ÌâÄ¿ÄÚÈÝ

¢ñ¡£Ò»¶¨ÎïÖʵÄÁ¿Å¨¶ÈÈÜÒºµÄÅäÖƺÍËá¼îÖк͵ζ¨ÊÇÖÐѧ»¯Ñ§ÖÐÁ½¸öµäÐ͵Ķ¨Á¿ÊµÑ顣ijÑо¿ÐÔѧϰС×éÔÚʵÑéÊÒÖÐÅäÖÆ1mol/LµÄÏ¡ÁòËá±ê×¼ÈÜÒº£¬È»ºóÓÃÆäµÎ¶¨Ä³Î´ÖªÅ¨¶ÈµÄNaOHÈÜÒº¡£ÏÂÁÐÓйØ˵·¨ÖÐÕýÈ·µÄÊÇ______________£¨´ð°¸¿ÉÄܲ»Î¨Ò»£©

A¡¢ÊµÑéÖÐËùÓõ½µÄµÎ¶¨¹Ü¡¢ÈÝÁ¿Æ¿£¬ÔÚʹÓÃÇ°¾ùÐèÒª¼ì©£»

B¡¢Èç¹ûʵÑéÖÐÐèÓÃ60mL µÄÏ¡ÁòËá±ê×¼ÈÜÒº£¬ÅäÖÆʱӦѡÓÃ100mLÈÝÁ¿Æ¿£»

C¡¢ÈÝÁ¿Æ¿Öк¬ÓÐÉÙÁ¿ÕôÁóË®£¬»áµ¼ÖÂËùÅä±ê×¼ÈÜÒºµÄŨ¶ÈƫС£»

D¡¢ËáʽµÎ¶¨¹ÜÓÃÕôÁóˮϴµÓºó£¬¼´×°Èë±ê׼Ũ¶ÈµÄÏ¡ÁòËᣬÔò²âµÃµÄNaOHÈÜÒºµÄŨ¶È½«Æ«´ó£»

E¡¢Óü׻ù³È×÷ָʾ¼Á£¬µÎ¶¨ÖÕµãʱ£¬ÈÜÒºÑÕÉ«´Ó³ÈÉ«±äΪºìÉ«£»

F¡¢ÅäÖÆÈÜÒººÍÖк͵樵ÄÁ½¸öʵÑéÖУ¬ÈôÔÚ×îºóÒ»´Î¶ÁÊý¾ù¸©ÊÓ¶ÁÊý£¬Ôòµ¼ÖÂʵÑé½á¹û¶¼Æ«´ó¡£

¢ò . ÏÖÓÃÈçͼËùʾµÄ×°ÖÃÖÆÈ¡½Ï¶àÁ¿µÄ±¥ºÍÂÈË®²¢²â¶¨±¥ºÍÂÈË®µÄpH¡£

»Ø´ðÓйØÎÊÌ⣺

£¨1£©Ð´³öÓйػ¯Ñ§·½³Ìʽ¡£

×°Öüףº____________________________________ £»

×°Öö¡£º           __________________        ¡£

£¨2£©Ö¤Ã÷ÂÈË®Òѱ¥ºÍµÄÏÖÏóÊÇ                                           ¡£

£¨3£©ÖÆÈ¡½áÊøʱ£¬×°ÖñûµÄÒºÌåÖк¬ÓеÄ΢Á£ÊÇ             £¨ÓÃ΢Á£·ûºÅ±íʾ £©£»

(4)Èô³·È¥×°ÖÃÒÒ£¬Ö±½Ó½«×°Öü׺ͱûÏàÁ¬¡£ÕâÑù×ö¶ÔʵÑé²â¶¨½á¹ûµÄÓ°ÏìÊÇ£º                      ¡£

²â¶¨±¥ºÍÂÈË®µÄpH·½·¨ÊÇ_______________________________________________¡£

¢ó.ʵÑéÉè¼Æ£ºÖ¤Ã÷NaOH¹ÌÌåÔÚ¿ÕÆøÖзÅÖò¿·Ö±äÖÊ

_______________________________________________________________

_______________________________________________________________________

¢ñ £¨  ¹²16·Ö£© £¨3·Ö£©  A¡¢B¡¢D¡£     (¶àÑ¡²»µÃ·Ö£¬ÉÙÑ¡1Ïî¿Û1·Ö )

¢ò. £¨8·Ö£©¡££¨1£©¼×£ºMnO2+4HClMnCl2+Cl2¡ü+2H2O£¨1·Ö£©

¶¡£ºCl2+2NaOH¡úNaCl+NaClO+H2O£¨1·Ö£©

£¨2£©±ûÖÐÒºÌå³Ê»ÆÂÌÉ«£¬±ûÉϲ¿¿Õ¼ä³Ê»ÆÂÌÉ«£¬ÓлÆÂÌÉ«ÆøÌå½øÈ붡ÖУ¨2·Ö£©

£¨3£© H2O¡¢Na+¡¢Cl-¡¢H+¡¢Cl¡¢HClO¡¢OH-£¨2·Ö£©£¨Ñ¡È«µÃ2·Ö£¬Â©Ñ¡µÃ1·Ö£¬¶àÑ¡²»µÃ·Ö£©

£¨4£©ÓÉÓÚÂÈ»¯Çâδ³ý¾¡»áµ¼ÖÂÖƵõÄÂÈË®ËáÐÔÔöÇ¿£¬²âµÃµÄpHƫС£¨1·Ö£©¡£

ÓÃpH¼ÆÖ±½Ó²â¶¨£¨ÓÃpHÊÔÖ½²â¶¨²»¸ø·Ö£©£¨1·Ö£©¡£

¢ó£º£¨3·Ö£©¡£È¥¹ÌÌåÑùÆ·ÉÙÁ¿ÈÜÓÚË®£¨1·Ö£©£¬ÏòÈÜÒºÖмÓÈë¹ýÁ¿CaCl2ÈÜÒº£¬ÈÜÒº²úÉú°×É«³Áµí˵Ã÷¸ÃÑùÆ·ÒѱäÖÊ£¨1·Ö£©£¬ÕâʱÏòÈÜÒºÖеÎÈë2µÎ·Ó̪£¬ÈôÈÜÒº³Êºìɫ˵Ã÷»¹Óв¿·ÖÇâÑõ»¯ÄÆû±äÖÊ£¨1·Ö£©£¬¼´²¿·ÖÑùÆ·±äÖÊ¡£


½âÎö:

ÂÔ

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
ÔÚ»¨Æ¿ÖмÓÈë¡°ÏÊ»¨±£ÏʼÁ¡±£¬ÄÜÑÓ³¤ÏÊ»¨µÄÊÙÃü£®Ï±íÊÇ1L¡°ÏÊ»¨±£ÏʼÁ¡±Öк¬Óеijɷ֣¬ÔĶÁºó»Ø´ðÏÂÁÐÎÊÌ⣺
³É·Ö ÖÊÁ¿£¨g£© Ħ¶ûÖÊÁ¿£¨g?mol-1£©
ÕáÌÇ 50.0 342
ÁòËá¼Ø 0.5 174
°¢Ë¾Æ¥ÁÖ 0.4 180
¸ßÃÌËá¼Ø 0.5 158
ÏõËáÒø 0.2 170
£¨1£©ÏÂÁС°ÏÊ»¨±£ÏʼÁ¡±µÄ³É·ÖÖУ¬ÊôÓڷǵç½âÖʵÄÊÇ
A
A
£®£¨ÌîÐòºÅ£©
A£®ÕáÌÇ       B£®ÁòËá¼Ø       C£®¸ßÃÌËá¼Ø      D£®ÏõËáÒø
£¨2£©¡°ÏÊ»¨±£ÏʼÁ¡±ÖÐK+£¨°¢Ë¾Æ¥ÁÖÖв»º¬K+£©µÄÎïÖʵÄÁ¿Å¨¶ÈΪ
2¡Á
0.5
174
+
0.5
158
2¡Á
0.5
174
+
0.5
158
mol?L-1£¨Ö»ÒªÇóд±í´ïʽ£¬²»ÓÃд³ö¼ÆËã½á¹û£©£®
£¨3£©ÔÚÅäÖÆÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈµÄÈÜҺʱ£¬Ò»°ã¿É·ÖΪÒÔϼ¸¸ö²½Ö裺
¢Ù³ÆÁ¿¢Ú¼ÆËã ¢ÛÈܽ⠢ÜÒ¡ÔÈ ¢ÝתÒÆ ¢ÞÏ´µÓ ¢ß¶¨ÈÝ ¢àÀäÈ´£®
ÆäÕýÈ·µÄ²Ù×÷˳ÐòΪ
¢Ú¢Ù¢Û¢à¢Ý¢Þ¢ß¢Ü
¢Ú¢Ù¢Û¢à¢Ý¢Þ¢ß¢Ü
£®
£¨4£©ÅäÖÆÉÏÊö1L¡°ÏÊ»¨±£ÏʼÁ¡±ËùÐèµÄÒÇÆ÷ÓУº1000mLÈÝÁ¿Æ¿¡¢ÉÕ±­¡¢²£Á§°ô¡¢Ò©³×¡¢
½ºÍ·µÎ¹Ü
½ºÍ·µÎ¹Ü
¡¢
ÍÐÅÌÌìƽ
ÍÐÅÌÌìƽ
£®£¨ÔÚºáÏßÉÏÌîдËùȱÒÇÆ÷µÄÃû³Æ£©
£¨5£©ÔÚÈÜÒºÅäÖƹý³ÌÖУ¬ÏÂÁвÙ×÷¶ÔÅäÖƽá¹ûûÓÐÓ°ÏìµÄÊÇ
BD
BD
£®
A£®½«Ò©Æ··ÅÈëÈÝÁ¿Æ¿ÖмÓÕôÁóË®ÖÁ»·Ðο̶ÈÏß
B£®ÈÝÁ¿Æ¿ÔÚʹÓÃǰδ¸ÉÔÀïÃæÓÐÉÙÁ¿ÕôÁóË®
C£®ÈÝÁ¿Æ¿ÔÚʹÓÃÇ°¸Õ¸ÕÅäÖÆÍêÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈµÄNaClÈÜÒº¶øδϴ¾»
D£®¶¨ÈÝÒ¡ÔȺó·¢ÏÖÒºÃæµÍÓÚÈÝÁ¿Æ¿µÄ¿Ì¶ÈÏߣ¬µ«Î´×öÈκδ¦Àí£®
¢ñ£®»¯Ñ§ÊÇÒ»ÃÅÒÔʵÑéΪ»ù´¡µÄѧ¿Æ£¬ÏÂÁÐʵÑé²Ù×÷µÄÃèÊöÖУ¬ÕýÈ·µÄÊÇ£¨ÌîÐòºÅ£©
¢Ù¢Û¢Ý
¢Ù¢Û¢Ý
£®
¢ÙÎÅÂÈÆøµÄÆøζʱ£¬ÓÃÊÖÇáÇáÔÚÆ¿¿ÚÉȶ¯£¬½öʹ¼«ÉÙÁ¿µÄÂÈÆøÆ®½ø±Ç¿×
¢ÚÓÃÍÐÅÌÌìƽ³ÆÈ¡10.4gʳÑÎʱ£¬½«Ê³ÑηÅÔÚÓÒÅÌÖеÄֽƬÉϳÆÁ¿
¢ÛÅäÖÆÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈµÄÈÜҺʱ£¬ÈÝÁ¿Æ¿Ï´µÓºóδ½øÐиÉÔï
¢ÜÒÔ·Ó̪×÷ָʾ¼Á£¬Óñê׼Ũ¶ÈµÄÑÎËáµÎ¶¨NaOHÈÜÒº£¬´ýÈÜÒº±ä³Édzºìɫʱ¼´Í£Ö¹µÎ¶¨
¢Ý²â¶¨ÈÜÒºµÄpHʱ£¬Óýྻ¡¢¸ÉÔïµÄ²£Á§°ôպȡ´ý²âÈÜÒº£¬µãÔÚÊÔÖ½Öв¿£¬´ý±äÉ«ºóÓë±ê×¼±ÈÉ«¿¨¶ÔÕÕ¶ÁÊý
¢ÞʵÑéÖÐÊ£ÓàµÄ¹ýÑõ»¯ÄÆ·ÛÄ©ÓÃÖ½°ü¹üºÃºó£¬·ÅÈëÀ¬»øÍ°ÄÚ
¢ò£®Ä³»¯Ñ§Ð¡×éÄâ²ÉÓÃÈçÏÂ×°Ö㨼гֺͼÓÈÈÒÇÆ÷ÒÑÂÔÈ¥£©À´µç½â±¥ºÍʳÑÎË®£¬²¢Óõç½â²úÉúµÄH2»¹Ô­CuO·ÛÄ©À´²â¶¨CuµÄÏà¶ÔÔ­×ÓÖÊÁ¿£¬Í¬Ê±ÑéÖ¤ÂÈÆøµÄÑõ»¯ÐÔ£®

£¨1£©Ð´³ö¼×Öз´Ó¦µÄ»¯Ñ§·½³Ìʽ
2NaCl+2H2O
 µç½â 
.
 
2NaOH+H2¡ü+Cl2¡ü
2NaCl+2H2O
 µç½â 
.
 
2NaOH+H2¡ü+Cl2¡ü
£®
£¨2£©ÎªÍê³ÉÉÏÊöʵÑ飬ÕýÈ·µÄÁ¬½Ó˳ÐòΪAÁ¬
E
E
£¬BÁ¬
C
C
£¨Ìî½Ó¿Ú×Öĸ£©£®
£¨3£©ÒÒ×°ÖÃÖÐXÊÔ¼Á¿ÉÒÔÊÇ
µí·ÛKIÈÜÒº
µí·ÛKIÈÜÒº
£¬±û×°ÖÃÖÐYÊÔ¼ÁµÄ×÷ÓÃÊÇ
ÎüÊÕÇâÆøÖеÄË®
ÎüÊÕÇâÆøÖеÄË®
£®
·½°¸
ÖÊÁ¿
·½°¸¼× ·½°¸ÒÒ
UÐ͹Ü+¹ÌÌå Ó²Öʲ£Á§¹Ü+¹ÌÌå
·´Ó¦Ç°ÖÊÁ¿/g a c
ÍêÈ«·´Ó¦ºóÖÊÁ¿/g b d
£¨4£©²â¶¨CuµÄÏà¶ÔÔ­×ÓÖÊÁ¿£º½«wg CuOÖÃÓÚÓ²Öʲ£Á§¹ÜÖУ¬°´ÒÔÏÂÁ½¸ö·½°¸²âµÃµÄÊý¾Ý¼ÆËãCuµÄÏà¶ÔÔ­×ÓÖÊÁ¿£®Çë»Ø´ð£º
ÄãÈÏΪ·½°¸
ÒÒ
ÒÒ
½Ï¼Ñ£¬°´½Ï¼Ñ·½°¸¼ÆË㣬µÃµ½CuµÄÏà¶ÔÔ­×ÓÖÊÁ¿±í´ïʽÊÇ
16(d+w-c)
c-d
16(d+w-c)
c-d
£®

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

¾«Ó¢¼Ò½ÌÍø