ÌâÄ¿ÄÚÈÝ

ʵÑéÊÒÖÐÓмס¢ÒÒÁ½Æ¿¶ªÊ§±êÇ©µÄÎåÉ«ÈÜÒº£¬ÆäÖÐһƿÊÇÑÎËᣬÁíһƿÊÇ̼ËáÄÆÈÜÒº¡£ÎªÈ·¶¨¼×¡¢ÒÒÁ½Æ¿ÈÜÒºµÄ³É·Ö¼°ÆäÎïÖʵÄÁ¿Å¨¶È£¬ÏÖ²Ù×÷ÈçÏ£º¢ÙÁ¿È¡25.00mL¼×ÈÜÒº£¬ÏòÆäÖлº»ºµÎ¼ÓÒÒÈÜÒº15.00mL£¬¹²ÊÕ¼¯µ½CO2ÆøÌå224 mL(±ê×¼×´¿ö)¡£¢ÚÁ¿È¡15.00 mLÒÒÈÜÒº£¬ÏòÆäÖлº»ºµÎ¼Ó¼×ÈÜÒº25.00 mL£¬¹²ÊÕ¼¯µ½CO2ÆøÌå112mL¡£(±ê×¼×´¿ö)

Çë»Ø´ð£º(1)¸ù¾ÝÉÏÊöÁ½ÖÖ²»Í¬²Ù×÷¹ý³Ì¼°ÊµÑéÊý¾Ý¿ÉÅжϼ×ÈÜÒºÊÇ___________£¬ÒÒÈÜÒºÊÇ___________¡£

(2)ÓÃÀë×Ó·½³Ìʽ±íʾÁ½´Î²Ù×÷µÃµ½²»Í¬ÆøÌåÌå»ýµÄÔ­Òò£º

¢Ù_______________________________________________________£»

¢Ú_______________________________________________________¡£

(3)¼×ÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈΪ___________mol¡¤L-1¡£(CO2ÔÚË®ÈÜÒºÖеÄÉÙÁ¿ÈܽâºöÂÔ²»¼Æ)

(4)½«nmLµÄ¼×ÈÜÒºÓëµÈÌå»ýµÄÒÒÈÜÒº°´¸÷ÖÖ¿ÉÄܵķ½Ê½»ìºÏ£¬²úÉúµÄÆøÌåÌå»ýΪVmL(±ê×¼×´¿ö)£¬ÔòVµÄȡֵ·¶Î§Îª__________________¡£

(1)ÑÎËá  Na2CO3ÈÜÒº

(2)¢Ù2H++£½CO2¡ü+H2O

¢ÚH++=

+H+====H2O+CO2¡ü

(3)0.8

(4)0£¾¡Ü8.96n

½âÎö£º´ËÌ⿼²éNa2CO3ÓëHClµÄÕýµÎÓë·´µÎ(µÎ¼Ó˳Ðò²»Í¬£¬²úÎﲻͬ)Èç¹ûÊÇÕýµÎ£¬¼´

2HCl+ Na2CO3=2NaCl+H2O +CO2¡ü

0.8mol/L¡Án¡Á10-3L=0.4n¡Á10-3mol

Èç¹ûÊÇ·´µÎ£¬Na2CO3+HCl=NaCl+NaHCO3

ÓÉÓÚNa2CO3¹ýÁ¿£¬ËùÒÔCO2Ϊ×îСÉú³ÉÁ¿£¬=0mol

¡àVµÄȡֵ·¶Î§:0£¼¡Ü8.96n


Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø