ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿µÍŨ¶ÈµÄÇâ·úËáÊÇÒ»ÔªÈõËᣬ´æÔÚÏÂÁÐÁ½¸öƽºâ£ºHFH++F-£¬HF+F-HF2- (½ÏÎȶ¨)¡£25¡æʱ£¬²»Í¬ËáÐÔÌõ¼þϵÄ2.0amol¡¤L-1HFÈÜÒºÖУ¬c(HF)¡¢c(F-)ÓëÈÜÒºpH(ºöÂÔÌå»ý±ä»¯)µÄ±ä»¯¹ØϵÈçͼËùʾ¡£ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨ £©

A.c(HF)+c(F-)=2.0amol¡¤L-1

B.c(F-)>c(HF)ʱ£¬ÈÜÒºÒ»¶¨³Ê¼îÐÔ

C.Ëæ×ÅÈÜÒºpHÔö´ó£¬²»¶ÏÔö´ó

D.25¡æʱ£¬HFµÄµçÀë³£ÊýKa=10-3.45

¡¾´ð°¸¡¿D

¡¾½âÎö¡¿

A£®¸ÃÈÜÒºÖк¬F΢Á£ÓУºF-¡¢HF¡¢£¬¸ù¾ÝÎïÁÏÊغã¿ÉÖª£¬£¬¹ÊA´íÎó£»

B£®ÓÉͼ¿ÉÖª£¬µ±ÈÜÒºpH>3.45ʱ£¬c(F-)>c(HF)£¬Òò´Ëµ±c(F-)>c(HF)ʱ£¬ÈÜÒº²»Ò»¶¨³Ê¼îÐÔ£¬¹ÊB´íÎó£»

C£®Ëæ×ÅÈÜÒºpHÔö´ó£¬c(F-)Öð½¥Ôö´ó£¬£¬Î¶Ȳ»±ä£¬Ka²»±ä£¬Òò´ËÖð½¥¼õС£¬¹ÊC´íÎó£»

D£®µ±pH=3.45ʱ£¬c(F-)=c(HF)£¬£¬¹ÊDÕýÈ·£»

¹Ê´ð°¸Îª£ºD¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿ÈýÂÈ»¯Áù°±ºÏîÜ(III)£¨[Co(NH3)6]Cl3 ÊǺϳÉÆäËûº¬îÜÅäºÏÎïµÄÖØÒªÔ­ÁÏ¡£ÔÚ»îÐÔÌ¿µÄ´ß»¯×÷ÓÃÏ£¬Í¨¹ýÑõ»¯¶þÂÈ»¯Áù°±ºÏîÜ(II)µÃµ½ÈýÂÈ»¯Áù°±ºÏîÜ(III)ÖƱ¸Á÷³ÌÈçÏ£º

×ÊÁÏ£º

¢ÙîÜÀë×Ó³£¼û¼Û̬ÓÐ+2(II)¼Û£¬+3(III)¼Û£¬Co(II)Àë×ÓÄÜÔÚË®ÈÜÒºÖÐÎȶ¨´æÔÚ£¬µ« Co(III)Àë×Ó²»ÄÜÎȶ¨´æÔÚ£¬Ö»ÄÜÒÔ¹Ì̬»òÂçºÏÎïÐÎʽ£¨Èç[Co(NH3)6]3+£©Îȶ¨´æÔÚÈÜÒºÖС£

¢Ú Co2+ÔÚ pH=9.4 ʱÍêÈ«³ÁµíΪ Co(OH)2

£¨1£© ʵÑéÖÐÐèÒª½« CoCl2¡¤6H2O ¾§ÌåÑÐϸ£¬ÆäÄ¿µÄÊÇ£º__________________¡£

£¨2£©ÔÚ¼ÓÈëŨ°±Ë®Ç°ÏȼÓÈë´óÁ¿ NH4ClÈÜÒº£¬Çë½áºÏƽºâÔ­Àí½âÊÍÔ­Òò______________________¡£

£¨3£©ÔÚ¡°Ñõ»¯¡±¹ý³ÌÖÐÐèˮԡ¿ØÎÂÔÚ 50¡«60¡æ£¬Î¶Ȳ»Äܹý¸ß£¬Ô­ÒòÊÇ______________________¡£

