£¨9·Ö£©Ä³Ñо¿ÐÔѧϰС×éÔÚÑо¿¶þÑõ»¯ÁòƯ°××÷ÓÃʱ£¬´Ó¡°ÂÈÆøµÄƯ°××÷ÓÃʵ¼ÊÉÏÊÇÂÈÆøÓëË®·´Ó¦Éú³ÉµÄ´ÎÂÈËáµÄƯ°××÷Óᱵõ½Æô·¢¡£ÎªÁË̽¾¿¶þÑõ»¯ÁòµÄƯ°××÷Óõ½µ×ÊǶþÑõ»¯Áò±¾Éí»¹ÊǶþÑõ»¯ÁòÓëË®×÷ÓõIJúÎ¸ÃС×éÉè¼ÆÁËÈçÏÂʵÑé¡£Çë»Ø´ðÏà¹ØÎÊÌâ¡£
£¨1£©ÎªÁË̽¾¿SO2ÄÜ·ñʹƷºìÍÊÉ«£¬¸ÃͬѧѡÔñÁËÕýÈ·µÄÒ©Æ·ºó£¬Éè¼ÆÁËÈçÏÂͼËùʾʵÑé×°Öã¬ÇëÖ¸³öʵÑé×°ÖÃͼÉè¼ÆÖеIJ»ºÏÀíÖ®´¦¡£

¢Ù                                                 £»  
¢Ú                                                ¡£
£¨2£©¸ÃͬѧѡÔñÁËÕýÈ·×°Öúó£¬ÊµÑéÖпØÖƶþÑõ»¯ÁòÒÔ´óԼÿÃë3¸öÆøÅݵÄËÙ¶Èͨ¹ýÆ·ºìµÄ¾Æ¾«ÈÜҺʱ£¬¾­¹ýһСʱºó£¬Æ·ºìÈÔ²»ÍÊÉ«¡£Îª´Ë£¬ÄãÈÏΪʹƷºìµÄË®ÈÜÒºÍÊÉ«µÄ΢Á£¿ÉÄÜÊÇ     ¡£
£¨3£©¸Ãͬѧ½øÒ»²½ÊµÑéÈçÏ£ºÈ¡µÈÁ¿ÏàͬŨ¶ÈµÄÆ·ºìË®ÈÜÒºÓÚÁ½Ö§ÊÔ¹ÜÖУ¬ÔÙ·Ö±ð¼ÓÈëÉÙÁ¿ÑÇÁòËáÄƹÌÌåºÍÑÇÁòËáÇâÄƹÌÌ壬Á½Ö§ÊÔ¹ÜÖеÄÆ·ºì¶¼ÍÊÉ«£¬ËûµÃ³ö½áÂÛ£ºÊ¹Æ·ºìÍÊÉ«µÄ΢Á£¿Ï¶¨ÊÇHSO3-»òSO32-¡£ÄãÈÏΪËûµÄ½áÂÛÊÇ·ñÕýÈ·     £¬ ÆäÀíÓÉÊÇ     ¡£
£¨4£©ÔÙÓÃÒÔÏÂ×°ÖÃ̽¾¿SO2µÄijЩ»¯Ñ§ÐÔÖÊ¡£

¢Ù×°ÖÃÒÒµÄ×÷ÓÃÊÇ     ¡£
¢ÚÈôXΪNa2SÈÜÒº£¬¹Û²ìµ½ÈÜÒºÖгöÏÖµ­»ÆÉ«»ë×Ç£¬ËµÃ÷SO2¾ßÓР    ¡£
a£®Ñõ»¯ÐÔ      b£®»¹Ô­ÐÔ      
c£®Æ¯°×ÐÔ      d.²»Îȶ¨ÐÔ
¢ÛÈôÊÔ¼ÁXΪCa(ClO)2ÈÜÒº£¬¿É¹Û²ìµ½°×É«³ÁµíÉú³É£¬Íê³É¸Ã¹ý³ÌµÄÀë×Ó·½³Ìʽ£º
Ca2£«+ClO£­+SO2+H2O£½     ¡ý+Cl£­+SO42£­+    ¡£

£¨1£©¢ÙȱÉÙ¶þÑõ»¯ÁòµÄ¸ÉÔï×°Öã¨1·Ö£©£» ¢Ú²»ÄÜʹÓ󤾱©¶·£¨1·Ö£©
£¨2£© H2SO3¡¢HSO3-¡¢SO32£­£¨1·Ö£©
£¨3£©²»ÕýÈ·£¨1·Ö£© ÒòΪÑÇÁòËá¸ùÀë×ÓºÍÑÇÁòËáÇâ¸ùÀë×Ó¶¼»áË®½âÉú³ÉÑÇÁòËᣨ1·Ö£©¡£
£¨4£©¢ÙβÆø´¦Àí£»£¨1·Ö£©¢Úa£»£¨1·Ö£©
¢ÛCa2++2ClO-+2SO2+2H2O=CaSO4¡ý+2Cl-+SO42-+4H+£®£¨2·Ö£©

½âÎöÊÔÌâ·ÖÎö£º£¨1£©¸ÃͬѧµÄʵÑéÄ¿µÄÊÇΪÁË̽¾¿¶þÑõ»¯ÁòµÄƯ°××÷Óõ½µ×ÊǶþÑõ»¯Áò±¾Éí»¹ÊǶþÑõ»¯ÁòÓëË®×÷ÓõIJúÎËùÒÔ¶þÑõ»¯Áò±ØÐëÊǸÉÔïµÄ£¬È±ÉÙ¸ÉÔï×°Öã»Æä´Î²»ÄÜÓ󤾱©¶·Ê¢·ÅÁòËᣬӦ¸ÃÓ÷ÖҺ©¶·£»
£¨2£©¶þÑõ»¯Áòͨ¹ýÆ·ºìµÄ¾Æ¾«ÈÜÒº£¬ÈÜÒº²»ÍÊÉ«£¬¶øͨ¹ýÆ·ºìµÄË®ÈÜÒºÍÊÉ«£¬¶þÑõ»¯ÁòͨÈë¾Æ¾«ÈÜÒºÓë¶þÑõ»¯ÁòµÄË®ÈÜÒºÖв»Í¬µÄÁ£×ÓÖ÷ÒªÓÐH2SO3¡¢HSO3-¡¢SO32£­£¬ÕâЩÁ£×Ó¶¼¿ÉÄÜÊÇʹƷºìÍÊÉ«µÄÁ£×Ó£»
£¨3£©ÑÇÁòËáÄƹÌÌåºÍÑÇÁòËáÇâÄƹÌÌåÈÜÓÚË®ºó£¬Ë®½â¶¼µÃµ½H2SO3£¬ËùÒÔ²»ÄÜÅжÏÊÇʹƷºìÍÊÉ«µÄ΢Á£¿Ï¶¨ÊÇHSO3-»òSO32-¡£¸ÃͬѧµÄÅжÏÊDz»ÕýÈ·µÄ£»
£¨4£©¢Ù×°ÖÃÒÒÓÃÓÚÎüÊÕÊ£ÓàµÄ¶þÑõ»¯Áò£¬·ÀÖ¹ÎÛȾ¿ÕÆø£»
¢ÚNa2SÓë¶þÑõ»¯Áò·´Ó¦Éú³ÉSµ¥ÖÊ£¬¶þÑõ»¯ÁòÖеÄSÔªËصĻ¯ºÏ¼Û½µµÍ£¬ËµÃ÷¶þÑõ»¯Áò¾ßÓÐÑõ»¯ÐÔ£¬´ð°¸Ñ¡a£»
¢ÛCa(ClO)2ÈÜÒºÓë¶þÑõ»¯Áò·¢ÉúÑõ»¯»¹Ô­·´Ó¦£¬Ê£Óà¸Ã·½³ÌʽÖеijÁµíÊÇCaSO4£¬¸ù¾ÝÔªËØÊغ㣬ÅжÏÁíÒ»ÎïÖÊÊÇH+£¬SO2ÖÐSÔªËØ»¯ºÏ¼ÛÉý¸ß2¼Û£¬Ca(ClO)2ÖÐClÔªËØÕûÌå½µµÍ2¼Û£¬ËùÒÔ·½³ÌʽÖÐSO2ÓëCa(ClO)2µÄϵÊý¾ùÊÇ1£¬¸ù¾ÝµçºÉÊغãºÍ¹Û²ì·¨ÅäƽÆäËûÎïÖʵÄϵÊý£¬Àë×Ó·½³ÌʽΪCa2++2ClO-+2SO2+2H2O=CaSO4¡ý+2Cl-+SO42-+4H+£®
¿¼µã£º¿¼²é¶þÑõ»¯ÁòµÄÖÆȡʵÑé¡¢ÐÔÖʵÄÑé֤ʵÑ飬¶Ô·´Ó¦µÄ·ÖÎö£¬Àë×Ó·½³ÌʽµÄÅäƽ

