ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿½ñÓÐÊÒÎÂÏÂËÄÖÖÈÜÒº£¬ÓйØÐðÊö²»ÕýÈ·µÄÊÇ

¢Ù

¢Ú

¢Û

¢Ü

pH

11

11

3

3

ÈÜÒº

°±Ë®

ÇâÑõ»¯ÄÆÈÜÒº

´×Ëá

ÑÎËá

A. ¢Ù¡¢¢ÜÁ½ÈÜÒºµÈÌå»ý»ìºÏ£¬ËùµÃÈÜÒºÖÐc(Cl£­)£¾c(NH4+)£¾c(OH£­)£¾c(H+)

B. ·Ö±ð¼ÓˮϡÊÍ10±¶£¬ËÄÖÖÈÜÒºµÄpH ¢Ù£¾¢Ú£¾¢Ü£¾¢Û

C. ¢Ù¡¢¢ÚÖзֱð¼ÓÈëÊÊÁ¿µÄÂÈ»¯ï§¾§Ìåºó£¬Á½ÈÜÒºµÄpH¾ù¼õС

D. VaL¢ÜÓëVbL¢ÚÈÜÒº»ìºÏºó,Èô»ìºÏºóÈÜÒºpH=4,ÔòVa ¡ÃVb= 11¡Ã9

¡¾´ð°¸¡¿A

¡¾½âÎö¡¿¢ÙÖÐc(NH3¡¤H2O)£¾0.001mol/L£¬¢ÚÖÐc(NaOH)=0.001mol/L£¬¢ÛÖÐc(CH3COOH)£¾0.001mol/L£¬¢ÜÖÐc(HCl)=0.001mol/L¡£A£®µÈÌå»ýµÄ¢Ù¡¢¢ÜÁ½ÈÜÒº»ìºÏ£¬°±Ë®¹ýÁ¿£¬ÈÜÒº³Ê¼îÐÔ£¬Ó¦Îªc(NH4+)£¾c(Cl-)£¾c(OH-)£¾c(H+)£¬·ñÔòµçºÉ²»Êغ㣬¹ÊA´íÎó£»B£®Ç¿Ëᡢǿ¼îÏ¡ÊÍ10±¶£¬pH±ä»¯1£¬ÔòÏ¡ÊÍ10±¶Ê±¢ÚµÄpH=10£¬¢ÜµÄpH=4£¬¶øÈõËá¡¢Èõ¼îÏ¡ÊÍ10±¶£¬pH±ä»¯Ð¡ÓÚ1£¬Ôò¢ÙµÄ10£¼pH£¼11£¬¢ÛµÄ3£¼pH£¼4£¬¼´·Ö±ð¼ÓˮϡÊÍ10±¶ËÄÖÖÈÜÒºµÄpHΪ¢Ù£¾¢Ú£¾¢Ü£¾¢Û£¬¹ÊBÕýÈ·£»C£®¢ÙÖдæÔÚNH3¡¤H2ONH4++OH-£¬¼ÓÈëÂÈ»¯ï§£¬c(NH4+)Ôö´ó£¬ÒÖÖÆNH3¡¤H2OµÄµçÀ룬ÔòpH¼õС£¬¢ÚÖмÓÈëÂÈ»¯ï§£¬·¢ÉúNH4++OH-=NH3¡¤H2O£¬c(OH-)¼õС£¬ÔòpH¼õС£¬¹ÊCÕýÈ·£»D£®Èô»ìºÏºóÈÜÒºpH=4£¬ÔòÑÎËá¹ýÁ¿£¬Ôò=0.0001mol/L£¬½âµÃVa£ºVb=11£º9£¬¹ÊDÕýÈ·£»¹ÊÑ¡A¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿Îíö²ÌìÆøÆµ·±³öÏÖ£¬ÑÏÖØÓ°ÏìÈËÃǵÄÉú»îºÍ½¡¿µ¡£ÆäÖÐÊ×ÒªÎÛȾÎïΪ¿ÉÎüÈë¿ÅÁ£ÎïPM2.5£¬ÆäÖ÷ÒªÀ´Ô´ÎªÈ¼Ãº¡¢»ú¶¯³µÎ²ÆøµÈ¡£Òò´Ë£¬¶ÔPM2.5¡¢SO2¡¢NOxµÈ½øÐÐÑо¿¾ßÓÐÖØÒªÒâÒå¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©½«PM2.5Ñù±¾ÓÃÕôÁóË®´¦ÀíÖÆ³É´ý²âÊÔÑù¡£

Èô²âµÃ¸ÃÊÔÑùËùº¬Ë®ÈÜÐÔÎÞ»úÀë×ӵĻ¯Ñ§×é·Ö¼°Æäƽ¾ùŨ¶ÈÈçÏÂ±í£º

Àë×Ó

K+

Na+

NH4+

SO42-

NO3-

Cl-

¶Èmol/L

4¡Á10-6

6¡Á10-6

2¡Á10-5

4¡Á10-5

3¡Á10-5

2¡Á10-5

¸ù¾Ý±íÖÐÊý¾ÝÅжÏÊÔÑùµÄpH=_________¡£

£¨2£©Æû³µÎ²ÆøÖÐNOxºÍCOµÄÉú³É£ºÒÑÖªÆû¸×ÖÐÉú³ÉNOµÄ·´Ó¦Îª£ºN2(g)+O2(g)2NO(g) ¡÷H>0¡£ºãΣ¬ºãÈÝÃܱÕÈÝÆ÷ÖУ¬ÏÂÁÐ˵·¨ÖÐÄÜ˵Ã÷¸Ã·´Ó¦´ïµ½»¯Ñ§Æ½ºâ״̬µÄÊÇ____

A.»ìºÏÆøÌåµÄÃܶȲ»Ôٱ仯 B.ÈÝÆ÷ÄÚµÄѹǿ²»Ôٱ仯

C.N2¡¢O2¡¢NOµÄÎïÖʵÄÁ¿Ö®±ÈΪ1¡Ã1¡Ã2 D.ÑõÆøµÄ°Ù·Öº¬Á¿²»Ôٱ仯

£¨3£©Îª¼õÉÙSO2µÄÅÅ·Å£¬³£²ÉÈ¡µÄ´ëÊ©ÓУº

¢Ù½«Ãº×ª»¯ÎªÇå½àÆøÌåȼÁÏ¡£

ÒÑÖª£ºH2(g)+ 1/2O2(g) =H2O(g) ¡÷H=£­241.8kJ¡¤mol-1

C(s)+1/2O2(g) =CO(g)¡÷H =-110.5kJ¡¤mol-1

¢ÚÏ´µÓº¬SO2µÄÑÌÆø¡£

д³ö½¹Ì¿ÓëË®ÕôÆø·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ£º_______________¡£

£¨4£©³µÁ¾ÅŷŵĵªÑõ»¯ÎúȼÉÕ²úÉúµÄ¶þÑõ»¯ÁòÊǵ¼ÖÂÎíö²ÌìÆøµÄ¡°×ï¿ý»öÊס±Ö®Ò»¡£»îÐÔÌ¿¿É´¦Àí´óÆøÎÛȾÎïNO¡£ÔÚ5LÃܱÕÈÝÆ÷ÖмÓÈëNOºÍ»îÐÔÌ¿£¨¼ÙÉèÎÞÔÓÖÊ£©¡£Ò»¶¨Ìõ¼þÏÂÉú³ÉÆøÌåEºÍF¡£µ±Î¶ȷֱðÔÚT1¡æºÍT2¡æÊ±£¬²âµÃ¸÷ÎïÖÊÆ½ºâʱÎïÖʵÄÁ¿£¨n/mol£©ÈçÏÂ±í£º

ÎïÖÊ

ζÈ/¡æ

»îÐÔÌ¿

NO

E

F

³õʼ

3.000

0.10

0

0

T1

2.960

0.020

0.040

0.040

T2

2.975

0.050

0.025

0.025

¢Ùд³öNOÓë»îÐÔÌ¿·´Ó¦µÄ»¯Ñ§·½³Ìʽ______________________£»

¢Ú¼ÆËãÉÏÊö·´Ó¦T1¡æÊ±µÄƽºâ³£ÊýK1=______________£»

ÈôT1£¼T2£¬Ôò¸Ã·´Ó¦µÄ¡÷H__________0(Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°£½¡±£©¡£

¢ÛÉÏÊö·´Ó¦T1¡æÊ±´ïµ½»¯Ñ§Æ½ºâºóÔÙͨÈë0.1molNOÆøÌ壬Ôò´ïµ½Ð»¯Ñ§Æ½ºâʱNOµÄת»¯ÂÊΪ________£»

¡¾ÌâÄ¿¡¿ÏÂÁÐÈÜÒºÖи÷΢Á£µÄŨ¶È¹ØÏµÕýÈ·µÄÊÇ

A. 0.2 mol¡¤L-1µÄNa2CO3ÈÜÒº£ºc(OH-)=c(HCO3-)+c(H+)+c(H2CO3)

B. pHÏàµÈµÄNaFÓëCH3COOKÈÜÒº£º[c(Na+)-c(F-)]>[c(K+)-c(CH3COO-)]

C. pHÏàµÈµÄ¢ÙNH4Cl ¢Ú(NH4)2SO4 ¢ÛNH4HSO4ÈÜÒº£¬NH4+´óС˳ÐòΪ¢Ù=¢Ú>¢Û

D. 0.2 mol¡¤L-1HClÓë0.1 mol¡¤L-1NaAlO2ÈÜÒºµÈÌå»ý»ìºÏ£ºc(Cl-)>c(Na+)>c(H+)>c(Al3+)>c(OH-)

¡¾´ð°¸¡¿C

¡¾½âÎö¡¿¸ù¾ÝÖÊ×ÓÊØºã£¬0.2 mol¡¤L-1µÄNa2CO3ÈÜÒº£ºc(OH-)=c(HCO3-)+c(H+)+2c(H2CO3)£¬¹ÊA´íÎ󣻸ù¾ÝµçºÉÊØºã£¬NaFÈÜÒºÖÐ[c(Na+)-c(F-)]= [c(OH-)-c(H+)]£¬CH3COOKÈÜÒº[c(K+)-c(CH3COO-)]= [c(OH-)-c(H+)]£¬PHÏàµÈ£¬ËùÒÔ[c(Na+)-c(F-)]=[c(K+)-c(CH3COO-)]£¬¹ÊB´íÎó£» NH4Cl¡¢(NH4)2SO4ÓÉÓÚNH4+Ë®½âÈÜÒº³ÊËáÐÔ£¬pHÏàµÈ£¬ËùÒÔNH4+Ũ¶ÈÏàµÈ£¬NH4HSO4ÈÜÒºÖ÷ÒªÓÉÓÚNH4HSO4µçÀë³ÊËáÐÔ£¬ËùÒÔNH4HSO4ÈÜÒºÖÐNH4+Ũ¶È×îС£¬¹ÊCÕýÈ·£»0.2 mol¡¤L-1HClÓë0.1 mol¡¤L-1NaAlO2ÈÜÒºµÈÌå»ý»ì£¬ÍêÈ«·´Ó¦Éú³ÉÂÈ»¯ÄÆ¡¢ÂÈ»¯ÂÁ¡¢ÇâÑõ»¯ÂÁ£¬ËùÒÔc(Cl-)>c(Na+)>c(Al3+)>c(H+)>c(OH-)£¬¹ÊD´íÎó¡£

