ÌâÄ¿ÄÚÈÝ

¸ù¾ÝÎïÖʵÄÁ¿µÄÓйع«Ê½¼ÆË㣨ע£º£¨2£©£¨3£©£¨4£©¼ÆËã½á¹û±£ÁôСÊýµãºóÁ½Î»£©

£¨1£©3.01¡Á1023¸öNH4+£¬ÆäÖÊÁ¿Îª g¡£

£¨2£©ÓûÅäÖÆ1L 0.2 mol.L-1µÄÑÎËáÈÜÒº£¬Ðè±ê×¼×´¿öÏÂHClµÄÌå»ýΪ L¡£

£¨3£©ÏÖ½«50 mLÃܶÈΪ1.84 g/cm3£¬ÖÊÁ¿·ÖÊýΪ98.0 %µÄŨÁòËᣬϡÊÍÖÁ250 mL£¬ÔòÏ¡Êͺó£¬ÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈΪ ¡£

£¨4£©19¿ËMgCl2Öк¬ÓÐÀë×ÓÊýÄ¿ ¡£

£¨5£©Ëùº¬Ô­×ÓÊýÏàµÈµÄ¼×ÍéºÍ°±Æø£¨NH3£©µÄÖÊÁ¿±ÈΪ ¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

£¨1£©¹Ì¶¨ºÍÀûÓÃCO2ÄÜÓÐЧµØÀûÓÃ×ÊÔ´£¬²¢¼õÉÙ¿ÕÆøÖеÄÎÂÊÒÆøÌ壮¹¤ÒµÉÏÓÐÒ»ÖÖÓÃCO2À´Éú²ú¼×´¼È¼Áϵķ¢·¨£ºCO2£¨g£©+3H2£¨g£©CH3OH£¨g£©+H2O£¨g£©¡÷H=-49.0kJ¡¤mol-1£®Ä³¿ÆѧʵÑ齫6mol CO2ºÍ8mol H2³äÈë2LµÄÃܱÕÈÝÆ÷ÖУ¬²âµÃH2µÄÎïÖʵÄÁ¿Ëæʱ¼ä±ä»¯Èçͼ1Ëùʾ£¨ÊµÏߣ©£®Í¼ÖÐÊý¾Ýa£¨1£¬6£©´ú±íµÄÒâ˼ÊÇ£ºÔÚ1minʱH2µÄÎïÖʵÄÁ¿ÊÇ6mol¡£

¢ÙaµãÕý·´Ó¦ËÙÂÊ____________£¨Ìî´óÓÚ¡¢µÈÓÚ»òСÓÚ£©Äæ·´Ó¦ËÙÂÊ¡£

¢ÚÏÂÁÐʱ¼ä¶Îƽ¾ù·´Ó¦ËÙÂÊ×î´óµÄÊÇ__________£¬×îСµÄÊÇ__________£®

A£®0¡«1min B£®1¡«3min C£®3¡«8min D£®8¡«11min

¢Û½ö¸Ä±äijһʵÑéÌõ¼þÔÙ½øÐÐÁ½´ÎʵÑ飬²âµÃH2µÄÎïÖʵÄÁ¿Ëæʱ¼ä±ä»¯ÈçͼÖÐÐéÏßËùʾ£¬ÇúÏßI¶ÔÓ¦µÄʵÑéÌõ¼þ¸Ä±äÊÇ__________£¬ÇúÏߢò¶ÔÓ¦µÄʵÑéÌõ¼þ¸Ä±äÊÇ__________¡£

£¨2£©ÀûÓùâÄܺ͹â´ß»¯¼Á£¬¿É½« CO2ºÍ H2O£¨g£©×ª»¯Îª CH4ºÍ O2¡£×ÏÍâ¹âÕÕÉäʱ£¬ÔÚ²»Í¬´ß»¯¼Á£¨¢ñ£¬¢ò£¬¢ó£©×÷ÓÃÏ£¬CH4²úÁ¿Ëæ¹âÕÕʱ¼äµÄ±ä»¯ÈçͼËùʾ¡£ÔÚ0¡«30СʱÄÚ£¬CH4µÄƽ¾ùÉú³ÉËÙÂÊ v¢ñ¡¢v¢òºÍv¢ó´Ó´óµ½Ð¡µÄ˳ÐòΪ____________£»·´Ó¦¿ªÊ¼ºóµÄ12СʱÄÚ£¬ÔÚµÚ__________ÖÖ´ß»¯¼Á×÷ÓÃÏ£¬ÊÕ¼¯µÄ CH4×î¶à¡£

£¨3£© 1100 ¡æʱ£¬Ìå»ýΪ2LµÄºãÈÝÈÝÆ÷Öз¢ÉúÈçÏ·´Ó¦£º

Na2SO4£¨s£©+4H2£¨g£©Na2S£¨s£©+4H2O£¨g£©£¬ÏÂÁÐÄÜÅжϷ´Ó¦´ïµ½Æ½ºâ״̬µÄÊÇ__________¡£

A£®ÈÝÆ÷ÄÚÆøÌåѹǿ²»Ôٱ仯

B£®H2µÄÌå»ý·ÖÊý²»Ôٱ仯

C£®lmolH--H¼ü¶ÏÁÑͬʱ2molH--O¼üÐγÉ

D£®Na2SµÄÖÊÁ¿²»Ôٱ仯

E£®vÕý£¨H2£©=vÄ棨H2O£©

F.ÈÝÆ÷ÄÚÆøÌåÃܶȲ»Ôٱ仯

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

¾«Ó¢¼Ò½ÌÍø