ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿Ä¿Ç°ÊÀ½çÉϺ£Ë®µ­»¯µÄÖ÷Òª·½·¨ÓÐÕôÁ󷨡¢µçÉøÎö·¨¡¢·´Éø͸·¨µÈ¡£µçÉøÎö·¨ÊÇÔÚÖ±Á÷µçÔ´×÷ÓÃÏÂͨ¹ýÀë×Ó½»»»Ä¤¶Ôº£Ë®½øÐд¦Àí(Ô­ÀíÈçͼËùʾ)£»·´Éø͸·¨ÊÇÀûÓÃѹǿ²îʹº£Ë®Ò»²àµÄË®·Ö×Óͨ¹ýÉø͸Ĥ½øÈ뵭ˮһ²à£¬´Ó¶øµÃµ½µ­Ë®ºÍŨËõµÄÑÎÈÜÒº(Ô­ÀíÈçͼËùʾ)¡£ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ

A. ÕôÁ󷨡¢µçÉøÎö·¨¡¢·´Éø͸·¨¾ù²»·¢Éú»¯Ñ§·´Ó¦

B. ·´Éø͸·¨ËùÓÃÉø͸ĤµÄ΢¿×Ö±¾¶·¶Î§ÊÇ1¡«100nm

C. ÕôÁ󷨾ßÓÐÉ豸¼òµ¥¡¢³É±¾µÍµÈÓŵã

D. ŨËõµÄÑÎÈÜÒº¿ÉÓÃÓÚÌáÈ¡»òÖƱ¸Ê³ÑΡ¢Ã¾¡¢äåµÈÎïÖÊ

¡¾´ð°¸¡¿D

¡¾½âÎö¡¿

A.µçÉøÎö·¨»á·¢Éúµç»¯Ñ§·´Ó¦£¬A´íÎó£»

B.Éø͸ĤֻÔÊÐíË®·Ö×Ó͸¹ý¶øÑÎÀàÀë×Ó²»ÄÜ͸¹ý£¬Òò´Ë΢¿×Ö±¾¶Ó¦Ð¡ÓÚ½ºÌå΢Á£µÄÖ±¾¶£¬¼´Ð¡ÓÚ1nm£¬B´íÎó£»

C.ÕôÁó·¨É豸²¢²»¼òµ¥ÇÒ»áÏûºÄ´óÁ¿ÄÜÔ´£¬³É±¾½Ï¸ß£¬C´íÎó¡£

D.µç½âŨËõºóµÄÑÎÈÜÒºº¬ÓдóÁ¿ÂÈ»¯ÄÆ¡¢ÂÈ»¯Ã¾ºÍ±»¯Î¿ÉÓÃÓÚÌáÈ¡»òÖƱ¸Ê³ÑΡ¢Ã¾¡¢äåµÈÎïÖÊ£¬¹ÊDÕýÈ·£»

´ð°¸Ñ¡D¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿°¢Ë¾Æ¥ÁÖ£¨ÒÒõ£Ë®ÑîËᣬ£©ÊÇÊÀ½çÉÏÓ¦ÓÃ×î¹ã·ºµÄ½âÈÈ¡¢ÕòÍ´ºÍ¿¹Ñ×Ò©£®ÒÒõ£Ë®ÑîËáÊÜÈÈÒ׷ֽ⣬·Ö½âζÈΪ128¡æ¡«135¡æ£®Ä³Ñ§Ï°Ð¡×éÔÚʵÑéÊÒÒÔË®ÑîËᣨÁÚôÇ»ù±½¼×ËᣩÓë´×Ëáôû[£¨CH3CO£©2O]ΪÖ÷ÒªÔ­ÁϺϳɰ¢Ë¾Æ¥ÁÖ£¬·´Ó¦Ô­ÀíÈçÏ£º

ÖƱ¸»ù±¾²Ù×÷Á÷³ÌÈçÏ£º

Ö÷ÒªÊÔ¼ÁºÍ²úÆ·µÄÎïÀí³£ÊýÈçϱíËùʾ£º

Ãû³Æ

Ïà¶Ô·Ö×ÓÖÊÁ¿

ÈÛµã»ò·Ðµã£¨¡æ£©

Ë®

Ë®ÑîËá

138

158£¨È۵㣩

΢ÈÜ

´×Ëáôû

102

139.4£¨·Ðµã£©

Ò×Ë®½â

ÒÒõ£Ë®ÑîËá

180

135£¨È۵㣩

΢ÈÜ

Çë¸ù¾ÝÒÔÉÏÐÅÏ¢»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©ÖƱ¸°¢Ë¾Æ¥ÁÖʱ£¬ÒªÊ¹ÓøÉÔïµÄÒÇÆ÷µÄÔ­ÒòÊÇ_____£®

£¨2£©ºÏ³É°¢Ë¾Æ¥ÁÖʱ£¬×îºÏÊʵļÓÈÈ·½·¨ÊÇ_____£®

£¨3£©Ìá´¿´Ö²úÆ·Á÷³ÌÈçÏ£¬¼ÓÈÈ»ØÁ÷×°ÖÃÈçͼ£º

¢ÙʹÓÃζȼƵÄÄ¿µÄÊÇ¿ØÖƼÓÈȵÄζȣ¬·ÀÖ¹_____£®

¢ÚÀäÄýË®µÄÁ÷½ø·½ÏòÊÇ_____£¨Ìî¡°a¡±»ò¡°b¡±£©£»

¢Û³ÃÈȹýÂ˵ÄÔ­ÒòÊÇ_____£®

¢ÜÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ_____£¨ÌîÑ¡Ïî×Öĸ£©£®

a£®´ËÖÖÌá´¿·½·¨ÖÐÒÒËáÒÒõ¥µÄ×÷ÓÃÊÇ×öÈܼÁ

b£®´ËÖÖÌá´¿´Ö²úÆ·µÄ·½·¨½ÐÖؽᾧ

c£®¸ù¾ÝÒÔÉÏÌá´¿¹ý³Ì¿ÉÒԵóö°¢Ë¾Æ¥ÁÖÔÚÒÒËáÒÒõ¥ÖеÄÈܽâ¶ÈµÍÎÂʱ´ó

d£®¿ÉÒÔÓÃ×ÏɫʯÈïÈÜÒºÅжϲúÆ·ÖÐÊÇ·ñº¬ÓÐδ·´Ó¦ÍêµÄË®ÑîËá

£¨4£©ÔÚʵÑéÖÐÔ­ÁÏÓÃÁ¿£º2.0gË®ÑîËá¡¢5.0mL´×Ëáôû£¨¦Ñ£½1.08g/cm3£©£¬×îÖճƵòúÆ·ÖÊÁ¿Îª2.2g£¬ÔòËùµÃÒÒõ£Ë®ÑîËáµÄ²úÂÊΪ_____£¨ÓðٷÖÊý±íʾ£¬Ð¡Êýµãºóһ룩£®

¡¾ÌâÄ¿¡¿PCl3ÊÇÁ׵ij£¼ûÂÈ»¯Î¿ÉÓÃÓÚ°ëµ¼ÌåÉú²úµÄÍâÑÓ¡¢À©É¢¹¤Ðò¡£ÓйØÎïÖʵIJ¿·ÖÐÔÖÊÈçÏ£º

ÈÛµã/¡æ

·Ðµã/¡æ

ÃܶÈ/ g¡¤mL£­1

ÆäËû

»ÆÁ×

44.1

280.5

1.82

2P£«3Cl2(ÉÙÁ¿) 2PCl3£»2P£«5Cl2(¹ýÁ¿) 2PCl5

PCl3

£­112

75.5

1.574

ÓöË®Éú³ÉH3PO3ºÍHCl£¬ÓöO2Éú³ÉPOCl3

(Ò»)ÖƱ¸

ÈçͼÊÇʵÑéÊÒÖƱ¸PCl3µÄ×°ÖÃ(²¿·ÖÒÇÆ÷ÒÑÊ¡ÂÔ)¡£

(1)ÒÇÆ÷ÒÒµÄÃû³ÆÊÇ________£»ÆäÖУ¬Óë×ÔÀ´Ë®½øË®¹ÜÁ¬½ÓµÄ½Ó¿Ú±àºÅÊÇ________¡£(Ìî¡°a¡±»ò¡°b¡±)

(2)ʵÑéÊÒÖƱ¸Cl2µÄÀë×Ó·½³Ìʽ___________________________¡£ÊµÑé¹ý³ÌÖУ¬Îª¼õÉÙPCl5µÄÉú³É£¬Ó¦¿ØÖÆ____________¡£

(3)¼îʯ»ÒµÄ×÷ÓãºÒ»ÊÇ·ÀÖ¹¿ÕÆøÖеÄË®ÕôÆø½øÈë¶øʹPCl3Ë®½â£¬Ó°Ïì²úÆ·µÄ´¿¶È£»¶þÊÇ_________¡£

(4)ÏòÒÇÆ÷¼×ÖÐͨÈë¸ÉÔïCl2֮ǰ£¬Ó¦ÏÈͨÈëÒ»¶Îʱ¼äCO2Åž¡×°ÖÃÖеĿÕÆø£¬ÆäÄ¿µÄÊÇ________¡£

(¶þ)·ÖÎö

²â¶¨²úÆ·ÖÐPCl3´¿¶ÈµÄ·½·¨ÈçÏ£ºÑ¸ËÙ³ÆÈ¡4.100 g²úÆ·£¬Ë®½âÍêÈ«ºóÅä³É500 mLÈÜÒº£¬È¡³ö25.00 mL¼ÓÈë¹ýÁ¿µÄ0.100 0 mol¡¤L£­1 20.00 mLµâÈÜÒº£¬³ä·Ö·´Ó¦ºóÔÙÓÃ0.100 0 mol¡¤L£­1 Na2S2O3ÈÜÒºµÎ¶¨¹ýÁ¿µÄµâ£¬ÖÕµãʱÏûºÄ12.00 mL Na2S2O3ÈÜÒº¡£

ÒÑÖª£ºH3PO3£«H2O£«I2===H3PO4£«2HI£»I2£«2Na2S2O3===2NaI£«Na2S4O6£»¼ÙÉè²â¶¨¹ý³ÌÖÐûÓÐÆäËû·´Ó¦¡£

(5)¸ù¾ÝÉÏÊöÊý¾Ý£¬¸Ã²úÆ·ÖÐPCl3(Ïà¶Ô·Ö×ÓÖÊÁ¿Îª137.5)µÄÖÊÁ¿·ÖÊýΪ________¡£ÈôµÎ¶¨ÖÕµãʱ¸©ÊÓ¶ÁÊý£¬ÔòPCl3µÄÖÊÁ¿·ÖÊý________(Ìî¡°Æ«´ó¡±¡°Æ«Ð¡¡±»ò¡°ÎÞÓ°Ï족)¡£

(Èý)̽¾¿

(6)Éè¼ÆʵÑéÖ¤Ã÷PCl3¾ßÓл¹Ô­ÐÔ£º_____________________________________¡£(ÏÞÑ¡ÊÔ¼ÁÓУºÕôÁóË®¡¢Ï¡ÑÎËá¡¢µâË®¡¢µí·Û)

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

¾«Ó¢¼Ò½ÌÍø