ÌâÄ¿ÄÚÈÝ

ÏÂÁÐ˵·¨»ò±íʾ·¨ÕýÈ·µÄÊÇ

A£®µÈÁ¿µÄ°×Á×ÕôÆøºÍ°×Á×¹ÌÌå·Ö±ðÍêȫȼÉÕ£¬ºóÕ߷ųöÈÈÁ¿¶à

B£®ÓÉC£¨Ê¯Ä«£©¡úC£¨½ð¸Õʯ£©£»¦¤H £½ £«1.19 kJ¡¤ mol¡ª1¿ÉÖª£¬½ð¸Õʯ±ÈʯīÎȶ¨

C£®ÔÚÏ¡ÈÜÒºÖУºH£«£¨aq£©£«OH£­£¨aq£©£½ H2O£¨l£©£»¦¤H£½£­57.3 kJ¡¤ mol¡ª1£¬Èô½«º¬0.5 mol H2SO4µÄŨÁòËáÓ뺬1 mol NaOHµÄÈÜÒº»ìºÏ£¬·Å³öµÄÈÈÁ¿´óÓÚ57.3 kJ

D£®ÔÚ101kPaʱ£¬2g H2ÍêȫȼÉÕÉú³ÉҺ̬ˮ£¬·Å³ö285.8 kJÈÈÁ¿£¬ÇâÆøȼÉÕµÄÈÈ»¯Ñ§·½³Ìʽ±íʾΪ2H2£¨g£©£«O2£¨g£©£½ 2H2O£¨l£©£»¦¤H£½£­285.8 kJ¡¤ mol¡ª1

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø