ÔÚÒ»¶¨Î¶ÈÏ£¬½«ÆøÌåXºÍÆøÌåY¸÷0.16mol³äÈë10LºãÈÝÃܱÕÈÝÆ÷ÖУ¬·¢Éú·´Ó¦X£¨g£©+Y£¨g£©¨T2Z£¨g£©¡÷H£¼0£¬Ò»¶Îʱ¼äºó´ïµ½Æ½ºâ£¬·´Ó¦¹ý³ÌÖвⶨµÄÊý¾ÝÈçÏÂ±í£º
t/min2479
n£¨Y£©/mol0.120.110.100.10
£¨1£©·´Ó¦Ç°2minµÄƽ¾ùËÙÂʦͣ¨Z£©=
 
£¬¸ÃζÈÏ´˷´Ó¦µÄƽºâ³£ÊýK=
 

£¨2£©ÆäËûÌõ¼þ²»±ä£¬Èô½µµÍζȣ¬·´Ó¦´ïµ½ÐÂƽºâÇ°¦Í£¨Ä棩
 
¦Í£¨Õý£© £¨Ìî¡°£¾¡±£¬¡°£¼¡±»ò¡°=¡±£©
£¨3£©ÆäËûÌõ¼þ²»±ä£¬ÈôÔÙ³äÈë0.2mol Z£¬Æ½ºâʱXµÄÌå»ý·ÖÊý
 
£¨Ìî¡°Ôö´ó¡±£¬¡°¼õС¡±»ò¡°²»±ä¡±£©
¿¼µã£º»¯Ñ§Æ½ºâµÄ¼ÆËã,»¯Ñ§Æ½ºâµÄÓ°ÏìÒòËØ
רÌ⣺»¯Ñ§Æ½ºâרÌâ
·ÖÎö£º£¨1£©2minÄÚYÎïÖʵÄÁ¿±ä»¯Îª0.16mol-0.12mol=0.04mol£¬¸ù¾Ýv=
¡÷n
V¡÷t
¼ÆËãv£¨Y£©£¬ÔÚÀûÓÃËÙÂÊÖ®±ÈµÈÓÚ»¯Ñ§¼ÆÁ¿ÊýÖ®±È¼ÆËãv£¨Z£©£»ÓɱíÖÐÊý¾Ý¿ÉÖª7minʱ£¬·´Ó¦µ½´ïƽºâ£¬¸ù¾ÝƽºâʱYµÄÎïÖʵÄÁ¿£¬ÀûÓÃÈý¶Îʽ¼ÆËãƽºâʱÆäËü×é·ÖµÄÎïÖʵÄÁ¿£¬ÓÉÓÚ·´Ó¦ÆøÌåÇâÆøµÄ»¯Ñ§¼ÆÁ¿ÊýÏàµÈ£¬ÓÃÎïÖʵÄÁ¿´úÌæŨ¶È´úÈëƽºâ³£Êý±í´ïʽ¼ÆË㣻
£¨2£©¸Ã·´Ó¦Õý·´Ó¦ÊÇ·ÅÈÈ·´Ó¦£¬½µµÍζÈƽºâÏòÕý·´Ó¦Òƶ¯£¬·´Ó¦´ïµ½ÐÂƽºâÇ°v£¨Ä棩£¼v£¨Õý£©£»
£¨3£©ÔÙͨÈë0.2 mol Z£¬µÈЧΪÔÚԭƽºâ»ù´¡ÉÏÔö´óѹǿ£¬·´Ó¦Ç°ºóÆøÌåµÄÌå»ý²»±ä£¬Æ½ºâ²»Òƶ¯£®
½â´ð£º ½â£º£¨1£©ÓɱíÖÐÊý¾Ý¿ÉÇóµÃÇ°2minÄÚÉú³ÉZΪ0.08mol£¬¹Ê¦Í£¨Z£©=
0.08mol
10L?2min
=0.004 mol/£¨L?min£©£¬ÓɱíÖÐÊý¾Ý¿ÉÖª7minʱ£¬·´Ó¦µ½´ïƽºâ£¬YµÄÎïÖʵÄÁ¿Îª0.10mol£¬´ËʱXµÄÎïÖʵÄÁ¿Ò²Îª0.10mol£¬ZµÄÎïÖʵÄÁ¿Ò²Îª0.12mol£¬X¡¢Y¡¢ZµÄÎïÖʵÄÁ¿Å¨¶È·Ö±ðΪ£º0.01mol?L-1¡¢0.01mol?L-1¡¢0.012mol?L-1£¬¹Êƽºâ³£ÊýK=
c2(Z)
c(X)c(Y)
=
0.0122
0.01¡Á0.01
=1.44£»
¹Ê´ð°¸Îª£º0.004 mol/£¨L?min£©£»1.44£»
£¨2£©¸Ã·´Ó¦µÄÕý·´Ó¦ÊÇ·ÅÈÈ·´Ó¦£¬½µµÍζÈƽºâÕýÏòÒƶ¯£¬·´Ó¦´ïµ½ÐÂƽºâÇ°v£¨Ä棩£¼v£¨Õý£©£¬¹Ê´ð°¸Îª£º£¼£»
£¨3£©Òò¸Ã·´Ó¦Ç°ºóÆøÌåµÄÌå»ý²»±ä£¬ÆäËûÌõ¼þ²»±ä£¬ÔÙ³äÈë0.2 mol Z£¬Æ½ºâ²»Òƶ¯£¬XµÄÌå»ý·ÖÊý²»±ä£¬¹Ê´ð°¸Îª£º²»±ä£®
µãÆÀ£º±¾Ì⿼²é»¯Ñ§·´Ó¦ËÙÂÊ¡¢»¯Ñ§Æ½ºâ³£Êý¡¢»¯Ñ§Æ½ºâµÄÓ°ÏìÒòËصȣ¬ÄѶÈÖеȣ¬×¢ÒâCÑ¡ÏîÖжԻ¯Ñ§Æ½ºâ³£ÊýµÄÀí½â£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÒÑÖª¸ÆµÄ½ðÊô»îÆÃÐÔ½éÓڼغÍÄÆÖ®¼ä£¬Æ仯ѧÐÔÖÊÓëÄÆÏàËÆ£®¸Æ¼°Æ仯ºÏÎïµÄÓйط´Ó¦ÈçÏ£º
¢ÙCa+2H2O¨TCa£¨OH£©2+H2¡ü             
¢ÚCaO+H2O¨TCa£¨OH£©2
¢ÛCaO2+H2O¨T
 
