ÌâÄ¿ÄÚÈÝ

ÏÂÁйØÓÚ·´Ó¦ÄÜÁ¿µÄ˵·¨ÕýÈ·µÄÊÇ

A£®Zn(s)£«CuSO4(aq)===ZnSO4(aq)£«Cu(s) ¦¤H£½£­216 kJ¡¤mol£­1£¬Ôò·´Ó¦Îï×ÜÄÜÁ¿£¾Éú³ÉÎï×ÜÄÜÁ¿

B£®ÏàͬÌõ¼þÏ£¬Èç¹û1 molÇâÔ­×ÓËù¾ßÓеÄÄÜÁ¿ÎªE1£¬1 molÇâ·Ö×ÓËù¾ßÓеÄÄÜÁ¿ÎªE2£¬Ôò2E1£½E2

C£®101 kPaʱ£¬2H2(g)£«O2(g)===2H2O(l) ¦¤H£½£­571.6 kJ¡¤mol£­1£¬ÔòH2µÄȼÉÕÈÈΪ571.6 kJ¡¤mol£­1

D£®H£«(aq)£«OH£­(aq)===H2O(l) ¦¤H£½£­57.3 kJ¡¤mol£­1£¬Ôò1 mol NaOHµÄÇâÑõ»¯ÄƹÌÌåÓ뺬0.5 mol H2SO4µÄÏ¡ÁòËá»ìºÏºó·Å³ö57.3 kJµÄÈÈÁ¿

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

ÓÐA¡¢B¡¢C¡¢DËÄÖÖÇ¿µç½âÖÊ£¬ËüÃÇÔÚË®ÖеçÀë²úÉúÏÂÁÐÀë×Ó(ÿÖÖÎïÖÊÖ»º¬Ò»ÖÖÒõÀë×ÓÇÒ»¥²»Öظ´)¡£

ÑôÀë×Ó

Na£«¡¢Ba2£«¡¢NH

ÒõÀë×Ó

CH3COO£­¡¢OH£­¡¢Cl£­¡¢SO

ÒÑÖª£º¢ÙA¡¢CÈÜÒºµÄpH¾ù´óÓÚ7£¬A¡¢BµÄÈÜÒºÖÐË®µÄµçÀë³Ì¶ÈÏàͬ£»¢ÚCÈÜÒººÍDÈÜÒºÏàÓöʱֻÉú³É°×É«³Áµí£¬BÈÜÒººÍCÈÜÒºÏàÓöʱֻÉú³É´Ì¼¤ÐÔÆøÎ¶µÄÆøÌ壬AÈÜÒººÍDÈÜÒº»ìºÏʱÎÞÏÖÏó¡£

£¨1£©AÊÇ______ ____£¬BÊÇ___ ___(Ìѧʽ) ¡£

ÓÃÀë×Ó·½³Ìʽ±íʾAµÄË®ÈÜÒºÖдæÔ򵀮½ºâ¹ØÏµ£º ¡£

£¨2£©Ð´³öCºÍD·´Ó¦µÄÀë×Ó·½³Ìʽ__________ _¡£

£¨3£©25 ¡æÊ±£¬0.1 mol¡¤L£­1 BÈÜÒºµÄpH£½a£¬ÔòBÈÜÒºÖУºc(H£«)£­c(OH¡ª)£½_____ ____£¨Ìî΢Á£Å¨¶È·ûºÅ£©=_____ ______(Óú¬ÓÐaµÄ¹ØÏµÊ½±íʾ)¡£

£¨4£©½«µÈÌå»ý¡¢µÈÎïÖʵÄÁ¿Å¨¶ÈµÄBÈÜÒººÍCÈÜÒº»ìºÏ£¬·´Ó¦ºóÈÜÒºÖи÷ÖÖÀë×ÓŨ¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòÊÇ _¡£

£¨5£©ÔÚÒ»¶¨Ìå»ýµÄ0.005 mol¡¤L£­1µÄCÈÜÒºÖУ¬¼ÓÈëÒ»¶¨Ìå»ýµÄ0.00125 mol¡¤L£­1µÄÑÎËᣬ»ìºÏÈÜÒºµÄpH£½11£¬Èô·´Ó¦ºóÈÜÒºµÄÌå»ýµÈÓÚCÈÜÒºÓëÑÎËáµÄÌå»ýÖ®ºÍ£¬ÔòCÈÜÒºÓëÑÎËáµÄÌå»ý±ÈÊÇ_ _____¡£

£¨6£©ÏÖʹÓÃËá¼îÖк͵ζ¨·¨²â¶¨ÊÐÊÛ°×´×µÄ×ÜËáÁ¿(g¡¤100mL£­1)¡£ÔÚ±¾ÊµÑéµÄµÎ¶¨¹ý³ÌÖУ¬ÏÂÁвÙ×÷»áʹʵÑé½á¹ûÆ«´óµÄÊÇ__ £¨ÌîдÐòºÅ)¡£

a£®¼îʽµÎ¶¨¹ÜÔڵζ¨Ê±Î´Óñê×¼NaOHÈÜÒºÈóÏ´

b£®¼îʽµÎ¶¨¹ÜµÄ¼â×ìÔڵζ¨Ç°ÓÐÆøÅÝ£¬µÎ¶¨ºóÆøÅÝÏûʧ

c£®×¶ÐÎÆ¿ÖмÓÈë´ý²â°×´×ÈÜÒººó£¬ÔÙ¼ÓÉÙÁ¿Ë®

d£®×¶ÐÎÆ¿Ôڵζ¨Ê±¾çÁÒÒ¡¶¯£¬ÓÐÉÙÁ¿ÒºÌ彦³ö

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

¾«Ó¢¼Ò½ÌÍø