¡¾ÌâÄ¿¡¿ÓÐÏÂÁл¯ºÏÎ¢ÙNaCl¢ÚNaOH¢ÛHCl¢ÜFeCl3¢ÝCH3COONa¢ÞCH3COOH¢ßNH3H2O¢àH2O
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©FeCl3ÈÜÒºÏÔ__________ÐÔ£¬ÓÃÀë×Ó·½³Ìʽ±íʾ______________________£»
CH3COOHÈÜÒºÏÔ________ÐÔ£¬ÓÃÀë×Ó·½³Ìʽ±íʾ__________________________£»
£¨2£©³£ÎÂÏ£¬pH=10µÄCH3COONaÈÜÒºÖУ¬Ë®µçÀë³öÀ´µÄc(OH-)=_________________£»
ÔÚpH=3HClµÄÈÜÒºÖУ¬Ë®µçÀë³öÀ´µÄc(H+)=___________________£»
£¨3£©ÒÑ֪ˮ´æÔÚÈçÏÂƽºâ£ºH2O+H2OH3O++OH-¡÷H>0£¬ÏÖÓûʹƽºâÏòÓÒÒƶ¯£¬ÇÒËùµÃÈÜÒºÏÔËáÐÔ£¬Ñ¡Ôñ·½·¨ÊÇ____________¡£
A.ÏòË®ÖмÓÈëNaHSO4¹ÌÌå
B.ÏòË®ÖмÓÈë(NH4)2SO4¹ÌÌå
C.¼ÓÈÈÖÁ100¡æ[ÆäÖÐc(H+)=1¡Á10-6molL-1]
D¡¢ÏòË®ÖмÓNa2CO3¹ÌÌå
£¨4£©ÈôµÈŨ¶È¡¢µÈÌå»ýµÄ¢ÚNaOHºÍ¢ßNH3H2O·Ö±ð¼ÓˮϡÊÍm±¶¡¢n±¶£¬Ï¡ÊͺóÁ½ÖÖÈÜÒºµÄpHÏàµÈ£¬Ôòm___________n£¨Ìî¡°<¡±¡¢¡°>¡±»ò¡°=¡±£©¡£
£¨5£©25¡æ£¬pH=aµÄÑÎËáVamLÓëpH=14-aµÄ°±Ë®VbmL»ìºÏ£¬ÈôÈÜÒºÏÔÖÐÐÔ£¬ÔòVa_____Vb(Ìî¡°>¡±¡¢¡°<¡±¡¢¡°=¡±¡¢¡°ÎÞ·¨È·¶¨¡±)
£¨6£©³ý¢àH2OÍ⣬ÈôÆäÓà7ÖÖÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈÏàͬ£¬ÔòÕâ7ÖÖÈÜÒº°´pHÓÉ´óµ½Ð¡µÄ˳ÐòΪ£º______________________________________________£¨ÌîÐòºÅ£©¡£
£¨7£©³£ÎÂÏÂpH=13µÄNaOHÈÜÒºÓëpH=2µÄÑÎËáÈÜÒº»ìºÏ£¬ËùµÃ»ìºÏÒºµÄpH£½11£¬ÔòNaOHÓëÑÎËáµÄÌå»ý±ÈΪ________
¡¾´ð°¸¡¿Ëá Fe 3+ + 3H2O Fe£¨OH£©3 + 3H + Ëá CH3COOH CH3COO- + H + 10-4 mol£¯L 10-11 mol£¯L B £¾ £¾ ¢Ú£¾¢ß£¾¢Ý£¾¢Ù£¾¢Ü£¾¢Þ£¾¢Û 1 :9
¡¾½âÎö¡¿
Ç¿ËáºÍÇ¿¼îÍêÈ«µçÀ룬ÈõËáºÍÈõ¼î²¿·ÖµçÀ룻FeCl3ÊôÓÚÇ¿ËáÈõ¼îÑΣ¬Ë®½âÏÔËáÐÔ£¬CH3COONaÊôÓÚÇ¿¼îÈõËáÑΣ¬Ë®½âÏÔ¼îÐÔ£»Ë®µÄµçÀëÊÇ΢ÈõµÄ£¬µçÀë¹ý³ÌÎüÈÈ£»ÓÉ´Ë·ÖÎö½â´ð¡£
(1)FeCl3ÊôÓÚÇ¿ËáÈõ¼îÑΣ¬ÔÚË®ÈÜÒºÖз¢ÉúË®½â·´Ó¦£ºFe3++3H2OFe(OH)3+3H+£¬Ê¹ÈÜÒºÖÐc(H+)>c(OH-)£¬ÈÜÒºÏÔËáÐÔ¡£CH3COOHÊôÓÚÈõËᣬÈÜÒºÏÔËáÐÔ£¬µçÀë·½³ÌʽΪ£ºCH3COOHCH3COO-+H+¡£
(2)CH3COONaÊôÓÚÇ¿¼îÈõËáÑΣ¬Ë®½âʹÈÜÒºÏÔ¼îÐÔ£¬CH3COO-+H2OCH3COOH+OH-£¬ÓÉÈÜÒºµÄpH=10Öª£¬ÈÜÒºÖÐc(H+)ÈÜÒº=10-10mol/L£¬ËùÒÔ¸ÃÈÜÒºÖÐÓÉË®µçÀë³öÀ´µÄc(OH-)Ë®=c(OH-)ÈÜÒº===10-4mol/L£»ÓÉpH=3µÄHClÈÜÒºÖª£¬¸ÃHClÈÜÒºÖÐc(H+)ÈÜÒº=10-3mol/L£¬HClÈÜÒºÖÐÓÉË®µçÀë³öÀ´µÄc(H+)Ë®=c(OH-)ÈÜÒº===10-11mol/L¡£
(3)A.NaHSO4ÔÚË®ÖеçÀëNaHSO4=Na++H++SO42-£¬Ôö´óÁËÈÜÒºÖÐc(H+)£¬Ë®µÄµçÀëƽºâÏò×óÒƶ¯¡£AÏî´íÎó£»B.(NH4)2SO4ÔÚË®ÖÐÄÜ·¢ÉúË®½â·´Ó¦£ºNH4++H2ONH3¡¤H2O+H+£¬Ê¹Ë®µÄµçÀëƽºâÏòÓÒÒƶ¯£¬ÇÒËùµÃÈÜÒºÏÔËáÐÔ£¬BÏîÕýÈ·£»C.ÒòË®µÄµçÀëÊÇÎüÈȹý³Ì£¬¼ÓÈÈÄÜʹˮµÄµçÀëƽºâÏòÓÒÒƶ¯£¬µ«ÈÜÒºÖÐc(H+)=c(OH-)£¬Ë®ÈÔÈ»ÏÔÖÐÐÔ£¬CÏî´íÎó£»D.Na2CO3ÔÚË®ÖÐÄÜ·¢ÉúË®½â·´Ó¦£ºCO32-+H2OHCO3-+OH-£¬Ê¹Ë®µÄµçÀëƽºâÏòÓÒÒƶ¯£¬µ«ËùµÃÈÜÒºÏÔ¼îÐÔ£¬DÏî´íÎó£»´ð°¸Ñ¡B¡£
(4) µÈŨ¶ÈµÄNaOHÈÜÒººÍNH3H2OÈÜÒº£¬Ï¡ÊÍÏàͬµÄ±¶Êýʱ£¬ÈÜÒºÖÐc(OH-)´óС¹ØϵÊÇ£ºNaOH>NH3H2O£¬ÈôʹϡÊͺóÈÜÒºpHÏàµÈ[¼´c(OH-)ÏàµÈ]£¬ÔòÓ¦½«NaOHÈÜÒº¼ÌÐøÏ¡ÊÍ£¬×îºó²Å¿ÉÄÜÓëNH3H2OµÄpHÏàµÈ£¬ËùÒÔÈôÏ¡ÊͺóÁ½ÈÜÒºµÄpHÏàµÈ£¬Ôòm>n¡£
(5)pH=aµÄÑÎËáÖÐc(H+)=10-amol/L£¬ÑÎËáÊÇÇ¿ËᣬÑÎËáÎïÖʵÄÁ¿Å¨¶ÈΪ10-amol/L£»pH=14-aµÄ°±Ë®ÖÐc(OH-)===10-amol/L£¬¶ø°±Ë®ÊÇÈõ¼î£¬°±Ë®ÎïÖʵÄÁ¿Å¨¶ÈÔ¶´óÓÚ10-amol/L£¬ÓÉÖкͷ´Ó¦·½³ÌʽHCl+NH3H2O=NH4Cl+H2O¿ÉÍÆÖª£¬»ìºÏºóÈôÈÜÒºÏÔÖÐÐÔ£¬ÔòÑÎËáÌå»ýÒªÔ¶´óÓÚ°±Ë®µÄÌå»ý¼´Va>Vb¡£
(6)HCl¡¢FeCl3¡¢CH3COOHÈÜÒºÏÔËáÐÔ£¬pHСÓÚ7£»NaClÏÔÖÐÐÔ£¬pHµÈÓÚ7£»NaOH¡¢CH3COONa¡¢NH3H2OÈÜÒºÏÔ¼îÐÔ£¬pH´óÓÚ7¡£Ò»°ãÀ´Ëµ£¬ÑÎÀàË®½â³Ì¶ÈСÓÚËá»ò¼îµÄµçÀë³Ì¶È£¬µÈŨ¶ÈFeCl3µÄËáÐÔÈõÓÚHClºÍCH3COOH£¬CH3COONaµÄ¼îÐÔÈõÓÚNaOHºÍNH3H2O¡£Ç¿ËáºÍÇ¿¼îÍêÈ«µçÀ룬ÈõËáºÍÈõ¼î²¿·ÖµçÀ룬ËùÒÔµÈŨ¶ÈʱHClµÄËáÐÔÇ¿ÓÚCH3COOH£¬NaOHµÄ¼îÐÔÇ¿ÓÚNH3H2O¡£×ܶøÑÔÖ®£¬µÈŨ¶ÈʱÆßÖÖÎïÖÊÈÜÒºµÄpHÓÉ´óµ½Ð¡µÄ˳ÐòΪ£ºNaOH>NH3H2O>CH3COONa>NaCl>FeCl3>CH3COOH>HCl¡£¼´¢Ú>¢ß>¢Ý>¢Ù>¢Ü>¢Þ>¢Û¡£
(7)pH=13µÄNaOHÈÜÒºÖÐc(OH-)===0.1mol/L£»pH=2µÄÑÎËáÈÜÒºÖÐc(H+)=10-2mol/L¡£ÒòΪËùµÃÈÜÒºpH=11£¬ËµÃ÷NaOH¹ýÁ¿£¬ÇÒc(OH-)»ìºÏ===10-3mol/L£¬ÓÖc(OH-)»ìºÏ=£¬ ´úÈëÊý¾ÝÓУº=10-3mol/L¡£½âµÃVNaOH(aq):VHCl(aq)=1:9¡£
Ä꼶 | ¸ßÖÐ¿Î³Ì | Ä꼶 | ³õÖÐ¿Î³Ì |
¸ßÒ» | ¸ßÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ | ³õÒ» | ³õÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ |
¸ß¶þ | ¸ß¶þÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õ¶þ | ³õ¶þÃâ·Ñ¿Î³ÌÍƼö£¡ |
¸ßÈý | ¸ßÈýÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õÈý | ³õÈýÃâ·Ñ¿Î³ÌÍƼö£¡ |
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÏÂÁз½³ÌʽÕýÈ·µÄÊÇ
A. ÓÃKIO3Ñõ»¯ËáÐÔÈÜÒºÖеÄKI£º 5I£+IO3£+3H2O =3I2+6OH£
B. ÓÃÏ¡ÏõËáÏ´µÓÊÔ¹ÜÄÚ±ÚµÄÒø¾µ£º Ag+2H£«+NO3£=Ag£«+NO¡ü+H2O
C. ½«¹ýÁ¿SO2ͨÈëÀ䰱ˮÖУº SO2+NH3¡¤H2O =HSO3£+NH4+
D. ÏòNH4HCO3ÈÜÒºÖмӹýÁ¿µÄNaOHÈÜÒº²¢¼ÓÈÈ£º NH4+ +OH£NH3¡ü+H2O
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÄÜÓ°ÏìË®µÄµçÀëƽºâ£¬²¢Ê¹ÈÜÒºÖÐc(H+)£¾c(OH-)µÄ´ëÊ©ÊÇ£¨ £©
A. ½«Ë®¼ÓÈÈÖó·Ð£¬²âµÃpH=6 B. ÏòË®ÖмÓÈëFeCl3¹ÌÌå
C. Ïò´¿Ë®ÖÐͶÈëһС¿é½ðÊôÄÆ D. ÏòË®ÖмÓÈëNa2CO3¹ÌÌå
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÔÚÁòËá;§Ìå½á¾§Ë®º¬Á¿²â¶¨µÄʵÑéÖУ¬»áÔì³ÉʵÑé½á¹ûÆ«µÍµÄÊÇ£¨ £©
A. ¼ÓÈÈʱ¼ä¹ý³¤£¬¹ÌÌ岿·Ö±äºÚB. ÛáÛöÄÚ¸½Óв»»Ó·¢ÔÓÖÊ
C. ¾§Ìå²»´¿£¬º¬Óв»»Ó·¢ÔÓÖÊD. ¼ÓÈȹý³ÌÖÐÓÐÉÙÁ¿¾§Ì彦³ö
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿(1)ΪÁËÑéÖ¤ Fe2+Óë Cu2+Ñõ»¯ÐÔÇ¿Èõ£¬ÏÂÁÐ×°ÖÃÄܴﵽʵÑéÄ¿µÄµÄÊÇ_______£¨Ìî×°ÖÃÐòºÅ£©£¬ ÆäÕý¼«µÄµç¼«·´Ó¦Ê½Îª_______£»Èô¹¹½¨¸ÃÔµç³ØʱÁ½¸öµç¼«µÄÖÊÁ¿ÏàµÈ£¬µ±µ¼ÏßÖÐͨ¹ý 0.4 mol µç×Óʱ£¬Á½¸öµç¼«µÄÖÊÁ¿²îΪ_______g¡£
(2)½« CH4 Éè¼Æ³ÉȼÁϵç³Ø£¬ÆäÀûÓÃÂʸü¸ß£¬×°ÖÃÈçͼËùʾ£¨A¡¢B Ϊ¶à¿×̼°ô£©¡£
ʵÑé²âµÃ OH- ¶¨ÏòÒÆÏò A µç¼«£¬Ôò_______£¨Ìî A »ò B£©´¦µç¼«Èë¿Úͨ CH4£¬Æäµç¼«·´Ó¦Ê½Îª_______¡£
(3)½ðÊôÒ±Á¶ºÍ´¦Àí³£Éæ¼°Ñõ»¯»¹Ô·´Ó¦¡£ÓÉÏÂÁÐÎïÖÊÒ±Á¶ÏàÓ¦½ðÊôʱ²ÉÓõç½â·¨µÄÊÇ_______£¨ÌîÑ¡ Ïî×Öĸ£©¡£
a£®Fe2O3 b£®NaCl c£®Cu2S d£®Al2O3
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÊµÑéÊÒÓûʹ1.6 g Fe2O3ÍêÈ«±»»¹Ô£¬ÊÂÏÈӦ׼±¸µÄCOÆøÌåµÄ±ê×¼×´¿öϵÄÌå»ýΪ(¡¡¡¡)
A. 672 mLB. 336 mLC. ´óÓÚ672 mLD. СÓÚ336 mL
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ºº»ÆÜËËØÊÇ´«Í³ÖвÝÒ©»ÆÜ˵ÄÓÐЧ³É·ÖÖ®Ò»£¬¶ÔÖ×Áöϸ°ûµÄɱÉËÓжÀÌØ×÷Óá£ÏÂÁÐÓйغº»ÆÜËËصÄÐðÊöÕýÈ·µÄÊÇ( )
A. ºº»ÆÜËËصķÖ×ÓʽΪC16H13O5
B. ¸ÃÎïÖÊÓöFeCl3ÈÜÒºÏÔÉ«
C. 1 mol¸ÃÎïÖÊÓëäåË®·´Ó¦£¬×î¶àÏûºÄ1 mol Br2
D. Óë×ãÁ¿H2·¢Éú¼Ó³É·´Ó¦,×î¶àÏûºÄH2 6mol
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿»¯Ñ§ÔÚÉú²úºÍÉú»îÖÐÓÐ×ÅÖØÒªµÄ×÷Óá£ÏÂÁÐÓйØ˵·¨²»ÕýÈ·µÄÊÇ( )
A. º¸½Ó½ðÊôʱ³£ÓÃÈÜÒº×ö³ýÐâ¼Á
B. æ϶ðϵÁÐÎÀÐÇÖÐʹÓõÄ̼ÏËά£¬ÊÇÒ»ÖÖÐÂÐÍÎÞ»ú·Ç½ðÊô²ÄÁÏ
C. Ö»Òª·ûºÏÏÞÁ¿£¬¡°Ê³ÓÃÉ«ËØ¡±¡¢¡°ÑÇÏõËáÑΡ±¿ÉÒÔ×÷ΪijЩʳƷµÄÌí¼Ó¼Á
D. PM2.5ÊÇÖ¸´óÆøÖÐÖ±¾¶Ð¡ÓÚ»òµÈÓÚ2.5΢Ã׵ĿÅÁ£ÎÊÇ·¢ÉúÎíö²ÌìÆøµÄÖ÷ÒªÔÒò£¬ÕâЩ¿ÅÁ£ÎïÀ©É¢ÔÚ¿ÕÆøÖж¼»áÐγɽºÌå
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÓÐһƿ³ÎÇåµÄÈÜÒº£¬ÆäÖпÉÄܺ¬ÓÐNH¡¢K£«¡¢Mg2£«¡¢Ba2£«¡¢Al3£«¡¢Fe3£«¡¢SO42-¡¢CO32-¡¢NO3-¡¢I£¡¢Cl££¬È¡¸ÃÈÜÒº½øÐÐÒÔÏÂʵÑ飺
¢ÙÓÃpHÊÔÖ½²âµÃ¸ÃÈÜÒº³ÊËáÐÔ£»
¢ÚÈ¡²¿·ÖÈÜÒº£¬¼ÓÈëÊýµÎÐÂÖƵÄÂÈË®¼°ÉÙÁ¿CCl4£¬¾Õñµ´¡¢¾²Öúó£¬CCl4²ã³Ê×ϺìÉ«£»
¢ÛÁíÈ¡²¿·ÖÈÜÒº£¬ÖðµÎ¼ÓÈëÏ¡NaOHÈÜÒº£¬Ê¹ÈÜÒº´ÓËáÐÔÖð½¥±äΪ¼îÐÔ£¬ÔÚÕû¸öµÎ¼Ó¹ý³ÌÖÐÎÞ³ÁµíÉú³É£»
¢ÜÈ¡²¿·ÖÉÏÊö¼îÐÔÈÜÒº£¬¼ÓÈëNa2CO3ÈÜÒº£¬Óа×É«³ÁµíÉú³É£»
¢Ý½«¢ÛµÃµ½µÄ¼îÐÔÈÜÒº¼ÓÈÈ£¬ÓÐÆøÌå·Å³ö£¬¸ÃÆøÌåÄÜʹʪÈóµÄºìɫʯÈïÊÔÖ½±äÀ¶¡£
¸ù¾ÝÉÏÊöʵÑéÊÂʵ»Ø´ðÎÊÌ⣺
£¨1£©Ð´³ö¢ÚËù·¢Éú·´Ó¦µÄÀë×Ó·½³Ìʽ_________¡£
£¨2£©¸ÃÈÜÒºÖп϶¨´æÔÚµÄÀë×ÓÊÇ________¡£
£¨3£©¸ÃÈÜÒºÖп϶¨²»´æÔÚµÄÀë×ÓÊÇ________¡£
£¨4£©¸ÃÈÜÒºÖл¹²»ÄÜÈ·¶¨ÊÇ·ñ´æÔÚµÄÀë×ÓÊÇ_________¡£
²é¿´´ð°¸ºÍ½âÎö>>
°Ù¶ÈÖÂÐÅ - Á·Ï°²áÁбí - ÊÔÌâÁбí
ºþ±±Ê¡»¥ÁªÍøÎ¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨Æ½Ì¨ | ÍøÉÏÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | µçÐÅթƾٱ¨×¨Çø | ÉæÀúÊ·ÐéÎÞÖ÷ÒåÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | ÉæÆóÇÖȨ¾Ù±¨×¨Çø
Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com