ÌâÄ¿ÄÚÈÝ

ÏÂÁÐÊÇÓйØÊµÑéµÄÐðÊöÖУ¬ºÏÀíµÄÊÇ                                    

¢ÙÓÃÍÐÅÌÌìÆ½³ÆÈ¡25.20 gNaCl¹ÌÌå  

¢ÚÓÃÌúÛáÛöׯÉÕÇâÑõ»¯ÄƹÌÌå

¢ÛʹÓÃÈÝÁ¿Æ¿µÄµÚÒ»²½²Ù×÷ÊÇÏȽ«ÈÝÁ¿Æ¿ÓÃÕôÁóˮϴµÓºóºæ¸É

¢ÜÓÃ25 mL¼îʽµÎ¶¨¹ÜÁ¿È¡14.80 mL 1 mol/L NaOHÈÜÒº

¢ÝÓÃʪÈóµÄpHÊÔÖ½²â¶¨Ä³ÈÜÒºµÄpH

¢ÞʵÑéÊÒÅäÖÆÂÈ»¯ÌúÈÜҺʱ£¬¿ÉÒÔÏȽ«ÂÈ»¯ÌúÈܽâÔÚÑÎËáÖУ¬ÔÙÅäÖÆµ½ËùÐèÒªµÄŨ¶È

A£®¢Ú¢Ü¢Þ           B£®¢Ú¢Û¢Þ           C£®¢Ù¢Ú¢Ü           D£®¢Ú¢Û¢Ü¢Ý¢Þ

 

¡¾´ð°¸¡¿

A

¡¾½âÎö¡¿

ÊÔÌâ·ÖÎö£ºÍÐÅÌÌìÆ½Ö»ÄܶÁÊýµ½0.1g£¬¢Ù²»ÕýÈ·£»½ðÊôÌúºÍÇâÑõ»¯ÄƲ»·´Ó¦£¬¿ÉÓÃÌúÛáÛöׯÉÕÇâÑõ»¯ÄƹÌÌ壬¢ÚÕýÈ·£»Ê¹ÓÃÈÝÁ¿Æ¿µÄµÚÒ»²½²Ù×÷ÊǼì©£¬¢Û²»ÕýÈ·£»ÇâÑõ»¯ÄÆÊÇÇ¿¼î£¬Ó¦¸ÃÓüîʽµÎ¶¨¹ÜÁ¿È¡£¬¢ÜÕýÈ·£»ÓÃpHÊÔÖ½²â¶¨Ä³ÈÜÒºµÄpHʱ£¬ÊÔÖ½²»ÄÜÊÂÏÈʪÈ󣬢ݲ»ÕýÈ·£»ÂÈ»¯ÌúÈÜÓÚË®£¬ÌúÀë×ÓË®½â£¬ÈÜÒºÏÔËáÐÔ£¬ËùÒÔʵÑéÊÒÅäÖÆÂÈ»¯ÌúÈÜҺʱ£¬¿ÉÒÔÏȽ«ÂÈ»¯ÌúÈܽâÔÚÑÎËáÖУ¬ÒÔÒÖÖÆÌúÀë×ÓË®½â£¬È»ºóÔÙÅäÖÆµ½ËùÐèÒªµÄŨ¶È£¬¢ÞÕýÈ·£¬´ð°¸Ñ¡A¡£

¿¼µã£º¿¼²é³£¼ûµÄ»ù±¾ÊµÑé²Ù×÷¡£

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×÷£¬ËùÒÔ¹ýÂËÊÔÌâÖ÷ÒªÊÇÒÔ³£¼ûÒÇÆ÷µÄÑ¡Óá¢ÊµÑé»ù±¾²Ù×÷ΪÖÐÐÄ£¬Í¨¹ýÊÇʲô¡¢ÎªÊ²Ã´ºÍÔõÑù×öÖØµã¿¼²éʵÑé»ù±¾²Ù×÷µÄ¹æ·¶ÐÔºÍ׼ȷÐÔ¼°Áé»îÔËÓÃ֪ʶ½â¾öʵ¼ÊÎÊÌâµÄÄÜÁ¦¡£

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø