10£®£¨1£©ÊµÑéÊÒÖУ¬ÖÆÈ¡Fe£¨OH£©3½ºÌåµÄ²Ù×÷·½·¨ÊÇÏò·ÐË®ÖÐÖðµÎ¼ÓÈëFeCl3ÈÜÒº£»¼ÌÐø¼ÓÈÈÖÁÈÜÒº³ÊºìºÖÉ«£¬Í£Ö¹¼ÓÈÈ£¬¼´µÃµ½Fe£¨OH£©3 ½ºÌ壬·´Ó¦·½³ÌʽÊÇFeCl3+3H2O$\frac{\underline{\;¼ÓÈÈ\;}}{\;}$Fe£¨OH£©3+3HCl£¬Ö¤Ã÷½ºÌåÒѾ­ÖƳɵķ½·¨ÊÇÓùâÊøÕÕÉäÈÜÒº£¬Èç¹ûÓж¡´ï¶ûЧӦ£¬Ôò˵Ã÷½ºÌåÒÑÖƳɣ¬Óô˷Öɢϵ½øÐÐÏÂÁÐʵÑ飺
£¨2£©½«Æä×°ÈëUÐιÜÄÚ£¬ÓÃʯīµç¼«½ÓֱͨÁ÷µç£¬Í¨µçÒ»¶Îʱ¼äºó£¬·¢ÏÖÒõ¼«¸½½üÑÕÉ«¼ÓÉÕâ±íÃ÷Fe£¨OH£©3½ºÁ£´øÕýµç£¬ÕâÖÖÏÖÏó³ÆΪµçÓ¾£®
£¨3£©ÏòÆäÖмÓÈë±¥ºÍ£¨NH4£©2SO4ÈÜÒº·¢ÉúµÄÏÖÏóÊÇÉú³É³Áµí£¬Ô­ÒòÊǵç½âÖÊÈÜÒºÄÜʹ½ºÌå·¢Éú¾Û³Á£®
£¨4£©ÏòÆäÖеÎÈë¹ýÁ¿Ï¡ÁòËᣬÏÖÏóÊÇÏÈÉú³É³Áµí£¬ºó³ÁµíÏûʧ£¬Ô­ÒòÊǽºÌåÏÈÓöÏ¡ÁòËá·¢Éú½ºÌåµÄ¾Û³Á£¬ºóÏ¡ÁòËὫÉú³ÉµÄÇâÑõ»¯Ìú³ÁµíÈܽ⣮
£¨5£©Ìá´¿´Ë·Öɢϵ³£Óõķ½·¨ÊÇÉøÎö£®

·ÖÎö £¨1£©Ïò·ÐË®ÖеÎÈ뼸µÎ±¥ºÍFeCl3ÈÜÒº£¬¼ÌÐøÖó·ÐÖÁÈÜÒº³ÊºìºÖÉ«£¬¼´ÖƵÃFe£¨OH£©3½ºÌ壻ÇâÑõ»¯Ìú½ºÌåµÄÖƱ¸ÊµÖÊÊÇÂÈ»¯ÌúµÄË®½â£»½ºÌåÌØÓеÄÐÔÖÊÊǶ¡´ï¶ûЧӦ£»
£¨2£©Fe£¨OH£©3½ºÁ£´øÕýµç£¬¹ÊÏòÒõ¼«Òƶ¯£¬Òõ¼«ÑÕÉ«±äÉ´ËΪ½ºÌåµÄµçÓ¾£»
£¨3£©µç½âÖÊÈÜÒºÄÜʹ½ºÌå·¢Éú¾Û³Á£»
£¨4£©µÎÈë¹ýÁ¿µÄÏ¡ÁòËᣬ»áÏȵ¼Ö½ºÌåµÄ¾Û³Á£¬Éú³ÉFe£¨OH£©3³Áµí£¬ºó¹ýÁ¿µÄÁòËáÄÜʹFe£¨OH£©3³ÁµíÈܽ⣻
£¨5£©½ºÌåµÄÌá´¿ÓÃÉøÎö£®

½â´ð ½â£º£¨1£©ÊµÑéÊÒÖƱ¸ÇâÑõ»¯Ìú½ºÌåÔ­ÀíÊÇÔÚ¼ÓÈÈÌõ¼þÏ£¬Èý¼ÛÌúÀë×ÓË®½âÉú³ÉÇâÑõ»¯Ìú½ºÌ壬Æä²Ù×÷Ϊ£ºÏò·ÐË®ÖÐÖðµÎ¼ÓÈëFeCl3ÈÜÒº£»¼ÌÐø¼ÓÈÈÖÁÈÜÒº³ÊºìºÖÉ«£¬Í£Ö¹¼ÓÈÈ£¬¼´µÃµ½Fe£¨OH£©3 ½ºÌ壻»¯Ñ§·½³ÌʽΪ£ºFeCl3+3H2O$\frac{\underline{\;¼ÓÈÈ\;}}{\;}$Fe£¨OH£©3+3HCl£»½ºÌåÌØÓеÄÐÔÖÊÊǶ¡´ï¶ûЧӦ£¬¹ÊÓùâÊøÕÕÉäÈÜÒº£¬Èç¹ûÓж¡´ï¶ûЧӦ£¬Ôò˵Ã÷½ºÌåÒÑÖƳɣ®¹Ê´ð°¸Îª£ºÏò·ÐË®ÖÐÖðµÎ¼ÓÈëFeCl3ÈÜÒº£»¼ÌÐø¼ÓÈÈÖÁÈÜÒº³ÊºìºÖÉ«£¬Í£Ö¹¼ÓÈÈ£¬¼´µÃµ½Fe£¨OH£©3 ½ºÌ壻FeCl3+3H2O$\frac{\underline{\;¼ÓÈÈ\;}}{\;}$Fe£¨OH£©3+3HCl£»ÓùâÊøÕÕÉäÈÜÒº£¬Èç¹ûÓж¡´ï¶ûЧӦ£¬Ôò˵Ã÷½ºÌåÒÑÖƳɣ»
£¨2£©Fe£¨OH£©3½ºÁ£´øÕýµç£¬¹ÊÓÃʯīµç¼«½ÓֱͨÁ÷µç£¬Í¨µçÒ»¶Îʱ¼äºó£¬Fe£¨OH£©3½ºÁ£ÏòÒõ¼«Òƶ¯£¬Òõ¼«ÑÕÉ«±äÉ´ËΪ½ºÌåµÄµçÓ¾£¬¹Ê´ð°¸Îª£º¼ÓÉFe£¨OH£©3½ºÁ£´øÕýµç£»µçÓ¾£»
£¨3£©Ïò½ºÌåÖмÓÈë±¥ºÍ£¨NH4£©2SO4ÈÜÒººó»áÉú³ÉFe£¨OH£©3³Áµí£¬Ô­ÒòÊǵç½âÖÊÈÜÒºÄÜʹ½ºÌå·¢Éú¾Û³Á£¬¹Ê´ð°¸Îª£ºÉú³É³Áµí£¬µç½âÖÊÈÜÒºÄÜʹ½ºÌå·¢Éú¾Û³Á£»
£¨4£©ÓÉÓÚÏ¡ÁòËáÊǵç½âÖÊÈÜÒº£¬¹Ê¿ªÊ¼µÎÈëÏ¡ÁòËᣬ»áµ¼Ö½ºÌåµÄ¾Û³Á£¬Éú³ÉFe£¨OH£©3³Áµí£¬ºó¹ýÁ¿µÄÁòËáÄÜʹFe£¨OH£©3³ÁµíÈܽ⣬¹Ê´ð°¸Îª£ºÏÈÉú³É³Áµí£¬ºó³ÁµíÏûʧ£»½ºÌåÏÈÓöÏ¡ÁòËá·¢Éú½ºÌåµÄ¾Û³Á£¬ºóÏ¡ÁòËὫÉú³ÉµÄÇâÑõ»¯Ìú³ÁµíÈܽ⣻
£¨5£©½ºÌåµÄÌá´¿ÓÃÉøÎö£¬¹Ê´ð°¸Îª£ºÉøÎö£®

µãÆÀ ±¾Ì⿼²éÁËFe£¨OH£©3 ½ºÌåµÄÖƱ¸ºÍ½ºÌåµÄÐÔÖÊ£¬ÌâÄ¿ÄѶȲ»´ó£¬Ò×´íµãΪ½ºÌåµÄÖƱ¸Ô­Àí£¬×¢ÒâÖƱ¸·½·¨£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

20£®Ï±íÖеÄʵÑé²Ù×÷²»ÄܴﵽʵÑéÄ¿µÄÊÇ£¨¡¡¡¡£©
Ñ¡ÏîʵÑé²Ù×÷ʵÑéÄ¿µÄ
AÏòº¬ÓÐKSCNÈÜÒºµÄFeSO4ÈÜÒºÖеμÓÁòËáËữµÄH2O2ÈÜÒº¼ìÑéH2O2µÄÑõ»¯ÐÔ´óÓÚFe3+
B½«ÆøÌåͨÈëÏ¡äåË®ÖмìÑéSO2 ÖÐÊÇ·ñ»ìÓÐÒÒÏ©
CµÎÈë×ÏɫʯÈïÊÔÒº¼ìÑé¾Æ¾«ÖÐÊÇ·ñ»ìÓд×Ëá
DÓñ¥ºÍ̼ËáÄÆÈÜҺϴµÓºó·ÖÒº³ýÈ¥ÒÒËáÒÒõ¥ÖÐÉÙÁ¿µÄÒÒ´¼ºÍÒÒËá
A£®AB£®BC£®CD£®D

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

1£®½ñÐèÅäÖÆpH=7.4µÄÁ×ËỺ³åÈÜÒº1000mL£¬ÎÊÐèÓÃ0.1mol/L Na2HPO4ºÍ0.1mol/L NaH2PO4ÈÜÒº¸÷¶àÉÙºÁÉý£¨H3PO4µÄKa2=6.2¡Á10-8£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

18£®¶þÑõ»¯Âȱ»ÊÀ½çÎÀÉú×éÖ¯£¨WHO£©È·ÈÏΪµÚËÄ´úA1ÐÍÏû¶¾¼Á£®
£¨1£©ÁòÌú¿ó·¨Éú²ú¶þÑõ»¯ÂȵÄÔ­ÀíΪ15ClO3-+FeS2+14H+¨T15ClO2¡ü+Fe3++2SO42-+7H2O£¬Ôòÿ1mol FeS2·¢Éú·´Ó¦Ê±£¬×ªÒƵç×ÓµÄÎïÖʵÄÁ¿Îª15mol£®
£¨2£©ÖÆÈ¡¶þÑõ»¯ÂÈÒ²¿ÉÓõç½â·¨£®
¢ÙÓÃDSA×÷Ñô¼«£¬ÓÃʯī×÷Òõ¼«£¬µç½âÂÈËáÄÆÓëÁòËáµÄ»ìºÏÈÜÒº£¬µç½â¹ý³Ì°üº¬ÈçÏ·´Ó¦£º
Òõ¼«·´Ó¦Ê½£º2ClO3-+4H++4e-¨T2ClO2-+2H2O£»
Òõ¼«ÈÜÒº·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º2ClO3-+2ClO2-+4H+¨T4ClO2¡ü+2H2O£»
Ñô¼«·´Ó¦Ê½£ºH2O-2e-¨T$\frac{1}{2}$O2¡ü+2H+£»
Ôòµç½âʱ×Ü·´Ó¦µÄÀë×Ó·½³ÌʽΪ4ClO3-+4H+=4ClO2¡ü+2H2O+O2¡ü£®
¢Ú½üÄêÀ´¿ª·¢µÄÓÃÌØÊâ½ðÊôÑõ»¯Îï×÷Ñô¼«µç½âÑÇÂÈËáÄÆ£¨NaClO2£©ÈÜÒºÖÆÈ¡¶þÑõ»¯ÂȵÄ×°ÖÃÈçͼ1Ëùʾ£®µç½âʱÑô¼«µÄµç¼«·´Ó¦Ê½ÎªClO2--e-=ClO2¡ü£»ÈÜÒºµÄpHÔö´ó£¨Ìî¡°Ôö´ó¡±»ò¡°¼õС¡±£©£®

£¨3£©NaClO2ÔÚÈÜÒºÖпÉÉú³ÉClO2¡¢HClO2¡¢ClO2-¡¢Cl-µÈ£¬ÆäÖÐHClO2ºÍClO2¶¼¾ßÓÐƯ°××÷Ó㬵«ClO2ÊÇÓж¾ÆøÌ壮ijζÈÏ£¬¸÷×é·Öº¬Á¿ËæpH±ä»¯Çé¿öÈçͼ2Ëùʾ£¨Cl-ûÓл­³ö£©£®ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇbd£®
a£®NaClO2ÈÜÒºÖУ¬$\frac{c£¨N{a}^{+}£©}{c£¨Cl{O}_{2}^{-}£©}$=1
b£®ÍùNaClO2ÈÜÒºÖеÎÈëÏ¡ÑÎËᣬµ±pH£¾6ʱ£¬ÈÜÒºÖÐÖ÷Òª·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ£ºClO2-+H+¨THClO2
c£®Ê¹ÓøÃƯ°×¼ÁµÄ×î¼ÑpHΪ1.0
d£®´ËζÈÏÂHClO2µÄµçÀëƽºâ³£ÊýµÄÊýÖµKa=10-6
£¨4£©ÔÚ³£ÎÂÏ£¬Ïò1LµÄÈÝÆ÷ÖгäÈë0.1mol ClO2£¬·¢ÉúµÄ·´Ó¦Îª2ClO2£¨g£©?Cl2£¨g£©+2O2£¨g£©£¬²âµÃƽºâʱO2µÄŨ¶ÈΪ0.02mol•L-1£¬Ôò¸Ã·´Ó¦ÔÚ³£ÎÂϵÄƽºâ³£ÊýΪ6.25¡Á10-4£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º¶àÑ¡Ìâ

