ÌâÄ¿ÄÚÈÝ

¸ù¾ÝÔªËØÖÜÆÚ±í֪ʶ¿ÉÖªBºÍÂÁµ¥Öʼ°»¯ºÏÎïÏàËÆ£»¸ù¾ÝÖÜÆÚ±íÖжԽÇÏß¹æÔò£¬BeÓëÂÁµ¥Öʼ°»¯ºÏÎïÐÔÖÊÒ²ÏàËÆ

¸ù¾ÝÔªËØÖÜÆÚ±í֪ʶ¿ÉÖªBºÍÂÁµ¥Öʼ°»¯ºÏÎïÏàËÆ£»¸ù¾ÝÖÜÆÚ±íÖжԽÇÏß¹æÔò£¬BeÓëÂÁµ¥Öʼ°»¯ºÏÎïÐÔÖÊÒ²ÏàËÆ£®»Ø´ðÏÂÁÐÎÊÌ⣺

(1)д³öBÓëNaOHÈÜÒº·´Ó¦µÄ»¯Ñ§·½³Ìʽ________£®

(2)д³öBeÓëNaOHÈÜÒº·´Ó¦µÄÀë×Ó·½³Ìʽ________£®

(3)¿ÉÓÃ________ÈÜÒº¼ø±ðBe(OH)2ºÍMg(OH)2£¬ÆäÀë×Ó·½³ÌʽΪ£º________£®

(4)ÒÑÖªÅðËá·Ö×ӽṹΪ£®ÅðËáΪ________Ëá(Ìî¡°Ç¿¡±¡¢¡°Èõ¡±)£»0.1molÅðËá¿É±»20mL 5mol¡¤L£­1 NaOHÈÜҺǡºÃÍêÈ«Öкͣ¬¾Ý´Ë¿ÉÖªËüΪ________ÔªËá(Ìî¡°Ò»¡±¡¢¡°¶þ¡±¡¢¡°Èý¡±)£»ÅðËáÖÐBÔ­×Ó×îÍâ²ãÓÐ________¸öµç×Ó£®

´ð°¸£º
½âÎö£º

¡¡¡¡(1)2B£«2NaOH£«2H2O£½2NaBO2£«3H2¡ü

¡¡¡¡(1)2B£«2NaOH£«2H2O£½2NaBO2£«3H2¡ü

¡¡¡¡(2)Be£«2OH£­£½£«H2¡ü

¡¡¡¡(3)NaOH,Be(OH)2£«2OH£­£½£«2H2O

¡¡¡¡(4)Èõ,Ò»,6


Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¸ù¾ÝÔªËØÖÜÆÚ±íÖеÚËÄÖÜÆÚÔªËØµÄÏà¹ØÖªÊ¶£¬»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©µÚËÄÖÜÆÚÔªËØµÄ»ù̬ԭ×ӵĵç×ÓÅŲ¼ÖÐ4s¹ìµÀÉÏÖ»ÓÐ1¸öµç×ÓµÄÔªËØÓÐ________ÖÖ£»Ð´³öCu+µÄºËÍâµç×ÓÅŲ¼Ê½________¡£

£¨2£©°´µç×ÓÅŲ¼£¬¿É½«ÖÜÆÚ±íÀïµÄÔªËØ»®·Ö³ÉÎå¸öÇøÓò£¬µÚËÄÖÜÆÚÔªËØÖÐÊôÓÚsÇøµÄÔªËØÓÐ________ÖÖ£¬ÊôÓÚdÇøµÄÔªËØÓÐ________ÖÖ¡£

£¨3£©CaO¾§°ûÈçÓÒͼËùʾ£¬CaO¾§ÌåÖÐCa2+µÄÅäλÊýΪ________£»CaOµÄÑæÉ«·´Ó¦Îª×©ºìÉ«£¬Ðí¶à½ðÊô»òËüÃǵϝºÏÎï¶¼¿ÉÒÔ·¢ÉúÑæÉ«·´Ó¦£¬ÆäÔ­ÒòÊÇ________________

£¨4£©Óɵþµª»¯¼Ø(KN3)ÈÈ·Ö½â¿ÉµÃ´¿N2£º2KN3(s)=2K(l)+3N2(g)£¬ÏÂÁÐÓйØËµ·¨ÕýÈ·µÄÊÇ________(ÌîÑ¡Ïî×Öĸ£©¡£

A£®NaN3ÓëKN3½á¹¹ÀàËÆ£¬Ç°Õß¾§¸ñÄܽÏС

B£®¾§Ì弨µÄ¾§°û½á¹¹ÈçͼËùʾ£º£¬Ã¿¸ö¾§°ûÖзÖ̯2¸ö¼ØÔ­×Ó

C£®µªµÄµÚÒ»µçÀëÄÜ´óÓÚÑõ

D£®µªÆø³£ÎÂϺÜÎȶ¨£¬ÊÇÒòΪµªµÄµç¸ºÐÔС

£¨5£©¶þÑõ»¯îÑ(TiO2)Êdz£Óõġ¢¾ßÓнϸߴ߻¯»îÐÔºÍÎȶ¨ÐԵĹâ´ß»¯¼Á¡£O2ÔÚÆä´ß»¯×÷ÓÃÏ£¬¿É½«CN-Ñõ»¯³ÉCNO-¡£CN-µÄµç×ÓʽΪ_______,CNO-µÄÖÐÐÄÔ­×ÓµÄÔÓ»¯·½Ê½Îª_______

£¨6£©ÔÚCrCl3ÈÜÒºÖУ¬Ò»¶¨Ìõ¼þÏ´æÔÚ×é³ÉΪ[CrCln(H2O)6£­n]x£« (nºÍx¾ùΪÕýÕûÊý£©µÄÅäÀë×Ó£¬½«Æäͨ¹ýÇâÀë×Ó½»»»Ê÷Ö¬£¨R-H)£¬¿É·¢ÉúÀë×Ó½»»»·´Ó¦£º[CrCln(H2O)6£­n]x£«+xR-HRx[CrCln(H2O)6£­n]+xH+¡£½«º¬0.0015 mol[CrCln(H2O)6£­n]x£«µÄÈÜÒº£¬ÓëR-HÍêÈ«½»»»ºó£¬ÖкÍÉú³ÉµÄH+ÐèŨ¶ÈΪ0.1200 mol¡¤L-1 NaOHÈÜÒº25.00 mL£¬Ôò¸ÃÅäÀë×ӵĻ¯Ñ§Ê½Îª_______¡£

 

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

¾«Ó¢¼Ò½ÌÍø