¹¤ÒµÉÏÒÔ»ÆÍ­¿ó(Ö÷Òª³É·ÖCuFeS2)ΪԭÁÏÖƱ¸½ðÊôÍ­£¬ÓÐÈçÏÂÁ½ÖÖ¹¤ÒÕ¡£
I.»ð·¨ÈÛÁ¶¹¤ÒÕ:½«´¦Àí¹ýµÄ»ÆÍ­¿ó¼ÓÈËʯӢ£¬ÔÙͨÈË¿ÕÆø½øÐбºÉÕ£¬¼´¿ÉÖƵôÖÍ­¡£
£¨1£©±ºÉÕµÄ×Ü·´Ó¦Ê½¿É±íʾΪ£º2CuFeS2 + 2SiO2+5O2£½2Cu+2FeSi03+4SO2¸Ã·´Ó¦µÄÑõ»¯¼ÁÊÇ_____¡£
£¨2£©ÏÂÁд¦ÀíSO2µÄ·½·¨£¬²»ºÏÀíµÄÊÇ_____

A£®¸ß¿ÕÅÅ·Å
B£®Óô¿¼îÈÜÒºÎüÊÕÖƱ¸ÑÇÁòËáÄÆ
C£®Óð±Ë®ÎüÊÕºó£¬ÔÙ¾­Ñõ»¯ÖƱ¸ÁòËáï§
D£®ÓÃBaCl2ÈÜÒºÎüÊÕÖƱ¸BaSO3
£¨3£©Â¯ÔüÖ÷Òª³É·ÖÓÐFeO ¡¢Fe2O3 ¡¢SiO2 ,Al2O3µÈ£¬ÎªµÃµ½Fe2O3¼ÓÑÎËáÈܽâºó£¬ºóÐø´¦Àí¹ý³ÌÖÐ,δÉæ¼°µ½µÄ²Ù×÷ÓÐ_____¡£
A¹ýÂË£»B¼Ó¹ýÁ¿NaOHÈÜÒº£»CÕô·¢½á¾§£»D×ÆÉÕ£»E¼ÓÑõ»¯¼Á
II.FeCl3ÈÜÒº½þÈ¡¹¤ÒÕ:ÆäÉú²úÁ÷³ÌÈçÏÂͼËùʾ

£¨4£©½þ³ö¹ý³ÌÖУ¬CuFeS2ÓëFeCl3ÈÜÒº·´Ó¦µÄÀë×Ó·½³ÌʽΪ _____________¡£
£¨5£©¸Ã¹¤ÒÕÁ÷³ÌÖУ¬¿ÉÒÔÑ­»·ÀûÓõÄÎïÖÊÊÇ_____(Ìѧʽ)¡£
£¨6£©ÈôÓÃʯīµç¼«µç½âÂËÒº£¬Ð´³öÑô¼«µÄµç¼«·´Ê½_____________¡£
£¨7£©»ÆÍ­¿óÖк¬ÉÙÁ¿Pb,µ÷½ÚC1һŨ¶È¿É¿ØÖÆÂËÒºÖÐPb2+µÄŨ¶È£¬µ±c(C1Ò»)£½2mo1¡¤L-1ʱÈÜÒºÖÐPb2+ÎïÖʵÄÁ¿Å¨¶ÈΪ__mol¡¤L-1¡£[ÒÑÖªKSP(PbCl2)£½1 x 10Ò»5]

£¨15·Ö£©£¨1£©CuFeS2¡¢O2£¨¸÷1·Ö£¬¹²2·Ö£©
£¨2£©A¡¢D£¨´ð¶Ô1¸ö¸ø1·Ö£¬´ð´í²»¸ø·Ö£¬¹²2·Ö£©
£¨3£©C£¨2·Ö£©
£¨4£©CuFeS2+4Fe3+£½Cu2++5Fe2++S£¨Ã»Åäƽ¿Û1·Ö£¬¹²3·Ö£©
£¨5£©FeCl3¡¡£¨2·Ö£©
£¨6£©Fe2+£­e£­£½Fe3+¡¡£¨2·Ö£©
£¨7£©2.5¡Á10£­6£¨2·Ö£©

