¡¶¶¨Á¿·ÖÎö¡·²âÊÔÌâ(A)(1)1Òªµã ÏÂÔØ±¾ÎÄ

¶¨ÖÁ·Ó̪¸ÕºÃÍÊÉ«£¬¼ÆËã¸ÖÑùÖÐP%ÒÔ¼°P2O5%¡££¨ÒÑÖªM(P)=30.97£¬M(P2O5)=141.94£©

¡¶¶¨Á¿·ÖÎö¡·ÆÚÖÕÄ£ÄâÊÔ¾í¶þ

УÃû___________ ϵÃû__________ רҵ____________ ÐÕÃû___________ ѧºÅ__________ ÈÕÆÚ____________

Ò»¡¢Ñ¡ÔñÌ⣺ÔÚÏÂÁи÷ÌâÖУ¬·Ö±ðÓдúÂëΪA¡¢B¡¢C¡¢DµÄ4¸ö±»Ñ¡´ð°¸£¬½«ÕýÈ·´ð°¸µÄ´úÂëÌîÈëÌâÄ©µÄ£¨ £©ÖС£ £¨±¾´óÌâΪ18СÌ⣬ÿСÌâ2·Ö£¬¹²36·Ö£©

1¡¢ÓÐËÄλѧÉú²â¶¨Í¬Ò»ÊÔÑù£¬±¨¸æ·ÖÎö½á¹ûµÄÏà¶ÔÎó²îÈçÏ£¬ÆäÖÐÕýÈ·µÄÊÇ£¨ £© A¡¢0.1020% B¡¢0.102% C¡¢0.10% D¡¢0.101%

2¡¢ÏÂÁÐÎïÖÊÔڵζ¨Ê±£¬Ó¦×°ÔÚ¼îʽµÎ¶¨¹ÜÖеÄÊÇ£¨ £© A¡¢K2Cr2O7 B¡¢AgNO3 C¡¢Na2H2Y D¡¢Na2S2O3

3 ¡¢ÏÂÁÐËÄ×éÊý¾Ý£¬°´Q0.9¼ìÑé·¨£¬Ö»ÓÐÒ»×éÓ¦½«ÒݳöÖµ0.2052ÉáÈ¥£¬Õâ×éÊý¾ÝÊÇ£¨ £© A¡¢0.2038 0.2040 0.2040 0.2047 0.2052 B¡¢0.2038 0.2041 0.2042 0.2042 0.2052 C¡¢0.2038 0.2042 0.2045 0.2048 0.2052 D¡¢0.2038 0.2042 0.2047 0.2048 0.2052

4¡¢ÒªÊ¹C(HCl)=0.1000mol¡¤L1µÄHCl¶ÔNaOHµÄµÎ¶¨¶ÈΪ0.002000g/ml , Ó¦Ïò1ÉýÈÜÒºÖмÓÈë___________mlµÄË®¡££¨ £©

£­

A¡¢600 B¡¢800 C¡¢1000 D¡¢1200

5¡¢ÏÂÁÐËá¼îµÎ¶¨²»ÄÜ׼ȷ½øÐеÄÊÇ£¨ £©

A¡¢0.1mol¡¤l1µÄHClµÎ¶¨0.1mol¡¤l1µÄNH3¡¤H2O (Kb=1.76¡Á105)

£­

£­

£­

B¡¢0.1mol¡¤l1µÄHClµÎ¶¨0.1mol¡¤l1µÄNaAc ( KHAc =1.8¡Á105)

£­

£­

£­

£­

£­

£­

C¡¢0.1mol¡¤l1µÄNaOHµÎ¶¨0.1mol¡¤l1µÄHCOOH (KHCOOH=1.8¡Á104) D¡¢0.1mol¡¤l1µÄHClµÎ¶¨0.1mol¡¤l1µÄNaCN (KHCN= 6.2¡Á10

£­

£­

£­10

)

6¡¢ÅäÖÃNaOH±ê×¼ÈÜÒº£¬Ó¦Ñ¡ÔñµÄÕýÈ··½·¨£º£¨ £©

A¡¢·ÖÎöÌìÆ½ÉϾ«È·³ÆÈ¡Ò»¶¨ÖØÁ¿µÄNaOHÓڽྻµÄÉÕ±­ÖУ¬¼ÓÕôÁóË®Èܽâºó£¬Ï¡ÊÍÖÁËùÐèÌå»ý£¬Ôٱ궨¡£

B¡¢Ì¨³ÓÉϾ«È·³ÆÈ¡Ò»¶¨ÖØÁ¿µÄNaOHÓڽྻµÄÉÕ±­ÖУ¬¼ÓÕôÁóË®Èܽâºó¶¨ÈÝÓÚËùÐèÌå»ýµÄÈÝÁ¿Æ¿ÖУ¬¼ÆËã³öÆä׼ȷŨ¶È¡£ C¡¢Ì¨³ÓÉϳÆÈ¡Ò»¶¨ÖØÁ¿µÄNaOHÓڽྻµÄÉÕ±­ÖУ¬¼ÓÕôÁóË®Èܽâºó£¬Ï¡ÊÍÖÁËùÐèÌå»ý£¬Ôٱ궨¡£

D¡¢·ÖÎöÌìÆ½ÉϾ«È·³ÆÈ¡Ò»¶¨ÖØÁ¿µÄÓÅÖÊ´¿NaOHÓڽྻµÄÉÕ±­ÖУ¬¼ÓÕôÁóË®Èܽâºó¶¨ÈÝÓÚËùÐèÌå»ýµÄÈÝÁ¿Æ¿ÖУ¬¼ÆËã³öÆä׼ȷŨ¶È¡£

7¡¢·ÖÎöʵÑéÖУ¬ÏÂÁÐÄÄ×éʵÑéÒÇÆ÷ÔÚʹÓÃǰ²»ÄÜÓñêÒº»òÊÔҺƯϴµÄÊÇ£¨ £© A¡¢µÎ¶¨¹Ü ÒÆÒº¹Ü ÈÝÁ¿Æ¿ ×¶ÐÎÆ¿ B¡¢µÎ¶¨¹Ü ÒÆÒº¹Ü

C¡¢ÈÝÁ¿Æ¿ ×¶ÐÎÆ¿ D¡¢ ÒÆÒº¹Ü ÈÝÁ¿Æ¿ ×¶ÐÎÆ¿

25

8¡¢ÒÔHCl±ê×¼ÈÜÒºµÎ¶¨Ä³¼îҺŨ¶È£¬ËáʽµÎ¶¨¹ÜÒòδϴ¸É¾»£¬µÎ¶¨Ê±¹ÜÄÚ¹ÒÓÐÒºµÎ£¬È´ÒÔ´íÎóµÄÌå»ý¶ÁÊý±¨³ö½á¹û£¬Ôò¼ÆËã¼îҺŨ¶È£¨ £©

A¡¢Æ«¸ß B¡¢Æ«µÍ C¡¢ÎÞÓ°Ïì D¡¢ÎÞ·¨ÅжÏ

9¡¢ÏÂÁÐÇé¿öʹ²â¶¨½á¹ûÆ«¸ßµÄÊÇ£¨ £© A¡¢ÏÔɫʱ¼ä¶Ì¡¢ÓÐÉ«Îïδ³ä·ÖÏÔÉ«£» B¡¢Ëá¶È¹ý¸ß£¬ÏÔÉ«¼ÁËáЧӦÑÏÖØ£» C¡¢¸ÉÈÅÀë×ÓÓëÏÔÉ«¼ÁÉú³ÉÓÐÉ«»¯ºÏÎï D¡¢¸ÉÈÅÀë×ÓÓëÏÔÉ«¼ÁÉú³ÉÎÞÉ«»¯ºÏÎï

10¡¢Ä³ÈÜÒº×ñÑ­¹âÎüÊÕ¶¨ÂÉ£¬µ±Å¨¶ÈΪCʱ͸¹âÂÊΪT£»µ±Å¨¶ÈΪC/2ʱ£¨Òº²ãºñ¶È²»±ä£©£¬Í¸¹âÂÊΪ( ) A¡¢T B¡¢T/2 C¡¢ T2 D¡¢

11¡¢ÒÑÖªTK2Cr2O7/Fe = 0.003284 g/ml ,ÇÒMFe = 55.85 g/mol ,ÔòC(( )mol/l

A¡¢0.05880 B¡¢0.3528 C¡¢0.5880 D¡¢0.005880

12¡¢Ò»Êøµ¥É«¹âͨ¹ýºñ¶ÈΪ1cmµÄÓÐÉ«ÈÜÒººó£¬Ç¿¶È¼õÈõ20%£¬µ±Ëüͨ¹ýºñ¶ÈΪ4cmµÄÏàͬÈÜҺʱ£¬Ç¿¶È½«¼õÈõ£¨ £© A¡¢59% B¡¢41% C¡¢ 36% D¡¢64%

13¡¢ÏÂÁбíÊö²»ÕýÈ·µÄÊÇ£¨ £©

A¡¢ÎüÊÕ¹âÆ×ÒÔ²¨³¤Îª×Ý×ø±ê£¬Îü¹â¶ÈΪºá×ø±ê£» B¡¢ÎüÊÕ¹âÆ×ÇúÏß±íÃ÷ÁËÎü¹â¶ÈËæ²¨³¤µÄ±ä»¯£» C¡¢ÎüÊÕ¹âÆ×ÇúÏß±íÃ÷ÁËÎü¹âÎïÖʵĹâÎüÊÕÌØÐÔ¡£

D¡¢ÎüÊÕ¹âÆ×ÇúÏßÖУ¬×î´óÎüÊÕ´¦µÄ²¨³¤Îª×î´óÎüÊÕ²¨³¤£»

14¡¢ÓÃAgNO3±ê×¼ÈÜÒºÖ±½ÓµÎ¶¨NaClÈÜÒº²â¶¨Clº¬Á¿Ê±£¬ËùʹÓõÄָʾ¼ÁΪ£¨ £© A¡¢K2Cr2O7 B¡¢NH4Fe(SO4)2¡¤12H2O C¡¢NH4SCN D¡¢K2CrO4 15 ¡¢ÅäλµÎ¶¨·¨µÄÖ±½Ó·¨ÖÕµãËù³ÊÏÖµÄÑÕɫΪ£¨ £© A¡¢MInÅäºÏÎïµÄÑÕÉ« B¡¢InµÄÑÕÉ«

C¡¢MYÅäºÏÎïµÄÑÕÉ« D¡¢ÉÏÊöA¡¢BµÄ»ìºÏÉ«

16¡¢ÓÃEDTAµÎ¶¨½ðÊôÀë×ÓMnʱ£¬¶Ô½ðÊôָʾ¼ÁµÄÒªÇ󣺣¨ £©

£«

£­

T

1K2Cr2O7)= 6A¡¢KB¡¢KC¡¢K

f?MIn/Kf?MY?100 f?MIn/Kf?MY?0.01 f?MIn£½Kf?MY

D¡¢Ö»ÒªMInÓëInÑÕÉ«ÓÐÃ÷ÏÔ²î±ð

17¡¢ÓÐÒ»¼îÒº£¬¿ÉÄÜÊÇNaOH»òNaHCO3»òNa2CO3»òËüÃǵĻìºÏÒº£¬ÈôÓñê×¼ËáµÎ¶¨ÖÁ·Ó̪ÖÕµãʱ£¬ºÄÈ¥ËáµÄÌå»ýΪV1ml , ¼ÌÒÔ¼×»ù³ÈΪָʾ¼Á£¬¼ÌÐøµÎ¶¨ÓÖºÄÈ¥V2 ml£¬µ±V1=V2ʱ£¬¸Ã¼îÒº×é³ÉΪ£¨ £©¡£

26

A¡¢Na2CO3ÓëNaOH B¡¢NaHCO3ÓëNa2CO3 C¡¢Na2CO3 D¡¢NaHCO3

18¡¢100ml0.1mol/LµÄNH3¡¤H20(PKb=4.74)ÖÐÂýÂýµÎÈëµÈŨ¶ÈµÄHCl±ê×¼ÈÜÒº100mlʱ£¬ÈÜÒºµÄPHֵΪ£¨ £© A¡¢5.13 B¡¢4.74 C¡¢3.74 D¡¢5.28

