ÌâÄ¿ÄÚÈÝ

£¨16·Ö£©Ä³Í¬Ñ§ÄâÓôÖÑõ»¯Í­£¨º¬ÉÙÁ¿Ñõ»¯ÑÇÌú¼°²»ÈÜÓÚËáµÄÔÓÖÊ£©ÖÆÈ¡ÎÞË®ÂÈ»¯Í­£¬Á÷³ÌÈçͼËùʾ£º

£¨1£©²½Öè¢ÙÖÐÉæ¼°µÄÀë×Ó·½³ÌʽÓУº                                    ¡£
£¨2£©²½Öè¢ÚÖмÓÈëH2O2µÄÄ¿µÄÊÇ£º                          ·´Ó¦µÄÀë×Ó·½³ÌʽΪ                               ¡£
£¨3£©ÒÑÖª£º
¡¡
ÇâÑõ»¯Î↑ʼ³ÁµíʱµÄpH
ÇâÑõ»¯Îï³ÁµíÍêȫʱµÄpH
Fe3£«
1.9
3.2
Cu2£«
4.7
6.7
Fe2£«
7
9
²½Öè¢ÛÖе÷½ÚpHµÄ×î¼Ñ·¶Î§Îª                                           £¬
²½Öè¢ÛÖпÉÒÔÓÃÓÚµ÷½ÚÈÜÒºpHµÄÊÔ¼ÁXÊÇ         £º
a£®Cu2(OH)2CO3        b£®CuO        c£®  Cu(OH)2     d£®NH3?H2O
£¨4£©²½Öè¢Ü½øÐеIJÙ×÷ÊÇ                     £¬¹ýÂË¡¢Ï´µÓ¡¢¸ÉÔï¡£            ÔÚ²½Öè¢ÝÖÐÒªµÃµ½ÎÞË®CuCl2£¬ÐèÒªÔÚ                         ¼ÓÈÈCuCl2¡¤2H2O¡£
£¨16·Ö£©£¨1£©CuO£«2H£«=Cu2£«£«H2O£¨2·Ö£©£¬FeO£«2H£«=Fe2£«£«H2O£¨2·Ö£©£»
£¨2£©½«Fe2£«Ñõ»¯³ÉFe3£«£¨2·Ö£©£¬ 2Fe2£«£«H2O2£«2H£«=2Fe3£«£«2H2O£¨2·Ö£©£»
£¨3£©3.2ÖÁ4.7£¨2·Ö£©£»  a b c £¨2·Ö£©£»
£¨4£©Õô·¢Å¨Ëõ£¬ÀäÈ´½á¾§£¨2·Ö£©¡£¸ÉÔïµÄHClÆøÁ÷ÖУ¨2·Ö£©
¡¿£¨1£©Ñõ»¯Í­ÊǼîÐÔÑõ»¯ÎºÍÑÎËá·´Ó¦µÄÀë×Ó·½³ÌʽΪCuO£«2H£«=Cu2£«£«H2O¡£
£¨2£©ÓÉÓÚÈÜÒºÖк¬ÓÐÑÇÌúÀë×Ó£¬¶øÑÇÌúÀë×ӵijÁµípH´óÓÚÍ­Àë×ӵģ¬ËùÒÔÓ¦¸Ã°ÑÑÇÌúÀë×Óת»¯ÎªÌúÀë×Ó¶ø²úÉúÇâÑõ»¯Ìú³Áµí£¬´Ó¶ø³ýÈ¥ÔÓÖÊ¡£
£¨3£©Òª³ÁµíÌúÀë×Ó¶ø±£ÁôÍ­Àë×Ó£¬Ôò¸ù¾ÝpH¿ÉÖª£¬Ó¦¸ÃÊÇ3.2ÖÁ4.7Ö®¼ä¡£ÓÉÓÚµ÷½ÚpHÊDz»ÄÜÒýÈëÔÓÖʵģ¬ËùÒÔ´ð°¸Ñ¡bc¡£
£¨4£©ÒªµÃµ½ÂÈ»¯Í­¾§Ì壬ÔòÓ¦¸ÃÊÇÕô·¢Å¨Ëõ£¬ÀäÈ´½á¾§£¬È»ºó¹ýÂË¡¢Ï´µÓ¸ÉÔï¼´¿É¡£ÂÈ»¯Í­ÊÇÇ¿ËáÈõ¼îÑΣ¬Ë®½âÉú³ÉÇâÑõ»¯Í­ºÍÂÈ»¯Çâ¡£ËùÒÔÔÚ¼ÓÈȹý³ÌÖУ¬Ó¦¸ÃÔÚÂÈ»¯ÎïµÄÆøÁ÷ÖнøÐÐÒÔ·ÀÖ¹CuCl2·¢ÉúË®½â¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
£¨16·Ö£©Ä³»¯Ñ§ÐËȤС×éΪ̽¾¿Í­ÓëŨÁòËáµÄ·´Ó¦£¬ÓÃÏÂͼËùʾװÖýøÐÐÓйØʵÑé¡£ ¼×ͬѧȡag Cu ƬºÍ12ml 18mol/LŨH2SO4·ÅÈëÔ²µ×ÉÕÆ¿ÖмÓÈÈ£¬Ö±µ½·´Ó¦Íê±Ï£¬×îºó·¢ÏÖÉÕÆ¿Öл¹ÓÐÒ»¶¨Á¿µÄH2SO4ºÍCuÊ£Óà¡£