½âÎöÊÔÌâ·ÖÎö£º£¨1£©¸ù¾Ý·½³Ìʽ2CuFeS2 + 2SiO2+5O2£½2Cu+2FeSiO3+4SO2¿ÉÖª£¬ÔÚ·´Ó¦ÖÐÌúÁòÔªËصĻ¯ºÏ¼Û´Ó£­2¼ÛÉý¸ßµ½+4¼Û£¬Ê§È¥µç×Ó£¬Òò´ËCuFeS2ÊÇ»¹Ô­¼Á£»ÑõÆøÖÐÑõÔªËصĻ¯ºÏ¼Û´Ó0¼Û½µµÍµ½£­2¼Û£¬Í­ÔªËصĻ¯ºÏ¼Û´Ó£«2¼Û½µµÍµ½0¼Û£¬Òò´Ë¸Ã·´Ó¦µÄÑõ»¯¼ÁÊÇCuFeS2¡¢O2¡£
£¨2£©A.SO2ÊÇ´óÆøÎÛȾÎ²»ÄÜËæÒâÅÅ·Å£¬A²»ÕýÈ·£»B.Óô¿¼îÈÜÒºÎüÊÕÖƱ¸ÑÇÁòËáÄÆ£¬¿ÉÒÔ·ÀÖ¹ÎÛȾ£¬BÕýÈ·£»C.Óð±Ë®ÎüÊÕºó£¬ÔÙ¾­Ñõ»¯ÖƱ¸ÁòËá泥¬Ò²¿ÉÒÔ·ÀÖ¹ÎÛȾ£¬CÕýÈ·£»D.SO2²»ÄÜÈÜÓÚÂÈ»¯±µÈÜÒºÖУ¬Òò´ËÓÃBaCl2ÈÜÒºÎüÊÕÖƱ¸BaSO3ÊDz»¿ÉÄܵģ¬D²»ÕýÈ·£¬´ð°¸Ñ¡AD¡£
£¨3£©Â¯ÔüÖ÷Òª³É·ÖÓÐFeO ¡¢Fe2O3 ¡¢SiO2 ,Al2O3µÈ£¬ÎªµÃµ½Fe2O3¼ÓÑÎËáÈܽâºó£¬ÐèÒª¹ýÂ˳ö¶þÑõ»¯¹è¡£È»ºó¼ÓÈëÑõ»¯¼Á½«ÈÜÒºÖеÄÑÇÌúÀë×ÓÑõ»¯Éú³ÉÌúÀë×Ó£¬ÔÙ¼ÓÈë¹ýÁ¿µÄÇâÑõ»¯ÄÆÈÜÒººóÉú³ÉÇâÑõ»¯Ìú³Áµí£¬¹ýÂË¡¢Ï´µÓ×ÆÉÕ¼´µÃµ½Ñõ»¯Ìú£¬Òò´ËºóÐø´¦Àí¹ý³ÌÖÐ,δÉæ¼°µ½µÄ²Ù×÷ÓÐÕô·¢½á¾§£¬´ð°¸Ñ¡C¡£
£¨4£©¹ýÂ˵õ½µÄÂËÖ½Öк¬ÓÐÁòµ¥ÖÊ£¬Õâ˵Ã÷CuFeS2ÓëFeCl3ÈÜÒº·¢ÉúµÄÊÇÑõ»¯»¹Ô­·´Ó¦£¬ÌúÀë×Ó½«ÁòÀë×ÓÑõ»¯£¬Òò´Ë·´Ó¦µÄÀë×Ó·½³ÌʽΪCuFeS2+4Fe3+£½Cu2++5Fe2++S¡£
£¨5£©µç½âʱ²úÉúµÄµç½âÒºÓÖ¿ÉÒÔÓë»ÆÍ­¿ó·´Ó¦£¬Òò´Ë¸Ã¹¤ÒÕÁ÷³ÌÖУ¬¿ÉÒÔÑ­»·ÀûÓõÄÎïÖÊÊÇFeCl3¡£
£¨6£©µç½â³ØÖÐÑô¼«Ê§È¥µç×Ó£¬Òò´ËÈôÓÃʯīµç¼«µç½âÂËÒº£¬ÔòÑÇÌúÀë×ÓÔÚÑô¼«Ê§È¥µç×Ó£¬ËùÒÔÑô¼«µÄµç¼«·´Ê½Fe2+£­e£­£½Fe3+¡£
£¨7£©ÒÑÖªKSP(PbCl2)£½1¡Á10Ò»5£¬Òò´Ëµ±ÈÜÒºÖÐc(C1Ò»)£½2mo1¡¤L-1ʱÈÜÒºÖÐPb2+ÎïÖʵÄÁ¿Å¨¶ÈΪ£½2.5¡Á10£­6mol¡¤L-1¡£
¿¼µã£º¿¼²éÑõ»¯»¹Ô­·´Ó¦Åжϡ¢¶þÑõ»¯ÁòβÆø´¦Àí¡¢»¯Ñ§ÊµÑé»ù±¾²Ù×÷ÒÔ¼°µç»¯Ñ§Ô­ÀíµÄÓ¦ÓÃÓë¼ÆËã