¶þ¡¢Ìî¿ÕÌ⣺£¨±¾´óÌâ¹²30£© 1¡¢£¨±¾Ð¡Ìâ2·Ö£©

ijѧÉú±ê¶¨KMnO4ËùµÃ·ÖÎö½á¹ûÊÇ£º0.1025¡¢0.1027¡¢0.1032£¬Ôٲⶨһ´ÎËùµÃ·ÖÎö½á¹û²»Ó¦ÉáÈ¥µÄµÄ½çÏÞÊÇ£º______________________ ¡££¨Q0.90=0.76£© 2¡¢£¨±¾Ð¡Ìâ2·Ö£©

20ml 0.1mol/lµÄHClºÍ20ml0.1mol/lµÄNa2CO3»ìºÏ£¬ÈÜÒºµÄÖÊ×ÓÌõ¼þʽ__________________________________________¡£ 3¡¢£¨±¾Ð¡Ìâ2·Ö£©

²â¶¨ÑùÆ·ÖÐP2O5º¬Á¿Ê±£¬ÑùÆ·¾­´¦ÀíÏȳÁµíΪMg3(PO4)2£¬ÔÙת»¯Îªµ¥ÖÊP½øÐгÆÁ¿£¬Ôò»¯Ñ§ÒòËØÎª________________________¡£ 4¡¢£¨±¾Ð¡Ìâ4·Ö£©

²ÉÓÃK2Cr2O7·¨²â¶¨ÂÌ·¯ÖÐÌúº¬Á¿Ê±£¬³£²ÉÓÃ_______________×÷ָʾ¼Á£¬ÖÕµãʱÑÕÉ«±ä»¯_____________________£¬³ýÁËÓÃH2SO4 ×÷ËáÐÔ½éÖÊÍ⣬»¹¼ÓÈëÒ»¶¨Á¿µÄH3PO4 £¬Ö÷ÒªÔ­Òò_______________________________________________¡£ 5¡¢£¨±¾Ð¡Ìâ4·Ö£©

ÔÚHCl½éÖÊÖУ¬ÓÃKMnO4 ²â¶¨Fe2+µÄº¬Á¿Ê±£¬½á¹ûÍùÍùÆ«______£¬ÆäÔ­Òò___________________________________________________________________ ¡£ 6¡¢£¨±¾Ð¡Ìâ3·Ö£©

ÓÃCe(SO4)2±ê×¼ÈÜÒºµÎ¶¨FeSO4(µÎ¶¨·´Ó¦Ce4+Fe2=Ce3+Fe3)£¬µ±µÎ¶¨ÖÁ99.9%ʱÈÜÒºµÄµçλÊÇ___________·ü£»µÎ¶¨

£«

£«

£«

£«

ÖÁ»¯Ñ§¼ÆÁ¿µãʱÊÇ__________·ü£»µÎ¶¨ÖÁ100.1%ÓÖÊÇ____________·ü¡£ £¨ÒÑÖªEFe3/Fe2=0.80V

£«

£«

0?E0?Ce4

£«

/Ce3=1.60V £©

£«

7¡¢£¨±¾Ð¡Ìâ3·Ö£©

ÓÃEDTAµÎ¶¨½ðÊôÀë×ÓÉú³ÉMYÅäºÏÎïʱ£¬µ±½éÖʵÄPHÔ½¸ß£¬Ôò?Y(H)Ô½______, µÄÎȶ¨ÐÔÔ½____________. 8¡¢(±¾Ð¡Ìâ4·Ö)

ÔÚPH=10ʱ£¬ÓÃEDTAµÎ¶¨Mg2, ÈÜÒºÖÐÊ×ÏȼÓÈë¸õºÚTָʾ¼Á£¬´ËʱÈÜÒº³Ê______É«£¬µ±µÎÈëEDTAºó£¬ÏÈÓëEDTA

£«

[Y?]±ÈÖµÔ½______£¬Éú³ÉÎïMY[Y]·´Ó¦µÄÊÇ_______________£¬´ïµ½¼ÆÁ¿µãʱ£¬ÈÜÒº³Ê_____É«£¬ÆäÔ­Òò

_________________________________________________________________¡£ 9¡¢£¨±¾Ð¡Ìâ4·Ö£©

¶ÔÓÚÒ»°ãµÎ¶¨·ÖÎö£¬ÒªÇóµ¥Ïî²âÁ¿µÄÏà¶ÔÎó²î¡Ü£°.£±%.³£Ó÷ÖÎöÌìÆ½¿ÉÒÔ³Æ×¼ÖÁ___________mg£¬Óòî¼õ·¨³ÆÈ¡ÊÔÑùʱ£¬Ò»°ãÖÁÉÙÓ¦³ÆÈ¡_______g£»

50mlµÎ¶¨¹ÜµÄ¶ÁÊýÒ»°ã¿ÉÒÔ¶Á×¼µ½________ml £¬¹ÊµÎ¶¨Ê±Ò»°ãµÎ¶¨ÈÝ»ý¿ØÖÆÔÚ________ml ÒÔÉÏ£¬ËùÒԵζ¨·ÖÎöÊÊÓÃÓÚ³£Á¿·ÖÎö¡£

10¡¢£¨±¾Ð¡Ìâ2·Ö£© ÓÃÇ¿¼îµÎ¶¨0.10mol¡¤L

£­1

µÄijÈõËᣬ׼ȷµÎ¶¨µÄǰÌáÌõ¼þ___________________ £»

£­1

ÓÃEDTAµÎ¶¨ 0.010mol¡¤LµÄij½ðÊôÀë×Ó£¬¿ÉÒÔ±»×¼È·µÎ¶¨µÄÌõ¼þ__________¡£

Èý¡¢¼ÆËãÌ⣺£¨±¾´óÌâ ¹²4СÌ⣬¹²¼Æ34·Ö£©

1¡¢£¨±¾Ð¡Ìâ8·Ö£© ³ÆÈ¡´¿NaHCO31.008gÈÜÓÚÊÊÁ¿Ë®ÖУ¬È»ºóÍù´ËÈÜÒºÖмÓÈë´¿¹ÌÌåNaOH0.3200g£¬×îºó½«´ËÈÜÒºÒÆÈë250mlÈÝÁ¿Æ¿ÖС£ÒÆÈ¡ÉÏÊöÈÜÒº50.00ml,ÒÔ0.1000mol¡¤L1HClÈÜÒºµÎ¶¨¡£¼ÆË㣨1£©ÒÔ·Ó̪Ϊָʾ¼ÁµÎ¶¨ÖÁÖÕµãʱ£¬ÏûºÄ

£­

27

HClÈÜÒº¶àÉÙºÁÉý£¿£¨2£©¼ÌÐø¼ÓÈë¼×»ù³Èָʾ¼ÁµÎ¶¨ÖÁÖÕµãʱ£¬ÓÖÏûºÄHClÈÜÒº¶àÉÙºÁÉý£¿£¨MNaHCO3?84.00 £¬

MNaOH = 40.00 £© \\

2¡¢£¨±¾Ð¡Ìâ8·Ö£©³ÆÈ¡º¬·ú¿óÑù0.5000g£¬Èܽ⣬ÔÚÈõ¼îÐÔ½éÖÊÖмÓÈë0.1000

mol¡¤L1 Ca2ÈÜÒº50.00ml £¬½«³Áµí¹ýÂË£¬ÊÕ¼¯ÂËÒººÍÏ´Òº£¬È»ºóÓÚPH=10.00ʱÓÃ0.05000 mol¡¤L1EDTA·µµÎ¶¨¹ýÁ¿µÄ

£­

£«

£­

Ca2ÖÁ»¯Ñ§¼ÆÁ¿µã£¬ÏûºÄ20.00ml.¼ÆËã¿óÑùÖзúµÄ°Ù·Öº¬Á¿¡£(MF = 19.00 )

£«

3¡¢£¨±¾Ð¡Ìâ10·Ö£© ׼ȷ³ÆÈ¡Ìú¿óÑù0.2000g , ÓÃHCl·Ö½âºó£¬¼ÓÔ¤´¦Àí¼ÁSnCl2½«Fe3»¹Ô­ÎªFe2,Ðè22.50mlKMnO4Óë

