¡¾ÌâÄ¿¡¿I.H2AÔÚË®ÖдæÔÚÒÔÏÂƽºâ£ºH2AH+ £«HA- £¬HA£H+£«A2- ¡£
£¨1£©NaHAÈÜÒºÏÔËáÐÔ£¬ÔòÈÜÒºÖÐÀë×ÓŨ¶ÈµÄ´óС˳ÐòΪ____________________¡£
£¨2£©³£ÎÂʱ£¬ÈôÏò0.1 mol/LµÄNaHAÈÜÒºÖÐÖðµÎµÎ¼Ó0.1mol/L KOHÈÜÒºÖÁÈÜÒº³ÊÖÐÐÔ¡£´Ëʱ¸Ã»ìºÏÈÜÒºµÄÏÂÁйØϵÖУ¬Ò»¶¨ÕýÈ·µÄÊÇ_______________¡£
A£®c(Na+ )£¾c(K+) B£®c(H +)c(OH)£½1¡Á10-14
C£®c(Na+ )£½c(K+) D£®c(Na+ )£«c(K+ )£½c(HA£)£«c(A2- )
£¨3£©ÒÑÖª³£ÎÂÏÂH2AµÄ¸ÆÑÎ(CaA)±¥ºÍÈÜÒºÖдæÔÚÒÔÏÂƽºâ£ºCaA(s)Ca2+ (aq)£«A2- (aq)£¬µÎ¼ÓÉÙÁ¿Na2A¹ÌÌ壬c(Ca2+ )_______________£¨Ìî¡°Ôö´ó¡±¡¢¡°¼õС¡±»ò¡°²»±ä¡±£©£¬ÔÒòÊÇ________________¡£
¢ò.º¬ÓÐCr2O72-µÄ·ÏË®¶¾ÐԽϴó£¬Ä³¹¤³§·ÏË®Öк¬4.00¡Á10-3 mol/LµÄCr2O72£¡£ÎªÊ¹·ÏË®ÄÜ´ï±êÅÅ·Å£¬×÷ÈçÏ´¦Àí£º
£¨1£©¸Ã·ÏË®ÖмÓÈëFeSO4¡¤7H2OºÍÏ¡ÁòËᣬ·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ£º_______________¡£
£¨2£©Óûʹ25 L¸Ã·ÏË®ÖÐCr2O7 ת»¯ÎªCr3+£¬ÀíÂÛÉÏÐèÒª¼ÓÈë__________g FeSO4¡¤7H2O¡£
£¨3£©Èô´¦ÀíºóµÄ·ÏË®ÖвÐÁôµÄc(Fe)£½1¡Á10-13mol/L£¬Ôò²ÐÁôµÄ Cr3+ µÄŨ¶ÈΪ__________¡££¨ÒÑÖª£ºKsp[Fe(OH)3]¡Ö1.0¡Á10-38mol/L£¬Ksp[Cr(OH)3]¡Ö1.0¡Á10-31 mol/L £©
¡¾´ð°¸¡¿I.£¨1£©c(Na£«)£¾c(HA£)£¾c(H£«)£¾c(A2£)£¾c(OH£) £¨2£©AB
£¨3£©¼õС£»¼ÓÈëNa2A¹ÌÌ壬c(A2- )Ôö´ó£¬´Ó¶øµ¼ÖÂÈܽâƽºâ×óÒÆ£¬c(Ca2+ )¼õС
¢ò.£¨1£©Cr2O72££«6Fe2£«£«14H£«£½2Cr3£«£«6Fe3£«£«7H2O £¨2£©166.8 £¨3£©1.0¡Á10£6mol/L
¡¾½âÎö¡¿
ÊÔÌâI.£¨1£©H2AÔÚË®ÖдæÔÚÒÔÏÂƽºâ£ºH2AH+ £«HA- £¬HA£H+£«A2- £¬ËùÒÔNaHAÖ»´æÔÚµçÀëƽºâ£¬ÈÜÒºÏÔËáÐÔ£¬ÈÜÒºÖÐÀë×ÓŨ¶ÈµÄ´óС˳ÐòΪc(Na£«)£¾c(HA£)£¾c(H£«)£¾c(A2£)£¾c(OH£)¡£
£¨2£©A¡¢NaHAÈÜÒº³ÊËáÐÔ£¬Ïò0.1mol/LµÄNaHAÈÜÒºÖÐÖðµÎµÎ¼Ó0.1mol/L KOHÈÜÒºÖÁÈÜÒº³ÊÖÐÐÔʱ£¬NaHAµÄÎïÖʵÄÁ¿Ó¦´óÓÚÇâÑõ»¯¼ØµÄÎïÖʵÄÁ¿£¬ËùÒÔͬһ»ìºÏÈÜÒºÖÐc(Na+)£¾c(K+)£¬AÕýÈ·£®B¡¢Ë®µÄÀë×Ó»ý³£ÊýÓëζÈÓйأ¬Î¶ÈÔ½¸ß£¬Ë®µÄÀë×Ó»ý³£ÊýÔ½´ó£¬³£ÎÂÏÂË®µÄÀë×Ó»ý³£ÊýÊÇ10£14£¬BÕýÈ·£»C¡¢¸ù¾ÝAÖзÖÎö¿ÉÖªC´íÎó£»D¡¢¸ù¾ÝµçºÉÊغã¿ÉÖªc(Na+)£«c(K+)£«c(H+)£½c(HA£)£«2c(A2- )£¬Ôòc(H£«)£½c(OH£)£¬ËùÒÔc£¨Na+£©+c£¨K+£©=c£¨HA-£©+2c£¨A2-£©£¬D´íÎ󣬴ð°¸Ñ¡AB¡£
£¨3£©ÓÉÓÚ¼ÓÈëNa2A¹ÌÌ壬c(A2- )Ôö´ó£¬´Ó¶øµ¼ÖÂÈܽâƽºâ×óÒÆ£¬c(Ca2+ )¼õС¡£
¢ò.£¨1£©ÑÇÌúÀë×Ó±»Ñõ»¯µÄÀë×Ó·½³ÌʽΪCr2O72££«6Fe2£«£«14H£«£½2Cr3£«£«6Fe3£«£«7H2O¡£
£¨2£©Ä³¹¤³§·ÏË®Öк¬4.00¡Á10-3 molL-1µÄCr2O72££¬n£¨Cr2O72££©£½25L¡Á4.00¡Á10-3mol/L£½0.1mol£»ÒÀ¾ÝÑõ»¯»¹Ô·´Ó¦Àë×Ó·½³ÌʽCr2O72££«6Fe2£«£«14H£«£½2Cr3£«£«6Fe3£«£«7H2O£¬µÃµ½n£¨Fe2+£©£½0.6mol£»ÐèÒªFeSO47H2OµÄÖÊÁ¿=0.6mol¡Á278g/mol£½166.8g£»
£¨3£©Èô´¦ÀíºóµÄ·ÏË®ÖвÐÁôµÄc£¨Fe3+£©£½1¡Á10-13molL-1£¬Ksp[Fe£¨OH£©3]£½c£¨Fe3+£©¡¤c3£¨OH-£©£½1.0¡Á10-38£¬¼ÆËãµÃµ½c3£¨OH-£©£½1¡Á10-25mol/L£¬Ôò²ÐÁôµÄCr3+µÄŨ¶ÈΪKsp[Cr£¨OH£©3]£½c£¨Cr3+£©c3£¨OH-£©£½1.0¡Á10-31 £¬c£¨Cr3+£©£½1¡Á10-6molL-1¡£
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¿¡¿ÏÂÁбíʾÎïÖʽṹµÄ»¯Ñ§ÓÃÓïÕýÈ·µÄÊÇ(¡¡¡¡)
A. ÒÒÏ©µÄ½á¹¹¼òʽ£ºCH2CH2 B. Òì±û»ùµÄ½á¹¹¼òʽ£º-CH(CH3)2
C. ôÇ»ùµÄµç×Óʽ£º D. ÐÂÎìÍéµÄ½á¹¹¼òʽ£º
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿NO2¡¢O2ºÍÈÛÈÚKNO3¿ÉÖÆ×÷ȼÁϵç³Ø£¬ÆäÔÀíÈçͼËùʾ¡£¸Ãµç³ØÔڷŵç¹ý³ÌÖÐʯī¢ñµç¼«ÉÏÉú³ÉÑõ»¯ÎïY£¬Y¿ÉÑ»·Ê¹Óá£ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ
A. ·Åµçʱ£¬NO3-Ïòʯī¢òµç¼«Ç¨ÒÆ
B. ʯī¢ò¸½½ü·¢ÉúµÄ·´Ó¦ÎªNO£«O2£«e£===NO3-
C. ¸Ãµç³Ø×Ü·´Ó¦Ê½Îª4NO2£«O2===2N2O5
D. µ±Íâµç·ͨ¹ý4 mol e£Ê±£¬¸º¼«ÉϹ²²úÉú2mol N2O5
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÒÑÖªNaClO2ÔÚË®ÈÜÒºÖÐÄÜ·¢ÉúË®½â¡£³£ÎÂʱ£¬ÓÐ1 mol/LµÄHClO2ÈÜÒººÍ1mol/LµÄHBF4(·úÅðËá)ÈÜÒºÆðʼʱµÄÌå»ý¾ùΪV0£¬·Ö±ðÏòÁ½ÈÜÒºÖмÓË®£¬Ï¡ÊͺóÈÜÒºµÄÌå»ýΪV£¬ËùµÃÇúÏßÈçͼËùʾ¡£ÏÂÁÐ˵·¨´íÎóµÄÊÇ
A. HClO2ΪÈõËᣬHBF4ΪǿËá
B. ³£ÎÂÏÂHClO2µÄµç¸ßƽºâ³£ÊýµÄÊýÁ¿¼¶Îª10¡ª4
C. ÔÚ0¡ÜpH¡Ü5ʱ£¬HBF4ÈÜÒºÂú×ãpH=lg(V/V0)
D. 25¡æʱ1L pH=2µÄHBF4ÈÜÒºÓë100¡æʱ1L pH=2µÄHBF4ÈÜÒºÏûºÄµÄNaOHÏàͬ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÎªÑé֤ͬÖ÷×åÔªËØÐÔÖʵĵݱä¹æÂÉ¡£Ä³Ð¡×éÓÃÈçͼËùʾµÄ×°ÖýøÐÐʵÑé(¼Ð³ÖÒÇÆ÷ÒÑÂÔÈ¥£¬×°ÖÃÆøÃÜÐÔÒѼìÑé)¡£
ʵÑé¹ý³Ì£º
¢ñ.´ò¿ªµ¯»É¼Ð£¬´ò¿ª»îÈûa£¬µÎ¼ÓŨÑÎËá¡£
¢ò.µ±×°ÖÃBºÍ×°ÖÃCÖеÄÈÜÒº¶¼±äΪ»Æɫʱ£¬¼Ð½ôµ¯»É¼Ð¡£
¢ó.µ±×°ÖÃBÖÐÈÜÒºÓÉ»ÆÉ«±äΪ×غìɫʱ£¬¹Ø±Õ»îÈûa¡£
¢ô.¡¡
£¨1£©½þÓÐNaOHÈÜÒºµÄÃÞ»¨µÄ×÷ÓÃ____________________________¡£
£¨2£©×°ÖÃAÖз¢ÉúµÄÖû»·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ___________________¡£
£¨3£©×°ÖÃBµÄÈÜÒºÖÐNaBrÍêÈ«±»Ñõ»¯£¬ÔòÏûºÄCl2µÄÎïÖʵÄÁ¿Îª__________¡£
£¨4£©ÎªÑéÖ¤äåÔªËصķǽðÊôÐÔÇ¿ÓÚµâÔªËØ£¬¹ý³Ì¢ôµÄ²Ù×÷ºÍÏÖÏóÊÇ__________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿³£ÎÂÏÂÏò20mL 0.1mol/L°±Ë®ÖÐͨÈëHClÆøÌ壬ÈÜÒºÖÐÓÉË®µçÀë³öµÄÇâÀë×ÓŨ¶ÈËæͨÈëHClÆøÌåµÄÌå»ý±ä»¯ÈçͼËùʾ¡£ÔòÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ
A. bµãͨÈëµÄHClÆøÌ壬ÔÚ±ê¿öÏÂΪ44.8mL
B. b¡¢cÖ®¼äÈÜÒºÖÐc(NH4+)>c(Cl-)
C. È¡10mLµÄcµãÈÜҺϡÊÍʱ£ºc(NH4+)/c(NH3¡¤H2O)¼õС
D. dµãÈÜÒº³ÊÖÐÐÔ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿Ä³ÓлúÎïÍêȫȼÉÕÉú³ÉCO2ºÍH2O¡£½«12.4 g¸ÃÓлúÎïµÄÍêȫȼÉÕ²úÎïͨ¹ýŨÁòËᣬŨÁòËáÔöÖØ10.8 g£¬ÔÙͨ¹ý¼îʯ»Ò£¬¼îʯ»ÒÔöÖØÁË17.6 g¡£ÏÂÁÐ˵·¨²»ÕýÈ·µÄÊÇ
A. ¸ÃÓлúÎïµÄ×î¼òʽΪCH3O
B. ¸ÃÓлúÎïµÄ·Ö×Óʽ¿ÉÄÜΪCH3O
C. ¸ÃÓлúÎïµÄ·Ö×Óʽһ¶¨ÎªC2H6O2
D. ¸ÃÓлúÎï¿ÉÄÜÊôÓÚ´¼Àà
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿º£Ë®ÖÐä庬Á¿Ô¼Îª 65 mg¡¤L-1,´Óº£Ë®ÖÐÌáÈ¡äåµÄ¹¤ÒÕÁ÷³ÌÈçÏÂ:
(1)ÒÔÉϲ½Öè¢ñÖÐÒÑ»ñµÃÓÎÀë̬µÄä壬²½Öè¢òÓÖ½«Ö®×ª±ä³É»¯ºÏ̬µÄä壬ÆäÄ¿µÄÊÇ____________¡£
(2)²½Öè¢òͨÈëÈÈ¿ÕÆø´µ³öBr2£¬ÀûÓÃÁËäåµÄ____¡£
A.Ñõ»¯ÐÔ ¡¡B.»¹ÔÐÔ¡¡ C.»Ó·¢ÐÔ D.¸¯Ê´ÐÔ
(3)²½Öè¢òÖÐÉæ¼°µÄÀë×Ó·´Ó¦ÈçÏÂ,ÇëÔÚÏÂÃæ·½¿òÄÚÌîÈëÊʵ±µÄ»¯Ñ§¼ÆÁ¿Êý:_____
(4)ÉÏÊöÁ÷³ÌÖдµ³öµÄäåÕôÆø£¬Ò²¿ÉÏÈÓöþÑõ»¯ÁòË®ÈÜÒºÎüÊÕ£¬ÔÙÓÃÂÈÆøÑõ»¯ºóÕôÁó¡£Ð´³öäåÓë¶þÑõ»¯ÁòË®ÈÜÒº·´Ó¦µÄ»¯Ñ§·½³Ìʽ:__________¡£
(5)ʵÑéÊÒ·ÖÀëä廹¿ÉÒÔÓÃÈܼÁÝÍÈ¡·¨,ÏÂÁпÉÒÔÓÃ×÷äåµÄÝÍÈ¡¼ÁµÄÊÇ____¡£
A.ÒÒ´¼ B.ËÄÂÈ»¯Ì¼¡¡¡¡C.ÉÕ¼îÈÜÒº D.±½
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿UO2ÓëÓ˵ª»¯ÎïÊÇÖØÒªµÄºËȼÁÏ£¬ÒÑÖª£º3(NH4)4[UO2(CO3)3]3UO2+10NH3¡ü+9CO2¡ü+N2¡ü+9H2O¡ü
»Ø´ðÏÂÁÐÎÊÌ⣺
(1)»ù̬µªÔ×Ó¼Ûµç×ÓÅŲ¼Í¼Îª______¡£
(2)·´Ó¦ËùµÃÆø̬»¯ºÏÎïÖÐÊôÓڷǼ«ÐÔ·Ö×ÓµÄÊÇ_______(Ìѧʽ)¡£
(3)ijÖÖÓ˵ª»¯ÎïµÄ¾§Ìå½á¹¹ÊÇNaClÐÍ¡£NaClµÄBom-HaberÑ»·ÈçͼËùʾ¡£ÒÑÖª£ºÔªËصÄÒ»¸öÆø̬Ô×Ó»ñµÃµç×Ó³ÉΪÆø̬ÒõÀë×ÓʱËù·Å³öµÄÄÜÁ¿³ÆΪµç×ÓÇ׺ÍÄÜ¡£ÏÂÁÐÓйØ˵·¨ÕýÈ·µÄÊÇ________(Ìî±êºÅ)¡£
a.Cl-Cl¼üµÄ¼üÄÜΪ119.6kJ/mol b.NaµÄµÚÒ»µçÀëÄÜΪ603.4kJ/mol
c.NaClµÄ¾§¸ñÄÜΪ785.6kJ/mol d.ClµÄµÚÒ»µç×ÓÇ׺ÍÄÜΪ348.3kJ/mol
(4)ÒÀ¾ÝVSEPRÀíÂÛÍƲâCO32-µÄ¿Õ¼ä¹¹ÐÍΪ_________¡£·Ö×ÓÖеĴóØ¢¼ü¿ÉÓ÷ûºÅØ¢±íʾ£¬ÆäÖÐm´ú±í²ÎÓëÐγɴóØ¢¼üµÄÔ×ÓÊý£¬n´ú±í²ÎÓëÐγɴóØ¢¼üµÄµç×ÓÊý(Èç±½·Ö×ÓÖеĴóØ¢¼ü¿É±íʾΪآ)£¬ÔòCO32-ÖеĴóØ¢¼üÓ¦±íʾΪ_____
(5)UO2¿ÉÓÃÓÚÖƱ¸UF4£º2UO2+5NH4HF22UF4¡¤2NH4F+3NH3¡ü+4H2O£¬ÆäÖÐHF2µÄ½á¹¹±íʾΪ[F¡ªH¡F]-£¬·´Ó¦ÖжÏÁѵĻ¯Ñ§¼üÓÐ_______ (Ìî±êºÅ)¡£
a.Çâ¼ü b.¼«ÐÔ¼ü c.Àë×Ó¼ü d.½ðÊô¼ü e.·Ç¼«ÐÔ¼ü
(6)Ó˵ª»¯ÎïµÄijÁ½ÖÖ¾§°ûÈçͼËùʾ£º
¢Ù¾§°ûaÖÐÓËÔªËصĻ¯ºÏ¼ÛΪ__________£¬ÓëU¾àÀëÏàµÈÇÒ×î½üµÄUÓÐ_______¸ö¡£
¢ÚÒÑÖª¾§°ûbµÄÃܶÈΪdg/cm3£¬UÔ×ӵİ뾶Ϊr1cm£¬NÔ×ӵİ뾶ΪΪr2cm£¬ÉèNAΪ°¢·ü¼ÓµÂÂÞ³£ÊýµÄÖµ£¬Ôò¸Ã¾§°ûµÄ¿Õ¼äÀûÓÃÂÊΪ___________(Áгö¼ÆËãʽ)¡£
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