ÌâÄ¿ÄÚÈÝ

15£®ÒÑÖª Na2SO4ºÍ NaCl »ìºÏÈÜÒºÖУ¬Cl-µÄÎïÖʵÄÁ¿Å¨¶ÈÊÇ Na+µÄÎïÖʵÄÁ¿Å¨¶ÈµÄ $\frac{3}{5}$ ±¶£¬ÏÂÁÐÐð ÊöÖÐÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®Na2SO4ºÍ NaCl µÄÎïÖʵÄÁ¿Ö®±ÈΪ 1£º3
B£®ÈÜÒºÖÐÒ»¶¨ÓР1molNa2SO4 ºÍ 3molNaCl
C£®Ëù´øµçºÉÊý SO42-ÊÇ Na+µÄ $\frac{3}{5}$ ±¶
D£®SO42-Óë Cl-µÄÎïÖʵÄÁ¿Ö®ºÍµÈÓÚ Na+µÄÎïÖʵÄÁ¿

·ÖÎö ÔÚ»ìºÏÈÜÒºÖдæÔÚµçºÉÊØºã£ºc£¨Cl-£©+2c£¨SO42-£©=c£¨Na+£©£¬Cl-µÄÎïÖʵÄÁ¿Å¨¶ÈÊÇNa+µÄÎïÖʵÄÁ¿Å¨¶ÈµÄ$\frac{3}{5}$±¶£¬Éèc£¨Na+£©=5mol/L£¬Ôòc£¨Cl-£©=3moL/L£¬c£¨SO42-£©=1mol/L£¬ÒԴ˽â´ð£®

½â´ð ½â£ºÔÚ»ìºÏÈÜÒºÖдæÔÚµçºÉÊØºã£ºc£¨Cl-£©+2c£¨SO42-£©=c£¨Na+£©£¬Cl-µÄÎïÖʵÄÁ¿Å¨¶ÈÊÇNa+µÄÎïÖʵÄÁ¿Å¨¶ÈµÄ$\frac{3}{5}$±¶£¬Éèc£¨Na+£©=5mol/L£¬Ôòc£¨Cl-£©=3moL/L£¬c£¨SO42-£©=1mol/L£¬
A£®Èçc£¨Cl-£©=3moL/L£¬Ôòc£¨SO42-£©=1mol/L£¬ÔòNa2SO4 ºÍNaClµÄÎïÖʵÄÁ¿Ö®±ÈΪ1£º3£¬¹ÊAÕýÈ·£»
B£®ÈÜÒºÌå»ýδ֪£¬ÒÀ¾Ýn=CV¿ÉÖª£¬ÎÞ·¨È·¶¨ÁòËáÄÆºÍÂÈ»¯ÄƵÄÎïÖʵÄÁ¿£¬¹ÊB´íÎó£»
C£®Éèc£¨Na+£©=5mol/L£¬Ôòc£¨SO42-£©=1mol/L£¬Ëù´øµçºÉÊýSO42-ÊÇNa+µÄ$\frac{2}{5}$±¶£¬¹ÊC´íÎó£»
D£®ÒÀ¾ÝµçºÉÊØºã£ºc£¨Cl-£©+2c£¨SO42-£©=c£¨Na+£©£¬SO42- ÓëCl-µÄÎïÖʵÄÁ¿Ö®ºÍСÓÚÓÚNa+µÄÎïÖʵÄÁ¿£¬¹ÊD´íÎó£»
¹ÊÑ¡£ºA£®

µãÆÀ ±¾Ì⿼²éÎïÖʵÄÁ¿Å¨¶ÈµÄÓйؼÆË㣬ÄѶȲ»´ó£¬¹Ø¼üÃ÷°×ÈÜÒº³ÊµçÖÐÐÔ£¬ÒõÑôÀë×ÓËù´øµçºÉ×ÜÊýÏàµÈ£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
5£®MgÄÜÔÚNO2ÖÐȼÉÕ£¬²úÎïΪMg3N2£¬MgOºÍN2£®Ä³¿ÆÑ§Ð¡×éͨ¹ýʵÑéÑéÖ¤²¿·Ö²úÎﲢ̽¾¿²úÎïµÄ±ÈÀý¹ØÏµ£®ÏÞÓÃÈçͼװÖÃʵÑ飨¼Ð³Ö×°ÖÃÊ¡ÂÔ£©
×ÊÁÏÐÅÏ¢£ºMg3N2+6H2O=3Mg£¨OH£©2+2NH3¡ü
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÀûÓÃÉÏÊö×°ÖÃÍê³ÉʵÑ飬Á¬½Ó˳ÐòΪB¡¢B¡¢A¡¢GF¡¢C£¨Ìî×ÖĸÐòºÅ£©£®
£¨2£©Á¬½ÓºÃÒÇÆ÷£¬×°ÈëҩƷǰÈçºÎ¼ìÑé×°ÖÃÆøÃÜÐԹرշÖҺ©¶·»îÈû£¬½«µ¼Æø¹ÜÄ©¶Ë²åÈëË®ÖУ¬¶ÔBÖÐ×¶ÐÎÆ¿Î¢ÈÈ£¬Èôµ¼¹Ü¿ÚÓÐÆøÅÝð³ö£¬Í£Ö¹¼ÓÈÈ£¬µ¼Æø¹ÜÖнøÈëÒ»¶ÎË®Öù£¬±íÊ¾ÆøÃÜÐÔÁ¼ºÃ
£¨3£©×°ÖÃAÖÐÊ¢×°µÄ¸ÉÔï¼Á¿ÉÒÔÊǢڢܣ¨ÌîÐòºÅ£©
¢ÙŨÁòËá   ¢ÚÎÞË®CaCl2   ¢Û¼îʯ»Ò    ¢ÜÎåÑõ»¯¶þÁ×
£¨4£©×°ÖÃFµÄ×÷ÓÃÊdzýÈ¥¹ýÁ¿µÄ¶þÑõ»¯µªÆøÌ壮
£¨5£©ÒªÈ·¶¨²úÎïÖÐÓÐN2Éú³É£¬¹Û²ìµ½µÄʵÑéÏÖÏóΪװÖÃCÖÐÒºÃæÏ½µ£¬DÖгöÏÖË®£¬È·¶¨²úÎïÖк¬ÓÐMg3N2µÄ¾ßÌåʵÑé²Ù×÷ΪȡÉÙÁ¿·´Ó¦ºóµÄ¹ÌÌå²úÎ¼ÓÈ뵽ˮÖвúÉúÓд̼¤ÐÔÆøÎ¶µÄÆøÌ壬¸ÃÆøÌåÄÜʹʪÈóµÄºìɫʯÈïÊÔÖ½±äÀ¶É«£®
£¨6£©ÒÑ֪װÖÃEÖгõʼ¼ÓÈëþ·ÛµÄÖÊÁ¿Îª13.2g£¬ÔÚ×ãÁ¿µÄNO2Öгä·ÖȼÉÕ£¬ÊµÑé½áÊøºó£¬Ó²Öʲ£Á§¹ÜÀäÈ´ÖÁÊÒΣ¬³ÆÁ¿£¬²âµÃÊ£Óà¹ÌÌåµÄÖÊÁ¿Îª21.0g£¬²úÉúN2µÄÌå»ýΪ1120mL£¨±ê×¼×´¿ö£©£®Ð´³ö²£Á§¹ÜÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º11Mg+4NO2=Mg3N2+8MgO+N2£®

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

¾«Ó¢¼Ò½ÌÍø