ÌâÄ¿ÄÚÈÝ

ÑÇÂÈËáÄÆ£¨NaClO2£©ÊÇÒ»ÖÖÖØÒªµÄº¬ÂÈÏû¶¾¼Á£¬Ö÷ÒªÓÃÓÚË®µÄÏû¶¾ÒÔ¼°É°ÌÇ¡¢ÓÍÖ¬µÄƯ°×Óëɱ¾ú¡£ÒÔÏÂÊǹýÑõ»¯Çâ·¨Éú²úÑÇÂÈËáÄƵŤÒÕÁ÷³Ìͼ£º

ÒÑÖª£º¢ÙNaClO2µÄÈܽâ¶ÈËæζÈÉý¸ß¶øÔö´ó£¬Êʵ±Ìõ¼þÏ¿ɽᾧÎö³öNaClO2?3H2O¡£
¢Ú´¿ClO2Ò׷ֽⱬը£¬Ò»°ãÓÃÏ¡ÓÐÆøÌå»ò¿ÕÆøÏ¡Ê͵½10£¥ÒÔÏ°²È«¡£
¢Û160 g/L NaOHÈÜÒºÊÇÖ¸160 gNaOH¹ÌÌåÈÜÓÚË®ËùµÃÈÜÒºµÄÌå»ýΪ1L¡£
£¨1£©160 g/L NaOHÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈΪ    ¡¡¡¡¡£ÈôÒª¼ÆËã¸ÃÈÜÒºµÄÖÊÁ¿·ÖÊý£¬
»¹ÐèÒªµÄÒ»¸öÌõ¼þÊÇ¡¡¡¡¡¡¡¡¡¡¡¡               ¡¡¡¡¡¡¡¡¡¡¡¡£¨ÓÃÎÄ×Ö˵Ã÷£©¡£
£¨2£©·¢ÉúÆ÷ÖйÄÈë¿ÕÆøµÄ×÷ÓÿÉÄÜÊÇ¡¡¡¡¡¡£¨Ñ¡ÌîÐòºÅ£©¡£
a£®½«SO2Ñõ»¯³ÉSO3£¬ÔöÇ¿ËáÐÔ£» b£®Ï¡ÊÍClO2ÒÔ·ÀÖ¹±¬Õ¨£»c£®½«NaClO3Ñõ»¯³ÉClO2
£¨3£©ÎüÊÕËþÄڵķ´Ó¦µÄ»¯Ñ§·½³ÌʽΪ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡£
ÎüÊÕËþµÄζȲ»Äܳ¬¹ý20¡æ£¬ÆäÄ¿µÄÊÇ¡¡¡¡                               ¡£
£¨4£©ÔÚ¼îÐÔÈÜÒºÖÐNaClO2±È½ÏÎȶ¨£¬ËùÒÔÎüÊÕËþÖÐӦά³ÖNaOHÉÔ¹ýÁ¿£¬ÅжÏNaOHÊÇ
·ñ¹ýÁ¿µÄ¼òµ¥ÊµÑé·½·¨ÊÇ                                              ¡£
£¨5£©ÎüÊÕËþÖÐΪ·ÀÖ¹NaClO2±»»¹Ô­³ÉNaCl£¬ËùÓû¹Ô­¼ÁµÄ»¹Ô­ÐÔÓ¦ÊÊÖС£³ýH2O2Í⣬»¹¿ÉÒÔÑ¡ÔñµÄ»¹Ô­¼ÁÊÇ¡¡    ¡¡¡¡£¨Ñ¡ÌîÐòºÅ£©¡£
a£®Na2O2         b£®Na2S          c£®FeCl2      
£¨6£©´ÓÂËÒºÖеõ½NaClO2?3H2O´Ö¾§ÌåµÄʵÑé²Ù×÷ÒÀ´ÎÊÇ¡¡¡¡¡¡¡¡¡¡£¨Ñ¡ÌîÐòºÅ£©¡£
a£®ÕôÁó    b£®Õô·¢   c£®×ÆÉÕ   d£®¹ýÂË    e£®ÀäÈ´½á¾§
ÒªµÃµ½¸ü´¿µÄNaClO2?3H2O¾§Ìå±ØÐë½øÐеIJÙ×÷ÊÇ¡¡¡¡¡¡¡¡¡¡  ¡¡¡¡£¨Ìî²Ù×÷Ãû³Æ£©¡£

£¨1£©¢Ù4mol/L£¨1·Ö£¬Î´Ð´µ¥Î»²»¸ø·Ö£©£¬¸ÃÈÜÒºµÄÃܶȣ¨1·Ö£©£»
£¨2£©b£¨2·Ö£©£»
£¨3£©2NaOH+2ClO2+H2O2¡ú2NaClO2+2H2O2+O2£¨2·Ö£©£»·ÀÖ¹H2O2·Ö½â£¨1·Ö£©£»
£¨4£©Á¬Ðø²â¶¨ÎüÊÕËþÄÚÈÜÒºµÄpHÖµ£¨2·Ö£©£»£¨5£©a £¨1·Ö£©£»
£¨6£©b¡¢e¡¢d£¨2·Ö£©£»Öؽᾧ£¨1·Ö£©
ÂÔ
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
ÎÒ¹ú»¯¹¤×¨¼ÒºîµÂ°ñµÄ¡°ºîÊÏÖƼ¡±ÔøΪÊÀ½çÖƼҵ×ö³öÁËÍ»³ö¹±Ïס£ËûÒÔNaCl¡¢NH3¡¢CO2µÈΪԭÁÏÏÈÖƵÃNaHCO3£¬½ø¶øÉú²ú³ö´¿¼î¡£Óйط´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º
NH3£«CO2£«H2O    NH4HCO3£»  NH4HCO3£«NaCl    NaHCO3¡ý£«NH4Cl £»
2NaHCO3Na2CO3£«CO2¡ü£«H2O
£¨1£©Ì¼ËáÇâï§Óë±¥ºÍʳÑÎË®·´Ó¦£¬ÄÜÎö³ö̼ËáÇâÄƾ§ÌåµÄÔ­ÒòÊÇ      £¨Ìî×Öĸ±êºÅ£©£»
a£®Ì¼ËáÇâÄÆÄÑÈÜÓÚË®            
b£®Ì¼ËáÇâÄÆÊÜÈÈÒ×·Ö½â
c£®Ì¼ËáÇâÄƵÄÈܽâ¶ÈÏà¶Ô½ÏС£¬ËùÒÔÔÚÈÜÒºÖÐÊ×ÏȽᾧÎö³ö
£¨2£©Ä³Ì½¾¿»î¶¯Ð¡×é¸ù¾ÝÉÏÊöÖƼîÔ­Àí£¬½øÐÐ̼ËáÇâÄƵÄÖƱ¸ÊµÑ飬ͬѧÃÇ°´¸÷×ÔÉè¼ÆµÄ·½°¸ÊµÑé¡£
¢Ù һλͬѧ½«¶þÑõ»¯Ì¼ÆøÌåͨÈ뺬°±µÄ±¥ºÍʳÑÎË®ÖÐÖƱ¸Ì¼ËáÇâÄÆ£¬ÊµÑé×°ÖÃÈçÓÒͼËùʾ£¨Í¼Öмг֡¢¹Ì¶¨ÓõÄÒÇÆ÷δ»­³ö£©¡£
ÊԻشðÏÂÁÐÓйØÎÊÌ⣺

£¨¢ñ£©ÒÒ×°ÖÃÖеÄÊÔ¼ÁÊÇ          
£¨¢ò£©¶¡×°ÖÃÖÐÏ¡ÁòËáµÄ×÷ÓÃÊÇ            
£¨¢ó£©ÊµÑé½áÊøºó£¬·ÖÀë³öNaHCO3¾§ÌåµÄ²Ù×÷ÊÇ         £¨Ìî·ÖÀë²Ù×÷µÄÃû³Æ£©¡£
£¨IV£©ÇëÄãÔÙд³öÒ»ÖÖʵÑéÊÒÖÆÈ¡ÉÙÁ¿Ì¼ËáÇâÄƵķ½·¨£º                  ¡£
£¨12·Ö£©ÁòËá±»ÈËÃÇÓþΪ¡°»¯Ñ§¹¤ÒµÖ®Ä¸¡±£¬ÔÚ¹úÃñÉú²úÖз¢»Ó×ÅÖØÒª×÷Óã¬ÁòËáµÄ²úÁ¿ÊǺâÁ¿Ò»¸ö¹ú¼Ò»¯Ñ§¹¤ÒµË®Æ½µÄ±êÖ¾¡£¹¤ÒµÉÏÉú²úÁòËá°üÀ¨Èý²½£º
£¨1£©µÚÒ»²½£¬ÔÚ·ÐÌÚ¯ÖÐìÑÉÕ»ÆÌú¿ó£¬·´Ó¦ÈçÏ£º4FeS2+11O28SO2+2Fe2O3£¬¸Ã·´Ó¦µÄÑõ»¯²úÎïÊÇ     £¬µ±Éú³É8 mol SO2ʱתÒƵç×ÓµÄÎïÖʵÄÁ¿Îª    ¡£
£¨2£©µÚ¶þ²½£¬ÔÚ½Ó´¥ÊÒÖз¢ÉúÈçÏ·´Ó¦£º2SO2+O2  2SO3¡£Ä³¿Æ¼¼Ð¡×éµÄͬѧÔÚÒ»¸öºãκãÈݵÄÈÝÆ÷ÖÐÄ£Äâ¸Ã·´Ó¦£¬ËûÃÇ·ÖÁ½´Î½øÐÐʵÑ飬µÚÒ»´ÎÏòÈÝÆ÷ÖмÓÈë2 mol SO2¡¢1 mol O2£¬·´Ó¦´ïƽºâºó²âµÃSO2µÄת»¯ÂÊΪ¦Á1£¬µÚ¶þ´ÎÏòÈÝÆ÷ÖмÓÈë3 mol SO2¡¢1.5 mol O2, ·´Ó¦´ïƽºâºó²âµÃSO2µÄת»¯ÂÊΪ¦Á2£¬Ôò¦Á1    ¦Á2£¨Ìî¡°´óÓÚ¡±¡¢¡°µÈÓÚ¡±»ò¡°Ð¡ÓÚ¡±£©¡£
£¨3£©µÚÈý²½£¬ÔÚÎüÊÕËþÖн«SO3ת»¯³ÉÁòËá¡£ÁòËáÊÇ»¯Ñ§ÊµÑéÖеij£ÓÃÊÔ¼Á£º
¢ÙŨÁòËá¿ÉÒÔÓë¶àÖÖÎïÖÊ·¢Éú»¯Ñ§·´Ó¦£¬ÊÔд³öŨÁòËáÓëÍ­·¢Éú·´Ó¦µÄ»¯Ñ§·½³Ìʽ                                  ¡£
¢ÚÓÐNa¡¢Mg¡¢Al¡¢CuËÄÖÖ½ðÊô£¬ÈôÁ½Á½»ìºÏºóÈ¡»ìºÏÎï14 gÓë×ãÁ¿Ï¡ÁòËá·´Ó¦£¬²úÉú±ê×¼×´¿öÏÂH25.6 L£¬Ôò´Ë»ìºÍÎïµÄ×éºÏ·½Ê½×î¶àÓР     ÖÖ¡£
£¨18·Ö£©ÁòËáÊÇ»¯Ñ§¹¤Òµ×îÖØÒªµÄ²úÆ·Ö®Ò»£¬¹¤ÒµÖÆ·¨ÈçÏ¡£
£¨1£©Éú²úÁòËá×î¹ÅÀϵķ½·¨ÊÇÒÔÂÌ·¯ÎªÔ­ÁÏ£¬ÔÚÕôÁó¸ªÖÐìÑÉÕ¡£·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º2FeSO4¡¤7H2OFe2O3+SO2+SO3¡ü+14H2O¡£ÆäÖÐÈýÑõ»¯ÁòÓëË®ÕôÆøͬʱÀäÄý±ãµÃµ½ÁòËá¡£ÓÃÈçͼËùʾװÖÃÄ£ÄâÓÃÂÌ·¯ÖÆÁòËáµÄʵÑ飬²¢¼ìÑéÉú³ÉµÄÁòËáºÍ¶þÑõ»¯Áò£¨¼ÓÈÈ×°ÖÃÒÑÂÔÈ¥£©¡£ÆäÖÐbΪ¸ÉÔïµÄÊԹܡ£