5£®ÏÂÁÐ˵·¨´íÎóµÄÊÇ£¨¡¡¡¡£©
A£®ÅäÖÆ0.1mol•L-1 NaOHÈÜҺʱ£¬Óó±ÊªµÄÉÕ±­³ÆÁ¿NaOH¹ÌÌ壬¶ÔÅäÖƽá¹ûÎÞÓ°Ïì
B£®ÅäÖÆÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈÈÜҺʱ£¬ÓÃÁ¿Í²Á¿È¡Å¨ÈÜÒºÌå»ýÑöÊÓ¶ÁÊý£¬ËùÅäÈÜÒºµÄŨ¶È½á¹ûÆ«¸ß
C£®98%µÄŨÁòËáÓõÈÌå»ýµÄˮϡÊͺó£¬ÁòËáµÄÖÊÁ¿·ÖÊýΪ49%
D£®Î¶ÈÒ»¶¨Ê±£¬Ïò±¥ºÍʳÑÎË®ÖмÓÈë¹ÌÌåÂÈ»¯ÄÆ£¬ÔòÈÜÒºµÄŨ¶ÈºÍÖÊÁ¿¾ù²»±ä

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

15£®ÎÙËáÄÆ£¨Na2WO4£¬WµÄ×î¸ß¼ÛΪ+6¼Û£©³£ÓÃÓÚÖÆÔì·À»ð£¬·ÀË®²ÄÁϵȣ¬ÏÖ³ÆÈ¡2.94gÎÙËáÄƼÓË®Èܽâºó£¬ÔÙ¼ÓÈëÒ»¶¨Á¿µÄпºÍÏ¡H2SO4£¬·´Ó¦Éú³ÉÒ»ÖÖÀ¶É«ÎÙµÄÑõ»¯ÎÎٵĻ¯ºÏ¼ÛΪ+5ºÍ+6µÄ»ìºÏ¼Û£©µ«ÎÞH2²úÉú£¬ÏÖÓÃ0.05mol•L-1µÄËáÐÔKMnO4ÈÜÒºµÎ¶¨ÉÏÊöÀ¶É«Ñõ»¯ÎÍêÈ«·´Ó¦ºó¹²ÏûºÄËáÐÔKMnO4ÈÜÒº24mL£¬Ôò¸ÃÀ¶É«ÎÙµÄÑõ»¯ÎïµÄ»¯Ñ§Ê½Îª£¨¡¡¡¡£©
A£®W10O21B£®W8O22C£®W10O27D£®W5O14

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

2£®ÖØÇâÊÇÇâÔªËصÄÒ»ÖÖÖØÒªµÄºËËØ£¬ËüµÄ»¯ºÏÎïÔÚÉú²úÖÐÓм«ÆäÖØÒªµÄÓ¦Óã¬ÈçÖØË®¿ÉÓ÷´Ó¦¶ÑÖеļõËÙ¼Á£®ÊԻشðÏÂÁÐÎÊÌ⣺
£¨1£©ÖØÇâÔ­×Ӻ˵çºÉÊýΪ1£¬ÖÐ×ÓÊýΪ1£®
£¨2£©Ð´³öµç½âÖØË®µÄ»¯Ñ§·½³Ìʽ2D2O$\frac{\underline{\;µç½â\;}}{\;}$2D2¡ü+O2¡ü£®
£¨3£©Ð´³öÖØÇâÐγɵÄÑÎËáºÍ̼Ëá¸Æ·´Ó¦µÄ»¯Ñ§·½³ÌʽCaCO3+2DCl=CaCl2+CO2¡ü+D2O£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

19£®ÏÂÁÐÓÉÌþµÄÑÜÉúÎïAÖƱ¸£¨¹ý³ÌÖÐÎÞ»ú²úÎïÒѺöÂÔ£©£®
£¨1£©Ð´³öÉÏÊö¹ý³ÌÖеÄÖмä²úÎïµÄ½á¹¹¼òʽ£ºACH3CH2Br£¬CCH3CHO£¬FBrCH2CH2Br£®
£¨2£©Ð´³öD+G¡úµÄ·´Ó¦·½³ÌʽCH3COOH+HOCH2CH2OH$¡ú_{¡÷}^{ŨH_{2}SO_{4}}$+2H2O£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

8£®CH2=CH-CH=C£¨CH3£©-CH=CH2ÓëµÈÎïÖʵÄÁ¿µÄBr2¼Ó³Éʱ¿ÉÄܵõ½µÄÒì¹¹ÌåÓУ¨¡¡¡¡£©
A£®5ÖÖB£®2ÖÖC£®3ÖÖD£®4ÖÖ

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