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

¶ÌÖÜÆÚÖ÷×åÔªËØA¡¢B¡¢C¡¢D¡¢EÔÚÔªËØÖÜÆÚ±íÖеÄλÖÃÈçÏÂͼËùʾ£¬ÆäÖÐAΪµØ¿ÇÖк¬Á¿×î¸ßµÄ½ðÊôÔªËØ¡£


ÇëÓû¯Ñ§ÓÃÓï»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©DÔªËØÔÚÖÜÆÚ±íÖеÄλÖ㺠              
£¨2£©A¡¢D ¡¢EÔªËؼòµ¥Àë×Ӱ뾶ÓÉ´óµ½Ð¡µÄ˳ÐòΪ_____£¾______ £¾______ (Ìî΢Á£·ûºÅ )
£¨3£©FÓëDͬÖ÷×åÇÒÏàÁÚ£¬ÆäÆø̬Ç⻯ÎïÎȶ¨ÐԵĴóС       £¾        £¨Ìî΢Á£·ûºÅ£©
£¨4£©ÓøßÄÜÉäÏßÕÕÉ京ÓÐ10µç×ÓµÄDÔªËØÇ⻯Îï·Ö×Óʱ£¬Ò»¸ö·Ö×ÓÄÜÊÍ·ÅÒ»¸öµç×Ó£¬Í¬Ê±²úÉúÒ»ÖÖ¾ßÓнϸßÑõ»¯ÐÔµÄÑôÀë×Ó£¬ÊÔд³ö¸ÃÑôÀë×ӵĵç×Óʽ             £¬¸ÃÑôÀë×ÓÖдæÔڵĻ¯Ñ§¼üÓР                  ¡£
£¨5£©CÔªËصļòµ¥Ç⻯ÎïÓëEÔªËصÄ×î¸ß¼ÛÑõ»¯ÎïµÄË®»¯Îï·´Ó¦£¬Éú³É»¯ºÏÎïK£¬ÔòKµÄË®ÈÜÒºÏÔ_____ÐÔ£¨Ìî¡°ËáÐÔ¡±¡¢¡°¼îÐÔ¡±»ò¡°ÖÐÐÔ¡±£©£¬ÓÃÀë×Ó·½³Ìʽ±íʾÆäÔ­Òò                £®
£¨6£©»¯ºÏÎïACµ¼ÈÈÐԺã¬ÈÈÅòÕÍϵÊýС£¬ÊÇÁ¼ºÃµÄÄÍÈȳå»÷²ÄÁÏ¡£ÆäÖÐÖƱ¸ACµÄÒ»ÖÖ·½·¨Îª£ºÓÃAÔªËصÄÑõ»¯Îï¡¢½¹Ì¿ºÍCµÄµ¥ÖÊÔÚ1600 ~ 1750¡æÉú³ÉAC£¬Ã¿Éú³É1 mol AC£¬ÏûºÄ18 g̼£¬ÎüÊÕb kJµÄÈÈÁ¿¡££¨ÈÈÁ¿Êý¾ÝΪ25¡æ¡¢101£®3 kPaÌõ¼þÏ£©Ð´³ö¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ                             ¡£
£¨7£©ÔÚFeºÍCuµÄ»ìºÏÎïÖмÓÈëÒ»¶¨Á¿CµÄ×î¸ß¼ÛÑõ»¯ÎïµÄË®»¯ÎïÏ¡ÈÜÒº£¬³ä·Ö·´Ó¦ºó£¬Ê£Óà½ðÊôm1 g£»ÔÙÏòÆäÖмÓÈëÏ¡ÁòËᣬ³ä·Ö·´Ó¦ºó£¬½ðÊôÊ£Óà m2 g ¡£ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ          ¡£
a£®¼ÓÈëÏ¡ÁòËáÇ°ºÍ¼ÓÈëÏ¡ÁòËáºóµÄÈÜÒºÖп϶¨¶¼ÓÐCu2+ 
b£®¼ÓÈëÏ¡ÁòËáÇ°ºÍ¼ÓÈëÏ¡ÁòËáºóµÄÈÜÒºÖп϶¨¶¼ÓÐFe2+ 
c£®m1Ò»¶¨´óÓÚm2
d£®Ê£Óà¹ÌÌåm1 g ÖÐÒ»¶¨Óе¥ÖÊÍ­£¬Ê£Óà¹ÌÌåm2 g ÖÐÒ»¶¨Ã»Óе¥ÖÊÍ­

