È¡µÈÌå»ý0.05mol/LµÄBa(OH)2ÈÜÒº£¬·Ö±ð×°Èë±êÓТ٢ڢܱۢàºÅµÄ4¸ö׶ÐÎÆ¿ÖУ®½«¢Ù¼ÓˮϡÊ͵½Ô­Ìå»ýµÄ2±¶£»ÔÚ¢Ú¢ÛÖзֱðͨÈëÉÙÁ¿µÄCO2£»¢Ü×÷¶ÔÕÕ£®·Ö±ðÔÚ¢Ù¢ÚÖеμӷÓ̪×öָʾ¼Á£»ÔÚ¢Û¢ÜÖеμӼ׻ù³È×öָʾ¼Á£¬ÓÃHClÈÜÒºµÎ¶¨ÉÏÊö4ÖÖÈÜÒº£¬ËùÏûºÄHClÈÜÒºµÄÌå»ýÊÇ

[¡¡¡¡]

A£®¢Ù£½¢Ú£½¢Û£½¢Ü
B£®¢Ú£½¢Û£¼¢Ù£½¢Ü
C£®¢Ù£¼¢Ú£¼¢Û£¼¢Ü
D£®¢Ú£¼¢Û£¼¢Ù£½¢Ü
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¢ñ£®ÊÒÎÂʱ1L 0.01mol?L-1µÄÁòËáÇâÄÆÈÜÒºµÄpHΪ
2
2
£¬ÔÚÆäÖеÎÈëµÈÌå»ýµÄ0.01mol?L-1µÄBa£¨OH£©2ÈÜÒººó£¬¼ÓˮϡÊ͵½10L£¬Ëù·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ
Ba2++OH-+H++SO42-¨TBaSO4¡ý+H2O
Ba2++OH-+H++SO42-¨TBaSO4¡ý+H2O
£¬pHΪ
11
11
£®
¢ò£®Ä³Ñ§ÉúÓÃ0.2000mol?L-1µÄ±ê×¼NaOHÈÜÒºµÎ¶¨Î´ÖªÅ¨¶ÈµÄÑÎËᣬÆä²Ù×÷ÈçÏ£º
¢ÙÓÃÕôÁóˮϴµÓ¼îʽµÎ¶¨¹Ü£¬²¢Á¢¼´×¢ÈëNaOHÈÜÒºÖÁ¡°0¡±¿Ì¶ÈÏßÒÔÉÏ£»
¢Ú¹Ì¶¨ºÃµÎ¶¨¹Ü²¢Ê¹µÎ¶¨¹Ü¼â×ì³äÂúÒºÌ壻
¢Ûµ÷½ÚÒºÃæÖÁ¡°0¡±»ò¡°0¡±¿Ì¶ÈÏßÉÔÏ£¬²¢¼Ç϶ÁÊý£»
¢ÜÒÆÈ¡20.00mL´ý²âҺעÈë½à¾»µÄ׶ÐÎÆ¿ÖУ¬²¢¼ÓÈë3µÎ·Ó̪ÈÜÒº£»
¢ÝÓñê×¼ÒºµÎ¶¨ÖÁÖյ㣬¼ÇÏµζ¨¹ÜÒºÃæ¶ÁÊý£®
Çë»Ø´ð£º
£¨1£©ÒÔÉϲ½ÖèÓдíÎóµÄÊÇ£¨Ìî±àºÅ£©
¢Ù
¢Ù
£¬¸Ã´íÎó²Ù×÷»áµ¼Ö²ⶨ½á¹û£¨Ìî¡°Æ«´ó¡±¡¢¡°Æ«Ð¡¡±»ò¡°ÎÞÓ°Ï족£©
Æ«´ó
Æ«´ó
£®
£¨2£©Åжϵζ¨ÖÕµãµÄÏÖÏóÊÇ£º
ÎÞÉ«±äΪ·Ûºì£¬°ë·ÖÖÓÄÚ²»±äÉ«
ÎÞÉ«±äΪ·Ûºì£¬°ë·ÖÖÓÄÚ²»±äÉ«
£®
£¨3£©ÈçÏÂͼÊÇij´ÎµÎ¶¨Ê±µÄµÎ¶¨¹ÜÖеÄÒºÃ棬Æä¶ÁÊýΪ
22.60
22.60
mL£®
£¨4£©¸ù¾ÝÏÂÁÐÊý¾Ý£ºÇë¼ÆËã´ý²âÑÎËáÈÜÒºµÄŨ¶È£º
0.2000
0.2000
mol/L£®
µÎ¶¨´ÎÊý ´ý²âÌå»ý£¨mL£© ±ê×¼ÉÕ¼îÌå»ý£¨mL£©
µÎ¶¨Ç°¶ÁÊý µÎ¶¨ºó¶ÁÊý
µÚÒ»´Î 20.00 0.40 20.40
µÚ¶þ´Î 20.00 2.00 24.10
µÚÈý´Î 20.00 4.00 24.00

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

³£ÎÂÏÂKsp£¨AgCl£©=1.8¡Á10-10£¬Ksp£¨AgI£©=1.0¡Á10-16£¬½«µÈÌå»ýµÄAgClºÍAgIµÄ±¥ºÍÈÜÒºµÄÇåÒº»ìºÏ£¬ÔÙÏòÆäÖмÓÈëÒ»¶¨Á¿µÄAgNO3¹ÌÌ壬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨¡¡¡¡£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÏÖÓÐŨ¶È¾ùΪ0.1mol/LµÄÏÂÁÐÈÜÒº£º¢ÙÁòËá¡¡¢Ú´×Ëá¡¡¢ÛÇâÑõ»¯ÄÆ¡¡¢ÜÂÈ»¯ï§
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©¢Ù¢Ú¢Û¢ÜËÄÖÖÈÜÒºÖÐÓÉË®µçÀë³öµÄH+Ũ¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòÊÇ£¨ÌîÐòºÅ£©
¢Ü¢Ú¢Û¢Ù
¢Ü¢Ú¢Û¢Ù
£®
£¨2£©½«¢ÛºÍ¢ÜµÈÌå»ý»ìºÏºó£¬»ìºÏÒºÖи÷Àë×ÓŨ¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòÊÇ
c£¨Na+£©=c£¨Cl-£©£¾c£¨OH-£©£¾c£¨NH4+£©£¾c£¨H+£©
c£¨Na+£©=c£¨Cl-£©£¾c£¨OH-£©£¾c£¨NH4+£©£¾c£¨H+£©
£®
£¨3£©ÒÑÖªt¡æ£¬KW=1¡Á10-13£¬Ôòt¡æ
£¾
£¾
£¨Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°=¡±£©25¡æ£®ÔÚt¡æʱ½«pH=11µÄNaOHÈÜÒºa LÓëpH=1µÄH2SO4ÈÜÒºb L»ìºÏ£¨ºöÂÔ»ìºÏºóÈÜÒºÌå»ýµÄ±ä»¯£©£¬ÈôËùµÃ»ìºÏÈÜÒºµÄpH=10£¬Ôòa£ºb=
101£º9
101£º9
£®
£¨4£©25¡æʱ£¬ÓÐpH=xµÄÑÎËáºÍpH=yµÄÇâÑõ»¯ÄÆÈÜÒº£¨x¡Ü6£¬y¡Ý8£©£¬È¡a L¸ÃÑÎËáÓëb L¸ÃÇâÑõ»¯ÄÆÈÜÒº·´Ó¦£¬Ç¡ºÃÍêÈ«Öкͣ¬Çó£º
¢ÙÈôx+y=14£¬Ôòa/b=
1
1
£¨ÌîÊý¾Ý£©£»
¢ÚÈôx+y=13£¬Ôòa/b=
0.1
0.1
£¨ÌîÊý¾Ý£©£»
¢ÛÈôx+y£¾14£¬Ôòa/b=
10x+y-14
10x+y-14
£¨Ìî±í´ïʽ£©£»
¢Ü¸ÃÑÎËáÓë¸ÃÇâÑõ»¯ÄÆÈÜÒºÍêÈ«Öкͣ¬Á½ÈÜÒºµÄpH£¨x¡¢y£©µÄ¹ØϵʽΪ
x+y=14+lg£¨
a
b
£©
x+y=14+lg£¨
a
b
£©
£¨Ìî±í´ïʽ£©£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â

»ðÁ¦·¢µç³§Êͷųö´óÁ¿µÄµªÑõ»¯ÎNOx£©¡¢¶þÑõ»¯ÁòºÍ¶þÑõ»¯Ì¼µÈÆøÌå»áÔì³É»·¾³ÎÛȾ£®¶Ôȼú·ÏÆø½øÐÐÍÑ̼ºÍÍÑÁòµÈ´¦Àí£¬¿ÉʵÏÖÂÌÉ«»·±£¡¢½ÚÄܼõÅÅ¡¢·ÏÎïÀûÓõÈÄ¿µÄ£®
£¨1£©ÍÑ̼£º½«CO2ת»¯Îª¼×´¼µÄÈÈ»¯Ñ§·½³ÌʽΪ£ºCO2£¨g£©+3H2£¨g£©?CH3OH£¨g£©+H2O£¨g£©¡÷H3
¢ÙÈ¡Îå·ÝµÈÌå»ýµÄCO2ºÍH2µÄ»ìºÏÆøÌ壨ÎïÖʵÄÁ¿Ö®±È¾ùΪ1£º3£©£¬·Ö±ð¼ÓÈëζȲ»Í¬¡¢ÈÝ»ýÏàͬµÄºãÈÝÃܱÕÈÝÆ÷ÖУ¬·¢ÉúÉÏÊö·´Ó¦£¬·´Ó¦Ïàͬʱ¼äºó£¬²âµÃ¼×´¼µÄÌå»ý·ÖÊý¦Õ£¨CH3OH£© Ó뷴ӦζÈTµÄ¹ØϵÇúÏßÈçͼ¢ñËùʾ£¬ÔòÉÏÊöCO2ת»¯Îª¼×´¼µÄ·´Ó¦µÄ¡÷H3
£¼
£¼
0£¨Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°=¡±£©

¢ÚÈçͼ¢òÊÇÔÚºãÎÂÃܱÕÈÝÆ÷ÖУ¬Ñ¹Ç¿ÎªP1ʱH2µÄÌå»ý·ÖÊýËæʱ¼ätµÄ±ä»¯ÇúÏߣ¬ÇëÔÚ¸ÃͼÖл­³ö¸Ã·´Ó¦ÔÚP2£¨P2£¾P1£©Ê±µÄH2Ìå»ý·ÖÊýËæʱ¼ätµÄ±ä»¯ÇúÏß
£®
£¨2£©ÔÚÒ»ºãκãÈÝÃܱÕÈÝÆ÷ÖгäÈë1mol CO2ºÍ3mol H2£¬½øÐÐÉÏÊö·´Ó¦£®²âµÃCO2ºÍCH3OH£¨g£©µÄÎïÖʵÄÁ¿Å¨¶ÈËæʱ¼ä±ä»¯Èçͼ¢óËùʾ£®
ÊԻشð£º0¡«10minÄÚ£¬ÇâÆøµÄƽ¾ù·´Ó¦ËÙÂÊΪ
0.225mol/£¨L?min£©
0.225mol/£¨L?min£©
£»¸ÃζÈÏ£¬·´Ó¦µÄƽºâ³£ÊýΪ
5.33
5.33
£»£¨±£ÁôÈý¸öÓÐЧÊý×Ö£©µÚ10minºó£¬Ïò¸ÃÈÝÆ÷ÖÐÔÙ³äÈë1mol CO2ºÍ3mol H2£¬ÔòÔٴδﵽƽºâʱCH3OH£¨g£©µÄÌå»ý·ÖÊý
±ä´ó
±ä´ó
£¨Ìî±ä´ó¡¢¼õÉÙ¡¢²»±ä£©£®
£¨3£©ÍÑÁò£ºÄ³ÖÖÍÑÁò¹¤ÒÕÖн«·ÏÆø¾­´¦Àíºó£¬ÓëÒ»¶¨Á¿µÄ°±Æø¡¢¿ÕÆø·´Ó¦£¬Éú³ÉÁòËá狀ÍÏõËá淋ĻìºÏÎï×÷Ϊ¸±²úÆ·»¯·Ê£®ÅäÖÆ100mlÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈµÄÁòËáï§ÈÜÒºËùÐè²£Á§ÒÇÆ÷³ýÉÕ±­¡¢½ºÍ·µÎ¹ÜÍ⻹Ð裺
²£Á§°ô¡¢100mLÈÝÁ¿Æ¿
²£Á§°ô¡¢100mLÈÝÁ¿Æ¿
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â