£«

£«

Ö®·´Ó¦¡£TKMnO£¨MNa2C

4/Na2C2O4?0.006700g/ml£¬ÇóÌú¿óÑùÖÐFeµÄ°Ù·Öº¬Á¿¡£

2O4?134.00 MFe?55.85£©

4¡¢£¨±¾Ð¡Ìâ8·Ö£© ÏÖȡijº¬Í­ÊÔÒº2.00ml¶¨ÈÝÖÁ100.00ml.´ÓÖÐÎüÈ¡2.00mlÏÔÉ«¶¨ÈÝÖÁ50.00ml ,ÓÃ1cmµÄÎüÊճزâµÃÎü¹â¶ÈΪ0.400¡£ÒÑÖªÏÔÉ«ÂçºÍÎïµÄĦ¶ûÎü¹âϵÊýΪ1.1¡Á104L¡¤mol1¡¤cm1¡£Çó¸Ãº¬Í­ÊÔÒºÖÐÍ­µÄº¬Á¿£¨ÒÔg¡¤L

£­

£­

£­1

¼Æ£©¡£ ( ÒÑ

ÖªMCu=63.50 )

²Î¿¼´ð°¸ ¡¶¶¨Á¿·ÖÎö¡·²âÊÔÌâA

Ò»¡¢Ñ¡ÔñÌ⣺1¡¢ C 2¡¢B 3¡¢B 4¡¢C 5¡¢B 6¡¢C 7¡¢C 8¡¢A 9¡¢ B 10¡¢D 11¡¢D 12¡¢C 13¡¢D 14¡¢C 15¡¢C 16¡¢D 17¡¢D 18¡¢B 19¡¢C 20¡¢C 21¡¢C 22¡¢A 23¡¢C 24¡¢B 25¡¢D 26¡¢B 27¡¢B 28¡¢C 29¡¢C 30¡¢C 31¡¢C 32¡¢D 33¡¢A 34¡¢A 35¡¢D 36¡¢C 37¡¢B 38¡¢A 39¡¢D 40¡¢A 41¡¢C 42¡¢B 43¡¢A 44¡¢D 45¡¢A 46¡¢D 47¡¢C 48¡¢D 49¡¢B 50¡¢C 51¡¢B 52¡¢B 53¡¢C 54¡¢B 55¡¢A 56¡¢A 58¡¢C 59¡¢A 60¡¢D 61¡¢B ¶þ¡¢Ìî¿ÕÌ⣺

1¡¢²»Ò»¶¨¸ß 2¡¢È¡¶à´Î²â¶¨µÄƽ¾ùÖµ 3¡¢£¨1£©20.1£¨2£©12.00£¨3£©0.1 4¡¢HCO3 H2O [H2CO3] + [H+] = [CO32

£­

£­

] +[OH]5¡¢(1) NaOH + Na2CO3 (2) Na2CO3 + NaHCO3(3) NaOH(4) NaHCO3(5) Na2CO3

£­

6¡¢£¨1£©¸ºÎó²î £¨2£©ÎÞÓ°Ïì £¨3£© ¸ºÎó²î 7¡¢1£º2 8¡¢£¨1£©12 £¨2£© 7.0 £¨3£© 7.0 9¡¢

5.3cb% 10¡¢5.8¡Á10a£­6

11¡¢Ka(Kb) ÒÔ¼°C , Ka¡Ý107 , ָʾ¼ÁµÄ±äÉ«·¶Î§Ó¦È«²¿»ò²¿·ÖλÓÚ

£­

µÎ¶¨Í»Ô¾·¶Î§ÄÚ 12¡¢ ¼×»ùºì 5.28 13¡¢ 0.05880 14¡¢

£«

£­

Ka1Ka2£­

£­

Ka1?C NaOH

15¡¢0.09910< x < 0.1036 16¡¢ ÎÞ Æ«¸ß17¡¢[ H] = [HCO3] £«2[CO32] £«[OH] 18¡¢0.007347 0.007465 19¡¢0.1 0.2 0.01 20 20¡¢6.0 5.0 ¼×»ùºì»ò¼×»ù 21¡¢30.00 ´ó

28