ¡ÑÎËá»á±»µ¹ÎüÈë¼×£¬Óë֮ǰ·´Ó¦Éú³ÉµÄ̼Ëá¸Æ³Áµí·´Ó¦£¬Éú³ÉÂÈ»¯¸Æ¡¢Ë®ºÍ¶þÑõ»¯Ì¼ÆøÌ壬¹Ê¿É¼ûÓÐÎÞÉ«ÆøÅݲúÉú£¬»ë×ǼõÉÙ¡£
µãÆÀ£º¶ÔÓÚÑéÖ¤ÐÍʵÑé̽¾¿£¬ÒªÊì¼ÇÎïÖʵÄÐÔÖÊ»ò±ä»¯¹æÂÉ£¬¸ù¾Ý¸ø³öµÄʵÑéÉè¼Æ·½°¸£¬½øÐÐʵÑé¡¢·ÖÎöºÍ̽¾¿£¬²¢Í¨¹ý¹Û²ì¡¢¼Ç¼ºÍ·ÖÎöµÄʵÑéÏÖÏó£¬À´ÑéÖ¤¸ÃÎïÖʵÄÐÔÖÊ»ò±ä»¯¹æÂɵȡ£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÊµÑéÌâ

ÒÔÌìÈ»ÆøΪԭÁϺϳÉÄòËصÄÖ÷Òª²½ÖèÈçÏÂͼËùʾ£¨Í¼ÖÐijЩת»¯²½Öè¼°Éú³ÉÎïδÁгö£©

(1)ÄòËØÓëÊìʯ»Ò»ìºÏÑÐÄ¥        (Ìî¡°ÓС±»ò¡°ÎÞ¡±£©´Ì¼¤ÐÔÆøζ¡£
(2)¼×ÍéºÍË®ÕôÆøÔÚ¸ßμ°´ß»¯¼ÁµÄÌõ¼þÏ·´Ó¦£¬Éú³ÉCOºÍH2£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ    
(3)·ÖÀë³ö°±ºó£¬¿ÉÑ­»·ÀûÓõÄÁ½ÖÖÎïÖÊÊÇ                ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÏÂÁÐʵÑéÖУ¬ÊµÑé·½°¸Óë½áÂ۵ĶÔÓ¦¹ØϵÕýÈ·µÄÊÇ
 
ʵÑé·½°¸
½áÂÛ
A
ÏòÊ¢×°CO£²µÄËÜÁÏÆ¿ÖмÓÈëNaOHÈÜÒº£¬¹Û²ìµ½Æ¿Éí±ä±ñ
˵Ã÷CO2ÄÜÓëNaOH·´Ó¦
B
ľ̿ÔÚ¿ÕÆøºÍÑõÆøÖоùÄÜȼÉÕ
˵Ã÷µªÆøÄÜÖ§³ÖȼÉÕ
C
ijºìÉ«·ÛÄ©·¢Éú·Ö½â·´Ó¦£¬Ö»Éú³Éµ¥Öʹ¯ºÍÑõÆø
¸Ã·ÛÄ©ÓÉÑõÔªËغ͹¯ÔªËØ×é³É
D
Ïò×ÏɫʯÈïÈÜÒºÖÐͨÈëCO2£¬Ê¯ÈïÈÜÒº±äºì
CO2ÊôÓÚËá
 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÊµÑéÌâ

¿ÎÍâС×éÑо¿¡°Ó°ÏìH2O2Éú³ÉO2·´Ó¦ËÙÂʵÄÒòËØ¡±µÄ¿ÎÌ⡣ͨ¹ý²éÔÄ×ÊÁÏ£¬ËûÃÇÌá³öÈçϲÂÏë¡£
¡¾Ìá³ö²ÂÏë¡¿´ß»¯¼ÁºÍ·´Ó¦ÎïŨ¶È¶¼»áÓ°ÏìH2O2Éú³ÉO2 µÄ·´Ó¦ËÙÂÊ
¡¾ÊµÑé¹ý³Ì¡¿ÊµÑé×°ÖÃÈçͼ£º

ʵÑé²Ù×÷£º¼ì²é×°ÖõÄÆøÃÜÐÔÁ¼ºÃ¡£½«·ÖҺ©¶·ÖеÄÒºÌå¼ÓÈë׶ÐÎÆ¿ÖУ¬Á¢¼´ÊÕ¼¯Ò»Æ¿·Å³öµÄÆøÌå¡£
ʵÑé¼Ç¼£º
ʵÑé±àºÅ
¢Ù
¢Ú
¢Û
·´Ó¦Îï
5%H2O2 50 mL
5%H2O2     
3%H2O250 mL
¼ÓÈë¹ÌÌå
0.5gCuO
0.5gMnO2
 0.5gMnO2       
ÊÕ¼¯µÈÌå»ýO2ËùÐèʱ¼ä
105 s
45s
78 s
£¨1£©ÊµÑé¢ÚÖмÓÈëH2O2ÈÜÒºµÄÌå»ýΪ                                mL£»
£¨2£©³ä·Ö·´Ó¦ºó£¬½«ÊµÑé¢ÚÖÐÊ£ÓàÎïÖÊÀïµÄMnO2ÌáÈ¡µÄ·½·¨ÊÇ            £»
¡¾½áÂÛ¡¿¸Ã̽¾¿¹ý³ÌµÃ³öµÄ½áÂÛÊÇ                                       £»
¡¾·´Ë¼¡¿ H2O2ÔÚ³£Î¼ÓÈëMnO2 ºó·´Ó¦ËÙÂʼӿ졣СÃôÌá³ö£¬ÎªÁ˸üºÃµÄÖ¤Ã÷¶þÑõ»¯ÃÌÊÇ·ñ¶ÔH2O2Éú³ÉO2µÄ·´Ó¦ËÙÂÊÓÐÓ°Ï죬»¹Ó¦¸ÃÔö¼ÓÒ»×é¶Ô±ÈʵÑé¡£¸ÃʵÑéÑ¡ÓõÄÒ©Æ·ºÍÓÃÁ¿Îª                                                              £»
¡¾ÊµÑéÍØÕ¹¡¿Ð¡ÃôÓÃÊÕ¼¯µÄÆøÌå½øÐÐÑõÆøµÄÐÔÖÊʵÑé¡£
£¨1£©ÌúË¿ÔÚÑõÆøÖÐȼÉÕ£¬Æä·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ                         £»
£¨2£©Ð¡Ãô·¢ÏÖÓÃÊÕ¼¯µÄÆøÌå½øÐÐÌú˿ȼÉÕʵÑéʱ£¬Ã»ÓвúÉúÃ÷ÏԵĻðÐÇËÄÉäÏÖÏó¡£Ôì³ÉÕâÖÖ½á¹ûµÄ¿ÉÄÜÔ­ÒòÊÇ                                              ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÊµÑéÌâ

ʵÑéÊÒÓÐÈýƿûÓбêÇ©µÄÎÞÉ«ÈÜÒº£¬ËüÃÇ·Ö±ðÊÇÏ¡ÁòËá¡¢ÁòËáÄÆÈÜÒº¡¢Ì¼ËáÄÆÈÜÒº¡£Éè¼ÆÒ»¸öʵÑé·½°¸£¬°ÑÕâÈýÆ¿ÈÜÒº½øÐмø±ð£¬²¢Íê³ÉʵÑ鱨¸æ¡££¨¿É¹©Ñ¡ÓõÄÊÔ¼ÁÓУº·Ó̪ÈÜÒº¡¢ÉúÐâÌú¶¤¡¢ÂÈ»¯±µÈÜÒº£©
ʵÑé²½Öè
Ô¤ÆÚÏÖÏóÓë½áÂÛ
£¨1£©
 
£¨2£©
 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÊµÑéÌâ

СÃôͬѧ°´ÒÔϲ½Öè¶ÔËýËùÓõÄÑÀ¸à½øÐÐ̽¾¿£º

¡¾×ÊÁÏ¡¿ÑÀ¸àÖг£ÓÃÇâÑõ»¯ÂÁ¡¢Ì¼Ëá¸ÆµÈÎïÖÊ×÷Ħ²Á¼Á¡£
Çë»Ø´ð£º
¢Å²Ù×÷AÊÇ                   ¡£
¢Æ´Ó̽¾¿¹ý³ÌÀ´¿´£¬ÄãÈÏΪÑÀ¸àÏÔ               ÐÔ¡£
¢ÇÄãÈÏΪÓÃ×÷ÑÀ¸àĦ²Á¼ÁµÄÎïÖÊÓ¦¾ß±¸Ê²Ã´ÌصãÄØ£¿£¨»Ø´ðÒ»Ìõ¼´¿É£©              ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÊµÑéÌâ

ͬѧÃÇÀ´µ½»¯Ñ§ÊµÑéÊÒ£¬ÒâÍâ¿´µ½ÁËÒ»¸ö²»ºÍгµÄ¡°Òô·û¡±£¨ÈçÏÂͼ£©¡£ÓÉ´Ë£¬¼¤·¢ÁËͬѧÃǵÄ̽¾¿ÓûÍû¡£

£¨1£©Í¬Ñ§ÃÇÈ¡¸ÃÆ¿ÖÐÉÙÁ¿ÈÜÒº µÎ¼ÓÏ¡ÑÎËᣬ¿´µ½Ã°ÆøÅÝ£¬ËµÃ÷Ò©Æ·ÒѱäÖÊ£¬±äÖÊ·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º                           ¡£
¡¾½øÐвÂÏ롿СÌìµÄ²ÂÏ룺NaOH ÈÜÒº²¿·Ö±äÖÊ£»
ÄãµÄ²ÂÏ룺               ¡£
¡¾ÊµÑé̽¾¿¡¿Ð¡ÌìÉè¼ÆÈçÏÂʵÑéÀ´ÑéÖ¤×Ô¼ºµÄ²ÂÏ룬ÇëÍê³ÉÏÂ±í£º
̽¾¿Ä¿µÄ
̽¾¿²½Öè
Ô¤¼ÆÏÖÏó
³ý¾¡ÈÜÒºÖеÄCO32-
¢Ù£ºÈ¡ÉÙÁ¿ÈÜÒºÓÚÊÔ¹ÜÖУ¬µÎ¼Ó×ãÁ¿µÄCaCl2ÊÔ¼Á
                        
Ö¤Ã÷ÈÜÒºÖÐÉдæNaOH 
¢Ú£ºÏòʵÑé¢Ù¹ýÂ˺óËùµÃÈÜÒºÖеμӷÓ̪ÊÔÒº
 
                       
¼ÙÉèÄãµÄ²ÂÏëÕýÈ·£¬²¢°´Ð¡ÌìµÄʵÑé·½°¸½øÐÐʵÑ飬ÔòÄã¹Û²ìµÄÏÖÏó             ¡£
¡¾ÊµÑ鷴˼¡¿£¨1£©ÏÂÁÐÎïÖÊ1.BaCl 2ÈÜÒº 2.Ca(NO3)2ÈÜÒº 3.Ca(OH)2ÈÜÒº4.Ba(OH)2ÈÜÒº£¬ ²»ÄÜÌæ´úСÌìʵÑéÖÐCaCl 2ÈÜÒºµÄÊÇ           (ÌîÐòºÅ)£¨ ¡£
£¨2£©Ð¡ÌìµÚ¶þ´ÎµÎ¼ÓµÄÊÔ¼Á³ýÓÃָʾ¼ÁÍ⣬»¹¿ÉÒÔÓà      Ìæ´ú¡£
¡¾ÍØÕ¹Ó¦Ó῱£´æNaOH ÈÜÒºµÄ·½·¨ÊÇ             ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÊµÑéÌâ

ÏÂͼÊÇÌá´¿CO¡¢CO2»ìºÏÆøÌåÖеÄCO²¢»¹Ô­Ñõ»¯ÌúµÄʾÒâͼ¡£Çë»Ø´ðÓйØÎÊÌ⣺

£¨1£©Îª±ÜÃâÒ»Ñõ»¯Ì¼ÎÛȾ¿ÕÆø£¬²¢»ØÊÕÀûÓÃÒ»Ñõ»¯Ì¼£¬·½¿òÖÐÁ¬½ÓµÄÊÇDºÍE£¬µ¼¹Ü½Ó¿ÚµÄÁ¬½Ó˳ÐòΪa¡ú (    )¡ú(    )¡ú (     )¡£D×°ÖÃÖÐNaOHµÄ×÷ÓÃÊÇ                 ¡£
Èç¹ûµ¼¹ÜÁ¬½Ó´íÎ󣬺ó¹ûÊÇ____________  ____________________________¡£
£¨2£©ÊµÑé½øÐÐÒ»¶Îʱ¼äºó£¬B×°ÖõIJ£Á§¹ÜÖеÄÏÖÏóΪ                              £¬·´Ó¦·½³ÌʽΪ                               ¡£ÓÃÕâÖÖ·½·¨¡°Á¶¡±µÃµÄÌúÓ빤ҵÉÏÁ¶³öµÄÉúÌúÔÚ×é³ÉÉϵÄ×î´óÇø±ðÊÇ___________             __         ____________¡£
£¨3£©ÊµÑé½áÊøºó£¬Òª´ÓA×°ÖõĻìºÏÈÜÒºÖлØÊյõ½½Ï´¿¾»µÄNaOH¹ÌÌå¡£
×ÊÁÏÏÔʾ£¬ÔÚ²»Í¬Î¶ÈÏÂNaOHµÄÈܽâ¶ÈÈçÏ£º
ζÈ(¡æ)
10
20
40
60
80
100
Èܽâ¶È(g/100gË®)
64
85
138
203
285
376
ÀûÓÃʵÑéÊÒµÄÊÔ¼ÁºÍÌõ¼þ£¬ÊµÑé²½ÖèÈçÏÂ(ÆäÖТñ¡¢¢ò¡¢¢óΪʵÑé²Ù×÷)£º

Çë¾ßÌåÌîдÊÔ¼ÁAµÄ»¯Ñ§Ê½ÒÔ¼°ÊµÑé²Ù×÷¢ñ¡¢¢ò¡¢¢óµÄÃû³Æ¡£
¢Ù»¯Ñ§Ê½£ºA_________  ____________£»
¢Ú²Ù×÷Ãû³Æ£º¢ñ_________  ________   ¢ò________  ___________  ¢ó________  __________¡£
¢ÛÈç¹ûÔÚ²Ù×÷¢òµÄ¹ý³ÌÖУ¬ÈÜÒºBÖгöÏÖÉÙÁ¿»ë×Ç£¬Ô­Òò¿ÉÄÜÊÇ                      ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÊµÑéÌâ

ÏÖÓк¬ÔÓÖʵÄÑõ»¯ÌúÑùÆ·(ÔÓÖʲ»²Î¼Ó·´Ó¦)£¬ÎªÁ˲ⶨ¸ÃÑùÆ·ÖÐÑõ»¯ÌúµÄÖÊÁ¿·ÖÊý£¬Ä³Í¬Ñ§³ÆÈ¡¸ÃÑùÆ·10g£¬²¢ÓÃÈçͼËùʾµÄ×°ÖýøÐÐʵÑ飬µÃµ½ÈçÏÂÁ½×éÊý¾Ý£º

 
        ·´Ó¦Ç°
    Ñõ»¯ÌúÍêÈ«·´Ó¦ºó
A×é
²£Á§¹ÜºÍÑõ»¯ÌúÑùÆ·µÄÖÊÁ¿43.7g
²£Á§¹ÜºÍ¹ÌÌåµÄÖÊÁ¿41.3g
B×é
ÉÕ±­ºÍ³ÎÇåʯ»ÒË®µÄÖÊÁ¿180g
ÉÕ±­ºÍÉÕ±­ÖÐÎïÖʵÄÖÊÁ¿186.8g
ÊԻشð£º
£¨1£©ÄãÈÏΪӦѡÔñ           ×éµÄʵÑéÊý¾ÝÀ´¼ÆËãÑùÆ·ÖÐÑõ»¯ÌúµÄÖÊÁ¿·ÖÊý£¬¼ÆËã½á¹ûΪ           ¡£
£¨2£©ÕâλͬѧËùÓõÄʵÑé×°ÖõIJ»×ãÖ®´¦                                       ¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