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÀûÓÃÏÂÁи÷×éÖеÄÎïÖÊÖƱ¸²¢ÊÕ¼¯ÉÙÁ¿ÏàÓ¦µÄÆøÌ壬ÄܲÉÓÃÈçͼװÖõÄÊÇ

¢ÙŨ°±Ë®ºÍ¹ÌÌåNaOHÖÆNH3  ¢Ú´óÀíʯºÍÏ¡ÑÎËáÖÆCO2
¢Û¹ýÑõ»¯ÇâÈÜÒººÍ¶þÑõ»¯ÃÌÖÆO2¢ÜÏ¡ÏõËáºÍͭƬÖÆNO
¢ÝŨÑÎËáºÍ¶þÑõ»¯ÃÌÖÆCl2      ¢ÞµçʯºÍË®ÖÆC2H2

20070404

 
¢ßпÁ£ºÍÏ¡ÁòËáÖÆH2           ¢àÒÒ´¼ºÍŨÁòËáÖÆC2H4

A£®¢Ú¢Û      B£®¢Ù¢Þ¢ß      C£®¢Ú¢Ý¢à      D£®¢Ù¢Ü¢Þ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ÔÚŨCaCl2ÈÜÒºÖÐͨÈëNH3ºÍCO2£¬¿ÉÒÔÖƵÃÄÉÃ×¼¶Ì¼Ëá¸Æ£¨Á£×ÓÖ±¾¶ÔÚ1~10nmÖ®¼ä£©£®ÏÂͼËùʾA~EΪʵÑéÊÒ³£¼ûµÄÒÇÆ÷×°Ö㨲¿·Ö¹Ì¶¨¼Ð³Ö×°ÖÃÂÔÈ¥£©£¬Çë¸ù¾ÝÒªÇó»Ø´ðÎÊÌ⣮

£¨1£©ÊµÑéÊÒÖƱ¸NH3µÄ·´Ó¦·½³Ìʽ£º_______________________________£»
£¨2£©ÊµÑéÊÒÖÆÈ¡¡¢ÊÕ¼¯¸ÉÔïµÄNH3£¬ÐèÑ¡ÓÃÉÏÊöÒÇÆ÷×°ÖõĽӿÚÁ¬½Ó˳ÐòÊÇ£¨Ñ¡Ìî×Öĸ£©£ºa½Ó    £¬    ½Ó    £¬    ½Óh£»
£¨3£©ÏòŨCaCl2ÈÜÒºÖÐͨÈëNH3ºÍCO2ÆøÌåÖÆÄÉÃ×¼¶Ì¼Ëá¸Æʱ£¬Ó¦ÏÈͨÈëµÄÆøÌåÊÇ           £¬ÊÔд³öÖÆÄÉÃ×¼¶Ì¼Ëá¸ÆµÄÀë×Ó·½³Ìʽ                               £»
£¨4£©ÔÚŨCaCl2ÈÜÒººÍNH3ÓÃÁ¿ÕýÈ·µÄÇé¿öÏ£¬CO2²»×ã»ò¹ýÁ¿¶¼»áµ¼ÖÂÄÉÃ×¼¶Ì¼Ëá¸Æ²úÁ¿Ï½µ£¬ÈôCO2¹ýÁ¿ÈÜÒºÖдóÁ¿´æÔÚµÄÀë×ÓÓУ¨²»¿¼ÂÇÈõµç½âÖʵĵçÀëºÍÑÎÀàË®½â²úÉúµÄÉÙÁ¿Àë×Ó£©________________,
£¨5£©È¡·´Ó¦ºóÈ¥³ýÁËCaCO3µÄÈÜÒº·Ö±ð×öÒÔÏÂʵÑ飬ÏÂÁÐʵÑéÅжϺÏÀíµÄÊÇ£º_________£®

A£®µÎ¼ÓÉÙÁ¿Na2CO3ÈÜÒº£¬ÈôÓгÁµí˵Ã÷CO2Ò»¶¨²»×ã
B£®µÎ¼ÓÉÙÁ¿ÑÎËᣬÈôÓÐÆøÅÝ£¬CO2Ò»¶¨¹ýÁ¿
C£®²âÁ¿ÈÜÒºpH£¬Èô´óÓÚ7£¬CO2Ò»¶¨²»×ãÁ¿
D£®µÎ¼ÓÉÙÁ¿BaCl2ÈÜÒº£¬ÈôÎÞ³Áµí£¬CO2Ò»¶¨Ã»ÓйýÁ¿
£¨6£©ÊÔÉè¼Æ¼òµ¥µÄʵÑé·½°¸£¬ÅжÏËùµÃ̼Ëá¸ÆÑùÆ·¿ÅÁ£ÊÇ·ñΪÄÉÃ×¼¶    

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÎÊ´ðÌâ


¹¤ÒµÉÏ¿ÉÓÃʳÑκÍʯ»ÒʯΪÖ÷ÒªÔ­ÁÏ£¬¾­²»Í¬µÄ·½·¨Éú²ú´¿¼î¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Â¬²¼À¼·¼ÊÇÒÔʳÑΡ¢Ê¯»Òʯ¡¢Å¨ÁòËá¡¢½¹Ì¿ÎªÔ­ÁÏ£¬ÔÚ¸ßÎÂϽøÐÐìÑÉÕ£¬ÔÙ½þÈ¡£¬½á¾§¶øÖƵô¿¼î¡£
¢ÙʳÑκÍŨÁòËá·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ___________£»
¢ÚÁòËáÄƺͽ¹Ì¿¡¢Ê¯»Òʯ·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ_________£¨ÒÑÖª²úÎï֮һΪCaS£©£»
£¨2£©°±¼î·¨µÄ¹¤ÒÕÈçÏÂͼËùʾ£¬µÃµ½µÄ̼ËáÇâÄƾ­ìÑÉÕÉú³É´¿¼î¡£

 
 
 

¢ÙͼÖеÄÖмä²úÎïCÊÇ_______£¬D_______¡££¨Ð´»¯Ñ§Ê½£©£»
¢Ú×°ÖÃÒÒÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ_______£»

£¨3£©ÁªºÏÖƼ¶Ô°±¼î·¨µÄ¸Ä½ø£¬ÆäÓŵãÊÇ______________£»
£¨4£©ÓÐÈËÈÏΪ̼ËáÇâ¼ØÓë̼ËáÇâÄƵĻ¯Ñ§ÐÔÖÊÏàËÆ£¬¹ÊÒ²¿ÉÓð±¼î·¨ÒÔÂÈ»¯¼ØºÍʯ»ÒʯΪԭÁÏÖÆ̼Ëá¼Ø¡£Çë½áºÏÏÂͼµÄÈܽâ¶È£¨S£©Ëæζȱ仯ÇúÏߣ¬·ÖÎö˵Ã÷ÊÇ·ñ¿ÉÐУ¿__________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÎÊ´ðÌâ

£¨15·Ö£©ÓÃÁ×»ÒʯÖÆÁ׷ʵĸ±²úÆ·Áù·ú¹èËáÄÆ(Na2SiF6)¿ÉÖƱù¾§Ê¯(Na3AlF6)£¬±ù¾§Ê¯Êǵç½âÂÁµÄÖúÈÛ¼Á,¿É½µµÍÑõ»¯ÂÁµÄÈ۵㡣ÏÂͼÊǹ¤ÒµÖÆÂÁµÄÁ÷³Ìͼ£º

£¨1£©¹¤ÒµÉÏ´ÓÂÁÍÁ¿óÖƱ¸½Ï¸ß´¿¶ÈAl2O3µÄÖ÷Òª¹¤ÒÕÁ÷³ÌÐè__________¸ö»·½Ú£¬µÚÒ»²½·´Ó¦µÄ·½³Ìʽ______________________________________________________________
£¨2£©¸ÃÖƱ¸¹¤ÒÕÖÐÓÐÁ½´Î¹ýÂ˲Ù×÷£¬¹ýÂ˲Ù×÷¢ÙµÄÂËÒºÊÇ________ÈÜÒº£¬ÂËÔüÊÇ________    ¡£
£¨3£©·Ö½âÍѹèºÍºÏ³É±ù¾§Ê¯»¯Ñ§·´Ó¦·½³Ìʽ·Ö±ðΪ£º_________________¡¢____________________¡£
£¨4£©¹¤ÒÕ¹ý³ÌÖТۺܵ͢ÄÄ¿µÄÊÇ_____________________£¬Ì¼ËáÄƺͶþÑõ»¯Ì¼ÊÇ·ñ¹»Óà              ¡£
£¨5£©µç½âAl2O3ÖÆAlʱ£¬I=200kA£¬Ò»ÌìÖÆAl 1£®430 t£¬µç½âЧÂÊÊǶàÉÙ£¿         

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÊµÑéÌâ

ÒÒËáÒÒõ¥ÊÇÎÞÉ«¾ßÓÐË®¹ûÏãζµÄÒºÌ壬·ÐµãΪ77.2 ¡æ£¬ÊµÑéÊÒij´ÎÖÆÈ¡ËüÓñù´×Ëá14.3 mL¡¢95%ÒÒ´¼ 23 mL¡£»¹Óõ½Å¨ÁòËá¡¢±¥ºÍ̼ËáÄÆÒÔ¼°¼«Ò×ÓëÒÒ´¼½áºÏ³ÉÁùË®ºÏÎïµÄÂÈ»¯¸ÆÈÜÒº¡£Ö÷ҪװÖÃÈçͼËùʾ£º

ʵÑé²½Ö裺
¢ÙÏÈÏòAÖеÄÕôÁóÉÕÆ¿ÖÐ×¢ÈëÉÙÁ¿ÒÒ´¼ºÍŨÁòËáºóÒ¡ÔÈ£¬ÔÙ½«Ê£ÏµÄËùÓÐÒÒ´¼ºÍ±ù´×Ëá×¢Èë·ÖҺ©¶·Àï´ýÓá£Õâʱ·ÖҺ©¶·Àï±ù´×ËáºÍÒÒ´¼µÄÎïÖʵÄÁ¿Ö®±ÈԼΪ5¡Ã7¡£
¢Ú¼ÓÈÈÓÍÔ¡±£ÎÂÔ¼135 ¡æ¡«145¡æ
¢Û½«·ÖҺ©¶·ÖеÄÒºÌ建»ºµÎÈëÕôÁóÉÕÆ¿Àµ÷½Ú¼ÓÈëËÙÂÊʹÕô³öõ¥µÄËÙÂÊÓë½øÁÏËÙÂÊ´óÌåÏàµÈ£¬Ö±µ½¼ÓÁÏÍê³É¡£
¢Ü±£³ÖÓÍԡζÈÒ»¶Îʱ¼ä£¬ÖÁ²»ÔÙÓÐÒºÌåÁó³öºó£¬Í£Ö¹¼ÓÈÈ¡£
¢ÝÈ¡ÏÂBÖеÄ׶ÐÎÆ¿£¬½«Ò»¶¨Á¿±¥ºÍNa2CO3ÈÜÒº·ÖÅúÉÙÁ¿¶à´ÎµØ¼Óµ½Áó³öÒºÀ±ß¼Ó±ßÕñµ´£¬ÖÁÎÞÆøÅݲúÉúΪֹ¡£
¢Þ½«¢ÝµÄÒºÌå»ìºÏÎï·ÖÒº£¬Æúȥˮ²ã¡£
¢ß½«±¥ºÍCaCl2ÈÜÒº£¨ÊÊÁ¿£©¼ÓÈëµ½·ÖҺ©¶·ÖУ¬Õñµ´Ò»¶Îʱ¼äºó¾²Ö㬷ųöË®²ã£¨·ÏÒº£©¡£
¢à·ÖҺ©¶·ÀïµÃµ½µÄÊdzõ²½Ìá´¿µÄÒÒËáÒÒõ¥´ÖÆ·¡£
ÊԻشð£º
£¨1£©ÊµÑéÖÐŨÁòËáµÄÖ÷Òª×÷ÓÃÊÇ__________                       _______¡£
£¨2£©ÓùýÁ¿ÒÒ´¼µÄÖ÷ҪĿµÄÊÇ________                            _________¡£
£¨3£©Óñ¥ºÍNa2CO3ÈÜҺϴµÓ´Öõ¥µÄÄ¿µÄÊÇ________                        ___¡£
Èç¹ûÓÃNaOHÈÜÒº´úÌæNa2CO3ÈÜÒº½«ÒýÆðµÄºó¹û___________________________________¡£
£¨4£©Óñ¥ºÍCaCl2ÈÜҺϴµÓ´Öõ¥µÄÄ¿µÄÊÇ_____                           ______¡£
£¨5£©ÔÚ²½Öè¢àËùµÃµÄ´Öõ¥Àﻹº¬ÓеÄÔÓÖÊÊÇ______  ___¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÊµÑéÌâ

(12·Ö)SnSO4ÊÇÒ»ÖÖÖØÒªµÄÁòËáÑΣ¬ÔÚ¹¤ÒµÉú²úÖÐÓÐ׏㷺µÄÓ¦Óá£ÆäÖƱ¸Â·ÏßÈçÏ£º

ÒÑÖª£ºÔÚËáÐÔÌõ¼þÏ£¬ÈÜÒºÖеÄSn2£«¿É±»¿ÕÆøÖеÄÑõÆøÑõ»¯³ÉSn4£«£» SnCl2ÄÜË®½âÉú³É¼îʽÂÈ»¯ÑÇÎý[Sn(OH)Cl]¡£
£¨1£© д³öÎïÖÊAµÄÃû³Æ£º________¡£
£¨2£© SnCl2ÓÃÑÎËá¶ø²»ÓÃË®ÈܽâµÄÔ­ÒòÊÇ____________________(Óû¯Ñ§·½³Ìʽ±íʾ)¡£
£¨3£© Îý·ÛµÄ×÷ÓÃÊdzýÈ¥ËáÈÜʱ²úÉúµÄÉÙÁ¿Sn4£«£¬Çëд³ö²úÉúSn4£«µÄÀë×Ó·½³Ìʽ£º______________________________¡£
£¨4£© ·´Ó¦¢ñÉú³ÉµÄ³ÁµíΪSnO£¬Ð´³ö¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º____¡£¸Ã·´Ó¦µÄζÈÐèÒª¿ØÖÆÔÚ75 ¡æ×óÓÒµÄÔ­ÒòÊÇ____¡£
£¨5£© ʵÑéÊÒÖС°Æ¯Ï´¡±³ÁµíµÄʵÑé²Ù×÷·½·¨ÊÇ____¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÊµÑéÌâ

¹ý̼ËáÄÆ£¨Na2CO3¡¤3H2O2£©£¬ÓйÌÌåË«ÑõË®µÄË׳ƣ¬¸Ã¾§Ìå¾ßÓÐNa2CO3ºÍH2O2µÄË«ÖØÐÔÖÊ£¬±»´óÁ¿Ó¦ÓÃÓÚÏ´µÓ¡¢Ó¡È¾¡¢·ÄÖ¯¡¢ÔìÖ½¡¢Ò½Ò©ÎÀÉúµÈÁìÓòÖУ¬¹ý̼ËáÄƵÄijÉú²úÁ÷³ÌÈçÏÂͼËùʾ¡£
ÒÑÖª£º2Na2CO3£«3H2O2£½2Na2CO3¡¤3H2O ¡÷H£¼0£»»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©ÏÂÁÐÎïÖÊ¿Éʹ¹ý̼ËáÄƽϿìʧЧµÄÊÇ£¨ÌîÐòºÅ£©_________¡£

A£®FeCl3ÈÜÒº B£®H2S C£®Ï¡ÁòËá D£®NaHCO3ÈÜÒº
£¨2£©·´Ó¦¢ÙÓ¦ÏÈͨÈëµÄÆøÌåÊÇ__________¡£
£¨3£©ÔÚÉÏÊöÁ÷³ÌÖУ¬Ïò·´Ó¦Ç°µÄH2O2ÖмÓÈëÎȶ¨¼ÁµÄ×÷ÓÃÊÇ____________________¡£
£¨4£©¸ÃÉú²úÁ÷³ÌÖпÉÑ­»·Ê¹ÓõÄÎïÖÊÊÇ______________________________£¨Ìѧʽ£©¡£
£¨5£©Éú²ú¹ý̼ËáÄƵÄÁ÷³ÌÖÐÒÅ©ÁËÒ»²½£¬Ôì³ÉËùµÃ²úÆ·´¿¶ÈÆ«µÍ£¬Çë¼òÊö¸Ã²½²Ù×÷¹ý³Ì___________¡£
£¨6£©ÊµÑéÊÒÀûÓÃÏÂͼװÖÃÖƱ¸¹ý̼ËáÄÆ£¬¸Ã×°ÖÃÖкãѹµÎҺ©¶·ÖÐÖ§¹ÜµÄ×÷ÓÃÊÇ_______£¬ÀäÄý¹ÜÓ¦´Ó__________´¦½øË®¡£
£¨7£©ÓÉʵÑé²â¶¨·´Ó¦Î¶ȶԲúÎïµÄÓ°ÏìÈçÏÂ±í£º¸ù¾ÝϱíÊý¾Ý£¬ÄãÈÏΪ·´Ó¦×î¼ÑµÄζÈÑ¡ÔñµÄ·¶Î§ÊÇ_______________¡£
T/¡æ
»îÐÔÑõ°Ù·Öº¬Á¿
²úÂÊ
5¡«10
13.94
85.49
10¡«15
14.02
85.78
15¡«20
15.05
88.38
20¡«25
14.46
83.01
¡¢

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÊµÑéÌâ

¼îʽ̼ËáÍ­ÊÇÒ»ÖÖ»¯¹¤Ô­ÁÏ£¬»¯Ñ§Ê½ÓÃmCu(OH)2¡¤nCuCO3±íʾ¡£ÊµÑéÊÒÒÔ·ÏͭмΪԭÁÏÖÆÈ¡¼îʽ̼ËáÍ­µÄ²½ÖèÈçÏ£º

¢ñ.·ÏͭмÖÆÏõËáÍ­
·½°¸1£º½«Í­Ð¼ÔÚ¿ÕÆøÖгä·Ö×ÆÉÕ£¬²ÐÁô¹ÌÌåÈÜÓÚÏ¡ÏõË᣻
·½°¸2£ºÈçͼ1(¼Ð³ÖÒÇÆ÷ÒÑÊ¡ÂÔ)¡£½«Å¨ÏõËỺÂý¼Óµ½·ÏͭмÖÐ(·Ïͭм¹ýÁ¿)£¬³ä·Ö·´Ó¦ºó¹ýÂË£¬µÃµ½ÏõËáÍ­ÈÜÒº¡£
·½°¸3£º½«·½°¸2ÖÐŨÏõËá»»³ÉÏ¡ÏõËᣬÆäËû²»±ä¡£
¢ò.¼îʽ̼ËáÍ­µÄÖƱ¸
¢ÙÏò´óÊÔ¹ÜÖмÓÈë̼ËáÄÆÈÜÒººÍÏõËáÍ­ÈÜÒº
¢Úˮԡ¼ÓÈÈÖÁ70 ¡æ×óÓÒ
¢ÛÓÃ0.4 mol¡¤L£­1µÄNaOHÈÜÒºµ÷½ÚpHÖÁ8.5£¬Õñµ´¡¢¾²ÖᢹýÂË
¢ÜÓÃÈÈˮϴµÓ¡¢ºæ¸É£¬µÃµ½¼îʽ̼ËáÍ­²úÆ·
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
(1)°´·½°¸1ʵÑ飬±ØÐëÑ¡ÔñµÄÏÂÁÐÒÇÆ÷ÓÐ________(ÌîÐòºÅ)¡£

(2)ͼ2ÖÐÄÜ·Ö±ðÌæ´úͼ1ÖÐBºÍC×°ÖõÄÊÇ________(Ìî×°ÖÃÐòºÅ)¡£
(3)ÒÑÖª£ºNO£«NO2£«2NaOH===2NaNO2£«H2O£»2NO2£«2NaOH===NaNO3£«NaNO2£«H2O£¬NO²»Äܵ¥¶ÀÓëNaOHÈÜÒº·´Ó¦£¬ÊµÑé½áÊøʱ£¬ÈçºÎ²Ù×÷²ÅÄÜʹװÖÃÖеÄÓж¾ÆøÌå±»NaOHÈÜÒºÍêÈ«ÎüÊÕ£¿__________________________¡£
(4)²½Öè¢ÜÖÐÏ´µÓµÄÄ¿µÄÊÇ______________________________________¡£
(5)²½Öè¢Û¹ýÂ˺óµÄÂËÒºÖк¬ÓÐCO32¡ª£¬¼ìÑéCO32¡ªµÄ·½·¨ÊÇ_________________________________________________________¡£
(6)²â¶¨¼îʽ̼ËáÍ­×é³ÉµÄ·½·¨Ö÷ÒªÓÐÁ½ÖÖ£º
·½·¨1¡¡×ÆÉÕ·¨£ºÈ¡34.6 g´¿¾»ÎïmCu(OH)2¡¤nCuCO3£¬ÔÚÓ²ÖÊÊÔ¹ÜÀï×ÆÉÕ£¬½«ÆøÌå²úÎïÒÀ´ÎͨÈë×ãÁ¿µÄŨÁòËá¡¢×ãÁ¿µÄ¼îʯ»ÒÖУ¬ÍêÈ«ÎüÊÕºóŨÁòËá¾»Ôö1.8 g£¬¼îʯ»Ò¾»Ôö8.8 g¡£
·½·¨2¡¡»¹Ô­·¨£ºÔÚÇâÆøÖмÓÇ¿ÈÈ£¬²â·´Ó¦Ç°ºó¹ÌÌåµÄÖÊÁ¿¡£
¢ÙÀûÓÃÉÏÊöÊý¾ÝÍÆËã¼îʽ̼ËáÍ­µÄ»¯Ñ§Ê½_____________________¡£
¢ÚÅäƽ»¯Ñ§·½³Ìʽ£ºmCu(OH)2¡¤nCuCO3£«________H2________Cu£«________CO2¡ü£«________H2O

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