ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿Ñо¿µç½âÖÊÔÚË®ÈÜÒºÖеÄƽºâÄÜÁ˽âËüµÄ´æÔÚÐÎʽ£¬ÓÐÖØÒªµÄʵ¼ÊÒâÒå¡£

£¨1£©³£ÎÂÏ£¬Ïò100mL0.01mol¡¤L-1HAµÄÈÜÒºÖÐÖðµÎ¼ÓÈë0.02mol¡¤L-1MOHÈÜÒº£¬ËùµÃÈÜÒºµÄpHËæMOHÈÜÒºµÄÌå»ý±ä»¯ÈçͼËùʾ£¨ÈÜÒºÌå»ý±ä»¯ºöÂÔ²»¼Æ£©¡£

¢Ù³£ÎÂÏ£¬0.01mol¡¤L-1HAÈÜÒºÖÐÓÉË®µçÀë³öµÄc£¨H+£©=_____mol¡¤L-1¡£

¢Ú³£ÎÂÏÂÒ»¶¨Å¨¶ÈµÄMAÏ¡ÈÜÒºµÄpH=a£¬Ôòa____7£¨Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°=¡±£©£¬ÓÃÀë×Ó·½³Ìʽ±íʾÆäÔ­ÒòΪ_____¡£

¢ÛXµãʱ£¬ÈÜÒºÖÐc£¨H+£©¡¢c£¨M+£©¡¢c£¨A-£©ÓÉ´óµ½Ð¡µÄ˳ÐòÊÇ_____¡£

¢ÜKµãʱ£¬ÈÜÒºÖÐc£¨H+£©+ c£¨M+£©- c£¨OH-£©=____molL-1¡£

£¨2£©25 ¡æʱ£¬ÒÑÖªKa£¨CH3COOH£©£½1.75¡Á10£­5£¬Kb£¨NH3¡¤H2O£©£½1.76¡Á10£­5¡£È¡Å¨¶È¾ùΪ0.100 0 mol¡¤L£­1µÄ´×ËáÈÜÒººÍ°±Ë®ÈÜÒº¸÷20.00 mLÓÚ׶ÐÎÆ¿ÖУ¬·Ö±ðÓÃ0.1000 mol¡¤L£­1NaOHÈÜÒº¡¢0.1000 mol¡¤L£­1ÑÎËá½øÐе樣¬µÎ¶¨¹ý³ÌÖÐpHËæµÎ¼ÓÈÜÒºÌå»ý±ä»¯¹ØϵÈçͼËùʾ¡£

¢Ù´ÓÇúÏßI¿ÉÖª£¬¸ÃµÎ¶¨²Ù×÷ʱ£¬Ó¦Ñ¡______×÷Ϊָʾ¼Á¡£

¢ÚÔÚÇúÏߢò£¬µ±µÎ¼ÓÈÜÒºµ½10.00 mLʱ£º

c£¨CH3COO£­£©£«c£¨OH£­£© ___ c£¨H£«£©£«c£¨CH3COOH£© £¨Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°=¡±£©

¡¾´ð°¸¡¿10-12 < M++ H2OMOH+H+ c(A-)> c(H+)> c(M+) 0.0005 ¼×»ù³È >

¡¾½âÎö¡¿

(1)¢ÙÓÉpHÖª£¬HAΪǿËᣬÈÜÒºÖÐÓÉË®µçÀë³öµÄc(H+)µÈÓÚÈÜÒºÖеÄOH-Ũ¶È¡£

¢Ú¶ÔÓÚÇ¿ËáÈõ¼îÑΣ¬½ðÊôÀë×Ó·¢ÉúË®½â£¬ÈÜÒºÏÔËáÐÔ¡£

¢ÛXµãʱ£¬´ÓÈÜÒºÖеÄÈÜÖʼ°µçÀëÄÜÁ¦½øÐзÖÎö£¬´Ó¶øµÃ³öÈÜÒºÖÐc(H+)¡¢c(M+)¡¢c(A-)ÓÉ´óµ½Ð¡µÄ˳Ðò¡£

¢Ü´ÓµçºÉÊغã½øÐзÖÎö£¬Çó³öc(A-)¡£

(2) ¢ÙÒòΪµÎ¶¨ÖÕµãʱ£¬ÈÜÒº³ÊËáÐÔ£¬Ó¦Ñ¡±äÉ«·¶Î§µãÔÚËáÐÔÇøÓòµÄָʾ¼Á¡£

¢ÚÀûÓøõãÈÜÒºµÄ×é³É¼°µçÀëÓëË®½âµÄÖ÷´Î£¬È·¶¨Î¢Á£Å¨¶ÈµÄÏà¶Ô´óС¡£

£¨1£©´ÓͼÖпÉÒÔ¿´³ö£¬0.01mol¡¤L-1HAµÄpH=2£¬ËµÃ÷ÆäΪǿËá¡£

¢Ù³£ÎÂÏ£¬0.01mol¡¤L-1HAÈÜÒºÖÐÓÉË®µçÀë³öµÄc(H+)= c(OH-)=mol¡¤L-1£»