£¨1£©CuÓëŨH2SO4·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ_____      ¡£×°ÖÃEÖÐÊÔ¹ÜDÄÚÊ¢ÓÐÆ·ºìÈÜÒº£¬µ±CÖÐÆøÌ弯Âúºó£¬DÖÐÓпÉÄܹ۲쵽µÄÏÖÏóÊÇ____________£¬Îª±ÜÃâʵÑé×°ÖÃDÓпÉÄÜÔì³É»·¾³ÎÛȾ?ÊÔÓÃ×î¼òµ¥·½·¨¼ÓÒÔ½â¾ö?ʵÑéÓÃÆ·×ÔÑ¡?____           _____¡£
£¨2£©×°ÖÃBµÄ×÷ÓÃÊÇÖü´æ¶àÓàµÄÆøÌå¡£BÖÐÓ¦·ÅÖõÄÒºÌåÊÇ_______£¨ÌîÐòºÅ£©¡£
a£®±¥ºÍNa2SO3ÈÜÒº    b£®ËáÐÔ KMnO4ÈÜÒº  c£®Å¨äåË®     d£®±¥ºÍNaHSO3ÈÜÒº
µ±D´¦ÓÐÃ÷ÏÔÏÖÏóºó£¬¹Ø±ÕÐýÈûK£¬ÒÆÈ¥¾Æ¾«µÆ£¬µ«ÓÉÓÚÓàÈȵÄ×÷Óã¬A´¦ÈÔÓÐÆøÌå²úÉú£¬´ËʱBÖÐÏÖÏóÊÇ_________                                         ¡£
£¨3£©·´Ó¦Íê±Ïºó£¬ÉÕÆ¿Öл¹ÓÐÒ»¶¨Á¿µÄÓàËᣬΪʲôȴ²»ÄÜʹCuÍêÈ«ÈܽâµÄÔ­ÒòÊÇ_____              ¡£Ê¹ÓÃ×ãÁ¿µÄÏÂÁÐÒ©Æ·²»ÄÜÓÃÀ´Ö¤Ã÷·´Ó¦½áÊøºóµÄÉÕÆ¿ÖеÄÈ·ÓÐÓàËáµÄÊÇ____  £¨ÌîÐòºÅ£©¡£
a£®Fe·Û b£®BaCl2ÈÜÒº       c£®CuO      d£®Na2CO3ÈÜÒº
ʵÑéÖÐijѧÉúÏòAÖз´Ó¦ºóÈÜÒºÖÐͨÈëÒ»ÖÖ³£¼ûÆøÌåµ¥ÖÊ£¬Ê¹Í­Æ¬È«²¿ÈܽâÇÒ½öÉú³ÉÁòËáÍ­ÈÜÒº£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ__________                          _  ¡£
£¨10£©°´ÒªÇóÍê³ÉÏõËá¼Ø¾§ÌåµÄÖƱ¸ÊµÑ飺
²½Öè
²Ù×÷
¾ßÌå²Ù×÷²½Öè
ÏÖÏó½âÊͽáÂÛµÈ
 
¢Ù
Èܽâ
È¡20¿ËNaNO3ºÍ17¿ËKClÈܽâÔÚ35mlË®ÖУ¬¼ÓÈÈÖÁ·Ð£¬²¢²»¶Ï½Á°è¡£
¹ÌÌåÈܽâ
¢Ú
Õô·¢
¼ÌÐø¼ÓÈȽÁ°è£¬Ê¹ÈÜÒºÕô·¢Å¨Ëõ¡£
ÓР   £á      ¾§ÌåÎö³ö¡£
¢Û
ÈȹýÂË
µ±ÈÜÒºÌå»ý¼õÉÙµ½Ô¼Ô­À´µÄÒ»°ëʱ£¬Ñ¸ËÙ³ÃÈȹýÂË
ÂËÒºÖеÄ×îÖ÷Òª³É·ÖΪ   £â    ¡£
¢Ü
ÀäÈ´
½«ÂËÒºÀäÈ´ÖÁÊÒΡ£
Óо§ÌåÎö³ö¡£
¢Ý
 
°´ÓйØÒªÇó½øÐвÙ×÷
µÃµ½³õ²úÆ·ÏõËá¼Ø¾§Ìå
¢Þ
 
½«µÃµ½µÄ³õ²úÆ·ÏõËá¼Ø¾§ÌåÈÜÓÚÊÊÁ¿µÄË®ÖУ¬¼ÓÈÈ¡¢½Á°è£¬´ýÈ«²¿ÈܽâºóÍ£Ö¹¼ÓÈÈ£¬Ê¹ÈÜÒºÀäÈ´ÖÁÊÒκó³éÂË¡£
 
µÃµ½´¿¶È½Ï¸ßµÄÏõËá¼Ø¾§Ìå
¢ß
 
¼ìÑé
·Ö±ðÈ¡¢Ý¡¢¢ÞµÃµ½µÄ²úÆ·£¬ÅäÖóÉÈÜÒººó·Ö±ð¼ÓÈë1µÎ1mol/lµÄHNO3ºÍ2µÎ0.1mol/lµÄAgNO3
¿É¹Û²ìµ½¢Ý¡¢¢Þ²úÆ·ÖгöÏÖµÄÏÖÏó·Ö±ðÊǢݲúÆ·ÖвúÉú°×É«³Áµí£¬¢Þ²úÆ·ÖÐÎÞÃ÷ÏÔÏÖÏó¡£
£¨1£©°ÑÉÏÊö¸÷²½ÖèÖеÄÄÚÈݲ¹³äÍêÕû£º£á£º¡¡¡¡¡¡¡¡¡¡£â£º¡¡¡¡¡¡¡¡ ¡¡
£¨2£©Ð´³ö²½Öè¢Û³ÃÈȹýÂ˵ÄÄ¿µÄ                  £¬²½Öè¢ÞµÄ²Ù×÷Ãû³Æ         £¬²½Öè¢ßÖвúÉú°×É«³ÁµíµÄÀë×Ó·½³Ìʽ                           
£¨3£©²½Öè¢ÞµÄ³éÂË×°ÖÃÈçͼËùʾ£¬ÒÇÆ÷AµÄÃû³Æ         £¬¸Ã×°ÖÃÖеĴíÎóÖ®´¦ÊÇ           £»³éÂËÍê±Ï»òÖÐ;ÐèÍ£³éÂËʱ£¬Ó¦ÏÈ               £¬È»ºó              ¡£

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

¾«Ó¢¼Ò½ÌÍø