ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿ÏÖÓÐÏÂÁÐÊ®ÖÖÎïÖÊ£º¢ÙH2 ¢ÚÂÁ ¢Û´×Ëá ¢ÜCO2 ¢ÝH2SO4 ¢ÞBa(OH)2¹ÌÌå ¢ß°±Ë® ¢àÏ¡ÏõËá¢áÈÛÈÚAl2(SO4)3 ¢âNaHSO4¡£

(1)°´ÎïÖʵķÖÀà·½·¨Ìîд±í¸ñµÄ¿Õ°×´¦£º

ÊôÓڷǵç½âÖʵÄÊÇ__________£»ÊôÓÚµç½âÖʵÄÊÇ__________£»Äܵ¼µçµÄÊÇ____________¡£

(2)ÉÏÊöÊ®ÖÖÎïÖÊÖÐÓÐÁ½ÖÖÎïÖÊÖ®¼ä¿É·¢ÉúÀë×Ó·´Ó¦£ºH++OH===H2O£¬¸ÃÀë×Ó·´Ó¦¶ÔÓ¦µÄ»¯Ñ§·½³ÌʽΪ_____________________________¡£

(3)д³ö¢ÛºÍ¢ß·´Ó¦µÄÀë×Ó·½³ÌʽΪ__________________£¬34.2 g ¢áÈÜÓÚË®Åä³É250 mLÈÜÒº£¬SO42-µÄÎïÖʵÄÁ¿Å¨¶ÈΪ_________________¡£

(4)¹ýÁ¿µÄ¢ÜͨÈë¢ÞµÄÈÜÒºÖз´Ó¦µÄÀë×Ó·½³ÌʽΪ______________________¡£

(5)½«¢âµÄÈÜÒº¼ÓÈë¢ÞµÄÈÜÒºÖÐÖÁÈÜÒºÖеÄBa2+Àë×ÓÇ¡ºÃÍêÈ«³Áµíʱ·´Ó¦µÄÀë×Ó·½³ÌʽΪ__________¡£

¡¾´ð°¸¡¿¢Ü¢Û¢Ý¢Þ¢á¢â¢Ú¢ß¢à¢áBa(OH)2+2HNO3=Ba(NO3)2+2H2OCH3COOH+NH3¡¤H2O=CH3COO+NH4++H2O1.20 mol/LCO2+OH=HCO3-2H++SO42-+Ba2++2OH= BaSO4¡ý+2H2O

¡¾½âÎö¡¿

(1). CO2 ÔÚÈÜÒºÖÐ×ÔÉí²»ÄܵçÀ룬ÊôÓڷǵç½âÖÊ£»´×Ëá¡¢ÁòËáÊôÓÚËᣬBa(OH)2ÊôÓڼAl2(SO4)3¡¢NaHSO4ÊôÓÚÑΣ¬ÔÚË®ÈÜÒºÖж¼¿ÉÒÔµçÀë³öÀë×ÓʹÈÜÒº¾ßÓе¼µçÐÔ£¬ÊôÓÚµç½âÖÊ£»ÂÁÊǽðÊô£¬¿ÉÒÔµ¼µç£¬ÔÚÏ¡ÏõËáÈÜÒºÖУ¬ÏõËáµçÀë³öÇâÀë×ÓºÍÏõËá¸ùÀë×Ó£¬¿ÉÒÔµ¼µç£¬ÔÚ°±Ë®ÈÜÒºÖУ¬Ò»Ë®ºÏ°±µçÀë³ö笠ùÀë×ÓºÍÇâÑõ¸ùÀë×Ó£¬¿ÉÒÔµ¼µç£¬ÈÛÈÚAl2(SO4)3µçÀë³öÂÁÀë×ÓºÍÁòËá¸ùÀë×Ó£¬¿ÉÒÔµ¼µç£¬¹Ê´ð°¸Îª£º¢Ü£»¢Û¢Ý¢Þ¢á¢â£»¢Ú¢ß¢à¢á

(2.)ÏõËáÓëÇâÑõ»¯±µ·´Ó¦µÄʵÖÊÊÇÇâÀë×ÓÓëÇâÑõ¸ùÀë×Ó·´Ó¦Éú³ÉË®£¬¿ÉÒÔÓÃÀë×Ó·½³ÌʽH++OH=H2O±íʾ£¬¶ÔÓ¦·´Ó¦·½³ÌʽΪ£ºBa(OH)2+2HNO3=Ba(NO3)2+2H2O£¬¹Ê´ð°¸Îª£ºBa(OH)2+2HNO3=Ba(NO3)2+2H2O£»

(3). ´×ËáºÍһˮºÏ°±ÔÚË®ÈÜÒºÖж¼²»ÄÜÍêÈ«µçÀ룬ËùÒÔ´×ËáºÍ°±Ë®·´Ó¦µÄÀë×Ó·½³ÌʽΪCH3COOH+NH3H2O=CH3COO+NH4++H2O£»34.2 g ÁòËáÂÁµÄÎïÖʵÄÁ¿Îª=0.10mol£¬ÓÉÁòËáÂÁµÄ»¯Ñ§Ê½¿ÉÖª£¬ÁòËá¸ùÀë×ÓµÄÎïÖʵÄÁ¿Îª0.30mol£¬ÈÜÓÚË®Åä³É250 mLÈÜÒº£¬SO42-µÄÎïÖʵÄÁ¿Å¨¶ÈΪc(SO42-)==1.20mol/L£¬¹Ê´ð°¸Îª£ºCH3COOH+NH3H2O=CH3COO+NH4++H2O£»1.20mol/L£»

(4).Ba(OH)2ÈÜÒºÓë¹ýÁ¿µÄCO2·´Ó¦Éú³É̼ËáÇâ±µ£¬¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽΪ£ºOH+CO2= HCO3-£¬¹Ê´ð°¸Îª£ºOH+CO2= HCO3-£»

(5).½«ÁòËáÇâÄƵÄÈÜÒº¼ÓÈëBa(OH)2ÈÜÒºÖÐÖÁÈÜÒºÖеÄBa2+Àë×ÓÇ¡ºÃÍêÈ«³Áµíʱ£¬Ba(OH)2ÓëNaHSO4µÄÎïÖʵÄÁ¿Ö®±ÈΪ1:2£¬¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽΪ£º2H++SO42+Ba2++2OH=BaSO4¡ý+2H2O£¬¹Ê´ð°¸Îª£º2H++SO42+Ba2++2OH=BaSO4¡ý+2H2O£»

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿ÓÃÖк͵ζ¨·¨²â¶¨Ä³ÉÕ¼îµÄ´¿¶È£¬ÊÔ¸ù¾ÝʵÑé»Ø´ð£º

£¨1£©³ÆÈ¡4.1gÉÕ¼îÑùÆ·¡£½«ÑùÆ·Åä³É250mL´ý²âÒº£¬ÐèÒªµÄÖ÷Òª²£Á§ÒÇÆ÷³ýÉÕ±­¡¢²£Á§°ôÍ⻹Ðè__________________ ¡¢______________________¡£

£¨2£©È¡10.00mL´ý²âÒº£¬ÓÃ___________________Á¿È¡¡£

£¨3£©ÓÃ0.2010mol¡¤L-1±ê×¼ÑÎËáµÎ¶¨´ý²âÉÕ¼îÈÜÒº£¬ÒÔ·Ó̪Ϊָʾ¼Á£¬µÎ¶¨Ê±×óÊÖÐýתËáʽµÎ¶¨¹ÜµÄ²£Á§»îÈû£¬ÓÒÊÖ²»Í£µØÒ¡¶¯×¶ÐÎÆ¿£¬Á½ÑÛ×¢ÊÓ_____________£¬Ö±µ½¿´µ½______________________________¼´¿ÉÅжϴﵽµÎ¶¨Öյ㡣

£¨4£©¸ù¾ÝÏÂÁÐÊý¾Ý£¬¼ÆËã´ý²âÉÕ¼îÈÜÒºµÄŨ¶ÈΪ£º_____________________£¨½á¹û±£ÁôËÄλÓÐЧÊý×Ö£©£¬ÑùÆ·ÉÕ¼îµÄÖÊÁ¿·ÖÊýΪ________________£¨½á¹û±£ÁôËÄλÓÐЧÊý×Ö£©¡£(¼ÙÉèÉÕ¼îÖв»º¬ÓÐÓëËá·´Ó¦µÄÔÓÖÊ)

µÎ¶¨´ÎÊý

´ý²âÒºÌå»ý

(mL)

±ê×¼ÑÎËáÌå»ý(mL)

µÎ¶¨Ç°¶ÁÊý(mL)

µÎ¶¨ºó¶ÁÊý(mL)

µÚÒ»´Î

10.00

0.50

20.40

µÚ¶þ´Î

10.00

4.00

24.10

£¨5£©µÎ¶¨¹ý³Ì£¬ÏÂÁÐÇé¿ö»áʹ²â¶¨½á¹ûÆ«¸ßµÄÊÇ_____________________________ÌîÐòºÅ£©¡£

¢ÙËáʽµÎ¶¨¹ÜÓÃˮϴºó±ã×°ÒºÌå½øÐе樣»¢Ú¼îʽµÎ¶¨¹Üˮϴºó£¬¾ÍÓÃÀ´Á¿È¡´ý²âÒº£»¢Û׶ÐÎÆ¿ÓÃÕôÁóˮϴµÓºó£¬ÓÖÓôý²âÒºÈóÏ´£»¢ÜµÎ¶¨¹ý¿ì³ÉϸÁ÷¡¢½«¼îÒº½¦µ½×¶ÐÎÆ¿±Ú¶øÓÖδҡÔÈÏ´Ï£»¢ÝÑÎËáÔڵζ¨Ê±½¦³ö׶ÐÎÆ¿Í⣻¢ÞµÎ¼ÓÑÎËᣬÈÜÒºÑÕÉ«ÍÊÈ¥µ«²»×ã°ë·ÖÖÓÓÖ»Ö¸´ºìÉ«£»¢ßµÎ¶¨Ç°£¬ËáʽµÎ¶¨¹ÜÓÐÆøÅÝ£¬µÎ¶¨ºóÏûʧ£»¢à¼Ç¼ÆðʼÌå»ýʱ£¬ÑöÊÓ¶ÁÊý£¬ÖÕµãʱ¸©ÊÓ¡£

¡¾ÌâÄ¿¡¿Ëæ×ŲÄÁÏ¿ÆѧµÄ·¢Õ¹£¬½ðÊô·°¼°Æ仯ºÏÎïµÃµ½ÁËÔ½À´Ô½¹ã·ºµÄÓ¦Ó㬲¢±»ÓþΪ¡°ºÏ½ðµÄάÉúËØ¡±¡£Îª»ØÊÕÀûÓú¬·°´ß»¯¼Á£¨º¬ÓÐV2O5¡¢VOSO4¼°²»ÈÜÐÔ²ÐÔü£©£¬¿ÆÑÐÈËÔ±×îÐÂÑÐÖÆÁËÒ»ÖÖÀë×Ó½»»»·¨»ØÊÕ·°µÄй¤ÒÕ£¬»ØÊÕÂÊ´ï91.7%ÒÔÉÏ¡£²¿·Öº¬·°ÎïÖÊÔÚË®ÖеÄÈܽâÐÔÈçϱíËùʾ£º

ÎïÖÊ

VOSO4

V2O5

NH4VO3

(VO2)2SO4

ÈܽâÐÔ

¿ÉÈÜ

ÄÑÈÜ

ÄÑÈÜ

Ò×ÈÜ

¸Ã¹¤ÒÕµÄÖ÷ÒªÁ÷³ÌÈçÏ¡£

Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©Çëд³ö¼ÓÈëNa2SO3ÈÜÒº·¢Éú·´Ó¦µÄÀë×Ó·½³Ìʽ_________¡£

£¨2£©´ß»¯Ñõ»¯ËùʹÓõĴ߻¯¼Á·°´¥Ã½(V2O5)Äܼӿì¶þÑõ»¯ÁòÑõ»¯ËÙÂÊ£¬´Ë¹ý³ÌÖвúÉúÁËÒ»Á¬´®µÄÖмäÌ壨ÈçÏÂͼ£©¡£ÆäÖÐa¡¢c¶þ²½µÄ»¯Ñ§·½³Ìʽ¿É±íʾΪ_________________£¬______________¡£

£¨3£©¸Ã¹¤ÒÕÖгÁ·¯ÂÊÊÇ»ØÊÕ·°µÄ¹Ø¼üÖ®Ò»£¬³Á·°ÂʵĸߵͳýÊÜÈÜÒºpHÓ°ÏìÍ⣬»¹ÐèÒª¿ØÖÆÂÈ»¯ï§ÏµÊý£¨NH4Cl¼ÓÈëÖÊÁ¿ÓëÁÏÒºÖÐV2O5µÄÖÊÁ¿±È£©ºÍζȡ£¸ù¾ÝÏÂͼÊÔ½¨Òé¿ØÖÆÂÈ»¯ï§ÏµÊýΪ_________£¬¿ØÖÆζȵķ½·¨Îª_________________¡£

£¨4£©½«ÂËÒº1ºÍÂËÒº2»ìºÏºóÓÃÂÈËá¼ØÑõ»¯£¬ÂÈÔªËر»»¹Ô­Îª×îµÍ¼Û£¬Æä·´Ó¦µÄÀë×Ó·´Ó¦·½³ÌʽΪ___¡£

