ÌâÄ¿ÄÚÈÝ
ÒÑÖªX¡¢YÊÇÁ½ÖÖÐÔÖÊÏàËÆµÄ¶ÌÖÜÆÚÔªËØ¡£
¢ñ£®ÈôX¡¢YÊÇÏàÁÚÏàËÆ
£¬ËüÃǵĵ¥Öʶ¼±ØÐë²ÉÓõç½â·¨ÖƱ¸£¬µ«¶¼ÎÞÐèÃÜ·â±£´æ£¬
£¨1£©XÀë×ӵĽṹʾÒâͼ_____________¡££¨2£©YÔªËØÔÚÖÜÆÚ±íÖÐλÖÃ__________¡£
¢ò£®ÈôX¡¢YÊÇͬ×åÏàËÆ
£¬XÊÇÐγɻ¯ºÏÎïÖÖÀà×î¶àµÄÔªËØ¡£
£¨3£©I2O3ÒÔÑõ»¯XO£¬³£ÓÃÓڲⶨXOº¬Á¿£¬ÒÑÖª£º¢Ù2I2(s)+5O2(g)=2I2O5(s)¡÷H=-75.66kJ¡¤mol-1
¢Ú2XO(g)+O2(g)=2XO2(g) ¡÷H=-566.0kJ¡¤mol-1¡£Çëд³öXO(g)ÓëI2O5(s)ºÍXO2(g)µÄÈÈ»¯Ñ§·½³Ìʽ£º______¡£
£¨4£©¹¤ÒµÉÏÓÃXµ¥ÖÊÓëYµÄÑõ»¯Îï·´Ó¦ÖÆÈ¡Yµ¥ÖʵĹý³ÌÖУ¬YOÊÇ·´Ó¦Öмä²úÎ¸ô¾ø¿ÕÆøÊ±YOºÍNaOHÈÜÒº·´Ó¦£¨²úÎïÖ®Ò»ÊÇNa2YO3£©µÄÀë×Ó·½³ÌʽÊÇ_________________________¡£
¢ó£®ÈôX¡¢YÊǶԽÇÏàËÆ
£¬X¡¢YµÄ×î¸ß¼Ûº¬ÑõËáµÄŨÈÜÒº¶¼ÓÐÇ¿Ñõ»¯ÐÔ¡£
£¨5£©ÏÂÁÐÊÔ¼Á¶¼¿ÉÒÔÖ¤Ã÷X¡¢YµÄ×î¸ß¼Ûº¬ÑõËáµÄŨÈÜÒº¶¼ÓÐÇ¿Ñõ»¯ÐÔµÄÊÇ_________¡£
A£®ÌúƬ B£®ÍƬ C£®¶þÑõ»¯Áò D£®Ä¾Ì¿
£¨6£©HAÊǺ¬ÓÐXÔªËØµÄÒ»ÔªËᣬ³£ÎÂÏ£¬½«0.2mol/LµÄHAÈÜÒºÓëµÈÌå»ý¡¢µÈŨ¶ÈµÄNaOHÈÜÒº»ìºÏ£¬ËùµÃÈÜÒº£¨¼ÙÉèÈÜÒºÌå»ý¿ÉÒÔÏà¼Ó£©Öв¿·Ö΢Á£×é³É¼°Å¨¶ÈÈçÓÒͼËùʾ£¬Í¼ÖÐN±íʾ______£¨Ìî΢Á£·ûºÅ£©¡£
![]()
![]()
£¨7£©Ä³»¯¹¤³§Éè¼ÆÒªÇó£º¿ÕÆøÖÐYO2º¬Á¿²»µÃ³¬¹ý0.02mg/L¡£Ä³Í¬Ñ§ÓÃÓÒͼËùʾ¼òÒ××°Öòⶨ¿ÕÆøÖеÄYO2º¬Á¿£º×¼È·ÒÆÈ¡10mL5¡Á10£4mol/LµÄ±ê×¼µâË®ÈÜÒº£¬×¢ÈëÊÔ¹ÜÖУ¬¼Ó2£3µÎµí·Ûָʾ¼Á£¬´ËʱÈÜÒº³ÊÀ¶É«£¬ÔÚÖ¸¶¨µÄ²â¶¨µØµã³éÆø£¬Ã¿´Î³éÆø100mL£¬Ö±µ½ÈÜÒºµÄÀ¶É«È«²¿Íʾ¡ÎªÖ¹£¬¼ÙÉè¸ÃͬѧµÄ²âÁ¿ÊÇ׼ȷµÄ£¬ÔòËû³éÆøµÄ´ÎÊýÖÁÉÙΪ_______´Îʱ·½¿É˵Ã÷¸Ã³§¿ÕÆøÖеÄYO2º¬Á¿´ï±ê¡£
T ¡ãCʱ£¬ÔÚÈÝ»ýΪ2 LµÄ3¸öºãÈÝÃܱÕÈÝÆ÷Öз¢Éú·´Ó¦:3A(g)+B(g)
XC(g)£¬°´²»Í¬·½Ê½Í¶Èë·´Ó¦Î²âµÃ·´Ó¦´ïµ½Æ½ºâʱµÄÓйØÊý¾ÝÈçÏ£º
ÈÝÆ÷ | ¼× | ÒÒ | ±û |
·´Ó¦ÎïµÄͶÈëÁ¿ | 3molA¡¢2mol B | 6molA¡¢4molB | 2molC |
´ïµ½Æ½ºâµÄʱ¼ä/ min | 5 | 8 | |
AµÄŨ¶È/mol • L-1 | C1 | C2 | |
CµÄÌå»ý·ÖÊý/% |
|
| |
»ìºÏÆøÌåµÄÃܶÈ/g • L-1 |
|
|
ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ
A£®Èô x <4,£¬2C1 £¼C2
B£®Èô x = 4£¬Ôò
=![]()
C£®ÎÞÂÛxµÄÖµÊǶàÉÙ±´¾ùÓÐ
=![]()
D£®ÈÝÆ÷¼×´ïµ½Æ½ºâËùÐèµÄʱ¼ä±ÈÈÝÆ÷ÒҴﵽƽºâËùÐèµÄʱ¼ä¶Ì