ÏÂÁиù¾ÝʵÑé²Ù×÷ºÍÏÖÏóËùµÃ³öµÄ½áÂÛÕýÈ·µÄÊÇ£¨¡¡¡¡£©
Ñ¡ÏîʵÑé²Ù×÷ʵÑéÏÖÏó½áÂÛ
AÓýྻµÄ²¬Ë¿ÕºÈ¡Ä³Ê³ÑÎÊÔÑùÔھƾ«µÆ»ðÑæÉÏ×ÆÉÕ»ðÑæÏÔ»ÆÉ«¸ÃʳÑÎÖв»º¬ÓÐKIO3
B½«SO2ÆøÌåͨÈëµ½Ba£¨NO3£©2ÈÜÒºÖÐÉú³É°×É«³Áµí´Ë³ÁµíÊÇBaSO3
CÈ¡¾ÃÖõÄNa2O2·ÛÄ©£¬ÏòÆäÖеμӹýÁ¿µÄÑÎËá²úÉúÎÞÉ«ÆøÌåNa2O2ûÓбäÖÊ
DÔÚCuSO4ÈÜÒºÖмÓÈëKIÈÜÒº£¬ÔÙ¼ÓÈë±½£¬Õñµ´¡¢¾²ÖÃÓа×É«³ÁµíÉú³É£¬±½²ã³Ê×ÏÉ«°×É«³Áµí¿ÉÄÜΪCuI
A¡¢AB¡¢BC¡¢CD¡¢D
¿¼µã£ºÄƵÄÖØÒª»¯ºÏÎï,¶þÑõ»¯ÁòµÄ»¯Ñ§ÐÔÖÊ,ÑæÉ«·´Ó¦
רÌ⣺
·ÖÎö£ºA£®¼ØµÄÑæÉ«·´Ó¦Äܱ»ÄƵÄÑæÉ«·´Ó¦ËùÑڸǣ¬¹Û²ì¼ØµÄÑæÉ«·´Ó¦ÒªÍ¸¹ýÀ¶É«îܲ£Á§£»
B£®ÏõËá¸ùÀë×ÓÔÚËáÐÔ»·¾³Ï¾ßÓÐÇ¿µÄÑõ»¯ÐÔ£¬Äܹ»Ñõ»¯¶þÑõ»¯Áò£»
C£®¾ÃÖõÄNa2O2·ÛÄ©»ìÓÐ̼ËáÄÆ£»
D£®·¢ÉúÑõ»¯»¹Ô­·´Ó¦Éú³ÉCuIºÍµâ£®
½â´ð£º ½â£ºA£®¼ØµÄÑæÉ«·´Ó¦Äܱ»ÄƵÄÑæÉ«·´Ó¦ËùÑڸǣ¬¹Û²ì¼ØµÄÑæÉ«·´Ó¦ÒªÍ¸¹ýÀ¶É«îܲ£Á§£¬Ôò¸ÃʵÑéÖв»ÄÜÈ·¶¨ÊÇ·ñº¬KIO3£¬¹ÊA´íÎó£»
B£®½«SO2ÆøÌåͨÈëµ½Ba£¨NO3£©2ÈÜÒºÖУ¬Éú³É°×É«³ÁµíΪÁòËá±µ£¬¹ÊB´íÎó£»
C£®¾ÃÖõÄNa2O2·ÛÄ©»ìÓÐ̼ËáÄÆ£¬¼ÓÑÎËáÉú³ÉÎÞÉ«ÆøÌåÖÐÓÐÑõÆøºÍ¶þÑõ»¯Ì¼£¬½áÂÛ²»ºÏÀí£¬¹ÊC´íÎó£»
D£®ÔÚCuSO4ÈÜÒºÖмÓÈëKIÈÜÒº£¬·¢ÉúÑõ»¯»¹Ô­·´Ó¦Éú³ÉCuIºÍµâ£¬ÔÙ¼ÓÈë±½£¬Õñµ´£¬Óа×É«³ÁµíÉú³É£¬±½²ã³Ê×ÏÉ«£¬¹ÊDÕýÈ·£»
¹ÊÑ¡£ºD£®
µãÆÀ£º±¾Ì⿼²é»¯Ñ§ÊµÑé·½°¸µÄÆÀ¼Û£¬Éæ¼°ÑæÉ«·´Ó¦¡¢¶þÑõ»¯ÁòµÄÐÔÖÊ¡¢¹ýÑõ»¯ÄƵÄÐÔÖÊ¡¢Ñõ»¯»¹Ô­·´Ó¦µÈ£¬ÌâÄ¿ÄѶȲ»´ó£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¼×Íé¹ã·º´æÔÚÓÚÌìÈ»Æø¡¢ÕÓÆø¡¢Ãº¿ó¿ÓÆøÖ®ÖУ¬ÊÇÓÅÖʵÄÆøÌåȼÁÏ£¬¸üÊÇÖÆÔìÐí¶à»¯¹¤²úÆ·µÄÖØÒªÔ­ÁÏ
¢ñ£®ÖÆÈ¡ÇâÆø
ÒÑÖª£ºCH4£¨g£©+H2O£¨g£©?CO£¨g£©+3H2£¨g£©¡÷H=+206.2kJ?mol-1
CH4£¨g£©+CO2£¨g£©?2CO£¨g£©+2H2£¨g£©¡÷H=+247.4kJ?mol-1
ÒÔ¼×ÍéΪԭÁÏÖÆÈ¡ÇâÆøÊǹ¤ÒµÉϳ£ÓõÄÖÆÇâ·½·¨
£¨1£©Çëд³öCH4£¨g£©ÓëH2O£¨g£©·´Ó¦Éú³ÉCO2£¨g£©ºÍH2£¨g£©µÄÈÈ»¯Ñ§·½³Ìʽ
 

£¨2£©Èô½«0.1molCH4ºÍ0.2molH2O£¨g£©Í¨ÈëÌå»ýΪ10LµÄÃܱÕÈÝÆ÷ÀÔÚÒ»¶¨Ìõ¼þÏ·¢Éú·´Ó¦£ºCH4£¨g£©+H2O?CO£¨g£©+3H2£¨g£©£¬CH4µÄƽºâת»¯ÂÊÓëζȡ¢Ñ¹Ç¿µÄ¹ØϵÈçͼ

¢ÙÒÑÖª100¡æʱ´ïµ½Æ½ºâËùÐèµÄʱ¼äΪ5min£¬ÔòÓÃH2±íʾµÄƽ¾ù·´Ó¦ËÙÂÊΪ
¢ÚͼÖеÄp1
 
p2£¨Ìî¡°£¼¡±¡¢¡°£¾¡±»ò¡°=¡±£©
£¨3£©ÈôÔÚ100¡æ£¬p1ѹǿÏ£¬½«xmolCH4£¬0.75molH2O£¨g£©¡¢0.025molCO¡¢0.075molH2ͨÈëÌå»ýΪ10LµÄÃܱÕÈÝÆ÷Àһ¶Îʱ¼äºó´ïµ½Æ½ºâʱ£¬CH4µÄת»¯ÂÊÈÔΪ£¨2£©Ïàͬ£¬Ôòx=
 
mol
¢ò£®ÖƱ¸¼×´¼
£¨4£©ÔÚѹǿΪ0.1MpaÌõ¼þÏ£¬½«amolCO Óë3molH2µÄ»ìºÏÆøÌåÔÚ´ß»¯×÷ÓÃÏÂÄÜ×Ô·¢·´Ó¦Éú³É¼×´¼£ºCO£¨g£©+2H2£¨g£©?CH3OH£¨g£©¡÷H£¼0£¬ÈôÈÝÆ÷ÈÝ»ý²»±ä£¬ÏÂÁдëÊ©¿ÉÔö¼Ó¼×´¼²úÂʵÄÊÇ
 

A£®³äÈëHe£¬Ê¹Ìåϵ×ÜѹǿÔö´ó 
B£®ÔÚ³äÈë1molCOºÍ3molH2
C£®Éý¸ßζÈ
D£®½«CH3OH£¨g£©´ÓÌåϵÖзÖÀë³öÀ´
E£®Ê¹Óøü¸ßЧµÄ´ß»¯¼Á
¢ó£®ºÏ³ÉÒÒËᣮ
£¨5£©¼×ÍéÖ±½ÓºÏ³ÉÒÒËá¾ßÓÐÖØÒªµÄÀíÂÛÒâÒåºÍÓ¦ÓüÛÖµ£¬¹â´ß»¯·´Ó¦¼¼ÊõʹÓÃCH4ºÍ
 
£¨Ìѧʽ£©Ö±½ÓºÏ³ÉÒÒËᣬÇÒ·ûºÏ¡°ÂÌÉ«»¯Ñ§¡±µÄÒªÇó£¨Ô­×ÓÀûÓÃÂÊ100%£©£¬ÈôÊÒÎÂϽµamol?L-1µÄÒÒËáÈÜÒººÍbmol?L-1£¬Ba£¨OH£©2ÈÜÒºµÈÌå»ý»ìºÏ£¬»Ö¸´ÊÒκó2c£¨Ba2+£©=c£¨CH3COO-£©£¬ÇëÓú¬aºÍbµÄ´úÊýʽ±íʾ»ìºÏÈÜÒºÖÐÒÒËáµÄµçÀëƽºâ³£ÊýKa=
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÏÂÁгýÔÓ·½°¸ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
Ñ¡Ïî±»Ìá´¿µÄÎïÖÊÔÓÖʳýÔÓÊÔ¼Á³ýÔÓ·½·¨
ACO2£¨g£©SO2£¨g£©±¥ºÍNaHSO3ÈÜÒº¡¢Å¨H2SO4Ï´Æø
BNH4Cl£¨aq£©Fe3+£¨aq£©NaOHÈÜÒº¹ýÂË
CCl2£¨g£©HCl£¨g£©±¥ºÍNaHCO3ÈÜÒº¡¢Å¨H2SO4Ï´Æø
DSO2£¨g£©SO3£¨g£©Å¨H2SO4Ï´Æø
A¡¢AB¡¢BC¡¢CD¡¢D

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ijÈÜÒº¿ÉÄÜ´æÔÚµÄÀë×ÓÈçÏ£ºNa+¡¢Ag +¡¢Ba2+¡¢Al3+¡¢AlO2-¡¢CO32-¡¢SO42-¡¢SO32-¡¢S2-£®ÏÖÏò¸ÃÈÜÒºÖмÓÈë¹ýÁ¿µÄÇâäåËᣬ²úÉúÆøÌåºÍµ­»ÆÉ«³Áµí£®ÏÂÁÐÅжÏÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A¡¢¸ÃÈÜÒºÖÐÒ»¶¨ÓÐAg+
B¡¢¸ÃÈÜÒºÖпÉÄÜûÓÐAlO2-¡¢CO32-
C¡¢¸ÃÈÜÒºÖÐÒ»¶¨ÓÐNa+
D¡¢¸ÃÈÜÒºÖпÉÄÜÓÐBa2+¡¢Al3+

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡°½á¹¹¾ö¶¨ÐÔÖÊ¡±ÊÇѧϰÓлú»¯Ñ§ÓÈΪÖØÒªµÄÀíÂÛ£¬²»½ö±íÏÖÔÚ¹ÙÄÜÍŶÔÎïÖÊÐÔÖʵÄÓ°ÏìÉÏ£¬»¹±íÏÖÔÚÔ­×Ó»òÔ­×ÓÍÅÏ໥µÄÓ°ÏìÉÏ£®ÒÔÏÂÊÂʵ²¢Î´Éæ¼°Ô­×Ó»òÔ­×ÓÍÅÏ໥ӰÏìµÄÊÇ£¨¡¡¡¡£©
A¡¢ÒÒ´¼ÊǷǵç½âÖʶø±½·ÓÓÐÈõËáÐÔ
B¡¢Â±´úÌþÄÑÈÜÓÚË®¶øµÍ¼¶´¼¡¢µÍ¼¶È©Ò×ÈÜÓÚË®
C¡¢¼×´¼Ã»ÓÐËáÐÔ£¬¼×Ëá¾ßÓÐËáÐÔ
D¡¢±½·ÓÒ×ÓëŨäåË®·´Ó¦Éú³É°×É«³Áµí¶ø±½ÓëÒºäåµÄ·´Ó¦ÐèÒªÌú·Û´ß»¯

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÔÚÒ»¶¨Å¨¶ÈµÄBaCl2ÈÜÒºÖÐͨÈëSO2ÆøÌ壬δ¼û³ÁµíÉú³É£¬ÈôÏÈͨÈëÁíÒ»ÖÖÆøÌ壬ÔÙͨÈëSO2£¬Ôò¿ÉÉú³É°×É«³Áµí£®ÁíÒ»ÖÖÆøÌå²»¿ÉÄÜÊÇ£¨¡¡¡¡£©
A¡¢NO2
B¡¢Cl2
C¡¢HCl
D¡¢NH3

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

Ë®ÈÜÒºÖÐÄÜ´óÁ¿¹²´æµÄÒ»×éÀë×ÓÊÇ£¨¡¡¡¡£©
A¡¢K+¡¢Cu2+¡¢OH-¡¢HCO3-
B¡¢Fe2+¡¢H+¡¢ClO-¡¢SiO32-
C¡¢Ca2+¡¢Fe3+¡¢Cl-¡¢CO32-
D¡¢Mg2+¡¢NH4+¡¢Br-¡¢SO42-

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

³£ÎÂÏ£¬Ïò10mL 0.1mol?L-1µÄH2C2O4ÈÜÒºÖÐÖðµÎ¼ÓÈë0.1mol?L-1 KOHÈÜÒº£¬ËùµÃµÎ¶¨ÇúÏßÈçͼËùʾ£®ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A¡¢KHC2O4ÈÜÒº³ÊÈõ¼îÐÔ
B¡¢Bµãʱ£ºc£¨K+£©£¾c£¨HC2O4-£©£¾c£¨C2O42-£©£¾c£¨H+£©£¾c£¨OH-£©
C¡¢Cµãʱ£ºc£¨K+£©£¾c£¨HC2O4-£©+c£¨C2O42-£©+c£¨H2C2O4£©
D¡¢Dµãʱ£ºc£¨H+£©+c£¨HC2O4-£©+c£¨H2C2O4£©=c£¨OH-£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¼×´¼Òѱ»¹ã·ºÓÃÓÚ¹¤ÒµÔ­ÁϺÍÆû³µÈ¼ÁÏ£®
£¨5£©ÓÃCOºÏ³É¼×´¼£¨CH3OH£©µÄ»¯Ñ§·½³ÌʽΪCO£¨g£©+2H2£¨g£©=CH3OH£¨g£©¡÷H£¼0£¬°´ÕÕÏàͬµÄÎïÖʵÄÁ¿Í¶ÁÏ£¬²âµÃCOÔÚ²»Í¬Î¶ÈϵÄƽºâת»¯ÂÊÓëѹǿµÄ¹ØϵÈçÓÒͼËùʾ£¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ
 
£®
a£®Î¶ȣºT1£¼T2£¼T3    b£®Õý·´Ó¦ËÙÂÊ£ºv£¨a£©£¾v£¨c£©£¬v£¨b£©£¾v£¨d£©
c£®Æ½ºâ³£Êý£ºK£¨a£©£¾K£¨c£©£¬K£¨b£©£¾K£¨d£©d£®Æ½¾ùĦ¶ûÖÊÁ¿£ºM£¨a£©£¾M£¨c£©£¬M£¨b£©£¾M£¨d£©

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