                 
¢ÜCaO+CO2¨TCaCO3
¢Ý2CaO2+2CO2¨T2CaCO3+O2
½áºÏËùѧ֪ʶÍê³ÉÏÂÁÐÎÊÌ⣺
£¨1£©Íê³ÉÉÏÊö¢ÛµÄ»¯Ñ§·½³Ìʽ£ºCaO2+H2O¨T
 
£»
£¨2£©ÓÃË«ÏßÇűê³ö·´Ó¦¢ÙµÄµç×ÓתÒÆ·½ÏòºÍÊýÄ¿£ºCa+2H2O¨TCa£¨OH£©2+H2¡ü
£¨3£©Ð´³öÒÔ¸ÆΪԭÁÏÖÆÈ¡¹ýÑõ»¯¸Æ£¨CaO2£©µÄ»¯Ñ§·½³Ìʽ£º
 
£»
£¨4£©ÔÚCaO2µÄË®ÈÜÒºÖУ¬µÎÈë·Ó̪ÈÜÒº£¬Ô¤ÆÚµÄʵÑéÏÖÏó¿ÉÄÜÊÇ
 
£»ÔÚ¸ÃÈÜÒºÖмÓÈëÉÙÁ¿µÄMnO2¹ÌÌ壬ÓÐÆøÅÝÉú³É£¬Ð´³ö¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÄÜÓÃäåË®¼ø±ðµÄÊÇ£¨¡¡¡¡£©
A¡¢¼×Íé¡¢ÒÒÏ©
B¡¢¼×Íé¡¢ÒÒÍé
C¡¢NaClÈÜÒº¡¢KClÈÜÒº
D¡¢MgSO4ÈÜÒº¡¢Mg£¨NO3£©2ÈÜÒº

