£¨1£©(2·Ö)ÏÂÁÐÓйØʵÑé²Ù×÷»òÅжϲ»ÕýÈ·µÄÊÇ ________£¨ÌîÐòºÅ£¬¶àÑ¡¿Û·Ö£©¡£
A£®ÅäÖÆÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈÈÜÒº£¬¶¨ÈÝʱ¸©Êӿ̶ÈÏ߻ᵼÖÂËùÅäÈÜҺŨ¶ÈƫС
B£®ÉÙÁ¿Å¨ÁòËáÕ´ÔÚƤ·ôÉÏ£¬Á¢¼´ÓÃÇâÑõ»¯ÄÆÈÜÒº³åÏ´
C£®ÒºäåÓж¾ÇÒÒ×»Ó·¢,ÐèÊ¢·ÅÔÚÄ¥¿ÚµÄϸ¿ÚÆ¿Àï,²¢Ë®·â±£´æ
D£®100 mLÈÝÁ¿Æ¿¿ÉÓÃÓÚÅäÖÆ95 mL 0.1 mol/L NaClÈÜÒº
E£®ÔÚÌìƽ×óÓÒÁ½ÅÌÖи÷·ÅÒ»ÕÅ°×Ö½ºó£¬¼´¿É½«NaOH¹ÌÌå·ÅÔÚ°×Ö½ÉϳÆÁ¿
(2) (2·Ö)Ë«ÑõË®£¨H2O2£©ÊǼ«ÈõµÄµç½âÖÊ£¬ H2O2ÈÜÒºÏÔÈõËáÐÔ¡£Èô°ÑH2O2¿´³É¶þÔªÈõËᣬÇëд³öËüÔÚË®ÖеĵçÀë·½³Ìʽ­­­­­­­­­­­­­­­­­­­­­                       ,
                         
(3) (4·Ö)ÓÉÇâÆøºÍÑõÆø·´Ó¦Éú³É1molË®ÕôÆø£¬·ÅÈÈ241.8kJ£¬Ð´³ö¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ­­                                             ¡£Èô1gË®ÕôÆøת»¯³ÉҺ̬ˮ·ÅÈÈ2.444 kJ£¬ÔòH2µÄȼÉÕÈȦ¤H=    kJ¡¤mol-1
£¨8·Ö£©ABE £¨2·Ö£©    H2O2H++HO2-   HO2-   H++O22-£¨2·Ö£©
2H2(g)+O2(g)="2H2O(g) " ¦¤H="-483.6" kJ¡¤mol-1     £¨2·Ö£©        -285.8£®£¨2·Ö£©
ÂÔ
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÏÂÁйØÓÚʵÑéÎÊÌâ´¦Àí·½·¨»ò²Ù×÷²»ÕýÈ·µÄÊÇ
A£®ÊµÑé½áÊøºó½«ËùÓеķÏÒºµ¹ÈëÏÂË®µÀÅųöʵÑéÊÒ£¬ÒÔÃâÎÛȾʵÑéÊÒ
B£®ÔÚʵÑéÊÒ£¬²»ÄÜÆ·³¢Ò©Æ·
C£®Ï¨ÃðʵÑé×ÀÉÏȼ×ŵľƾ«£¬¼ò±ãºÏÀíµÄ·½·¨¾ÍÊÇÓÃʪĨ²¼¸ÇÃð
D£®ÅäÖÆÏ¡ÁòËáʱ£¬ÏÈÔÚÉÕ±­ÖмÓÒ»¶¨Ìå»ýµÄË®£¬Ôٱ߽Á°è±ßÑر­±Ú¼ÓÈëŨÁòËá

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

ʵÑéÊÒÓÃÏÂÁз½·¨²â¶¨Ä³Ë®ÑùÖÐO2µÄº¬Á¿¡£

(1)ÓÃÈçͼËùʾװÖã¬Ê¹Ë®ÖÐÈܽâµÄO2ÔÚ¼îÐÔÌõ¼þϽ«Mn2£«Ñõ»¯³ÉMnO(OH)2µÄ·´Ó¦µÄÀë×Ó·½³ÌʽΪ______________________________________¡£
ʵÑé²½Ö裺
¢Ù´ò¿ªÖ¹Ë®¼ÐaºÍb£¬´ÓA´¦Ïò×°ÖÃÄÚ¹ÄÈë¹ýÁ¿N2£¬´Ë²Ù×÷µÄÄ¿µÄÊÇ______________________________________¡£
¢ÚÓÃ×¢ÉäÆ÷³éÈ¡20.00 mLijˮÑù´ÓA´¦×¢Èë׶ÐÎÆ¿¡£
¢ÛÔÙ·Ö±ð´ÓA´¦×¢È뺬m mol NaOHµÄÈÜÒº¼°¹ýÁ¿µÄMnSO4ÈÜÒº¡£
¢ÜÍê³ÉÉÏÊö²Ù×÷ºó£¬ÏÂÃæÓ¦½øÐеIJÙ×÷ÊÇ_______________¡£cÊdz¤Ï𽺹ܣ¬Ê¹Óýϳ¤Ï𽺹ܵÄÄ¿µÄÊÇ__________________________________¡£
(2)ÓÃI£­½«Éú³ÉµÄMnO(OH)2ÔÙ»¹Ô­ÎªMn2£«£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪMnO(OH)2£«2I£­£«4H£«===Mn2£«£«I2£«3H2O¡£
ʵÑé²½Ö裺
¢Ý´ò¿ªÖ¹Ë®¼Ða¡¢b£¬·Ö±ð´ÓA´¦×¢Èë×ãÁ¿NaIÈÜÒº¼°º¬n mol H2SO4µÄÁòËáÈÜÒº¡£
¢ÞÖظ´¢ÜµÄ²Ù×÷¡£
(3)ÓÃNa2S2O3±ê×¼ÈÜÒºµÎ¶¨²½Öè(2)ÖÐÉú³ÉµÄI2£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪI2£«2Na2S2O3===2NaI£«Na2S4O6¡£
ʵÑé²½Ö裺
¢ßÈ¡ÏÂ׶ÐÎÆ¿£¬ÏòÆäÖмÓÈë2¡«3µÎ__________×÷ָʾ¼Á¡£
¢àÓÃ0.005 mol¡¤L£­1µÄNa2S2O3µÎ¶¨ÖÁÖյ㡣
(4)¼ÆËã¡£µÎ¶¨¹ý³ÌÖУ¬µÎ¶¨Ç°ºóµÎ¶¨¹ÜÖеÄÒºÃæ¶ÁÊýÈçͼËùʾ¡£

¢á¾­¼ÆË㣬´ËË®ÑùÖÐÑõ(O2)µÄº¬Á¿Îª________mg¡¤L£­1¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

£¨18·Ö£©ÊµÑéÊÒÓÃÂÈ»¯ÄƹÌÌåÅäÖÆ1.00mol/LµÄNaClÈÜÒº0.5 L£¬»Ø´ðÏÂÁÐÎÊÌâ
£¨1£©¼òÊö¸ÃʵÑéµÄÖ÷ҪʵÑé²½Öè          
£¨2£©ËùÐèÒÇÆ÷Ϊ£ºÈÝÁ¿Æ¿ £¨¹æ¸ñ£º      £©¡¢ÍÐÅÌÌìƽ¡¢»¹ÐèÒªÄÇЩʵÑéÒÇÆ÷²ÅÄÜÍê³É¸ÃʵÑ飬Çëд³ö£º                        ¡£
£¨3£©ÊÔ·ÖÎöÏÂÁвÙ×÷¶ÔËùÅäÈÜÒºµÄŨ¶ÈÓкÎÓ°Ïì¼°Ôì³É¸ÃÓ°ÏìµÄÔ­Òò¡£
¢ÙÅäÖÆÇ°ÈÝÁ¿Æ¿Ï´µÓºóûºæ¸É£¬¶ÔËùÅäÈÜҺŨ¶ÈµÄÓ°Ïì          ÊÇ        £¨Æ«´ó¡¢Æ«Ð¡¡¢ÎÞÓ°Ïì¡£ÒÔÏÂͬ£©
¢ÚδϴµÓÉÕ±­ºÍ²£Á§°ô£¬¶ÔËùÅäÈÜҺŨ¶ÈµÄÓ°Ï죺        ¡£
¢ÛΪ¼ÓËÙ¹ÌÌåÈܽ⣬¿ÉÉÔ΢¼ÓÈȲ¢²»¶Ï½Á°è¡£ÔÚδ½µÖÁÊÒÎÂʱ£¬Á¢¼´½«ÈÜҺתÒÆÖÁÈÜÁ¿Æ¿¶¨ÈÝ¡£¶ÔËùÅäÈÜҺŨ¶ÈµÄÓ°Ï죺        ¡£
¢Ü¶¨Èݺ󣬼Ӹǵ¹×ªÒ¡ÔȺ󣬷¢ÏÖÈÜÃæµÍÓڿ̶ÈÏߣ¬ÓֵμÓÕôÁóË®ÖÁ¿Ì¶È¡£¶ÔËùÅäÈÜҺŨ¶ÈµÄÓ°Ïì        ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÏÂÁÐʵÑé²Ù×÷Ó밲ȫʹʴ¦Àí´íÎóµÄÊÇ                
A£®Ê¹ÓÃË®ÒøζȼƲâÁ¿ÉÕ±­ÖÐˮԡζÈʱ£¬²»É÷´òÆÆË®ÒøÇò£¬Óõιܽ«Ë®ÒøÎü³ö·ÅÈëË®·âµÄСƿÖУ¬²ÐÆƵÄζȼƲåÈë×°ÓÐÁò·ÛµÄ¹ã¿ÚÆ¿ÖÐ
B£®ÝÍÈ¡·ÖÒººó£¬ÉϲãÈÜÒº´ÓÉÏ¿Úµ¹³ö£¬Ï²ãÈÜÒº´ÓÏ¿ڷųö
C£®ÖƱ¸ÒÒËáÒÒõ¥Ê±£¬½«ÒÒ´¼ºÍÒÒËáÒÀ´Î¼ÓÈ뵽ŨÁòËáÖÐ
D£®µÎ¶¨¹ÜʹÓÃ֮ǰÊ×ÏÈÒª¼ì²éÊÇ·ñ©ˮ£¬Ï´µÓÖ®ºó»¹ÒªÓÃÏàÓ¦µÄÈÜÒºÈóÏ´²ÅÄÜÓÃÀ´µÎ¶¨

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÔÚʹÓÃÈÝÁ¿Æ¿ÅäÖÆÈÜҺʱ£¬ÏÂÁвÙ×÷ÕýÈ·µÄÊÇ £¨ £©  
A£®Ê¹ÓÃÈÝÁ¿Æ¿Ç°¶¼±ØÐë¼ì²éÈÝÁ¿Æ¿ÊÇ·ñ©ˮ
B£®ÈÝÁ¿Æ¿ÓÃÕôÁóˮϴ¾»ºó£¬ÔÙÓôýÅäÈÜÒºÈóÏ´
C£®³ÆºÃµÄ¹ÌÌåÊÔÑùÐèÓÃÖ½ÌõСÐĵØËÍÈëÈÝÁ¿Æ¿ÖÐ
D£®Ò¡ÔȺó·¢ÏÖ°¼ÒºÃæϽµ£¬ÔÙ¼ÓË®ÖÁ¿Ì¶ÈÏß

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÏÂͼËùʾʵÑé»ù±¾²Ù×÷ÕýÈ·µÄÊÇ£¨   £©

A                 B               C                  D

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÏÂÁÐÒÇÆ÷ʹÓÃÇ°±ØÐë¼ì²éÊÇ·ñ©ˮµÄÊÇ                              
A£®·ÖҺ©¶·B£®Â©¶·C£®Õô·¢ÃóD£®³¤¾±Â©¶·

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÏÂÁÐʵÑé²Ù×÷ÖдíÎóµÄÊÇ£¨   £©
A£®ÇâÑõ»¯ÄÆÓи¯Ê´ÐÔ£¬ËùÒÔÒ»°ã·ÅÔÚСÉÕ±­ÖгÆÁ¿
B£®Ê¹ÓÃÊԹܼУ¬Ó¦´ÓÊԹܵײ¿ÍùÉÏÌ×£¬¼ÐÔÚÊÔ¹ÜÖÐÉϲ¿
C£®×öÍêʵÑéºó£¬ÓÃÊ£µÄÈκÎÒ©Æ·²»µÃ¶ªÆú£¬Ò²²»Òª·Å»ØÔ­Æ¿
D£®ÊµÑéÊÒÖÆÈ¡ÑõÆø֮ǰ£¬Ïȼì²é×°ÖõÄÆøÃÜÐÔ£¬ÔÙ×°ÈëÒ©Æ·

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