ÌâÄ¿ÄÚÈÝ
Á׺ÍÆ仯ºÏÎïÔÚ¿ÆÑм°Éú²úÖоùÓÐ×ÅÖØÒªµÄ×÷Óá£
£¨1£©ºìÁ×P(s)ºÍCl2(g)·¢Éú·´Ó¦Éú³ÉPCl3(g)PCl5(g)£¬·´Ó¦¹ý³ÌÈçÏ£º
2P(s) + 3Cl2(g) = 2PCl3(g) ¡÷H=£612kJ/mol
2P(s) + 5Cl2(g) = 2PCl5(g) ¡÷H=£798kJ/mol
Æø̬ PCl5Éú³ÉÆø̬PCl3ºÍCl2µÄÈÈ»¯Ñ§·½³ÌʽΪ________¡£
£¨2£©Ò»¶¨Î¶ÈÏ£¬ÔÚÈý¸öºãÈÝÌå»ý¾ùΪ2.0LµÄÈÝÆ÷Öз¢Éú·´Ó¦£ºPCl5£¨g£© PCl3£¨g£©+Cl2£¨g£©
±àºÅ | ζȣ¨¡æ£© | Æðʼ/mol | ƽºâ/mol | ´ïµ½Æ½ºâËùÐèʱ¼ä/s | |
PCl5£¨g£© | PCl3£¨g£© | Cl2£¨g£© | |||
I | 320 | 0.40 | 0.10 | 0.10 | t1 |
II | 320 | 0.80 | t2 | ||
III | 410 | 0.40 | 0.15 | 0.15 | t3 |
¢Ùƽºâ³£ÊýK£ºÈÜÒºII____ÈÜÒºIII£¨Ìî¡°>¡±»ò¡°=¡±»ò¡°<¡±£©
¢Ú·´Ó¦´ïµ½Æ½ºâʱ£¬PCl3µÄת»¯ÂÊ£ºÈÝÆ÷II_____ÈÝÆ÷I£¨Ìî¡°>¡±»ò¡°=¡±»ò¡°<¡±£©
£¨3£©ÑÇÁ×ËᣨH2PO5£©Óë×ãÁ¿µÄNaOHÈÜÒº·´Ó¦Éú³ÉNa2HPO3¡£µç½âNa2HPO3ÈÜÒºÒ²¿ÉµÃµ½ÑÇ÷[ËᣬװÖÃʾÒâͼÈçͼ£º
Òõ¼«µÄµç¼«·´Ó¦Ê½Îª_______£»²úÆ·ÊÒÖз´Ó¦µÄÀë×Ó·½³ÌʽΪ__________¡£
£¨4£©Ò»¶¨Î¶ÈÏ£¬Ksp[Mg3(PO4)2]=6.0¡Á10-29£¬Ksp[Ca3(PO4)2]=6.0¡Á10-26ÏòŨ¶È¾ùΪ0.20 mol¡¤L-1µÄMgCl2ºÍCaCl2»ìºÏÈÜÒºÖÐÖðµÎ¼ÓÈëNa3PO3£¬ÏÈÉú³É________³Áµí£¨Ìѧʽ£©£»µ±²âµÃÈÜÒºÆäÖÐÒ»ÖÖ½ðÊôÑôÀë×Ó³ÁµíÍêÈ«£¨Å¨¶ÈСÓÚ105mol/L£©Ê±£¬ÈÜÒºÖеÄÁíÒ»ÖÖ½ðÊôÑôÀë×ÓµÄÎïÖʵÄÁ¿Å¨¶Èc= ______mol/L