´ð°¸£º10-12¡£

¢Úµ±pH=7ʱ£¬V(MOH)=51mL£¬n(HA)=0.001mol£¬n(MOH)=0.00102mol£¬MOH¹ýÁ¿£¬´Ó¶ø˵Ã÷MAΪǿËáÈõ¼îÑΣ¬pH=a£¬Ôòa<7£»

Àë×Ó·½³Ìʽ±íʾÆäÔ­ÒòΪM++ H2OMOH+H+£»

´ð°¸£º<£»M++ H2OMOH+H+¡£

¢ÛXµãʱ£¬n(HA)=0.001mol£¬n(MOH)=0.0005mol£¬·´Ó¦Éú³Én(MA)=0.0005mol£¬Ê£Óàn(HA)=0.0005mol£¬ÈÜÒºÖÐc(H+)¡¢c(M+)¡¢c(A-)ÓÉ´óµ½Ð¡µÄ˳ÐòÊÇc(A-)> c(H+)> c(M+)£»

´ð°¸£ºc(A-)> c(H+)> c(M+)¡£

¢ÜKµãʱ£¬n(HA)=0.001mol£¬n(MOH)=0.002mol£¬·´Ó¦Éú³Én(MA)=0.001mol£¬Ê£Óàn(HA)=0.001mol£¬ÈÜÒºÖÐc(H+)+ c(M+)- c(OH-)= c(A-)=0.0005molL-1¡£

´ð°¸£º0.0005¡£

£¨2£©´Ó´×ËáºÍ°±Ë®µÄËá¼îÐÔÅжϣ¬ÇúÏßIΪ°±Ë®£¬ÇúÏߢòΪ´×Ëá¡£

¢Ù´ÓÇúÏßI¿ÉÖª£¬¸ÃµÎ¶¨²Ù×÷ʱ£¬µÎ¶¨ÖÕµãÔÚËáÐÔÇøÓòÄÚ£¬Ó¦Ñ¡¼×»ù³È×÷Ϊָʾ¼Á£»

´ð°¸£º¼×»ù³È¡£

¢ÚÔÚÇúÏߢò£¬µ±µÎ¼ÓÈÜÒºµ½10.00 mLʱ£¬n(CH3COOH)=0.002mol£¬n(NaOH)=0.001mol£¬

·´Ó¦ºóÈÜÒºÖУ¬n(CH3COOH)=0.001mol£¬n(CH3COONa)=0.001mol£¬ÒÔ´×ËáµÄµçÀëΪÖ÷£¬

c(CH3COO£­)£«c(OH£­) >c(H£«)£«c(CH3COOH) £»

´ð°¸£º>¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿2018Ä꣬ÃÀ¹úÍ˳öÁË¡¶°ÍÀèЭ¶¨¡·ÊµÐÐÔÙ¹¤Òµ»¯Õ½ÂÔ£¬¶øÖйúÈ´¼Ó´óÁË»·±£Á¦¶È£¬Éú¶¯Ú¹ÊÍÁËÎÒ¹ú¸ºÔðÈεĴó¹úÐÎÏó¡£½üÄêÀ´£¬ÎÒ¹ú´óÁ¦¼ÓÇ¿ÎÂÊÒÆøÌåCO2´ß»¯Ç⻯ºÏ³É¼×´¼¼¼ÊõµÄ¹¤Òµ»¯Á¿²úÑо¿£¬ÊµÏֿɳÖÐø·¢Õ¹¡£

£¨1£©ÒÑÖª£ºCO2£¨g£©+ H2£¨g£©H2O£¨g£© +CO£¨g£© ¦¤H1 = + 41.1 kJmol-1

CO£¨g£©+2H2£¨g£©CH3OH£¨g£© ¦¤H2= £­90.0 kJmol-1

д³öCO2´ß»¯Ç⻯ºÏ³É¼×´¼µÄÈÈ»¯Ñ§·½³Ìʽ_________¡£

£¨2£©ÎªÌá¸ßCH3OH²úÂÊ£¬ÀíÂÛÉÏÓ¦²ÉÓõÄÌõ¼þÊÇ_________£¨Ìî×Öĸ£©¡£

a ¸ßθßѹ b µÍεÍѹ c ¸ßεÍѹ d µÍθßѹ

£¨3£©250¡æ¡¢ÔÚÌå»ýΪ2.0 LµÄºãÈÝÃܱÕÈÝÆ÷ÖмÓÈë6 mol H2¡¢2 mol CO2ºÍ´ß»¯¼Á£¬10 minʱ·´Ó¦´ïµ½Æ½ºâ£¬²âµÃc£¨CH3OH£© = 0.75 mol¡¤ L£­1¡£

