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͸Ĥ½øÈëµË®Ò»²à£¬´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¡£
¡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)