£¨5£©¾­¹ýÈÈÖØ·ÖÎö²âµÃ£ºNH4VO3ÔÚ±ºÉÕ¹ý³ÌÖУ¬¹ÌÌåÖÊÁ¿µÄ¼õÉÙÖµ£¨×Ý×ø±ê£©Ëæζȱ仯µÄÇúÏßÈçͼËùʾ¡£ÔòNH4VO3ÔÚ·Ö½â¹ý³ÌÖÐ_________¡£

A£®ÏÈ·Ö½âʧȥH2O£¬ÔÙ·Ö½âʧȥNH3 B£®ÏÈ·Ö½âʧȥNH3£¬ÔÙ·Ö½âʧȥH2O

C£®Í¬Ê±·Ö½âʧȥH2OºÍNH3D£®Í¬Ê±·Ö½âʧȥH2¡¢N2ºÍH2O

£¨6£©È«·°µç³ØµÄµç½âÖÊÈÜҺΪVOSO4ÈÜÒº£¬µç³ØµÄ¹¤×÷Ô­ÀíΪVO2+ + V2++2H+ VO2+ +H2O +V3+¡£µç³Ø³äµçʱÑô¼«µÄµç¼«·´Ó¦Ê½Îª___________¡£

¡¾ÌâÄ¿¡¿¢ñ\(1)¡°ÄÉÃײÄÁÏ¡±Êǵ±½ñ²ÄÁÏ¿ÆѧÑо¿µÄÇ°ÑØ,ÆäÑо¿³É¹û¹ã·ºÓ¦ÓÃÓÚ´ß»¯¼Á¼°¾üÊ¿ÆѧÖС£Ëùν¡°ÄÉÃײÄÁÏ¡±ÊÇÖ¸Ñо¿¡¢¿ª·¢³öµÄ΢Á£Ö±¾¶´Ó¼¸ÄÉÃ×µ½¼¸Ê®ÄÉÃ׵IJÄÁÏ,È罫ÄÉÃײÄÁÏ·ÖÉ¢µ½·ÖÉ¢¼ÁÖÐ,ËùµÃ»ìºÏÎï¿ÉÄܾßÓеÄÐÔÖÊÊÇ ____¡£

A.ÄÜÈ«²¿Í¸¹ýÂËÖ½ B.Óж¡´ï¶ûЧӦ C.ËùµÃÒºÌå³Ê½º×´ D.ËùµÃÎïÖÊÒ»¶¨ÊÇÐü×ÇÒº

(2)°Ñµí·ÛÈÜÒºÈÜÓÚ·ÐË®ÖÐ,ÖƳɵí·Û½ºÌå,¼ø±ðÈÜÒººÍµí·Û½ºÌå¿ÉÒÔÀûÓõķ½·¨ÊÇ__________¡£

(3)°ÑÉÙÁ¿µÄFeCl3±¥ºÍÈÜÒºµÎÈë·ÐË®ÖÐ,ÖƳÉFe(OH)3½ºÌåºóµÎÈëÉÙÁ¿Ï¡ÁòËá¿É¹Û²ìµ½µÄÏÖÏóÊÇ£º__£¬¼ÌÐøµÎ¼ÓÖÁÏ¡ÁòËá¹ýÁ¿¿É¹Û²ìµ½µÄÏÖÏóÊÇ_____£¬Ð´³öÉÏÊö·´Ó¦µÄ»¯Ñ§·½³Ìʽ____¡£

¢ò\¸øÏÂͼ¢Ù~¢ÝÑ¡ÔñÊʵ±µÄÎïÖÊ£¬Ê¹ÓÐÁ¬ÏßµÄÁ½ÎïÖÊÄÜ·¢Éú·´Ó¦¡£¹©Ñ¡ÔñµÄÊÔ¼ÁÓÐÏ¡ÁòËá¡¢¶þÑõ»¯Ì¼¡¢Í­Æ¬¡¢Ê³ÑΡ¢Éúʯ»Ò¡¢Ò»Ñõ»¯Ì¼¡¢´¿¼î¡¢ÌúƬºÍľ̿·Û¡£

£¨1£©ÇëÍƶÏËüÃǵĻ¯Ñ§Ê½·Ö±ðΪ£º¢Ú______£¬¢Û ______£¬¢Ý_______¡£

£¨2£©Ð´³öÏÂÁÐÐòºÅÖ®¼äµÄ»¯Ñ§·½³Ìʽ£º¢ÙºÍ¢Ú£º_______________£¬¢ÚºÍ¢Ü£º____________£¬¢ÛºÍ¢Ü£º______________¡£

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

¾«Ó¢¼Ò½ÌÍø