£¨1£©¼ºÖª£ºFe(s)+
1
2
O2(g)=FeO(s)
¡÷H=-272.0kJ?mol-1
2Al(s)+
3
2
O2(g)=Al2O3(s)
¡÷H=-1675.7kJ?mol-1
AlºÍFeO·¢ÉúÂÁÈÈ·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽÊÇ
2Al£¨s£©+3FeO£¨s£©¨TAl2O3£¨s£©+3Fe£¨s£©¡÷H=-859.7 kJ?mol-1
2Al£¨s£©+3FeO£¨s£©¨TAl2O3£¨s£©+3Fe£¨s£©¡÷H=-859.7 kJ?mol-1
£®
£¨2£©Ä³¿ÉÄæ·´Ó¦ÔÚ²»Í¬Ìõ¼þϵķ´Ó¦Àú³Ì·Ö±ðΪA¡¢B£¬ÈçͼËùʾ£®
¢Ù¾ÝͼÅжϸ÷´Ó¦ÊÇ
Ï©
Ï©
£¨Ìî¡°Îü¡±»ò¡°·Å¡±£©ÈÈ·´Ó¦£¬µ±·´Ó¦´ïµ½Æ½ºâºó£¬ÆäËûÌõ¼þ²»±ä£¬Éý¸ßζȣ¬·´Ó¦ÎïµÄת»¯ÂÊ
¼õС
¼õС
£¨Ìî¡°Ôö´ó¡±¡¢¡°¼õС¡±»ò¡°²»±ä¡±£©£»
¢ÚÆäÖÐBÀú³Ì±íÃ÷´Ë·´Ó¦²ÉÓõÄÌõ¼þΪ
D
D
£®£¨Ñ¡ÌîÐòºÅ£©£®
A£®Éý¸ßζÈ
B£®Ôö´ó·´Ó¦ÎïµÄŨ¶È
C£®½µµÍζÈ
D£®Ê¹Óô߻¯¼Á
£¨3£©1000¡æʱ£¬ÁòËáÄÆÓëÇâÆø·¢ÉúÏÂÁз´Ó¦£º
Na2SO4£¨S£©+4H2£¨g£©¨TNa2S£¨s£©+4H2O£¨g£©
¸Ã·´Ó¦µÄƽºâ³£Êý±í´ïʽΪ
c4(H2O)
c4(H2)
c4(H2O)
c4(H2)
£»
ÒÑÖªK1000¡æ£¼K1200¡æ£¬Èô½µµÍÌåϵζȣ¬»ìºÏÆøÌåµÄƽ¾ùÏà¶Ô·Ö×ÓÖÊÁ¿½«
Ôö´ó
Ôö´ó
£¨Ìî¡°Ôö´ó¡±¡¢¡°¼õС¡±»ò¡°²»±ä¡±£©£®
£¨4£©³£ÎÂÏ£»Èç¹ûÈ¡0.1mol?L-1 HAÈÜÒºÓë0.1mol?L-1 NaOHÈÜÒºµÈÌå»ý»ìºÏ£¨»ìºÏºóÈÜÒºÌå»ýµÄ±ä»¯ºöÂÔ²»¼Æ£©£¬²âµÃ»ìºÏÒºµÄpH=8£®
¢Ù»ìºÏÒºÖÐÓÉË®µçÀë³öµÄOH-Ũ¶ÈÓë0.1mol?L-1 NaOHÈÜÒºÖÐÓÉË®µçÀë³öµÄOH-Ũ¶ÈÖ®±ÈΪ
107£º1
107£º1
£»
¢ÚÒÑÖªNH4AÈÜҺΪÖÐÐÔ£¬ÓÖÖª½«HAÈÜÒº¼Óµ½Na2CO3ÈÜÒºÖÐÓÐÆøÌå·Å³ö£¬ÊÔÍƶϣ¨NH4£©2CO3ÈÜÒºµÄpH
£¾
£¾
7£¨Ìî¡°£¼¡±¡¢¡°£¾¡±»ò¡°=¡±£©£»
ÏàͬζÈÏ£¬µÈÎïÖʵÄÁ¿Å¨¶ÈµÄÏÂÁÐËÄÖÖÑÎÈÜÒº°´pHÓÉ´óµ½Ð¡µÄÅÅÁÐ˳ÐòΪ
c£¾a£¾b£¾d
c£¾a£¾b£¾d
£¨ÌîÐòºÅ£©
a£®NH4HCO3    b£®NH4A    c£®£¨NH4£©2CO3    d£®NH4Cl£®

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