¢Ù Ç°10 minµÄƽ¾ù·´Ó¦ËÙÂÊv£¨H2£©£½______ mol¡¤L£­1¡¤min£­1¡£

¢Ú »¯Ñ§Æ½ºâ³£ÊýµÄÖµ ______¡£

£¨4£©ÒÔCO2£¨g£©ºÍH2£¨g£©ÎªÔ­ÁϺϳɼ״¼£¬·´Ó¦µÄÄÜÁ¿±ä»¯ÈçÏÂͼËùʾ¡£

¢Ù ²¹È«ÉÏͼ£ºÍ¼ÖÐA´¦Ó¦ÌîÈë_____¡£

¢Ú ¸Ã·´Ó¦ÐèÒª¼ÓÈëÍ­£­Ð¿»ù´ß»¯¼Á¡£¼ÓÈë´ß»¯¼Áºó£¬¸Ã·´Ó¦µÄ¦¤ H_____£¨Ìî¡°±ä´ó¡±¡°±äС¡±»ò¡°²»±ä¡±£©¡£

£¨5£©CO2»¹¿ÉÒÔÓëCH4·´Ó¦Éú³ÉCH3COOH¡£·´Ó¦Àú³ÌÈçÏÂͼ1¡££¨ÖмäÌåµÄÄÜÁ¿¹ØϵÈçÐé¿òÖÐÇúÏßËùʾ£© ÊÒÎÂÏÂijÈÜÒºÖÐCH3COOHºÍCH3COO£­Á½ÖÖ΢Á£Å¨¶ÈËæpH±ä»¯µÄÇúÏߣ¬ÈçÏÂͼ2¡£
¢ÙÖмäÌå¢ÙµÄÄÜÁ¿______ÖмäÌå¢ÚµÄÄÜÁ¿¡££¨Ì¡°>¡±»ò¡°=¡±»ò¡°<¡±£©

¢Ú´Óͼ2Öеóö£¬ÊÒÎÂÏ£¬CH3COOHµÄµçÀë³£ÊýֵΪ____¡£

¡¾ÌâÄ¿¡¿Ä³ÊµÑéС×éѧϰÁË·´Ó¦CO2+2Mg2MgO+CµÄÏà¹Ø֪ʶºó£¬Óû̽¾¿Ã¾ÊÇ·ñÓëSO2ÆøÌå·´Ó¦£¬Éè¼ÆÁËÏÂÁÐʵÑé×°ÖÃͼ(×¢£ºÊ¯ÃÞÈÞÊÇÒ»ÖÖÄÍ»ð²ÄÁÏ£¬²»²ÎÓë·´Ó¦)¡£

Çë»Ø´ðÏÂÁÐÎÊÌ⣺

(1)Èô¿É¹©Ñ¡Ôñ¹ÌÌåÊÔ¼ÁÓУºÄ¾Ì¿¡¢Í­Æ¬¡¢ÑÇÁòËáÄƹÌÌ壬Ôò×°ÖÃAÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ____________________________________________¡£

(2)×°ÖÃBÖÐËùÊ¢ÈëµÄÒ©Æ·ÊÇ______________________¡£

(3)ÒÇÆ÷aµÄÃû³ÆÊÇ___________£¬Æä×÷ÓÃÊÇ______________________¡£

(4)ʵÑéÍê±Ïºó¹Û²ìµ½Ê¯ÃÞÈÞÉÏÓлÆÉ«ÎïÖÊÉú³É£¬½«Ê¯ÃÞÈÞ¼°Æ丽×ÅÎïͶÈ뵽ϡÑÎËáÖУ¬»á²úÉú¾ßÓгô¼¦µ°ÆøζµÄÆøÌ壬ÔòþÓëSO2³ý·¢ÉúSO2+2Mg2MgO+S·´Ó¦Í⣬»¹·¢ÉúµÄ·´Ó¦ÊÇ____________¡£

(5)ÔÚ×°ÖÃC¡¢D¼ä²åÈëͼËùʾװÖ㬿ÉÒÔ»ñµÃÑÇÁòËá(H2SO3)ÈÜÒº¡£

¢ÙÒºÌåX¿ÉÒÔÊÇÏÂÁÐÊÔ¼ÁÖеÄ___________(ÌîÐòºÅ)¡£

A.±½ B.ÆûÓÍ C.ËÄÂÈ»¯Ì¼ D.¾Æ¾«

¢ÚʵÑéÍê±Ïºó£¬Í¨¹ý___________(ʵÑé²Ù×÷Ãû³Æ)¿ÉÒÔ½«ÉÏ¡¢ÏÂÁ½²ãÒºÌå·Ö¿ª¡£

¢ÛÏòÑÇÁòËáÈÜÒºÖÐͨÈë¿ÕÆø£¬ÈÜÒºµÄpH_______(Ìî¡°Ôö´ó¡±¡¢¡°¼õС¡±»ò¡°²»±ä¡±)£¬ÆäÔ­ÒòÊÇ______(Óû¯Ñ§·½³Ìʽ±íʾ)¡£

¡¾ÌâÄ¿¡¿ÎªÁ˸üÉî¿ÌµØÈÏʶ±ËصÄÐÔÖÊ£¬Ä³»¯Ñ§Ð¡×é¶Ô±Ëؼ°Æ仯ºÏÎïµÄÖƱ¸ºÍÐÔÖʽøÐÐÈçÏÂ̽¾¿ÊµÑ飬¸ù¾ÝʵÑé»Ø´ðÎÊÌâ¡£

[ʵÑéÒ»]ÂÈÆøµÄÖƱ¸

£¨1£©¸ÃС×éÄâÓÃͼ¼×ʵÑé×°ÖÃÀ´ÖƱ¸´¿¾»¡¢¸ÉÔïµÄÂÈÆø£¬²¢Íê³ÉÓë½ðÊôÌúµÄ·´Ó¦(¼Ð³ÖÒÇÆ÷ÂÔÈ¥)¡£Ã¿¸öÐéÏß¿ò±íʾһ¸öµ¥Ôª×°Öã¬ÇëÓÃÎÄ×ÖÃèÊö½«ÏÂÁÐ×°ÖõĴíÎóÖ®´¦¸ÄÕý£º___¡£

[ʵÑé¶þ]̽¾¿ÂÈ»¯ÑÇÌúÓëÑõÆø·´Ó¦µÄ²úÎï

ÒÑÖªÂÈ»¯ÑÇÌúµÄÈÛµãΪ674¡æ£¬·ÐµãΪ1023¡æ£»ÈýÂÈ»¯ÌúÔÚ100¡æ×óÓÒʱÉý»ª£¬¼«Ò×Ë®½â¡£ÔÚ500¡æÌõ¼þÏÂÂÈ»¯ÑÇÌúÓëÑõÆø¿ÉÄÜ·¢ÉúÏÂÁз´Ó¦£º12FeCl2+3O22Fe2O3+8FeCl3¡¢4FeCl2+3O22Fe2O3 +4Cl2£¬¸Ã»¯Ñ§Ð¡×éÑ¡ÓÃͼÒÒ²¿·Ö×°ÖÃ(×°ÖÿÉÒÔÖظ´Ñ¡ÓÃ)½øÐÐÂÈ»¯ÑÇÌúÓëÑõÆø·´Ó¦²úÎïµÄ̽¾¿¡£

£¨2£©ÊµÑé×°ÖõĺÏÀíÁ¬½Ó˳ÐòΪA¡ú___¡úE¡£

£¨3£©¼òÊö½«×°ÖÃFÖеĹÌÌåÅä³ÉÈÜÒºµÄ²Ù×÷·½·¨£º___¡£

[ʵÑéÈý]±ËØ»¯ºÏÎïÖ®¼ä·´Ó¦µÄʵÑéÌõ¼þ¿ØÖÆ̽¾¿

£¨4£©ÔÚ²»Í¬ÊµÑéÌõ¼þÏÂKClO3¿É½«KIÑõ»¯ÎªI2»òKIO3¡£¸ÃС×éͬѧÉè¼ÆµÄÒ»×éʵÑéµÄÊý¾Ý¼Ç¼Èçϱí(ʵÑé¿ØÖÆÔÚÊÒÎÂϽøÐÐ)£º

ÊԹܱêºÅ

1

2

3

4

0.20molL-1KIÈÜÒº/mL

1.0

1.0

1.0

1.0

KClO3(s)/g

0.10

0.10

0.10

0.10

6.0molL-1H2SO4ÈÜÒº/mL

0

3.0

6.0

9.0

ÕôÁóË®/mL

9.0

6.0

3.0

0

ʵÑéÏÖÏó

¢Ù¸Ã×éʵÑéµÄÄ¿µÄÊÇ___¡£

¢Ú2ºÅÊԹܷ´Ó¦ÍêÈ«ºó£¬È¡ÉÙÁ¿ÊÔ¹ÜÖеÄÈÜÒº£¬µÎ¼Óµí·ÛÈÜÒººóÏÔÀ¶É«£¬¼ÙÉ軹ԭ²úÎïÖ»ÓÐKCl£¬Ð´³ö·´Ó¦µÄÀë×Ó·½³Ìʽ£º___¡£

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø