½üÄêÀ´£¬¸ßÃÌËá¼ØÔÚÒûÓÃË®ºÍ¹¤ÒµÎÛË®´¦ÀíÁìÓòµÄÏû·ÑÐèÇóÔö³¤½Ï¿ì¡£ÊµÑéÊÒ¿ÉÓöþÑõ»¯ÃÌΪÖ÷ÒªÔÁÏÖƱ¸¸ßÃÌËá¼Ø¡£Æ䲿·ÖÁ÷³ÌÈçÏ£º
£¨1£©µÚ¢Ù²½ÖвÉÓÃÌúÛáÛö¶ø²»ÓôÉÛáÛöµÄÔÒòÊÇ£¨Óû¯Ñ§·½³Ìʽ±íʾ£©_______________________________________________________________¡£
£¨2£©KOH¡¢KClO3ºÍMnO2¹²ÈÛ·´Ó¦Éú³ÉÄ«ÂÌÉ«K2MnO4µÄ»¯Ñ§·½³ÌʽΪ________________________________________________________________¡£
£¨3£©µÚ¢Ü²½Í¨ÈëCO2£¬¿ÉÒÔʹMnO42-·¢Éú·´Ó¦£¬Éú³ÉMnO4-ºÍMnO2¡£ÔòK2MnO4ÍêÈ«·´Ó¦Ê±£¬×ª»¯ÎªKMnO4µÄ°Ù·ÖÂÊԼΪ____________________£¨¾«È·µ½0.1%£©¡£
£¨4£©µÚ¢Ý²½³ÃÈȹýÂ˵ÄÄ¿µÄÊÇ________________________________¡£
£¨5£©µÚ¢Þ²½¼ÓÈÈŨËõÖÁÒºÃæÓÐϸС¾§ÌåÎö³öʱ£¬Í£Ö¹¼ÓÈÈ£¬ÀäÈ´½á¾§¡¢___________¡¢Ï´µÓ¡¢¸ÉÔï¡£¸ÉÔï¹ý³ÌÖУ¬Î¶Ȳ»Ò˹ý¸ß£¬ÒòΪ_________________¡£
Ä꼶 | ¸ßÖÐ¿Î³Ì | Ä꼶 | ³õÖÐ¿Î³Ì |
¸ßÒ» | ¸ßÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ | ³õÒ» | ³õÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ |
¸ß¶þ | ¸ß¶þÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õ¶þ | ³õ¶þÃâ·Ñ¿Î³ÌÍƼö£¡ |
¸ßÈý | ¸ßÈýÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õÈý | ³õÈýÃâ·Ñ¿Î³ÌÍƼö£¡ |
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
´ß»¯¼ÁÊÇ»¯¹¤¼¼ÊõµÄºËÐÄ£¬¾ø´ó¶àÊýµÄ»¯¹¤Éú²ú¾ùÐè²ÉÓô߻¯¹¤ÒÕ¡£
¢ÅÈËÃdz£Óô߻¯¼ÁÀ´Ñ¡Ôñ·´Ó¦½øÐеķ½Ïò¡£ÈçͼËùʾΪһ¶¨Ìõ¼þÏÂ1mol CH3OHÓëO2·¢Éú·´Ó¦Ê±£¬Éú³ÉCO¡¢CO2»òHCHOµÄÄÜÁ¿±ä»¯Í¼[·´Ó¦ÎïO2(g)ºÍÉú³ÉÎïH2O(g)ÂÔÈ¥]¡£
¢ÙÔÚÓд߻¯¼Á×÷ÓÃÏ£¬CH3OHÓëO2·´Ó¦Ö÷ÒªÉú³É £¨Ìî¡°CO¡¢CO2»òHCHO¡±£©¡£
¢Ú2HCHO(g)£«O2(g)=2CO(g)£«2H2O(g) ¡÷H= ¡£