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

¹¤ÒµÉú²ú´¿¼îµÄ¹¤ÒÕÁ÷³ÌʾÒâͼÈçÏ£º

Íê³ÉÏÂÁÐÌî¿Õ£º
(1)´ÖÑÎË®¼ÓÈë³Áµí¼ÁA¡¢B³ýÔÓÖÊ(³Áµí¼ÁAÀ´Ô´ÓÚʯ»ÒÒ¤³§)£¬Ð´³öA¡¢BµÄ»¯Ñ§Ê½¡£
A______________¡¡B____________¡£
(2)ʵÑéÊÒÌá´¿´ÖÑεÄʵÑé²Ù×÷ÒÀ´ÎΪ£º
È¡Ñù¡¢__________¡¢³Áµí¡¢__________¡¢__________¡¢ÀäÈ´½á¾§¡¢__________¡¢ºæ¸É¡£
(3)¹¤ÒµÉú²ú´¿¼î¹¤ÒÕÁ÷³ÌÖУ¬Ì¼Ëữʱ²úÉúµÄÏÖÏóÊÇ__________________¡£
̼ËữʱûÓÐÎö³ö̼ËáÄƾ§Ì壬ÆäÔ­ÒòÊÇ______________________£®
(4)̼Ëữºó¹ýÂË£¬ÂËÒºD×îÖ÷ÒªµÄ³É·ÖÊÇ______________(Ìîд»¯Ñ§Ê½)£¬¼ìÑéÕâÒ»³É·ÖµÄÒõÀë×ӵľßÌå·½·¨ÊÇ£º______________________¡£
(5)°±¼î·¨Á÷³ÌÖа±ÊÇÑ­»·Ê¹Óõģ¬Îª´Ë£¬ÂËÒºD¼ÓÈëʯ»ÒË®²úÉú°±¡£¼Óʯ»ÒË®ºóËù·¢ÉúµÄ·´Ó¦µÄÀë×Ó·½³ÌʽΪ£º____________________________¡£
(6)²úÆ·´¿¼îÖк¬ÓÐ̼ËáÇâÄÆ¡£Èç¹ûÓüÓÈÈ·Ö½âµÄ·½·¨²â¶¨´¿¼îÖÐ̼ËáÇâÄƵÄÖÊÁ¿·ÖÊý£¬´¿¼îÖÐ̼ËáÇâÄƵÄÖÊÁ¿·ÖÊý¿É±íʾΪ£º__________(×¢Ã÷ÄãµÄ±í´ïʽÖÐËùÓõÄÓйطûºÅµÄº¬Òå)¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

Í­ÊÇÖØÒªµÄ½ðÊô²ÄÁÏ¡£
(1)¹¤ÒµÉÏ¿ÉÓÃCu2SºÍO2·´Ó¦ÖÆÈ¡´ÖÍ­£¬¸Ã·´Ó¦ÖÐÑõ»¯¼ÁΪ________¡£µç½â´ÖÍ­ÖÆÈ¡¾«Í­£¬µç½âʱ£¬Ñô¼«²ÄÁÏÊÇ________£¬µç½âÒºÖбØÐ뺬ÓеÄÑôÀë×ÓÊÇ________¡£
(2)ÔÚ100 mL 18 mol¡¤L£­1ŨÁòËáÖмÓÈë¹ýÁ¿µÄͭƬ£¬¼ÓÈÈʹ֮³ä·Ö·´Ó¦£¬·´Ó¦Öб»»¹Ô­µÄH2SO4Ϊ________mol¡£
(3)µç×Ó¹¤ÒµÔøÓÃÖÊÁ¿·ÖÊýΪ30%µÄFeCl3ÈÜÒº¸¯Ê´·óÓÐÍ­²­µÄ¾øÔµ°åÖÆÓ¡Ë¢µç·°å£¬ÎªÁË´ÓʹÓùýµÄ·Ï¸¯Ê´ÒºÖлØÊÕÍ­£¬²¢ÖØеõ½FeCl3ÈÜÒº£¬Éè¼ÆÈçÏÂʵÑéÁ÷³Ì¡£

ÉÏÊöÁ÷³ÌÖУ¬Ëù¼ÓÊÔ¼ÁµÄ»¯Ñ§Ê½Îª£ºX________£¬Y________£¬Z________£»µÚ¢Þ²½·´Ó¦µÄÀë×Ó·½³ÌʽΪ
___________________________________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

Ìú¼°ÌúµÄ»¯ºÏÎïÓ¦Óù㷺£¬ÈçFeCl3¿ÉÓÃ×÷´ß»¯¼Á¡¢Ó¡Ë¢µç·ͭ°å¸¯Ê´¼ÁºÍÍâÉËֹѪ¼ÁµÈ¡£
£¨1£©Ð´³öFeCl3ÈÜÒº¸¯Ê´Ó¡Ë¢µç·ͭ°åµÄÀë×Ó·½³Ìʽ                  ¡£

 
£¨2£©Èô½«£¨1£©Öеķ´Ó¦Éè¼Æ³ÉÔ­µç³Ø£¬Çë»­³öÔ­µç³ØµÄ×°ÖÃͼ£¬±ê³öÕý¡¢¸º¼«£¬²¢Ð´³öµç¼«·´Ó¦Ê½¡£
Õý¼«·´Ó¦                          ¸º¼«·´Ó¦                      ¡£
£¨3£©¸¯Ê´Í­°åºóµÄ»ìºÏÈÜÒºÖУ¬ÈôCu2£«¡¢Fe3£«ºÍFe2£«µÄŨ¶È¾ùΪ
0.10mol¡¤L£­1£¬Çë²ÎÕÕϱí¸ø³öµÄÊý¾ÝºÍÒ©Æ·£¬¼òÊö³ýÈ¥CuCl2ÈÜÒºÖÐFe3£«ºÍFe2£«µÄʵÑé²½Öè             ¡£
 
ÇâÑõ»¯Î↑ʼ³ÁµíʱµÄpH
ÇâÑõ»¯Îï³ÁµíÍêȫʱµÄpH
Fe3£«
Fe2£«
Cu2£«
1.9
7.0
4.7
3.2
9.0
6.7
ÌṩµÄÒ©Æ·£ºCl2 Å¨H2SO4  NaOHÈÜÒº  CuO  Cu
 
£¨4£©Ä³¿ÆÑÐÈËÔ±·¢ÏÖÁÓÖʲ»Ðâ¸ÖÔÚËáÖи¯Ê´»ºÂý£¬µ«ÔÚijЩÑÎÈÜÒºÖи¯Ê´ÏÖÏóÃ÷ÏÔ¡£Çë´ÓÉϱíÌṩµÄÒ©Æ·ÖÐÑ¡ÔñÁ½ÖÖ£¨Ë®¿ÉÈÎÑ¡£©£¬Éè¼Æ×î¼ÑʵÑ飬ÑéÖ¤ÁÓÖʲ»Ðâ¸ÖÒ×±»¸¯Ê´¡£
Óйط´Ó¦µÄ»¯Ñ§·½³Ìʽ                                                   
ÁÓÖʲ»Ðâ¸Ö¸¯Ê´µÄʵÑéÏÖÏó                                               

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

I  ¡°Ì×¹ÜʵÑ顱Êǽ«Ò»Ö§½ÏСµÄ²£Á§ÒÇÆ÷×°ÈëÁíÍâÒ»¸ö²£Á§ÒÇÆ÷ÖУ¬¾­×é×°À´Íê³ÉÔ­À´ÐèÒª¸ü¶àÒÇÆ÷½øÐеÄʵÑé¡£ÒòÆä¾ßÓÐÐí¶àÓŵ㣬±»¹ã·ºÓ¦ÓÃÓÚ»¯Ñ§ÊµÑéÖУ¬ÈçͼʵÑéΪ¡°Ì×¹ÜʵÑ顱£¬Ð¡ÊÔ¹ÜÄÚÈûÓÐÕ´ÓÐÎÞË®ÁòËáÍ­·ÛÄ©µÄÃÞ»¨Çò¡£Çë¹Û²ìʵÑé×°Ö㬷ÖÎöʵÑéÔ­Àí£¬»Ø´ðÏÂÁÐÎÊÌ⣺
   
