ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿Ä¿Ç°ï®µç³ØµÄÓ¦ÓÃÈÕÒæ¹ã·º£¬¶ø·Ï¾É﮵ç³ØµÄ»ØÊÕÀûÓÃÊÇÊ®·ÖÖØÒªµÄ¿ÎÌâ¡£»ØÊÕÀûÓ÷ϾÉ﮵ç³ØµÄÁ÷³ÌÈçÏ£º

ÒÑÖª£º¢Ù﮵ç³Ø·ÏÁϵÄÖ÷Òª³É·ÖÊÇLiCoO2¡¢ÂÁ¡¢Ì¿ºÚ¼°ÆäËûÔÓÖÊ¡£

¢Ú¡°ÈÜÒºA¡±ÖÐÖ÷Òª½ðÊôÀë×ÓÊÇCo2+¡¢Li+£¬»¹º¬ÓÐÉÙÁ¿Fe3+¡¢Al3+¡¢Cu2+¡£

Çë»Ø´ð£º

(1)²½Öè¢ñÖÐÂÁÈܽâµÄÀë×Ó·½³ÌʽΪ_____£»

²½Öè¢óÖÐLiCoO2¹ÌÌåÈܽâµÄ»¯Ñ§·½³ÌʽΪ_______£»

(2)¹ØÓÚ²½Öè¢ò£¬ÏÂÁÐʵÑé²Ù×÷»ò˵·¨ºÏÀíµÄÊÇ_________

A£®×ÆÉÕÇ°£¬×ÆÉÕʹÓõÄÕô·¢ÃóÏ´¾»ºó²»ÐèÒª²Á¸É£¬È»ºó¼ÓÈë¹ÌÌåX½øÐÐ×ÆÉÕ

B£®×ÆÉÕʱÐèÒªÓò£Á§°ô²»¶Ï½Á°è

C£®×ÆÉÕÖÁºãÖØÊÇָǰºóÁ½´Î³ÆÁ¿ËùµÃÖÊÁ¿Ö®²î²»µÃ³¬¹ýÒ»¶¨µÄÔÊÐíÎó²î£¬Õâ¸öÔÊÐíÎó²îÒ»¶¨Îª0.01g

D£®ÔÚµç½âÈÛÈÚµÄAl2O3ÖƱ¸½ðÊôÂÁʱ£¬Í¨³£Ðè¼ÓÈë±ù¾§Ê¯(NaAlF6)ÒÔÔöÇ¿Æäµ¼µçÐÔ

(3)²½Öè¢ôÖгýÔÓʱ¿É¼ÓÈ백ˮµ÷½ÚÈÜÒºµÄpH£¬ÊµÑé±íÃ÷ÈÜÒºAÖи÷ÖÖ½ðÊôÀë×ӵijÁµíÂÊËæpHµÄ±ä»¯ÈçÏÂͼËùʾ£¬Ôò²½Öè¢ôÖпɳýÈ¥µÄÔÓÖÊÀë×ÓÊÇ_________¡£

(4)²½Öè¢õ¿É²ÉÓÃËá¼îµÎ¶¨·¨²â¶¨Ì¼Ëá﮵Ĵ¿¶È¡£µÎ¶¨Ô­ÀíÈçÏ£ºÌ¼Ëáï®Äܹ»ÓëÑÎËá·´Ó¦Éú³ÉÂÈ»¯ï®ºÍ¶þÑõ»¯Ì¼£¬ÔÚ̼Ëáï®Î´ÍêÈ«·´Ó¦Ê±£¬ÈÜÒº±£³ÖÖÐÐÔ(pH=7)¡£·´Ó¦ÍêÈ«ºó£¬Ëæ×ÅÑÎËáµÄ¼ÌÐøµÎÈ룬ÈÜÒºpHϽµ£¬ÒÔ¼×»ùºì¡ªäå¼×·ÛÂÌΪָʾ¼Á£¬ÓÃÑÎËá±ê×¼ÒºµÎ¶¨ÊÔÑù£¬ÓÃÏûºÄÑÎËá±ê×¼µÎ¶¨ÈÜÒºµÄÁ¿À´¼ÆËã̼Ëá﮵ĺ¬Á¿¡£

¢ÙÅäÖÆÔ¼0.30molL-1ÑÎËáÈÜÒº²»ÐèÒªÓõ½µÄÒÇÆ÷ÓÐ________(Ìî±àºÅ)

a.Á¿Í²b.µç×ÓÌìƽc.©¶·d.ÉÕ±­e.ÈÝÁ¿Æ¿f.²£Á§°ôg.½ºÍ·µÎ¹Ü

¢ÚÓÃ0.3000molL-1ÑÎËá±ê×¼ÈÜÒºµÎ¶¨£¬ÆäÖÐÕýÈ·²Ù×÷²½ÖèµÄ˳ÐòΪ________

a£®¼ÓÈë0.1-0.2mL¼×»ùºì¡ªäå¼×·ÛÂÌ×÷ָʾ¼Á£»

b£®Öó·ÐÈ¥³ýCO2£¬ÔÙÀäÈ´µ½ÊÒΣ»

c£®½«ÊÔÑùÖÃÓÚ250mL׶ÐÎÆ¿ÖУ¬¼ÓÈë20mLË®Èܽ⣻

d£®ÓÃÑÎËá±ê×¼ÒºµÎ¶¨ÖÁÊÔÒºÓÉÂÌÉ«±ä³É¾ÆºìÉ«£»

e£®¼ÌÐøµÎ¶¨ÖÁ¾ÆºìÉ«(µÎ¶¨Í»Ô¾ÇøÓò)¼´ÎªÖÕµã

¢Û̼Ëá﮵Ĵ¿¶È¿ÉÓÃÏÂʽ¼ÆË㣺¦Ø=£¬ÆäÖУº¦Ø¡ªÌ¼Ëáï®ÊÔÑùµÄ´¿¶È£»m¡ªÌ¼Ëáï®ÊÔÑùµÄÖÊÁ¿(g)£»c¡ªÑÎËá±ê×¼ÒºµÄŨ¶È(molL-1)V1¡ªÊÔÑùµÎ¶¨Ê±ÏûºÄÑÎËáµÄÌå»ý(mL)£»V0¡ªµÎ¶¨¿Õ°×ÈÜÒº(Ö¸²»¼ÓÊÔÑù½øÐеζ¨)ʱ±ê×¼ÑÎËáÏûºÄµÄÌå»ý(mL)£¬0.03694¡ªÓë1.00mL±ê×¼ÑÎËá(c=1.000molL-1)Ï൱µÄ̼Ëá﮵ÄÖÊÁ¿(g)¡£

µÎ¶¨¿Õ°×ÈÜÒºµÄÄ¿µÄÊÇ_________£¬ÉÏÊöµÎ¶¨²Ù×÷ÖУ¬È±ÉÙ¡°Öó·ÐÈ¥³ýCO2£¬ÔÙÀäÈ´µ½ÊÒΡ±Õâ¸ö²½Ö裬²â¶¨½á¹û½«_________(Ìî¡°Æ«´ó¡±¡¢¡°Æ«Ð¡¡±¡¢¡°ÎÞÓ°Ï족)¡£

¡¾´ð°¸¡¿2Al+2OH-+2H2O£½2+3H2¡ü»ò 2Al+2OH-+6H2O==2[Al(OH)4]- +3H2¡ü 2LiCoO2+H2O2+3H2SO4£½Li2SO4+2CoSO4+O2¡ü+4H2O B Fe3+¡¢Al3+ bc cadbe ÒÔÏû³ýʵÑéÖеÄÊÔ¼Á¡¢²Ù×÷µÈ×ÛºÏÒòËØÔì³ÉµÄµÎ¶¨Îó²î ƫС

¡¾½âÎö¡¿

﮵ç³Ø·ÏÁÏÖк¬ÓÐLiCoO2¡¢ÂÁ¡¢Ì¿ºÚ£¬½«·ÏÁÏÏÈÓüîÒº½þÅÝ£¬½«Al³ä·ÖÈܽ⣬¹ýÂ˺óµÃµ½µÄÂËÒºÖк¬ÓÐÆ«ÂÁËáÄÆ£¬ÔÚ½þ³öÒºÖмÓÈëÏ¡ÁòËáÉú³ÉÇâÑõ»¯ÂÁ£¬×ÆÉÕ¡¢µç½â¿ÉµÃµ½ÂÁ£¬ÂËÔüÖк¬ÓÐLiCoO2£¬½«ÂËÔüÓÃË«ÑõË®¡¢ÁòËá´¦ÀíºóÉú³ÉLi2SO4¡¢CoSO4£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪ£º2LiCoO2+H2O2+3H2SO4=Li2SO4+2CoSO4+O2¡ü+4H2O£¬Ìâ¸øÐÅÏ¢¿ÉÖªAÈÜÒºÖ÷ÒªµÄ½ðÊôÀë×ÓÊÇCo2+¡¢Li+£¬»¹º¬ÓÐÉÙÁ¿Fe3+¡¢Al3+¡¢Cu2+£¬¾­³ýÔÓºó¼ÓÈë²ÝËá泥¬¿ÉµÃµ½CoC2O4¹ÌÌ壬ĸҺÖк¬ÓÐLi+£¬¼ÓÈë±¥ºÍ̼ËáÄÆÈÜÒººó¹ýÂË£¬×îºóµÃµ½Ì¼Ëá﮹ÌÌ壬ÒԴ˽â´ð¸ÃÌâ¡£