£¨4£©Ð´³ö¡°Ñõ»¯¡±¹ý³ÌÖз´Ó¦µÄÀë×Ó·½³Ìʽ_______________¡£

£¨5£©Îª²â¶¨²úÆ·ÖÐîܵĺ¬Á¿£¬½øÐÐÏÂÁÐʵÑ飺

¢Ù³ÆÈ¡ÑùÆ· 4.000 g ÓÚÉÕÆ¿ÖУ¬¼ÓË®Èܽ⣬¼ÓÈë×ãÁ¿µÄ NaOH ÈÜÒº£¬¼ÓÈÈÖÁ·Ð 15¡«20 min£¬½« [Co£¨NH3£©6]Cl3 Íêȫת»¯Îª Co£¨OH£©3£¬ÀäÈ´ºó¼ÓÈë×ãÁ¿ KI ¹ÌÌåºÍ HCl ÈÜÒº£¬³ä·Ö·´Ó¦Ò»¶Îʱ¼äºó£¬½«ÉÕÆ¿ÖеÄÈÜҺȫ²¿×ªÒÆÖÁ 250.00 mL ÈÝÁ¿Æ¿ÖУ¬¼ÓË®¶¨ÈÝ£¬È¡ÆäÖÐ 25.00 mL ÊÔÑù¼ÓÈ뵽׶ÐÎÆ¿ÖУ»

¢ÚÓà 0.100 0 mol¡¤L £­1 Na2S2O3 ±ê×¼ÈÜÒºµÎ¶¨£¬ÈÜÒº±äΪdz»ÆÉ«ºó£¬¼ÓÈëµí·ÛÈÜÒº×÷ָʾ¼Á¼ÌÐøµÎ¶¨ÖÁÖյ㣬Öظ´ 2 ´ÎʵÑ飬²âµÃÏûºÄ Na2S2O3 ÈÜÒºµÄƽ¾ùÌå»ýΪ 15.00 mL¡££¨ÒÑÖª£º2Co3£«£«2I£­=2Co2£«£«I2 £¬I2 £«2S2O32£­=2I£­£«S4O62£­£©.ͨ¹ý¼ÆËãÈ·¶¨¸Ã²úÆ·ÖÐîܵĺ¬Á¿___________________¡£

¡¾ÌâÄ¿¡¿I£®ÏÂÃæÁгöÁ˼¸×éÎïÖÊ£¬ÇëÓÃÎïÖʵÄ×éºÅÌîдÏÂ±í¡£

ÀàÐÍ

ͬλËØ

ͬËØÒìÐÎÌå

ͬ·ÖÒì¹¹Ìå

ͬϵÎï

×éºÅ

__

__

__

__

¢Ù Óë ¢Ú Óë ¢Û½ð¸ÕʯºÍʯī ¢Ü12C¡¢13C¡¢14C ¢ÝºÍ

II£®ÏÂͼÖÐA¡¢B¡¢C ·Ö±ðÊÇÈýÖÖÌþµÄ½á¹¹Ä£ÐÍ£º

Çë»Ø´ðÏÂÁÐÎÊÌ⣺

(1)A µÄµç×Óʽ______________£¬B µÄ½á¹¹¼òʽ________________¡£

(2)A¼°ÆäͬϵÎïµÄ·Ö×Óʽ·ûºÏͨʽ_____________(Óà n ±íʾ)¡£µ± n£½____________ʱ£¬ÍéÌþ¿ªÊ¼³öÏÖͬ·ÖÒì¹¹Ì壻µ± n=6 ʱ£¬Í¬·ÖÒì¹¹ÌåÓÐ__________ÖÖ¡£

(3)A¡¢B¡¢C ÈýÖÖÓлúÎïÖУ¬ËùÓÐÔ­×Ó¾ù¹²ÃæµÄÊÇ___________(ÌîÃû³Æ)¡£½á¹¹¼òʽΪ µÄÓлúÎïÖУ¬´¦ÓÚͬһƽÃæÄÚµÄÔ­×ÓÊý×î¶àΪ__________£¬´¦ÓÚ Í¬Ò»Æ½ÃæÄÚµÄ̼ԭ×ÓÊýÖÁÉÙΪ____________¡£

(4)ÓлúÎï C ²»¾ßÓеĽṹ»òÐÔÖÊÊÇ_____________(Ìî×ÖĸÐòºÅ)¡£

a£®ÊÇ̼̼˫¼üºÍ̼̼µ¥¼ü½»ÌæµÄ½á¹¹ b£®Óж¾¡¢²»ÈÜÓÚË®¡¢ÃܶȱÈˮС

c£®²»ÄÜʹËáÐÔ KMnO4 ÈÜÒººÍäåË®·´Ó¦ÍÊÉ« d£®Ò»¶¨Ìõ¼þÏÂÄÜÓëÇâÆø»òÑõÆø·´Ó¦

(5)µÈÖÊÁ¿µÄÈýÖÖÓлúÎïÍêȫȼÉÕÉú³É H2O ºÍ CO2£¬ÏûºÄÑõÆøµÄÌå»ý(Ïàͬ״¿öÏÂ)×î´óµÄÊÇ_______(ÌîA »òB »ò C)¡£

III£®Ä³ÓлúÎïµÄ½á¹¹¼òʽÈçͼ£¬1mol ¸ÃÓлúÎï×î¶à¿ÉÒÔºÍ______molÇâÆø·´Ó¦£¬×î¶à¿ÉÒÔºÍ_____molNaOH ·´Ó¦¡£

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

¾«Ó¢¼Ò½ÌÍø