ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿ÒÑÖª25 ¡æʱ²¿·ÖÈõµç½âÖʵĵçÀëƽºâ³£ÊýÊý¾ÝÈçÏÂ±í£º

ÈõËữѧʽ

HSCN

CH3COOH

HCN

H2CO3

µçÀëƽºâ³£Êý

1.3¡Á101

1.7¡Á105

6.2¡Á1010

K1=4.3¡Á107

K2=5.6¡Á1011

»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©Ð´³ö̼ËáµÄµÚÒ»¼¶µçÀëƽºâ³£Êý±í´ïʽ£ºK1=_______________________¡£

£¨2£©µÈÎïÖʵÄÁ¿Å¨¶ÈµÄa.CH3COONa¡¢b.NaCN¡¢c.Na2CO3¡¢d.NaHCO3ÈÜÒºµÄpHÓÉ´óµ½Ð¡µÄ˳ÐòΪ________(Ìî×Öĸ)¡£

£¨3£©³£ÎÂÏ£¬0.1 mol¡¤L1µÄCH3COOHÈÜÒº¼ÓˮϡÊ͹ý³ÌÖУ¬ÏÂÁбí´ïʽµÄÊý¾Ý±ä´óµÄÊÇ________(ÌîÐòºÅ)

A£®[H£«] B£®[H£«]/[CH3COOH]

C£®[H£«]¡¤[OH] D£®[OH]/[H£«]

£¨4£©Ìå»ý¾ùΪ100 mL pH=2µÄCH3COOHÓëÒ»ÔªËáHX£¬¼ÓˮϡÊ͹ý³ÌÖÐpHÓëÈÜÒºÌå»ýµÄ¹ØϵÈçͼËùʾ£¬ÔòHXµÄµçÀëƽºâ³£Êý________(Ìî¡°´óÓÚ¡±¡¢¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡±)CH3COOHµÄµçÀëƽºâ³£Êý¡£

£¨5£©Ð´³öÉÙÁ¿CO2ͨÈë´ÎÂÈËáÄÆÈÜÒºÖеÄÀë×Ó·½³Ìʽ£º_____________________________¡£

¡¾´ð°¸¡¿ cbdaBDСÓÚCO2£«H2O£«ClO===HCO3-£«HClO

¡¾½âÎö¡¿

£¨1£©Ì¼ËáµÄµÚÒ»²½µçÀë·½³ÌʽÊÇ £¬¸ù¾ÝµçÀë·½³ÌʽÊéдµçÀëƽºâ³£Êý±í´ïʽ£»£¨2£©ÈõËáµçÀëƽºâ³£ÊýԽС£¬Ëá¸ùÀë×ÓË®½â³Ì¶ÈÔ½´ó£¬Ë®½â³Ì¶È£ºCO32-£¾CN-£¾HCO3-£¾CH3COO-£¬Ë®½â¾ùÏÔ¼îÐÔ£¬Ë®½â³Ì¶ÈÔ½´ó£¬¼îÐÔԽǿ£»£¨3£©CH3COOHÈÜÒº¼ÓˮϡÊ͹ý³Ì£¬´Ù½øµçÀ룬c(H+)¼õС£¬c(CH3COO-)¼õС£» Kw²»±ä£¬c(OH-)Ôö´ó£»£¨4£©ÓÉͼ¿ÉÖª£¬Ï¡ÊÍÏàͬµÄ±¶Êý£¬HXµÄpH±ä»¯³Ì¶È±È´×ËáС£¬ÔòHX±È´×ËáµÄËáÐÔÈõ£¬µçÀëƽºâ³£ÊýС£»£¨5£©ËáÐÔHCO3-£¼HClO£¼H2CO3£¬ÉÙÁ¿CO2ͨÈë´ÎÂÈËáÄÆÈÜÒºÖÐÉú³É̼ËáÇâÄƺʹÎÂÈËá¡£

£¨1£©Ì¼ËáµÄµÚÒ»²½µçÀë·½³ÌʽÊÇ £¬Ì¼ËáµÄµÚÒ»¼¶µçÀëƽºâ³£Êý±í´ïʽÊÇK1=£»£¨2£©ÈõËáµçÀëƽºâ³£ÊýԽС£¬Ëá¸ùÀë×ÓË®½â³Ì¶ÈÔ½´ó£¬Ë®½â³Ì¶È£ºCO32-£¾CN-£¾HCO3-£¾CH3COO-£¬Ë®½â¾ùÏÔ¼îÐÔ£¬Ë®½â³Ì¶ÈÔ½´ó£¬¼îÐÔԽǿ£¬ËùÒÔpHÓÉ´óµ½Ð¡µÄ˳ÐòΪNa2CO3>NaCN>NaHCO3>CH3COONa£»£¨3£©CH3COOHÈÜÒº¼ÓˮϡÊ͹ý³Ì£¬´Ù½øµçÀ룬c£¨H+£©¼õС£¬µçÀëƽºâ³£Êý £¬c£¨CH3COO-£©¼õС£¬ËùÒÔ[H£«]/[CH3COOH]Ôö´ó£» Kw²»±ä£¬c(H+)¼õС£¬ËùÒÔc(OH-)Ôö´ó£¬[OH]/[H£«]Ôö´ó£¬¹ÊÑ¡BD¡££¨4£©ÓÉͼ¿ÉÖª£¬Ï¡ÊÍÏàͬµÄ±¶Êý£¬HXµÄpH±ä»¯³Ì¶È±È´×ËáС£¬ÔòHX±È´×ËáµÄËáÐÔÈõ£¬µçÀëƽºâ³£ÊýСÓÚ´×Ë᣻£¨5£©ËáÐÔHCO3-£¼HClO£¼H2CO3£¬ÉÙÁ¿CO2ͨÈë´ÎÂÈËáÄÆÈÜÒºÖÐÉú³É̼ËáÇâÄƺʹÎÂÈËᣬÀë×Ó·½³ÌʽÊÇCO2£«H2O£«ClO = HCO3-£«HClO¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿¼×°·Ç¦µâ£¨CH3NH3PbI3£©ÓÃ×÷È«¹Ì̬¸ÆîÑ¿óÃô»¯Ì«ÑôÄܵç³ØµÄÃô»¯¼Á£¬¿ÉÓÉCH3NH2¡¢PbI2¼°HIΪԭÁϺϳɣ¬»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©ÖÆÈ¡¼×°·µÄ·´Ó¦ÎªCH3OH£¨g£©£«NH3£¨g£©CH3NH2£¨g£©£«H2O£¨g£©¡¡¦¤H¡£ÒÑÖª¸Ã·´Ó¦ÖÐÏà¹Ø»¯Ñ§¼üµÄ¼üÄÜÊý¾ÝÈçÏ£º

¹²¼Û¼ü

C¡ªO

H¡ªO

N¡ªH

C¡ªN

C¡ªH

¼üÄÜ/kJ¡¤mol£­1

351

463

393

293

414

Ôò¸Ã·´Ó¦µÄ¦¤H£½____kJ¡¤mol£­1¡£

£¨2£©ÉÏÊö·´Ó¦ÖÐËùÐèµÄ¼×´¼¹¤ÒµÉÏÀûÓÃˮúÆøºÏ³ÉCO£¨g£©£«2H2£¨g£©CH3OH£¨g£©¡¡¦¤H<0¡£ÔÚÒ»¶¨Ìõ¼þÏ£¬½«1 mol COºÍ2 mol H2ͨÈëÃܱÕÈÝÆ÷ÖнøÐз´Ó¦£¬µ±¸Ä±äijһÍâ½çÌõ¼þ£¨Î¶Ȼòѹǿ£©Ê±£¬CH3OHµÄÌå»ý·ÖÊý¦Õ£¨CH3OH£©±ä»¯Ç÷ÊÆÈçͼËùʾ£º

¢Ùƽºâʱ£¬MµãCH3OHµÄÌå»ý·ÖÊýΪ10%£¬ÔòCOµÄת»¯ÂÊΪ___¡£

¢ÚXÖáÉÏaµãµÄÊýÖµ±Èbµã____ £¨Ìî¡°´ó¡±»ò¡°Ð¡¡±£©¡£Ä³Í¬Ñ§ÈÏΪÉÏͼÖÐYÖá±íʾζȣ¬ÄãÈÏΪËûÅжϵÄÀíÓÉÊÇ_________________¡£

£¨3£©¹¤ÒµÉϿɲÉÓÃCH3OHCO+2H2 À´ÖÆÈ¡¸ß´¿¶ÈµÄCOºÍH2¡£ÎÒ¹úѧÕß²ÉÓÃÁ¿×ÓÁ¦Ñ§·½·¨£¬Í¨¹ý¼ÆËã»úÄ£Ä⣬Ñо¿ÁËÔÚîÙ»ù´ß»¯¼Á±íÃæÉϼ״¼ÖÆÇâµÄ·´Ó¦Àú³Ì£¬ÆäÖÐÎü¸½ÔÚîÙ´ß»¯¼Á±íÃæÉϵÄÎïÖÖÓÃ*±ê×¢¡£

¼×´¼£¨CH3OH£©ÍÑÇâ·´Ó¦µÄµÚÒ»²½Àú³Ì£¬ÓÐÁ½ÖÖ¿ÉÄÜ·½Ê½£º

·½Ê½ A£ºCH3OH* ¡úCH3O* £«H* Ea= +103.1kJ¡¤mol-1

·½Ê½ B£ºCH3OH* ¡úCH3* £«OH* Eb= +249.3kJ¡¤mol-1

ÓɻÄÜEÖµÍƲ⣬¼×´¼Áѽâ¹ý³ÌÖ÷ÒªÀú¾­µÄ·½Ê½Ó¦Îª___£¨ÌîA¡¢B£©¡£

ÏÂͼΪ¼ÆËã»úÄ£ÄâµÄ¸÷²½·´Ó¦µÄÄÜÁ¿±ä»¯Ê¾Òâͼ¡£

¸ÃÀú³ÌÖУ¬·ÅÈÈ×î¶àµÄ²½ÖèµÄ»¯Ñ§·½³ÌʽΪ________¡£

£¨4£©PbI2Óë½ðÊôï®ÒÔLiI-Al2O3¹ÌÌåΪµç½âÖÊ×é³É﮵âµç³Ø£¬Æä½á¹¹Ê¾ÒâͼÈçÏ£¬µç³Ø×Ü·´Ó¦¿É±íʾΪ£º2Li+PbI2=2LiI+Pb£¬Ôòb¼«Éϵĵ缫·´Ó¦Ê½Îª£º_____¡£

£¨5£©CH3NH2µÄµçÀë·½³ÌʽΪCH3NH2+H2OCH3NH3++OH-µçÀë³£ÊýΪkb£¬ÒÑÖª³£ÎÂÏÂpkb=-lgkb=3.4£¬Ôò³£ÎÂÏÂÏòCH3NH2ÈÜÒºµÎ¼ÓÏ¡ÁòËáÖÁc£¨CH3NH2£©=c£¨CH3NH3+£©Ê±£¬ÈÜÒºpH=______¡£

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

¾«Ó¢¼Ò½ÌÍø