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÏÂÁÐÓлúÎïÖУ¬ÊôÓڸ߷Ö×Ó»¯ºÏÎïµÄÊÇ£¨¡¡¡¡£©
A¡¢ÓÍÖ¬B¡¢ÕáÌÇ
C¡¢ÆÏÌÑÌÇD¡¢µ°°×ÖÊ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÏÂÁйØÓÚÓлú»¯ºÏÎïµÄÈÏʶÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A¡¢µí·ÛºÍÏËάËØÊÇͬ·ÖÒì¹¹Ìå
B¡¢µí·Û¡¢µ°°×ÖÊ¡¢ÓÍÖ¬¶¼ÊôÓÚÌìÈ»¸ß·Ö×Ó»¯ºÏÎï
C¡¢Ö²ÎïÓͲ»ÄÜʹäåµÄËÄÂÈ»¯Ì¼ÈÜÒºÍÊÉ«
D¡¢ÓÍÖ¬ÔÚËáÐÔÌõ¼þÏÂË®½âÓëÔÚ¼îÐÔÌõ¼þÏÂË®½â£¬²úÎﲻͬ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÓÃÊʵ±ÈÜÒº°Ñ3.31gijÌú¿óʯÑùÆ·Èܽ⣬Ȼºó¼Ó¹ýÁ¿¼îÈÜÒº£¬Éú³É³Áµí£¬ÔÙ×ÆÉÕ³Áµí£¬µÃ2.40g Fe2O3£®ÒÑÖª¸ÃÌú¿óʯÖÐÌúµÄÑõ»¯ÎïµÄÖÊÁ¿·ÖÊýΪ70%£®ÊÔ¼ÆË㣺
£¨1£©¸ÃÌú¿óʯÖÐÌúµÄÖÊÁ¿·ÖÊý
 
£®
£¨2£©¸ÃÌú¿óʯÖÐÌúµÄÑõ»¯ÎïµÄ»¯Ñ§Ê½
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÅðÔªËØÔÚ»¯Ñ§ÖÐÓкÜÖØÒªµÄµØ룬Åð¼°Æ仯ºÏÎï¹ã·ºÓ¦ÓÃÓÚÓÀ´Å²ÄÁÏ¡¢³¬µ¼²ÄÁÏ¡¢¸»È¼ÁϲÄÁÏ¡¢¸´ºÏ²ÄÁϵȸßвÄÁÏÁìÓò£®
£¨1£©Èý·ú»¯ÅðÔÚ³£Î³£Ñ¹ÏÂΪ¾ßÓд̱Ƕñ³ôºÍÇ¿´Ì¼¤ÐÔµÄÎÞÉ«Óж¾¸¯Ê´ÐÔÆøÌ壬Æä·Ö×ÓµÄÁ¢Ìå¹¹ÐÍΪ
 
£¬BÔ­×ÓµÄÔÓ»¯ÀàÐÍΪ
 
£®
£¨2£©Á×»¯ÅðÊÇÒ»ÖÖÊܵ½¸ß¶È¹Ø×¢µÄÄÍĥͿÁÏ£¬Ëü¿ÉÓÃ×÷½ðÊôµÄ±íÃæ±£»¤²ã£®Í¼1ÊÇÁ×»¯Åð¾§ÌåµÄ¾§°ûʾÒâͼ£¬ÔòÁ×»¯ÅðµÄ»¯Ñ§Ê½Îª
 
£¬¸Ã¾§ÌåµÄ¾§ÌåÀàÐÍÊÇ
 
£®

£¨3£©ÅðËᣨH3B03£©ÊÇÒ»ÖÖƬ²ã×´½á¹¹µÄ°×É«¾§Ì壬²ãÄÚµÄH3B03·Ö×Ó¼äͨ¹ýÇâ¼üÏàÁ¬£¨Èçͼ2£©£®
¢ÙÅðËá·Ö×ÓÖÐB×îÍâ²ãÓÐ
 
¸öµç×Ó£¬1molH3B03µÄ¾§ÌåÖÐÓÐ
 
 molÇâ¼ü£®
¢ÚÅðËáÈÜÓÚË®Éú³ÉÈõµç½âÖÊһˮºÏÅðËáB£¨OH£©3?H2 O£¬ËüµçÀëÉú³ÉÉÙÁ¿[B£¨OH£©4]Ò»ºÍH+£¬ÔòÅðËáΪ
 
 ÔªËᣬ[B£¨OH£©4]Ò»º¬ÓеĻ¯Ñ§¼üÀàÐÍΪ
 
£®
£¨4£©H3P04µÄK1¡¢K2¡¢K3·Ö±ðΪ7.6¡Á10-3¡¢6.3¡Á10-8¡¢4.4¡Á10-13£¬ÏõËáÍêÈ«µçÀ룬¶øÑÇÏõËáK=5.1¡Á10-4£¬Çë¸ù¾Ý½á¹¹ÓëÐÔÖʵĹØϵ½âÊÍ£º
¢ÙH3P04µÄK1Ô¶´óÓÚK2µÄÔ­Òò
 
£» ¢ÚÏõËá±ÈÑÇÏõËáËáÐÔÇ¿µÄÔ­Òò
 
£®
£¨5£©NiO¾§Ìå½á¹¹ÓëNaCl¾§ÌåÀàËÆ£¬Æ侧°ûµÄÀⳤΪa cm£¬Ôò¸Ã¾§ÌåÖоàÀë×î½üµÄÁ½¸öÑôÀë×Ӻ˼äµÄ¾àÀëΪ
 
£¨Óú¬ÓÐaµÄ´úÊýʽ±íʾ£©£®ÔÚÒ»¶¨Î¶ÈÏ£¬NiO¾§Ìå¿ÉÒÔ×Ô·¢µØ·ÖÉ¢²¢Ðγɡ°µ¥·Ö×Ӳ㡱£¨Èçͼ3£©£¬¿ÉÒÔÈÏΪÑõÀë×Ó×÷ÃÜÖµ¥²ãÅÅÁУ¬ÄøÀë×ÓÌî³äÆäÖУ¬ÁÐʽ²¢¼ÆËãÿƽ·½Ã×Ãæ»ýÉÏ·ÖÉ¢µÄ¸Ã¾§ÌåµÄÖÊÁ¿Îª
 
g£¨ÑõÀë×ӵİ뾶Ϊ1.40¡Á10-10m£¬
3
¡Ö1.732£©£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÏÂÁÐÖеÄʵÑéÒ»¶¨²»»á³öÏÖ³ÁµíÏÖÏóµÄÊÇ£¨¡¡¡¡£©
A¡¢CO2ÆøÌåͨÈëNaAlO2ÈÜÒºÖÐ
B¡¢SO2ÆøÌåͨÈëBaCl2ÈÜÒºÖÐ
C¡¢CO2ÆøÌåͨÈë±¥ºÍNa2CO3ÈÜÒºÖÐ
D¡¢SO2ÆøÌåͨÈëBa£¨OH£©2ÈÜÒºÖÐ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÓëˮһÑù£¬¼×´¼Ò²ÄÜ΢ÈõµçÀ룺2CH3OH£¨l£©¨TCH3OH2++CH3O-£¬25¡æʱ£¬K=2.0¡Á10-17£®ÈôÍùÒ»¶¨Á¿µÄ¼×´¼ÖмÓÈë½ðÊôÄÆ£¬ÔòÓйØÐðÊöÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A¡¢½ðÊôÄÆÓë¼×´¼·´Ó¦±È½ðÊôÄÆÓëË®·´Ó¦¸ü¾çÁÒ
B¡¢½áºÏH+µÄÄÜÁ¦CH3O-£¾OH-
C¡¢ËùµÃµ½ÈÜÒºÖÐK=c£¨CH3O-£©¡Ác£¨CH3OH2+£©
D¡¢ËùµÃµ½ÈÜÒºÖÐc£¨Na+£©=c£¨CH3OH2+£©+c£¨CH3O-£©

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