ÌâÄ¿ÄÚÈÝ

£¨1£©¸ù¾ÝÓÒͼ»Ø´ðÏÂÁÐÎÊÌ⣺
¢Ù¸ÃÒÇÆ÷µÄ׼ȷÃû³ÆÊÇ______£®
¢Ú¹ØÓÚ¸ÃÒÇÆ÷µÄÐðÊö£¬²»ÕýÈ·µÄÊÇ______£®
A£®Æä¡°0¡±¿Ì¶ÈÔÚÉÏB£®¾«È·¶È±ÈÁ¿Í²¸ß
C£®¿ÉÁ¿È¡KMnO4ÈÜÒºD£®¿ÉÁ¿È¡NaOHÈÜÒº
¢ÛÏÖÒªÓøÃÒÇÆ÷Á¿È¡VmLÒºÌ壬Èô¿ªÊ¼Ê±Æ½Êӿ̶ȶÁÊý£¬½áÊøʱ¸©Êӿ̶ȶÁÊý£¬Ôòʵ¼ÊËùµÃÒºÌåÌå»ý±ÈVmL______£¨Ìî¡°Æ«´ó¡±¡¢¡°Æ«Ð¡¡±£©£®
£¨2£©ÓÃ0.100mol/LNaOHÈÜÒºµÎ¶¨Î´ÖªÅ¨¶ÈµÄÑÎËᣬ½øÐÐÒÔÏµζ¨²Ù×÷£º
A£®ÓÃÕôÁóˮϴ¾»¼îʽµÎ¶¨¹Ü£¬È»ºóÓñê×¼NaOHÈÜÒºÈóÏ´£¬ÔÙ½«±ê×¼NaOHÈÜҺװÈë¼îʽµÎ¶¨¹ÜÖУ¬µ÷ÕûÒºÃæÖÁÁã¿Ì¶ÈÏߣ®
B£®ÓÃËáʽµÎ¶¨¹ÜÁ¿È¡25.00mLÑÎËáÓÚ׶ÐÎÆ¿ÖУ¬²¢µÎ¼Ó¼¸µÎ·Óָ̪ʾ¼Á£®
C£®ÔÚ׶ÐÎƿϵæÒ»ÕÅ°×Ö½£¬µÎ¶¨ÖÁÖյ㣬¼Ç϶ÁÊýΪ20.00mL£®
¾Í´ËʵÑéÍê³ÉÏÂÁÐÌî¿Õ£º
¢ÙµÎ¶¨Ê±ÑÛ¾¦Ó¦¹Û²ì______£®
¢ÚµÎµ½Öյ㣬ÈÜÒºµÄÑÕÉ«ÓÉ______É«±äΪ______É«£¬ÇÒ°ë·ÖÖÓÄÚ²»ÍÊÉ«£®
¢Û¸ù¾ÝÉÏÊöÊý¾ÝÇóµÃÑÎËáµÄÎïÖʵÄÁ¿Å¨¶ÈΪ______mol/L£®
£¨1£©¢Ù¸ÃÒÇÆ÷ÏÂÃæÊDz£Á§ÐýÈû£¬ËùÒÔÊÇËáʽµÎ¶¨¹Ü£¬
¹Ê´ð°¸Îª£ºËáʽµÎ¶¨¹Ü£»
¢ÚA£®ËáʽµÎ¶¨¹ÜµÄ¡°0¡±¿Ì¶ÈÔÚÉÏ·½£¬¹ÊAÕýÈ·£»
B£®µÎ¶¨¹Ü׼ȷ¶ÈΪ0.01mL£¬Á¿Í²×¼È·¶ÈΪ0.1mL£¬µÎ¶¨¹Ü¾«È·¶È±ÈÁ¿Í²¸ß£¬¹ÊBÕýÈ·£»
C£®È¡KMnO4ÈÜÒº¾ßÓÐÇ¿Ñõ»¯ÐÔ£¬Äܹ»Ñõ»¯Ï𽺹ܣ¬ËùÒÔ±ØÐëʹÓÃËáʽµÎ¶¨¹ÜÁ¿È¡£¬¹ÊCÕýÈ·£»
D£®Á¿È¡NaOHÈÜÒº£¬±ØÐëʹÓüîʽµÎ¶¨¹Ü£¬¹ÊD´íÎó£»
¹Ê´ð°¸Îª£ºD£»
¢Û¿ªÊ¼Ê±Æ½Êӿ̶ȶÁÊý£¬½áÊøʱ¸©Êӿ̶ȶÁÊý£¬µ¼Ö¶ÁÊýƫС£¬Á¿È¡µÄÈÜÒºµÄÌå»ýÆ«´ó£¬
¹Ê´ð°¸Îª£ºÆ«´ó£®
£¨2£©¢ÙµÎ¶¨¹ý³ÌÖУ¬ÑÛ¾¦¹Û²ì׶ÐÎÆ¿ÖÐÈÜÒºµÄÑÕÉ«±ä»¯£¬±ÜÃâµÎÈë¹ýÁ¿µÄÇâÑõ»¯ÄÆÈÜÒº£¬
¹Ê´ð°¸Îª£º×¶ÐÎÆ¿ÖÐÈÜÒºµÄÑÕÉ«±ä»¯£»
¢Ú׶ÐÎÆ¿ÖеÄÑÎËáµÎÈë·Ó̪£¬ÈÜҺΪÎÞÉ«£¬·´Ó¦½áÊøʱ£¬ÈÜÒºÏÔʾ·ÛºìÉ«£¬
¹Ê´ð°¸Îª£ºÎÞ£»·Ûºì£¨Ç³ºì£©
¢ÛÏûºÄ±ê×¼ÒºÇâÑõ»¯ÄÆÈÜÒºµÄÌå»ýΪ20mL£¬¼´0.02L£¬ÇâÑõ»¯ÄƵÄÎïÖʵÄÁ¿Îª£º0.100mol/L¡Á0.02L=0.002mol£¬ÓÉÓÚn£¨NaOH£©=n£¨HCl£©£¬ËùÒÔ´ý²âÒºÑÎËáµÄŨ¶ÈΪ£º
0.002mol
0.025L
=0.0800mol/L£¬
¹Ê´ð°¸Îª£º0.0800£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø