(Ò»)£¨1£©¼×ÍéÒ²ÊÇÒ»ÖÖÇå½àȼÁÏ£¬µ«²»ÍêȫȼÉÕʱÈÈЧÂʽµµÍ²¢»á²úÉúÓж¾ÆøÌåÔì³ÉÎÛȾ¡£
ÒÑÖª£º CH4(g) + 2O2(g) £½ CO2(g) + 2H2O(l)   ¦¤H1£½¨D890.3 kJ/mol
2CO (g) + O2(g) £½ 2CO2(g)            ¦¤H2£½¨D566.0 kJ/mol
Ôò¼×Íé²»ÍêȫȼÉÕÉú³ÉÒ»Ñõ»¯Ì¼ºÍҺ̬ˮʱµÄÈÈЧÂÊÖ»ÊÇÍêȫȼÉÕʱµÄ________±¶£¨¼ÆËã½á¹û±£Áô1λСÊý£©¡£
£¨2£©¼×ÍéȼÁϵç³Ø¿ÉÒÔÌáÉýÄÜÁ¿ÀûÓÃÂÊ¡£ÏÂͼÊÇÀûÓü×ÍéȼÁϵç³Øµç½â50 mL 2 mol/LµÄÂÈ»¯Í­ÈÜÒºµÄ×°ÖÃʾÒâͼ£º

Çë»Ø´ð£º
¢Ù¼×ÍéȼÁϵç³ØµÄ¸º¼«·´Ó¦Ê½ÊÇ________¡£
¢Úµ±Ïß·ÖÐÓÐ0.1 molµç×Óͨ¹ýʱ£¬________£¨Ìî¡°a¡±»ò¡°b¡±£©¼«ÔöÖØ________g¡£
£¨¶þ£©Ï±íÊǼ¸ÖÖÈõµç½âÖʵĵçÀëƽºâ³£Êý¡¢ÄÑÈܵç½âÖʵĠ  
ÈܶȻýKsp (25¡æ)¡£
µç½âÖÊ
ƽºâ·½³Ìʽ
ƽºâ³£ÊýK
Ksp
CH3COOH
CH3COOHCH3COO-£«H+
1.76¡Á10-5
 
H2CO3
H2CO3H+£«HCO3-
HCO3-H+£«CO32-
K1£½4.31¡Á10-7
K2£½5.61¡Á10-11
 
C6H5OH
C6H5OH  C6H5O-£«H+
1.1¡Á10-10
 
H3PO4
H3PO4H+£«H2PO4-
H2PO4-H+£«HPO32-
HPO42-H+£«PO43-
K1£½7.52¡Á10-3
K2£½6.23¡Á10-8
K3£½2.20¡Á10-13
 
NH3¡¤H2O
NH3¡¤H2O NH4+£«OH-
1.76¡Á10-5
 
BaSO4
BaSO4 Ba2+£«SO42-
 
1.07¡Á10-10
BaCO3
BaCO3 Ba2+£«CO32-
 
2.58¡Á10-9
 
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÓÉÉϱí·ÖÎö£¬Èô¢ÙCH3COOH ¢ÚHCO3-¢ÛC6H5OH ¢ÜH2PO4- ¾ù¿É¿´×÷ËᣬÔòËüÃÇËáÐÔÓÉÇ¿µ½ÈõµÄ˳ÐòΪ__________________________(Ìî±àºÅ)£»
(2)25¡æʱ£¬½«µÈÌå»ýµÈŨ¶ÈµÄ´×ËáºÍ°±Ë®»ìºÏ£¬»ìºÏÒºÖУºc(CH3COO-)______c(NH4+)£»(Ìî¡°£¾¡±¡¢¡°£½¡±»ò¡°£¼¡±)
£¨3£©25¡æʱ£¬Ïò10ml 0.01mol/L±½·ÓÈÜÒºÖеμÓVml 0.01mol/L°±Ë®£¬»ìºÏÈÜÒºÖÐÁ£×ÓŨ¶È¹ØϵÕýÈ·µÄÊÇ(     )£»
A£®Èô»ìºÏÒºpH£¾7£¬ÔòV¡Ý10
B£®Èô»ìºÏÒºpH£¼7£¬Ôòc((NH4+) £¾c (C6H5O-) £¾c (H+)£¾c (OH£­)
C£®V=10ʱ£¬»ìºÏÒºÖÐË®µÄµçÀë³Ì¶ÈСÓÚ10ml 0.01mol/L±½·ÓÈÜÒºÖÐË®µÄµçÀë³Ì¶È
D£®V=5ʱ£¬2c(NH3¡¤H2O)+ 2 c (NH4+)=" c" (C6H5O-)+ c (C6H5OH)
£¨4£©ÈçÏÂͼËùʾ£¬ÓÐT1¡¢T2Á½ÖÖζÈÏÂÁ½ÌõBaSO4ÔÚË®ÖеijÁµíÈܽâƽºâÇúÏߣ¬»Ø´ðÏÂÁÐÎÊÌ⣺
ÌÖÂÛT1ζÈʱBaSO4µÄ³ÁµíÈܽâƽºâÇúÏߣ¬ÏÂÁÐ˵·¨²»ÕýÈ·µÄÊÇ(   )

A£®¼ÓÈëNa2SO4¿ÉʹÈÜÒºÓÉaµã±äΪbµã
B£®ÔÚT1ÇúÏßÉÏ·½ÇøÓò(²»º¬ÇúÏß)ÈÎÒâÒ»µãʱ£¬ ¾ùÓÐBaSO4³ÁµíÉú³É
C£®Õô·¢ÈܼÁ¿ÉÄÜʹÈÜÒºÓÉdµã±äΪÇúÏßÉÏa¡¢ bÖ®¼äµÄijһµã(²»º¬a¡¢b)
D£®ÉýοÉʹÈÜÒºÓÉbµã±äΪdµã
¢å £¨1£©0.7   £¨2·Ö£©  £¨2£©¢ÙCH4-8e-+2H2O=CO2+8H+ ¢Ú  b    3.2    £¨¸÷2·Ö£© 
¢æ£¨1£©¢Ù¢Ü¢Û¢Ú  £¨2·Ö£© £¨2£©  =   £¨1·Ö£©   £¨3£© D  £¨2·Ö£© £¨4£© D £¨2·Ö£©

ÊÔÌâ·ÖÎö£º¢å£¨1£©¼×Íé²»ÍêȫȼÉÕµÄÈÈ»¯Ñ§·½³ÌʽΪ: CH4(g) + O2(g) £½ CO(g) + 2H2O(l) ¦¤H£½¨D607.3 kJ/mol,Ôò¼×Íé²»ÍêȫȼÉÕÉú³ÉÒ»Ñõ»¯Ì¼ºÍҺ̬ˮʱµÄÈÈЧÂÊÖ»ÊÇÍêȫȼÉÕʱµÄ=0.7±¶ £¨2£©¢Ùµç½âÖÊΪËáÐԹʸº¼«·´Ó¦Ê½Îª£ºCH4-8e-+2H2O=CO2+8H+ ¢ÚCu2+ÔÚÒõ¼«·Åµç¼´b¼«£¬Ã¿Í¨¹ý0.1molµç×Ó£¬ÓÐ0.5molCu2+·Åµç£¬¼´3.2g ¢æ£¨1£© KÖµÔ½´óËáÐÔԽǿ £¨2£©25¡æʱ´×ËáºÍ°±Ë®µÄµçÀë³Ì¶ÈÏàͬ£¬¹Ê½«µÈÌå»ýµÈŨ¶ÈµÄ´×ËáºÍ°±Ë®»ìºÏºóÈÜÒº³ÊÖÐÐÔ£¬¹Ê»ìºÏÒºÖÐc(CH3COO-)=c(NH4+)£»£¨3£©ÒòΪNH3¡¤H2OµÄµçÀë³Ì¶ÈÔ¶´óÓÚC6H5OH£¬¹ÊÏò10ml 0.01mol/L±½·ÓÈÜÒºÖеμÓVml 0.01mol/L°±Ë®£¬°±Ë®µÄÌå»ý²»ÐèÒªµ½10mLÈÜÒºµÄpH>7,A ´í£»B²»×ñÑ­µçºÉÊغ㣬´í£»V=10ʱ£¬»ìºÏҺΪ±½·Óï§ÈÜÒº£¬±½·Óï§Ë®½â´Ù½øË®µÄµçÀ룬¹Ê»ìºÏÒºÖÐË®µÄµçÀë³Ì¶È´óÓÚ10ml 0.01mol/L±½·ÓÈÜÒºÖÐË®µÄµçÀë³Ì¶È£¬C´í£¨4£©Éý¸ßζȣ¬BaSO4µÄµçÀëƽºâÏòÓÒÒƶ¯£¬c£¨SO42-£©ºÍc(Ba2+)¶¼Ôö´ó£¬D´í¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÔÚÏÂÁÐÈÜÒºÖУ¬¸÷×éÀë×ÓÒ»¶¨Äܹ»´óÁ¿¹²´æµÄÊÇ£¨    £©
A£®25¡æÓÉË®µçÀëµÄc£¨H+£©= 10-12mol/L µÄÈÜÒº£ºFe3+ Cl£­NO3£­ K+
B£®Ê¹·Ó̪±äºìµÄÈÜÒº£ºNa+ Cl£­ SO42£­AlO2£­
C£®Ä³ÎÞÉ«ÈÜÒº£ºHCO3£­NO3£­ Al3+ Ba2+
D£®25¡æʱ£¬pH=1µÄÈÜÒº£º Ba2+ NO3£­ K+ I£­