¡¾ÌâÐÍ¡¿µ¥Ñ¡Ìâ
¡¾½áÊø¡¿
8

¡¾ÌâÄ¿¡¿Ö±½ÓÅŷź¬SO2µÄÑÌÆø»áÐγÉËáÓ꣬Σº¦»·¾³¡£Ä³»¯Ñ§ÐËȤС×é½øÐÐÈçÏÂÓйØSO2ÐÔÖʺͺ¬Á¿²â¶¨µÄ̽¾¿»î¶¯¡£

£¨1£©×°ÖÃAÖÐÒÇÆ÷aµÄÃû³ÆÎª________¡£Ð´³öAÖз¢Éú·´Ó¦µÄÀë×Ó·½³Ìʽ_________________________¡£ÈôÀûÓÃ×°ÖÃAÖвúÉúµÄÆøÌåÖ¤Ã÷+4¼ÛµÄÁòÔªËØ¾ßÓÐÑõ»¯ÐÔ£¬ÊÔÓû¯Ñ§·½³Ìʽ±íʾ¸ÃʵÑé·½°¸µÄ·´Ó¦Ô­Àí_______________________________________¡£

£¨2£©Ñ¡ÓÃͼÖеÄ×°ÖúÍҩƷ̽¾¿ÑÇÁòËáÓë´ÎÂÈËáµÄËáÐÔÇ¿Èõ£º

¢Ù¼×ͬѧÈÏΪ°´A¡úC¡úF¡úÎ²Æø´¦Àí˳ÐòÁ¬½Ó×°ÖÿÉÒÔÖ¤Ã÷ÑÇÁòËáºÍ´ÎÂÈËáµÄËáÐÔÇ¿Èõ£¬ÒÒͬѧÈÏΪ¸Ã·½°¸²»ºÏÀí£¬ÆäÀíÓÉÊÇ__________________________________¡£

¢Ú±ûͬѧÉè¼ÆµÄºÏÀíʵÑé·½°¸Îª£º°´ÕÕA¡úC¡ú______¡úF¡úÎ²Æø´¦Àí(Ìî×Öĸ)˳ÐòÁ¬½Ó×°Öá£Ö¤Ã÷ÑÇÁòËáµÄËáÐÔÇ¿ÓÚ´ÎÂÈËáµÄËáÐÔµÄʵÑéÏÖÏóÊÇ____________________________¡£

¢ÛÆäÖÐÑbÖÃCµÄ×÷ÓÃÊÇ____________________________¡£

£¨3£©ÎªÁ˲ⶨװÖÃA²ÐÒºÖÐSO2µÄº¬Á¿£¬Á¿È¡10.00mL²ÐÒºÓÚÔ²µ×ÉÕÆ¿ÖУ¬¼ÓÈÈʹSO2È«²¿Õô³ö£¬ÓÃ20.00mL0.0500mol¡¤L-1µÄKMnO4ÈÜÒºÎüÊÕ¡£³ä·Ö·´Ó¦ºó£¬ÔÙÓÃ0.2000 mol¡¤L-1µÄKI±ê×¼ÈÜÒºµÎ¶¨¹ýÁ¿µÄKMnO4£¬ÏûºÄKIÈÜÒº20.00mL¡£

ÒÑÖª£º5SO2+2MnO4-+2H2O=2Mn2++5SO42-+4H+ 10I-+2MnO4-+16H+=2Mn2++5I2+8H2O

¢Ù²ÐÒºÖÐSO2µÄº¬Á¿Îª____g¡¤L-1¡£

¢ÚÈôµÎ¶¨Ç°¶ÁÊýʱƽÊÓ£¬µÎ¶¨ÖÕµã¶ÁÊýʱ¸©ÊÓÔò²â¶¨½á¹û________£¨Ìî¡°Æ«¸ß¡±¡¢¡°Æ«µÍ¡±»ò¡°ÎÞÓ°Ï족£©¡£

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

¾«Ó¢¼Ò½ÌÍø