¢ÛÔÚÏ¡ÁòËá´ß»¯Ï£¬HCHO¿ÉÒÔͨ¹ý·´Ó¦Éú³É·Ö×ÓʽΪC3H6O3µÄ»·×´Èý¾Û¼×È©·Ö×Ó£¬Æä·Ö×ÓÖÐͬÖÖÔ×ӵĻ¯Ñ§»·¾³¾ùÏàͬ¡£Ð´³öÈý¾Û¼×È©µÄ½á¹¹¼òʽ£º ¡£
¢Ü¼×´¼ÖÆÈ¡¼×È©¿ÉÓÃAg×÷´ß»¯¼Á£¬º¬ÓÐAgCl»áÓ°ÏìAg´ß»¯¼ÁµÄ»îÐÔ£¬Óð±Ë®¿ÉÒÔÈܽâ³ýÈ¥ÆäÖеÄAgCl£¬Ð´³ö¸Ã·´Ó¦µÄÀë×Ó·½³Ìʽ£º ¡£
¢ÆÒ»ÖÖÒÔÍ×÷´ß»¯¼ÁÍÑÁòÓÐÈçÏÂÁ½¸ö¹ý³Ì£º
¢ÙÔÚ͵Ä×÷ÓÃÏÂÍê³É¹¤ÒµÎ²ÆøÖÐSO2µÄ²¿·Ö´ß»¯Ñõ»¯£¬Ëù·¢Éú·´Ó¦Îª£º
2SO2£«2n Cu£«(n£«1)O2£«(2£2 n) H2O="2n" CuSO4£«(2£2n) H2SO4
´Ó»·¾³±£»¤µÄ½Ç¶È¿´£¬´ß»¯ÍÑÁòµÄÒâÒåΪ £»Ã¿ÎüÊÕ±ê×¼×´¿öÏÂ11.2L SO2£¬±»SO2»¹ÔµÄO2µÄÖÊÁ¿Îª g¡£
¢ÚÀûÓÃÏÂͼËùʾµç»¯Ñ§×°ÖÃÎüÊÕÁíÒ»²¿·ÖSO2£¬²¢Íê³ÉCuµÄÔÙÉú¡£Ð´³ö×°ÖÃÄÚËù·¢Éú·´Ó¦µÄÀë×Ó·½³Ìʽ ¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
£¨1£©½«·Ï·°´ß»¯¼Á(Ö÷Òª³É·ÖV2O5)ÓëÏ¡ÁòËá¡¢ÑÇÁòËá¼ØÈÜÒº»ìºÏ£¬³ä·Ö·´Ó¦£¬ËùµÃÈÜÒºÏÔËáÐÔ£¬º¬VO2£«¡¢K£«¡¢SO42-µÈ¡£Ð´³ö¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ_________________________________¡£
£¨2£©ÏòÉÏÊöËùµÃÈÜÒºÖмÓÈëKClO3ÈÜÒº£¬³ä·Ö·´Ó¦ºó£¬ÈÜÒºÖÐÐÂÔö¼ÓÁËVO2+¡¢Cl£¡£Ð´³ö²¢Åäƽ¸Ã·´Ó¦µÄÀë×Ó·½³Ìʽ£¬²¢±ê³öµç×ÓתÒƵÄÊýÄ¿ºÍ·½Ïò______________________¡£
£¨3£©ÔÚ20.00 mLµÄ0.1 mol¡¤L£1 VO2+ÈÜÒºÖУ¬¼ÓÈë0.195 gп·Û£¬Ç¡ºÃÍê³É·´Ó¦£¬Ôò»¹Ô²úÎï¿ÉÄÜÊÇ______________________________________________________________¡£
a£®V b£®V2£« c£®VO2+ d£®VO2£«
£¨4£©ÒÑÖªV2O5ÄܺÍÑÎËá·´Ó¦Éú³ÉÂÈÆøºÍVO2£«¡£ÇëÔÙдһ¸öÀë×Ó·´Ó¦·½³Ìʽ£¬ËµÃ÷»¹ÔÐÔ£ºSO32-£¾Cl££¾VO2£«__________________________________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
(13·Ö)ÌúºÍÌúµÄ»¯ºÏÎïÔÚ¹¤ÒµÉú²úºÍÈÕ³£Éú»îÖж¼Óй㷺µÄÓÃ;¡£
£¨1£©ÔÚ¶¨Ïò±¬ÆÆÖУ¬³£ÀûÓÃÑõ»¯ÌúÓëÂÁ·´Ó¦·Å³öµÄÈÈÁ¿À´Çиî¸Ö½î£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£ß£ß¡£
£¨2£©ÒÑÖª£º2Fe2O3(s)£«3C(s)£½3CO2(g)£«4Fe(s) ¡÷H£½+468.2 kJ¡¤mol-1
C(s)+O2(g)£½CO2(g) ¡÷H="-393.5" kJ¡¤mol-1¡£
ÔòFe(s)ÓëO2 (g)·´Ó¦Éú³ÉFe2 O3 (s)µÄÈÈ»¯Ñ§·½³ÌʽΪ£ß_____________________¡£
£¨3£©¿ÉÓÃKMnO4ÈÜÒºµÎ¶¨Fe2+µÄŨ¶È£¬·´Ó¦µÄÀë×Ó·½³ÌʽÈçÏ£º5Fe2£«£«MnO4££«8H£«£½5Fe3£«£«Mn2£«£«4H2O
¢ÙKMnO4ÈÜҺӦʢ·ÅÔڣߣߣߣߣߵζ¨¹ÜÖУ»
¢ÚÅжϴﵽµÎ¶¨ÖÕµãµÄÏÖÏóÊǣߣߣߣߣߣ»
¢ÛÓÃÁòËáËữµÄ0.020 00 mol¡¤L-1¡£KMnO4ÈÜÒºµÎ¶¨Ä³FeSO4ÈÜÒºÖÁÖյ㣬ʵÑéÊý¾Ý¼Ç¼ÈçÏÂ±í£º
Çë·ÖÎöÊý¾Ý²¢¼ÆË㣬¸ÃFeSO4ÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈΪ£ß£ß£ß£ß£ß¡£
£¨4£©ÐÂÐÍÄÉÃײÄÁÏZnFe2Ox£¬¿ÉÓÃÓÚ³ýÈ¥¹¤Òµ·ÏÆøÖеÄijЩÑõ»¯Îï¡£ÖÆȡвÄÁϺͳýÈ¥·ÏÆøµÄת»¯¹ØϵÈçÏÂͼ£º
¢ÙÒÑÖªZnFe2O4ÓëH2·´Ó¦µÄÎïÖʵÄÁ¿Ö®±ÈΪ2:1£¬ÔòZnFe2OxÖÐx=£ß£ß£ß£ß£ß£»
¢ÚÓÃZnFe2Ox³ýÈ¥SO2µÄ¹ý³ÌÖУ¬Ñõ»¯¼ÁÊǣߣߣߣߣߡ£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
ÍÑÁò¼¼ÊõÄÜÓÐЧ¿ØÖÆSO2¶Ô¿ÕÆøµÄÎÛȾ¡£
(1)ÏòúÖмÓÈëʯ»Òʯ¿É¼õÉÙȼÉÕ²úÎïÖÐSO2µÄº¬Á¿£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ
_______________________________¡£
(2)º£Ë®³ÊÈõ¼îÐÔ£¬Ö÷Òªº¬ÓÐNa£«¡¢K£«¡¢Ca2£«¡¢Mg2£«¡¢Cl£¡¢SO42¡ª¡¢Br£¡¢HCO3¡ªµÈ¡£º¬SO2µÄÑÌÆø¿ÉÀûÓú£Ë®ÍÑÁò£¬Æ乤ÒÕÁ÷³ÌÈçͼËùʾ£º
¢ÙÏòÆØÆø³ØÖÐͨÈë¿ÕÆøµÄÄ¿µÄÊÇ_____________________________________¡£
¢ÚͨÈë¿ÕÆøºóÆØÆø³ØÖк£Ë®ÓëÌìÈ»º£Ë®Ïà±È£¬Å¨¶ÈÓÐÃ÷ÏÔ²»Í¬µÄÀë×ÓÊÇ________¡£
a£®Cl£¡¡¡¡¡¡ b£®SO42¡ª¡¡¡¡¡¡ c£®Br£¡¡¡¡¡¡ d£®HCO3¡ª
(3)ÓÃNaOHÈÜÒºÎüÊÕÑÌÆøÖеÄSO2£¬½«ËùµÃµÄNa2SO3ÈÜÒº½øÐеç½â£¬¿ÉµÃµ½NaOH£¬Í¬Ê±µÃµ½H2SO4£¬ÆäÔÀíÈçͼËùʾ(µç¼«²ÄÁÏΪʯī)¡£
¢ÙͼÖÐa¼«ÒªÁ¬½ÓµçÔ´µÄ________(Ìî¡°Õý¡±»ò¡°¸º¡±)¼«£¬C¿ÚÁ÷³öµÄÎïÖÊÊÇ________¡£
¢ÚSO32¡ª·ÅµçµÄµç¼«·´Ó¦Ê½Îª____________________________¡£
¢Ûµç½â¹ý³ÌÖÐÒõ¼«Çø¼îÐÔÃ÷ÏÔÔöÇ¿£¬ÓÃƽºâÒƶ¯µÄÔÀí½âÊÍÔÒò£º
__________________________________________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
£¨1£©Ç뽫ÏÂÁÐÎåÖÖÎïÖÊ£ºKBr¡¢Br2¡¢I2¡¢KI¡¢K2SO4·Ö±ðÌîÈëÏÂÁкáÏßÉÏ£¬×é³ÉÒ»¸öδÅäƽµÄ»¯Ñ§·½³Ìʽ£º
KBrO3£«________£«H2SO4¨D¡ú________£«________£«________£«________£«H2O¡£
£¨2£©Èç¹û¸Ã»¯Ñ§·½³ÌʽÖÐI2ºÍKBrµÄ»¯Ñ§¼ÆÁ¿Êý·Ö±ðÊÇ8ºÍ1£¬Ôò
¢ÙBr2µÄ»¯Ñ§¼ÆÁ¿ÊýÊÇ________£»
¢ÚÇ뽫·´Ó¦ÎïµÄ»¯Ñ§Ê½¼°ÅäƽºóµÄ»¯Ñ§¼ÆÁ¿ÊýÌîÈëÏÂÁÐÏàÓ¦µÄλÖÃÖУº
________KBrO3£«________£«________H2SO4¨D¡ú¡¡£»
¢ÛÈôתÒÆ10 molµç×Ó£¬Ôò·´Ó¦ºóÉú³ÉI2µÄÎïÖʵÄÁ¿Îª________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
ij̽¾¿Ð¡×齫һÅúµç×Ó·ÏÆúÎï¼òµ¥´¦Àíºó£¬µÃµ½º¬Cu¡¢Al¡¢Fe¼°ÉÙÁ¿Au¡¢PtµÈ½ðÊôµÄ»ìºÏÎ²¢Éè¼ÆÈçÏÂÖƱ¸ÁòËá;§ÌåºÍÎÞË®ÂÈ»¯ÌúµÄ·½°¸£º
ÒÑÖª£ºCu2+ + 4NH3¡¤H2O£½[Cu(NH3)4]2+ + 4H2O
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©²½Öè¢ÙCuÓëËá·´Ó¦µÄÀë×Ó·½³ÌʽΪ ¡£
£¨2£©²½Öè¢Ú¼ÓH2O2µÄ×÷ÓÃÊÇ £¬ÂËÔü2Ϊ(Ìѧʽ) ¡£
£¨3£©²½Öè¢Ý²»ÄÜÖ±½Ó¼ÓÈÈÍÑË®µÄÀíÓÉÊÇ ¡£
£¨4£©ÈôÂËÒº1ÖÐCu2+µÄŨ¶ÈΪ0£®02mol¡¤L-1£¬ÔòÇâÑõ»¯Í¿ªÊ¼³ÁµíʱµÄpH =
(ÒÑÖª£ºKsp[Cu(OH)2]£½2£®0¡Á10-20)¡£
£¨5£©ÒÑÖª£º2Cu2+£«4I-£½ 2CuI¡ý£«I2 I2£«2S2O32-£½ 2I-£«S4O62-
ijͬѧΪÁ˲ⶨCuSO4¡¤5H2O²úÆ·µÄÖÊÁ¿·ÖÊý¿É°´ÈçÏ·½·¨£ºÈ¡3£®00g²úÆ·£¬ÓÃË®Èܽâºó£¬¼ÓÈë×ãÁ¿µÄKIÈÜÒº£¬³ä·Ö·´Ó¦ºó¹ýÂË¡¢Ï´µÓ£¬½«ÂËҺϡÊÍÖÁ250mL£¬È¡50mL¼ÓÈëµí·ÛÈÜÒº×÷ָʾ¼Á£¬ÓÃ0£®080 mol¡¤L-1 Na2S2O3±ê×¼ÈÜÒºµÎ¶¨£¬´ïµ½µÎ¶¨ÖÕµãµÄÒÀ¾ÝÊÇ ¡£
ËÄ´ÎƽÐÐʵÑéºÄÈ¥Na2S2O3±ê×¼ÈÜÒºÊý¾ÝÈçÏ£º
ʵÑéÐòºÅ | 1 | 2 | 3 | 4 |
ÏûºÄNa2S2O3±ê×¼ÈÜÒº(mL) | 25£®00 | 25£®02 | 26£®20 | 24£®98 |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
»¯Ñ§ÔÚ»·¾³±£»¤ÖÐÆð×ÅÊ®·ÖÖØÒªµÄ×÷Óᣴ߻¯·´Ïõ»¯·¨ºÍµç»¯Ñ§½µ½â·¨¿ÉÓÃÓÚÖÎÀíË®ÖÐÏõËáÑεÄÎÛȾ¡£
£¨1£©´ß»¯·´Ïõ»¯·¨ÖУ¬H2Äܽ«NO3-»¹ÔΪN2¡£25¡æʱ£¬·´Ó¦½øÐÐl0 min£¬ÈÜÒºµÄpHÓÉ7±äΪl 2¡£
¢ÙN2µÄ½á¹¹Ê½Îª
¢ÚÉÏÊö·´Ó¦µÄÀë×Ó·½³ÌʽΪ £¬Æäƽ¾ù·´Ó¦ËÙÂÊv(NO3-)Ϊ mol¡¤L-1¡¤min-1¡£
¢Û»¹Ô¹ý³ÌÖпÉÉú³ÉÖмä²úÎïNO2-£¬Ð´³ö2ÖÖ´Ù½øNO2-Ë®½âµÄ·½·¨ ¡£
£¨2£©µç»¯Ñ§½µ½âNO3-µÄÔÀíÈçÏÂͼËùʾ¡£
¢ÙµçÔ´Õý¼«Îª (Ìî¡°A¡±»ò¡°B¡±)£¬Òõ¼«·´Ó¦Ê½Îª£º
¢ÚÈôµç½â¹ý³ÌÖÐתÒÆÁË2 molµç×Ó£¬ÔòĤÁ½²àµç½âÒºµÄÖÊÁ¿±ä»¯²î(¡÷m×ó-¡÷mÓÒ)Ϊ g¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
ÔÚÑõ»¯»¹Ô·´Ó¦ÖУ¬µç×ÓתÒƵıíʾ·½·¨Í¨³£Óõ¥ÏßÇÅ·¨ºÍË«ÏßÇÅ·¨¡£ÊÔ·ÖÎöÕâÁ½ÖÖ±íʾ·½·¨µÄÌØÕ÷£¬²¢Ìî¿Õ¡£
£¨1£©±êµ¥ÏßÇÅ£¨4·Ö£©
KIO3£«6HI£½KI£«3I2£«3H2OÑõ»¯²úÎïÓ뻹ԲúÎïÎïÖʵÄÁ¿Ö®±È£½ ¡£
£¨2£©Åäƽ»¯Ñ§·½³Ìʽ¼°±êË«ÏßÇÅ£¨5·Ö£©
Fe + HNO3(Ï¡)£½ Fe(NO3)2+ NO¡ü+ H2O·´Ó¦ÖУ¬Ñõ»¯¼ÁÓ뻹ԼÁµÄÎïÖʵÄÁ¿Ö®±È ¡£
²é¿´´ð°¸ºÍ½âÎö>>
°Ù¶ÈÖÂÐÅ - Á·Ï°²áÁбí - ÊÔÌâÁбí
ºþ±±Ê¡»¥ÁªÍøÎ¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨Æ½Ì¨ | ÍøÉÏÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | µçÐÅթƾٱ¨×¨Çø | ÉæÀúÊ·ÐéÎÞÖ÷ÒåÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | ÉæÆóÇÖȨ¾Ù±¨×¨Çø
Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com