(1)²½Öè¢ñÖУ¬ÂÁÓëÇâÑõ»¯ÄÆ·´Ó¦Éú³ÉÆ«ÂÁËáÄÆ£¬Àë×Ó·½³ÌʽΪ2Al+2OH-+2H2O£½2+3H2¡ü»ò2Al+2OH-+6H2O==2[Al(OH)4]-+3H2¡ü£»²½Öè¢óÖУ¬·´Ó¦ÎïÓÐLiCoO2¡¢Ë«ÑõË®¡¢ÁòËᣬÉú³ÉÎïÓÐLi2SO4ºÍCoSO4£¬LiCoO2ÖÐCo»¯ºÏ¼Û½µµÍ£¬ÔòH2O2»¯ºÏ¼ÛÉý¸ßÉú³ÉO2£¬ËùÒÔ»¯Ñ§·½³ÌʽΪ2LiCoO2+H2O2+3H2SO4£½Li2SO4+2CoSO4+O2¡ü+4H2O£»

(2)²½Öè¢ò½«ÇâÑõ»¯ÂÁ£¬×ÆÉÕ¡¢µç½âºóµÃµ½ÂÁ£¬

A£®Õô·¢ÃóÒ»°ãÓÃÓÚÕô·¢ÈÜÒº£¬ÇâÑõ»¯ÂÁΪ¹ÌÌ壬һ°ãÓÃÛáÛö×ÆÉÕ£¬A´íÎó£»

B£®×ÆÉÕʱÐèÒªÓò£Á§°ô²»¶Ï½Á°è£¬±ÜÃâ¾Ö²¿¹ýÈÈ£¬ÒýÆð±Ä½¦£¬BÕýÈ·£»

C£®ÔÚÖØÁ¿·ÖÎö·¨ÖУ¬¾­ºæ¸É»ò×ÆÉÕµÄÛáÛö»ò³Áµí£¬Ç°ºóÁ½´Î³ÆÖØÖ®²îСÓÚ0.2mg£¬C´íÎó£»

D£®ÔÚµç½âÈÛÈÚµÄAl2O3ÖƱ¸½ðÊôÂÁʱ£¬Í¨³£Ðè¼ÓÈë±ù¾§Ê¯(NaAlF6)£¬Ä¿µÄÊǽµµÍÑõ»¯ÂÁµÄÈ۵㣬D´íÎó¡£

´ð°¸Ñ¡B£»

(3)Èçͼ¿ÉÖªpH<5.0ʱ£¬Co2+²»Éú³É³Áµí£¬¶øFe3+¡¢Al3+Ò×Éú³É³Áµí¶ø³ýÈ¥£¬ËùÒԿɳýÈ¥µÄÔÓÖÊÀë×ÓÊÇFe3+¡¢Al3+£»

(4)¢ÙÅäÖÆÔ¼0.30molL-1ÑÎËáÈÜÒºÐèÒªµÄÒÇÆ÷ÓÐÁ¿Í²¡¢ÉÕ±­¡¢ÈÝÁ¿Æ¿¡¢²£Á§°ô¡¢½ºÍ·µÎ¹Ü£¬Ôò²»ÐèÒªÓõ½µÄÒÇÆ÷ÓÐbc£»

¢ÚÓÃ0.3000molL-1ÑÎËá±ê×¼ÈÜÒºµÎ¶¨£¬´óÌå²½ÖèΪ£ºÈ¡ÑùÓÚ׶ÐÎÆ¿ÖУ¬¼ÓË®Èܽ⣻¼ÓÈëָʾ¼Á£»ÓÃÑÎËá±ê×¼ÒºµÎ¶¨ÖÁָʾ¼Á±äÉ«£»Öó·ÐÈ¥³ýCO2£¬ÔÙÀäÈ´£»¼ÌÐøµÎ¶¨ÖÁָʾ¼Á±äÉ«¼´ÎªµÎ¶¨Öյ㣬ËùÒÔ˳ÐòΪcadbe£»

¢ÛµÎ¶¨¿Õ°×ÈÜÒºµÄÄ¿µÄÊÇÏû³ýʵÑéÖеÄÊÔ¼Á¡¢²Ù×÷µÈ×ÛºÏÒòËØÔì³ÉµÄµÎ¶¨Îó²î£»ÉÏÊöµÎ¶¨²Ù×÷ÖУ¬ÈçȱÉÙÖó·ÐÈ¥³ýCO2£¬ÔòCO2ÈÜÓÚˮʹÈÜÒº³ÊËáÐÔ£¬Ëù¼ÓÑÎËáÌå»ýÆ«ÉÙ£¬ËùÒԲⶨ½á¹û½«Æ«Ð¡¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿¶þÑõ»¯ÂÈ(ClO2)ÊÇ»ÆÂÌÉ«Ò×ÈÜÓÚË®µÄÆøÌ壬È۵㣭59¡æ¡¢·Ðµã11¡æ£¬µ«ÆäŨ¶È¹ý¸ßʱÒ×·¢Éú·Ö½â£¬Òò´Ë³£½«ÆäÖƳÉNaClO2¹ÌÌå±ãÓÚÔËÊäºÍÖü´æ¡£»Ø´ðÏÂÁÐÎÊÌ⣺

(1)ʵÑéÊÒÓÃNH4Cl¡¢ÑÎËá¡¢NaClO2(ÑÇÂÈËáÄÆ)ΪԭÁÏ£¬Í¨¹ýÒÔϹý³ÌÖƱ¸ClO2£º

¢Ùµç½âʱ·¢Éú·´Ó¦·½³ÌʽΪ__________¡£

¢ÚÈÜÒºXÖдóÁ¿´æÔÚµÄÈÜÖÊÓÐ__________(Ìѧʽ)¡£

¢Û³ýÈ¥ClO2ÖеÄNH3¿ÉÑ¡ÓõÄÊÔ¼ÁÊÇ_________(Ìî±êºÅ)¡£

a£®Ë® b£®¼îʯ»Ò c£®Å¨ÁòËá d£®±¥ºÍʳÑÎË®

(2)ʵÑéÊÒÖÐÓùýÑõ»¯Çâ·¨½«ClO2ÖƱ¸³ÉNaClO2¹ÌÌ壬ÆäʵÑé×°ÖÃÈçͼËùʾ¡£

¢ÙAÖз¢ÉúµÄ·´Ó¦Îª2NaClO3+H2O2+H2SO4=2ClO2¡ü+O2¡ü+Na2SO4+2H2O£¬ÒÇÆ÷AµÄÃû³ÆÊÇ_______¡£

¢Úд³öÖƱ¸NaClO2¹ÌÌåµÄ»¯Ñ§·½³Ìʽ£º__________¡£±ùˮԡÀäÈ´µÄÄ¿µÄÊÇ________¡£

¢Û¿ÕÆøÁ÷ËÙ¹ý¿ì»ò¹ýÂý£¬¾ù»á½µµÍNaClO2µÄ²úÂÊ£¬ÆäÔ­ÒòÊÇ______________¡£

¢ÜΪÁ˲ⶨNaClO2´ÖÆ·µÄ´¿¶È£¬È¡ÉÏÊö´Ö²úÆ·10.0gÈÜÓÚË®Åä³É1LÈÜÒº£¬È¡³ö10mL£¬ÈÜÒºÓÚ׶ÐÎÆ¿ÖУ¬ÔÙ¼ÓÈë×ãÁ¿ËữµÄKIÈÜÒº£¬³ä·Ö·´Ó¦(NaClO2±»»¹Ô­ÎªCl-£¬ÔÓÖʲ»²Î¼Ó·´Ó¦)£¬¼ÓÈë2¡«3µÎµí·ÛÈÜÒº£¬ÓÃ0.20mol¡¤L-1Na2S2O3±ê×¼ÒºµÎ¶¨£¬´ïµ½µÎ¶¨´ïÖÕµãʱÓÃÈ¥±ê×¼Òº20.00mL£¬ÊÔ¼ÆËãNaClO2´ÖÆ·µÄ´¿¶È_______¡£(ÒÑÖª£º2Na2S2O3+I2=Na2S4O6+2NaI)

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

¾«Ó¢¼Ò½ÌÍø