ÏÂÁÐÏà¹ØÀë×Ó·½³ÌʽÊéдÕýÈ·µÄÊÇ   (¡¡¡¡)¡£
A£®NaClOÈÜÒºÓëFeCl2ÈÜÒº»ìºÏ£º6Fe2£«£«3ClO£­£«3H2O=2Fe(OH)3¡ý£«3Cl£­£«4Fe3£«
B£®ÓÃʳ´×¼ìÑéÑÀ¸àÖеÄ̼Ëá¸Æ£ºCaCO3£«2H£«=Ca2£«£«CO2¡ü£«H2O
C£®FeCl2ËáÐÔÈÜÒº·ÅÔÚ¿ÕÆøÖбäÖÊ£º2Fe2£«£«4H£«£«O2=2Fe3£«£«2H2O
D£®µç½âMgCl2Ë®ÈÜÒºµÄÀë×Ó·½³Ìʽ£º2Cl£­£«2H2OH2¡ü£«Cl2¡ü£«2OH£­
A
Ñ¡ÏîAÖÐClO£­½«Fe2£«Ñõ»¯ÎªFe3£«£¬AÏîÕýÈ·¡£Ñ¡ÏîBÖÐʳ´×ÓÐЧ³É·ÖΪÒÒËᣬÒÒËáÓ¦¸Ãд·Ö×Óʽ£¬ÕýÈ·µÄÀë×Ó·½³ÌʽΪCaCO3£«2CH3COOH=Ca2£«£«2CH3COO£­£«CO2¡ü£«H2O¡£Ñ¡ÏîCÖеÃʧµç×Ӻ͵çºÉ²»Êغ㣬ÕýÈ·µÄÀë×Ó·½³ÌʽΪ4Fe2£«£«4H£«£«O2=4Fe3£«£«2H2O¡£Ñ¡ÏîDÖеç½âÂÈ»¯Ã¾Ë®ÈÜÒºÉú³ÉµÄOH£­ÓëMg2£«·´Ó¦Éú³ÉÇâÑõ»¯Ã¾³Áµí£¬ÕýÈ·µÄÀë×Ó·½³ÌʽΪMg2£«£«2Cl£­£«2H2OMg(OH)2¡ý£«H2¡ü£«Cl2¡ü¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

Ñо¿ÈËÔ±ÑÐÖÆÀûÓÃµÍÆ·Î»ÈíÃÌ¿ó½¬(Ö÷Òª³É·ÖÊÇMnO2)ÎüÊÕÁòËá³§µÄÎ²ÆøSO2£¬ÖƱ¸ÁòËáÃ̵ÄÉú²úÁ÷³ÌÈçÏ£º

ÒÑÖª£º½þ³öÒºµÄpH£¼2£¬ÆäÖеĽðÊôÀë×ÓÖ÷ÒªÊÇMn2£«£¬»¹º¬ÓÐÉÙÁ¿µÄFe2£«¡¢Al3£«¡¢Ca2£«¡¢Pb2£«µÈÆäËû½ðÊôÀë×Ó¡£PbO2µÄÑõ»¯ÐÔ´óÓÚMnO2¡£PbSO4ÊÇÒ»ÖÖ΢ÈÜÎïÖÊ¡£ÓйؽðÊôÀë×ӵİ뾶¡¢ÐγÉÇâÑõ»¯Îï³ÁµíʱµÄpH¼ûÏÂ±í£¬ÑôÀë×ÓÎü¸½¼ÁÎü¸½½ðÊôÀë×ÓµÄЧ¹û¼ûÏÂͼ¡£
Àë×Ó
Àë×Ó°ë¾¶(pm)
¿ªÊ¼³Áµí
ʱµÄpH
ÍêÈ«³Áµí
ʱµÄpH
Fe2£«
74
7.6
9.7
Fe3£«
64
2.7
3.7
Al3£«
50
3.8
4.7
Mn2£«
80
8.3
9.8
Pb2£«
121
8.0
8.8
Ca2£«
99
£­
£­
 

Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Ð´³ö½þ³ö¹ý³ÌÖÐÉú³ÉMn2+·´Ó¦µÄ»¯Ñ§·½³Ìʽ                            ¡£
£¨2£©Ñõ»¯¹ý³ÌÖÐÖ÷Òª·´Ó¦µÄÀë×Ó·½³Ìʽ                                 ¡£
£¨3£©ÔÚÑõ»¯ºóµÄÒºÌåÖмÓÈëʯ»Ò½¬£¬ÓÃÓÚµ÷½ÚpHÖµ£¬´Ë´¦µ÷½ÚpHÖµÓõ½µÄÒÇÆ÷ÊÇ       £¬Ó¦µ÷½ÚpHµÄ·¶Î§Îª                     ¡£
£¨4£©ÑôÀë×ÓÎü¸½¼Á¿ÉÓÃÓÚ³ýÈ¥ÔÓÖʽðÊôÀë×Ó¡£ÇëÒÀ¾Ýͼ¡¢±íÐÅÏ¢»Ø´ð£¬¾ö¶¨ÑôÀë×ÓÎü¸½¼ÁÎü¸½Ð§¹ûµÄÒòËØÓР                ¡¢                 µÈ£»Îü¸½²½Öè³ýÈ¥µÄÖ÷ÒªÀë×ÓΪ£º               ¡£
£¨5£©CaSO4ÊÇÒ»ÖÖ΢ÈÜÎïÖÊ£¬ÒÑÖªKsp(CaSO4)=9.10¡Á10¡ª6¡£ÏÖ½«c mol¡¤L¡ª1CaCl2ÈÜÒºÓë2.00¡Á10¡ª2 mol¡¤L¡ª1Na2SO4ÈÜÒºµÈÌå»ý»ìºÏ£¨ºöÂÔÌå»ýµÄ±ä»¯£©£¬ÔòÉú³É³Áµíʱ£¬cµÄ×îСֵÊÇ¡¡¡¡¡¡¡¡¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÏÂÁÐÓйصÄÀë×Ó·½³ÌʽÕýÈ·µÄÊÇ
A£®ÂÈ»¯ÂÁÈÜÒºÖмÓÈë¹ýÁ¿°±Ë®£ºAl3£«£«4NH3¡¤H2O=£«4
B£®Í­Æ¬½ÓµçÔ´Õý¼«£¬Ì¼°ô½ÓµçÔ´¸º¼«£¬µç½âÁòËáÈÜÒº£º
C£®Á×ËáÒ»ÇâÄÆÈÜҺˮ½â£º£«H2O£«H3O£«
D£®ÊµÑéÊÒÅäÖÆµÄÑÇÌúÑÎÈÜÒºÔÚ¿ÕÆøÖб»Ñõ»¯£º4Fe2£«£«O2£«2H2O=4Fe3£«£«4OH£­

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÏÂÁбíʾ¶ÔÓ¦·´Ó¦µÄÀë×Ó·½³ÌʽÖУ¬²»ÕýÈ·µÄÊÇ£¨  £©¡£
A£®ÏòMg(HCO3)2ÈÜÒºÖмÓÈë¹ýÁ¿µÄNaOHÈÜÒº£ºMg2£«£«2HCO3£­£«4OH£­=Mg(OH)2¡ý£«2CO32£­£«2H2O
B£®ÏòNH4Al(SO4)2ÈÜÒºÖеÎÈëBa(OH)2ÈÜҺǡºÃʹSO42£­·´Ó¦ÍêÈ«£º2Ba2£«£«4OH£­£«Al3£«£«2SO42£­=2BaSO4¡ý£«AlO2£­£«2H2O
C£®FeBr2ÈÜÒºÖÐͨÈë¹ýÁ¿µÄCl2£º2Fe2£«£«4Br£­£«3Cl2=2Fe3£«£«2Br2£«6Cl£­
D£®ÏòFe(NO3)2ÈÜÒºÖмÓÈëÏ¡ÑÎË᣺3Fe2£«£«4H£«£«NO3£­=3Fe3£«£«NO¡ü£«2H2O

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

Ïòº¬ÓÐ0.2 mol NaOHºÍ0.1 mol Ba(OH)2µÄÈÜÒºÖгÖÐøÎȶ¨µØÍ¨ÈëCO2ÆøÌ壬µ±Í¨ÈëÆøÌåΪ8.96 L(0 ¡æ£¬1.01¡Á105Pa)ʱÁ¢¼´Í£Ö¹£¬ÔòÕâÒ»¹ý³ÌÖУ¬ÈÜÒºÖÐÀë×ÓµÄÎïÖʵÄÁ¿ÓëͨÈëCO2ÆøÌåµÄÌå»ý¹ØÏµÍ¼ÏñÕýÈ·µÄÊÇ(ÆøÌåµÄÈܽâºÍÀë×ÓµÄË®½âºöÂÔ²»¼Æ)(  )¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

½âÊÍÏÂÁÐÊÂʵµÄ·½³Ìʽ²»ÕýÈ·µÄÊÇ
A£®´ÎÂÈËá¸ÆÈÜÒºÖÐͨÈë¹ýÁ¿¶þÑõ»¯Áò£ºCa2+ + 2ClO£­+ H2O + SO2£½CaSO3¡ý+ 2HClO
B£®ÁòËáÐÍËáÓê·ÅÖÃÒ»¶Îʱ¼äÈÜÒºµÄpHϽµ£º2H2SO3£«O2£½2H2SO4
C£®´¿¼îÒº¿ÉÒÔÇåÏ´ÓÍÎÛµÄÔ­Òò£ºCO32¯+H2OHCO3¯+OH¯
D£®ÏòK2Cr2O7ÈÜÒºÖмÓÈëÉÙÁ¿NaOHŨÈÜÒº£¬ÈÜÒºÓɳÈÉ«±äΪ»ÆÉ«£ºCr2O72¡ª+H2O2CrO42¡ª+2H+

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÏÂÁбíʾ¶ÔÓ¦»¯Ñ§·´Ó¦µÄÀë×Ó·½³ÌʽÕýÈ·µÄÊÇ
A£®Na2O2ÈÜÓÚË®²úÉúO2£º2O22¡ª£«2H2O=O2¡ü£«4OH£­
B£®Ïò°±Ë®Í¨Èë×ãÁ¿SO2£ºSO2£«2NH3¡¤H2O=2NH4£«£«SO32£­£«H2O
C£®´ÎÂÈËáÄÆÓëŨÑÎËá·´Ó¦²úÉúCl2£ºClO£­£«Cl¡ª£«H2O=Cl2¡ü£«2OH¡ª
D£®NaHCO3ÈÜÒºÓëÉÙÁ¿Ba(OH)2ÈÜÒº·´Ó¦£ºBa2£«£«2OH£­£«2HCO3£­=BaCO3¡ý£«CO32¡ª£«2H2O

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÏÂÁÐÀë×Ó·½³Ìʽ²»ÕýÈ·µÄÊÇ  £¨     £©
A£®äåÒÒÍéÔÚÇâÑõ»¯ÄÆË®ÈÜÒºÖеķ´Ó¦£ºCH3CH2Br + OH-CH3CH2OH +Br-
B£®´×ËáÈÜÒºÓëÇâÑõ»¯Í­·´Ó¦£º2CH3COOH + Cu(OH)2Cu2+ + 2CH3COO-+ 2H2O
C£®±½·ÓÄÆÈÜÒºÖÐͨÈëÉÙÁ¿¶þÑõ»¯Ì¼£º2C6H5O£­+ CO2 + H2O2C6H5OH + CO32-
D£®¼×È©ÈÜÒºÓë×ãÁ¿µÄÒø°±ÈÜÒº¹²ÈÈHCHO+4[Ag(NH3)2]++4OH£­CO32-+2NH4+ +4Ag¡ý+6NH3+2H2O

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÏÂÁÐÀë×Ó·½³Ìʽ²»ÕýÈ·µÄÊÇ
A£®´×ËáÈÜÒºÖмÓÈëÉÙÁ¿ÇâÑõ»¯Ã¾¹ÌÌ壺Mg(OH)2+2CH3COOH=Mg2++2CH3COO-+2H2O
B£®H2O2ÈÜÒºