ÌâÄ¿ÄÚÈÝ

£¨1£©50 mL 0.50 mol/LÑÎËáÓë50 mL 0.55 mol/L NaOHÈÜÒºÔÚÈçͼËùʾµÄ×°ÖÃÖнøÐÐÖкͷ´Ó¦¡£Í¨¹ý²â¶¨·´Ó¦¹ý³ÌÖÐËù·Å³öµÄÈÈÁ¿¿É¼ÆËãÖкÍÈÈ£¬»Ø´ðÏÂÁÐÎÊÌ⣺

¢Ù´ÓʵÑé×°ÖÃÉÏ¿´£¬Í¼ÖÐÉÐȱÉÙµÄÒ»ÖÖ²£Á§ÓÃÆ·ÊÇ      ¡£
¢Ú´óÉÕ±­ÉÏÈç²»¸ÇÓ²Ö½°å£¬ÇóµÃµÄÖкÍÈÈÊýÖµ½«________ (Ìî¡° Æ«´ó¡±¡°Æ«Ð¡¡±»ò¡°ÎÞÓ°Ï족)¡£
£¨2£©Ëá¼îÖк͵ζ¨ÊÇÖÐѧ»¯Ñ§³£¼ûʵÑé¡£
ijѧУ»¯Ñ§¿ÎÍâС×éÓÃ0.2000 mol¡¤L-1ÑÎËáµÎ¶¨Î´ÖªÅ¨¶ÈµÄÇâÑõ»¯ÄÆÈÜÒº£¬ÊԻشðÏÂÁÐÎÊÌâ¡£
¢ÙµÎ¶¨¹ý³ÌÖУ¬ÑÛ¾¦Ó¦×¢ÊÓ                                     ¡£
¢ÚÔÚÌú¼Ų̈ÉϵæÒ»ÕŰ×Ö½£¬ÆäÄ¿µÄÊÇ                             ¡£
¢Û¸ù¾ÝϱíÊý¾Ý£¬¼ÆËã±»²âÉÕ¼îÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈÊÇ    mol¡¤L£­1¡££¨±£ÁôËÄλÓÐЧÊý×Ö£©

¢ÜÏÂÁÐʵÑé²Ù×÷¶ÔµÎ¶¨½á¹û²úÉúʲôӰÏì(Ìî¡°Æ«¸ß¡±¡¢¡°Æ«µÍ¡±»ò¡°ÎÞÓ°Ï족)?
a.¹Û²ìËáʽµÎ¶¨¹ÜÒºÃæÊ±£¬¿ªÊ¼¸©ÊÓ£¬µÎ¶¨ÖÕµãÆ½ÊÓ£¬ÔòµÎ¶¨½á¹û________¡£
b.Èô½«×¶ÐÎÆ¿Óôý²âÒºÈóÏ´£¬È»ºóÔÙ¼ÓÈë10.00 mL´ý²âÒº£¬ÔòµÎ¶¨½á¹û_______¡£

£¨1£©¢Ù»·Ðβ£Á§½Á°è°ô   ¢ÚƫС
£¨2£©¢Ù ×¶ÐÎÆ¿ÄÚÈÜÒºÑÕÉ«µÄ±ä»¯  ¢Ú±ãÓÚ¹Û²ì×¶ÐÎÆ¿ÄÚÒºÌåÑÕÉ«µÄ±ä»¯£¬¼õСµÎ¶¨Îó²î ¢Û0.4000£¨±£ÁôÓÐЧÊý×Ö£¬·ñÔòÎÞ·Ö£© ¢ÜÆ«¸ß¡¢Æ«¸ß

½âÎöÊÔÌâ·ÖÎö£º£¨1£©¢ÙʵÑéÖÐÐèÒª½Á°è£¬ËùÒÔ´ÓʵÑé×°ÖÃÉÏ¿´£¬Í¼ÖÐÉÐȱÉÙµÄÒ»ÖÖ²£Á§ÓÃÆ·ÊÇ»·Ðβ£Á§½Á°è°ô¡£
¢Ú´óÉÕ±­ÉÏÈç²»¸ÇÓ²Ö½°å£¬Ôò»áÔì³ÉÈÈÁ¿µÄËðʧ£¬ÔòÇóµÃµÄÖкÍÈÈÊýÖµ½«Æ«Ð¡¡£
£¨2£©¢ÙµÎ¶¨¹ý³ÌÖУ¬ÑÛ¾¦Ó¦×¢ÊÓ×¶ÐÎÆ¿ÄÚÈÜÒºÑÕÉ«µÄ±ä»¯¡£
¢ÚÔÚÌú¼Ų̈ÉϵæÒ»ÕŰ×Ö½£¬ÆäÄ¿µÄÊDZãÓÚ¹Û²ì×¶ÐÎÆ¿ÄÚÒºÌåÑÕÉ«µÄ±ä»¯£¬¼õСµÎ¶¨Îó²î¡£
¢Û¸ù¾Ý±íÖеÄÊý¾Ý¿ÉÖª£¬Á½´ÎʵÑéÖÐÏûºÄÑÎËáµÄÌå»ýÊÇ20.10mlºÍ19.90ml£¬ÔòÏûºÄÑÎËáÌå»ýµÄƽ¾ùÖµÊÇ20.00ml£¬ËùÒÔÉÕ¼îÈÜÒºµÄŨ¶ÈÊÇ£½0.4000mol/L¡£
¢Ü¹Û²ìËáʽµÎ¶¨¹ÜÒºÃæÊ±£¬¿ªÊ¼¸©ÊÓ£¬µÎ¶¨ÖÕµãÆ½ÊÓ¡£ÓÉÓÚ¸©ÊÓ¶ÁÊýƫС£¬ËùÒÔÔòµÎ¶¨½á¹ûÆ«¸ß£»Èô½«×¶ÐÎÆ¿Óôý²âÒºÈóÏ´£¬È»ºóÔÙ¼ÓÈë10.00 mL´ý²âÒº£¬ÔòÏûºÄÑÎËáµÄÌå»ýÔö¼Ó£¬ËùÒԵ樽á¹ûÆ«¸ß¡£
¿¼µã£º¿¼²éÖкÍÈȺÍÖк͵ζ¨ÊµÑéµÄÓйØÅжÏ
µãÆÀ£º»¯Ñ§ÊµÑé³£ÓÃÒÇÆ÷µÄʹÓ÷½·¨ºÍ»¯Ñ§ÊµÑé»ù±¾²Ù×÷ÊǽøÐл¯Ñ§ÊµÑéµÄ»ù´¡£¬¶Ô»¯Ñ§ÊµÑéµÄ¿¼²éÀë²»¿ª»¯Ñ§ÊµÑéµÄ»ù±¾²Ù×÷£¬ËùÒÔ¸ÃÀàÊÔÌâÖ÷ÒªÊÇÒÔ³£¼ûÒÇÆ÷µÄÑ¡Óá¢ÊµÑé»ù±¾²Ù×÷ΪÖÐÐÄ£¬Í¨¹ýÊÇʲô¡¢ÎªÊ²Ã´ºÍÔõÑù×öÖØµã¿¼²éʵÑé»ù±¾²Ù×÷µÄ¹æ·¶ÐÔºÍ׼ȷ¼°Áé»îÔËÓÃ֪ʶ½â¾öʵ¼ÊÎÊÌâµÄÄÜÁ¦¡£±¾ÌâµÄÄѵãÊÇÎó²î·ÖÎö£¬ÐèҪעÒâʵÑéÔ­Àí£¬È»ºó½áºÏ²Ù×÷Áé»îÔËÓü´¿É¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

£¨1£©50 mL 0.50 mol/LÑÎËáÓë50 mL 0.55 mol/L NaOHÈÜÒºÔÚÈçͼËùʾµÄ×°ÖÃÖнøÐÐÖкͷ´Ó¦¡£Í¨¹ý²â¶¨·´Ó¦¹ý³ÌÖÐËù·Å³öµÄÈÈÁ¿¿É¼ÆËãÖкÍÈÈ£¬»Ø´ðÏÂÁÐÎÊÌ⣺

¢Ù´ÓʵÑé×°ÖÃÉÏ¿´£¬Í¼ÖÐÉÐȱÉÙµÄÒ»ÖÖ²£Á§ÓÃÆ·ÊÇ      ¡£

¢Ú´óÉÕ±­ÉÏÈç²»¸ÇÓ²Ö½°å£¬ÇóµÃµÄÖкÍÈÈÊýÖµ½«________ (Ìî¡° Æ«´ó¡±¡°Æ«Ð¡¡±»ò¡°ÎÞÓ°Ï족)¡£

£¨2£©Ëá¼îÖк͵ζ¨ÊÇÖÐѧ»¯Ñ§³£¼ûʵÑé¡£

ijѧУ»¯Ñ§¿ÎÍâС×éÓÃ0.2000 mol¡¤L-1ÑÎËáµÎ¶¨Î´ÖªÅ¨¶ÈµÄÇâÑõ»¯ÄÆÈÜÒº£¬ÊԻشðÏÂÁÐÎÊÌâ¡£

¢ÙµÎ¶¨¹ý³ÌÖУ¬ÑÛ¾¦Ó¦×¢ÊÓ                                     ¡£

¢ÚÔÚÌú¼Ų̈ÉϵæÒ»ÕŰ×Ö½£¬ÆäÄ¿µÄÊÇ                             ¡£

¢Û¸ù¾ÝϱíÊý¾Ý£¬¼ÆËã±»²âÉÕ¼îÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈÊÇ    mol¡¤L£­1¡££¨±£ÁôËÄλÓÐЧÊý×Ö£©

¢ÜÏÂÁÐʵÑé²Ù×÷¶ÔµÎ¶¨½á¹û²úÉúʲôӰÏì(Ìî¡°Æ«¸ß¡±¡¢¡°Æ«µÍ¡±»ò¡°ÎÞÓ°Ï족)?

a.¹Û²ìËáʽµÎ¶¨¹ÜÒºÃæÊ±£¬¿ªÊ¼¸©ÊÓ£¬µÎ¶¨ÖÕµãÆ½ÊÓ£¬ÔòµÎ¶¨½á¹û________¡£

b.Èô½«×¶ÐÎÆ¿Óôý²âÒºÈóÏ´£¬È»ºóÔÙ¼ÓÈë10.00 mL´ý²âÒº£¬ÔòµÎ¶¨½á¹û_______¡£

 

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

¾«Ó¢¼Ò½ÌÍø