£¨1£©¸ÃʵÑéµÄÄ¿µÄÊÇ_____________________
£¨2£©ÊµÑ鿪ʼǰ΢ÈÈÊԹܣ¬ËµÃ÷×°Öò»Â©ÆøµÄÏÖÏóÊÇ                               
£¨3£©Ò»¶Îʱ¼äºó½áÊøʵÑ飬´ý×°ÖÃÀäÈ´£¬È¡³öСÊÔ¹ÜÖйÌÌåÈÜÓÚË®£¬È»ºóµÎ¼Ó1mol/LÑÎËᣬ²úÉúCO2µÄÁ¿ÓëÑÎËáµÄÁ¿µÄ¹ØϵÈçͼËùʾ¡£ÆäÖкÏÀíµÄÊÇ________________

IIȼÁϵç³ØÊÇÒ»ÖÖÁ¬ÐøµÄ½«È¼ÁϺÍÑõ»¯¼ÁµÄ»¯Ñ§ÄÜÖ±½Óת»¯ÎªµçÄܵĻ¯Ñ§µç³Ø¡£ÇâÆø¡¢Ìþ¡¢ë¡¢¼×´¼µÈÒºÌå»òÆøÌ壬¾ù¿ÉÒÔ×÷ȼÁϵç³ØµÄȼÁÏ¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÒÔ¼×ÍéºÍÑõÆøΪԭÁÏ£¬ÇâÑõ»¯ÄÆÈÜҺΪµç½âÖÊÈÜÒº¹¹³Éµç³Ø¡£Ð´³öÆäÕý¼«·´Ó¦Ê½   
£¨2£©ÒÔÉÏÊöµç³ØΪµçÔ´£¬Ê¯Ä«Îªµç¼«µç½â1L0. 1mol/LµÄÂÈ»¯¼ØÈÜÒº¡£»Ø´ðÏÂÁÐÎÊÌ⣺
д³öµç½â×Ü·´Ó¦µÄÀë×Ó·½³Ìʽ                                     

£¨3£©ÊÒÎÂʱ£¬µç½âÒ»¶Îʱ¼äºó£¬È¡25mLÉÏÊöµç½âºóÈÜÒº£¬µÎ¼Ó0.2mol/L´×Ëᣬ¼ÓÈë´×ËáµÄÌå»ýÓëÈÜÒºµÄpHµÄ¹ØϵÈçͼËùʾ£¨²»¿¼ÂÇÄÜÁ¿ËðʧºÍÆøÌåÈÜÓÚË®£¬ÈÜÒºÌå»ý±ä»¯ºöÂÔ²»¼Æ£©
¢Ù¼ÆËãÏûºÄ±ê×¼×´¿öϼ×Íé        mL
¢ÚÈôͼÖеÄBµãpH=7£¬ÔòËá¼îÇ¡ºÃÍêÈ«·´Ó¦µÄµãÔÚ       Çø¼ä£¨Ìî¡°AB¡±¡¢¡°BC¡±»ò¡°CD¡±£©
¢ÛABÇø¼äÈÜÒºÖи÷Àë×ÓŨ¶È´óС¹ØϵÖпÉÄÜÕýÈ·µÄÊÇ                 
A£® c(K+)£¾c(OH£­)£¾c (CH3COO£­) £¾c(H+)   
B£® c(K+)£¾c(CH3COO£­)£¾c(OH£­) £¾c(H+)
C£® c(K+)£¾c(CH3COO£­)=c(OH£­) £¾c(H+)