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

ÒÔʳÑÎΪԭÁϽøÐÐÉú²ú²¢×ÛºÏÀûÓõÄijЩ¹ý³ÌÈçÏÂͼËùʾ£º

£¨1£©Îª³ýÈ¥´ÖÑÎÖеÄCa2+¡¢Mg2+ºÍSO£¬µÃµ½´¿¾»µÄNaCl¾§Ì壬Ðè¼ÓÈëÒÔÏÂÊÔ¼Á£º
A£®¹ýÁ¿µÄNaOHÈÜÒº£»B£®¹ýÁ¿µÄNa2CO3ÈÜÒº£»C£®ÊÊÁ¿µÄÑÎË᣻D£®¹ýÁ¿µÄBaCl2ÈÜÒº¡£
ÕýÈ·µÄ¼ÓÈë˳ÐòÊÇ_______£¬»¹È±ÉٵIJÙ×÷²½ÖèÊÇ__________ºÍ___________¡£
£¨2£©½«ÂËÒºµÄpHµ÷ÖÁËáÐÔ³ýÈ¥µÄÀë×ÓÊÇ________£¬·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ________¡£
£¨3£©ÈôÓöèÐԵ缫µç½â200mL1.5mol/LʳÑÎË®£¬µ±µç½â2minʱ£¬Á½¼«¹²ÊÕ¼¯µ½448mLÆøÌ壨±ê×¼×´¿öÏ£©¡£¼ÙÉèµç½âÇ°ºóÈÜÒºµÄÌå»ý²»±ä£¬Ôòµç½âºó¸ÃÈÜÒºµÄpHΪ______¡£
£¨4£©ÈôÏò·ÖÀë³öNaHCO3¾§ÌåºóµÄĸҺÖмÓÈë¹ýÁ¿Éúʯ»Ò£¬Ôò¿É»ñµÃÒ»ÖÖ¿ÉÒÔÑ­»·Ê¹ÓõÄÎïÖÊ£¬Æ仯ѧʽÊÇ___________¡£
£¨5£©´¿¼îÔÚÉú²úÉú»îÖÐÓй㷺µÄÓ¦Óá£
¢Ù´¿¼î¿ÉÓÃÓÚ³ýÔį̂ÓÍÎÛ¡£ÆäÔ­ÒòÊÇ£¨½áºÏÀë×Ó·½³Ìʽ±íÊö£©_____________¡£
¢Ú³£ÎÂÏ£¬ÏòijpH=11µÄNa2CO3ÈÜÒºÖмÓÈë¹ýÁ¿Ê¯»ÒÈ飬¹ýÂ˺óËùµÃÈÜÒºpH=13¡£Ôò·´Ó¦Ç°µÄÈÜÒºÖÐÓë·´Ó¦ºóµÄÂËÒºÖÐË®µçÀë³öµÄ£¨OH-£©µÄ±ÈÖµÊÇ_________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÏÂÁÐÀë×ÓÔÚÖ¸¶¨ÈÜÒºÖÐÒ»¶¨ÄÜ´óÁ¿¹²´æµÄÊÇ
A£®¼îÐÔÈÜÒºÖУºNa+¡¢K+¡¢SO32£­¡¢S2£­
B£®ÎÞÉ«ÈÜÒºÖУºNa+¡¢Fe3+¡¢Cl£­¡¢SO42£­
C£®ÎÞÉ«ÈÜÒºÖУºNa+¡¢K+¡¢HCO3£­¡¢AlO2£­
D£®ËáÐÔÈÜÒºÖУºNa+¡¢K+¡¢NO3£­¡¢ClO£­

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÔÚÇ¿ËáÐÔÎÞɫ͸Ã÷ÈÜÒºÖУ¬ÏÂÁи÷×éÀë×ÓÄÜ´óÁ¿¹²´æµÄÊÇ£¨   £©¡£
A£®Fe3+¡¢K+¡¢Cl-¡¢NO3-B£®Ag+¡¢Na+¡¢NO3-¡¢Cl-
C£®Zn2+¡¢Al3+¡¢SO42-¡¢Cl-D£®Ba2+¡¢NH4+¡¢Cl-¡¢HCO3-