¢ÙÊÔ¹ÜbÖеõ½µÄÖ÷Òª²úÎïÊÇ           £¬¼ìÑé¸Ã²úÎïµÄ·½·¨ÊÇ£¨½áºÏÀë×Ó·½³Ìʽ¼òҪ˵Ã÷£©                                                                      ¡£
¢ÚΪ¼ìÑé·´Ó¦µÄÁíÒ»ÖÖÉú³ÉÎÊÔ¹ÜcÖÐÓ¦¼ÓÈëµÄÊÔ¼ÁÊÇ            £¬¼òÊöÏàÓ¦µÄʵÑéÏÖÏó¼°½áÂÛ                                          ¡£
¢ÛËùµÃÁòËáµÄÀíÂÛŨ¶È£¨ÈÜÖʵÄÖÊÁ¿·ÖÊý£©Îª                                ¡£
£¨2£©Ä¿Ç°£¬ÎÒ¹ú²ÉÓá°½Ó´¥·¨¡±ÖÆÁòËᣬÉú²úÉ豸ÈçͼËùʾ£º

¢ÙͼÖÐÉ豸AµÄÃû³ÆÊÇ                      £¬a¡¢bÁ½´¦Ëùº¬ÆøÌåµÄ»¯Ñ§Ê½·Ö±ðΪ                               ¡¢                              ¡£
¢ÚÓйؽӴ¥·¨ÖÆÁòËáµÄÏÂÁÐ˵·¨ÖУ¬²»ÕýÈ·µÄÊÇ           ¡£
A£®¶þÑõ»¯ÁòµÄ½Ó´¥Ñõ»¯ÔںϳÉËþÖз¢Éú
B£®ÎüÊÕËþµÃµ½µÄÁòËáŨ¶ÈΪ98£¥
C£®ìÑÉÕº¬Áò48£¥µÄ»ÆÌú¿óʱ£¬ÈôFeS2ËðʧÁË2£¥£¬ÔòSËðʧ4£¥
D£®B×°ÖÃÖз´Ó¦µÄÌõ¼þ֮һΪ½Ï¸ßζÈÊÇΪÁËÌá¸ßSO2µÄת»¯ÂÊ
£¨3£©Ëæ×Å»¯Ñ§¹¤ÒµµÄ·¢Õ¹£¬¡°½Ó´¥·¨¡±È«ÃæÌæ´úÁË¡°ÂÌ·¯Èȷֽⷨ¡±£¬ÇëÄã´Ó×ÛºÏЧÒæµÄ½Ç¶ÈÖ¸³ö¡°½Ó´¥·¨¡±ÖÆÁòËáµÄÓÅÊÆ£º¢ÙÉú²ú³É±¾µÍ¡¢¢ÚÔ­ÁÏÒ׵ᢢ۠               ¡¢¢Ü                ¡¢¢Ý               ¡¢¢Þ               £¨¿É²»ÌîÂú£©¡£

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

¾«Ó¢¼Ò½ÌÍø