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

¶þ¼×ÃÑ(CH3OCH3)ÊÇÎÞÉ«ÆøÌ壬¿É×÷ΪһÖÖÐÂÐÍÄÜÔ´¡£ÓɺϳÉÆø(×é³ÉΪH2¡¢COºÍÉÙÁ¿µÄCO2)Ö±½ÓÖƱ¸¶þ¼×ÃÑ£¬ÆäÖеÄÖ÷Òª¹ý³Ì°üÀ¨ÒÔÏÂËĸö·´Ó¦£º
¼×´¼ºÏ³É·´Ó¦£º
(¢¡)CO(g)£«2H2(g)=CH3OH(g)    ¦¤H1£½£­90.1 kJ¡¤mol£­1
(¢¢)CO2(g)£«3H2(g)=CH3OH(g)£«H2O(g)    ¦¤H2£½£­49.0 kJ¡¤mol£­1
ˮúÆø±ä»»·´Ó¦£º
(¢£)CO(g)£«H2O(g)=CO2(g)£«H2(g)   ¦¤H3£½£­41.1 kJ¡¤mol£­1
¶þ¼×ÃѺϳɷ´Ó¦£º
(¢¤)2CH3OH(g)=CH3OCH3(g)£«H2O(g)    ¦¤H4£½£­24.5 kJ¡¤mol£­1
»Ø´ðÏÂÁÐÎÊÌ⣺
(1)Al2O3ÊǺϳÉÆøÖ±½ÓÖƱ¸¶þ¼×ÃÑ·´Ó¦´ß»¯¼ÁµÄÖ÷Òª³É·ÖÖ®Ò»¡£¹¤ÒµÉÏ´ÓÂÁÍÁ¿óÖƱ¸½Ï¸ß´¿¶ÈAl2O3µÄÖ÷Òª¹¤ÒÕÁ÷³ÌÊÇ____________________________________________(ÒÔ»¯Ñ§·½³Ìʽ±íʾ)¡£
(2)·ÖÎö¶þ¼×ÃѺϳɷ´Ó¦(¢¤)¶ÔÓÚCOת»¯ÂʵÄÓ°Ïì
________________________________________________________________________¡£
(3)ÓÐÑо¿ÕßÔÚ´ß»¯¼Á(º¬Cu­Zn­Al­OºÍAl2O3)¡¢Ñ¹Ç¿Îª5.0 MPaµÄÌõ¼þÏ£¬ÓÉH2ºÍCOÖ±½ÓÖƱ¸¶þ¼×ÃÑ£¬½á¹ûÈçͼËùʾ¡£ÆäÖÐCOת»¯ÂÊËæζÈÉý¸ß¶ø½µµÍµÄÔ­ÒòÊÇ_________________________________________________¡£

(4)¶þ¼×ÃÑÖ±½ÓȼÁϵç³Ø¾ßÓÐÆô¶¯¿ì¡¢Ð§ÂʸߵÈÓŵ㣬ÆäÄÜÁ¿ÃܶȸßÓÚ¼×´¼Ö±½ÓȼÁϵç³Ø(5.93 kW¡¤h¡¤kg£­1)¡£Èôµç½âÖÊΪËáÐÔ£¬¶þ¼×ÃÑÖ±½ÓȼÁϵç³ØµÄ¸º¼«·´Ó¦Îª__________
_____________________£¬Ò»¸ö¶þ¼×ÃÑ·Ö×Ó¾­¹ýµç»¯Ñ§Ñõ»¯£¬¿ÉÒÔ²úÉú________________¸öµç×ӵĵçÁ¿£»¸Ãµç³ØµÄÀíÂÛÊä³öµçѹΪ1.20 V£¬ÄÜÁ¿ÃܶÈE£½_______________(ÁÐʽ¼ÆËã¡£ÄÜÁ¿Ãܶȣ½µç³ØÊä³öµçÄÜ/ȼÁÏÖÊÁ¿£¬1 kW¡¤h£½3.6¡Á106 J)¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ij»¯Ñ§ÐËȤС×é°´ÕÕÏÂÁÐÁ÷³Ì½øÐС°ÓÉþÂÁºÏ½ðÖƱ¸ÁòËáÂÁ¾§Ì塱µÄʵÑé¡£

£¨1£©Ã¾ÂÁºÏ½ðÖмÓNaOHÈÜÒºµÄÀë×Ó·´Ó¦·½³ÌʽΪ       £¬»­³ö½ðÊôXµÄÔ­×ӽṹʾÒâͼ       £¬¹ÌÌåAµÄ»¯Ñ§Ê½Îª       £»
£¨2£©Ð´³öÁòËáÂÁÔÚË®ÖеĵçÀë·½³Ìʽ        £¬²Ù×÷¢Ú°üº¬µÄ²½ÖèÓÐÕô·¢Å¨Ëõ¡¢       ¡¢¹ýÂË¡¢¸ÉÔï¡£
£¨3£©¸ÃÐËȤС×éΪ²â¶¨Ã¾ÂÁºÏ½ðÖи÷×é³ÉµÄÖÊÁ¿·ÖÊý£¬Éè¼ÆÈçͼװÖã¬ÔòÐèÒª²â¶¨µÄÊý¾ÝÓР          ¡¡¡¢            ¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