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

³£ÎÂÏ£¬ÏÂÁи÷×éÀë×Ó»ò·Ö×ÓÔÚÖ¸¶¨ÈÜÒºÖÐÄÜ´óÁ¿¹²´æµÄÊÇ£¨   £©
A£®°±Ë®ÖУºNH4£«¡¢Fe2£«¡¢SO42£­¡¢HSO3£­
B£®pH=lµÄÈÜÒºÖУº¡¢Mg2£«¡¢Fe3£«¡¢Cl£­¡¢NO3£­
C£®Í¨Èë×ãÁ¿CO2µÄÈÜÒºÖУºCa2£«¡¢Na£«¡¢SiO32£­¡¢CO32£­
D£®ÓÉË®µçÀë³öµÄc(OH£­)£½1¡Á10¡ª12mol¡¤L¡ª1µÄÈÜÒºÖУºNH4+¡¢HCO3£­¡¢Na+¡¢Cl£­

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÏÂÁÐÀë×Ó×éÔÚ¸ø¶¨Ìõ¼þÏÂÄÜ·ñ´óÁ¿¹²´æµÄÅжÏÕýÈ·£¬Ëù·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽҲÕýÈ·µÄÊÇ
Ñ¡Ïî
Ìõ¼þ
Àë×Ó×é
Àë×Ó¹²´æÅжϼ°·´Ó¦µÄÀë×Ó·½³Ìʽ
A
µÎ¼Ó°±Ë®
Na£«¡¢Al3£«¡¢Cl£­¡¢NO3£­
²»Äܹ²´æ£¬Al3£«£«3OH£­£½Al(OH)3¡ý
B
pH£½1µÄÈÜÒº
Fe2£«¡¢Al3£«¡¢SO42£­¡¢MnO4£­
²»Äܹ²´æ£¬
5Fe2£«£«MnO4£­£«8H£«£½Mn2£«£«5Fe3£«£«4H2O
C
ÓÉË®µçÀë³öµÄH£«Å¨¶ÈΪ1¡Á10£­12mol/L
NH4£«¡¢Na£«¡¢NO3£­¡¢Cl£­
Ò»¶¨¹²´æ£¬NH4£«£«H2ONH3¡¤H2O£«H£«
D
ͨÈëÉÙÁ¿SO2ÆøÌå
K£«¡¢Na£«¡¢ClO£­¡¢SO42£­
²»Äܹ²´æ£¬
2ClO£­£«SO2£«H2O£½2HClO£«SO32£­
 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÏÖÓÐŨ¶È¾ùΪ0.10mol¡¤L£­1µÄ´×ËáV1 mLºÍÇâÑõ»¯ÄÆÈÜÒºV2 mL£¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ
A£®³£ÎÂÏ£¬ÉÏÊöÈÜÒºÖУ¬´×ËáÈÜÒºµÄpH=1£¬ÇâÑõ»¯ÄÆÈÜÒºµÄpH=13
B£®³£ÎÂÏ£¬ÈôÁ½ÈÜÒº»ìºÏºópH=7£¬Ôò»ìºÏÒº£ºc(Na£«)£½c(CH3COO£­)
C£®Èô V1= V2£¬½«Á½ÈÜÒº»ìºÏ£¬ËùµÃ»ìºÏÒº£ºc(CH3COO£­)£¾c(Na£«)£¾c(H£«)£¾c(OH£­)
D£®V1ÓëV2ÈÎÒâ±È»ìºÏ£¬ËùµÃ»ìºÏÒº£ºc (Na£«)+ c(H+) £½c(OH£­)+ c (CH3COO£­)

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

³£ÎÂÏ£¬ÏÂÁи÷×éÀë×ÓÔÚÖ¸¶¨ÈÜÒºÖÐÒ»¶¨ÄÜ´óÁ¿¹²´æµÄÊÇ£¨  £©
A£®1.0 mol¡¤L£­1µÄKNO3ÈÜÒº£ºH£«¡¢Fe2£«¡¢Cl£­¡¢SO42£­
B£®±¥ºÍÂÈË®ÖУºNH4£«¡¢SO32£­¡¢AlO2£­¡¢Cl£­
C£®ÓëÂÁ·´Ó¦²úÉú´óÁ¿ÇâÆøµÄÈÜÒº£ºNa£«¡¢K£«¡¢CO32£­¡¢NO3£­
D£®c(H£«)=1.0¡Á10£­13mol/LÈÜÒºÖУºK£«¡¢Na£«¡¢CH3COO£­¡¢Br£­

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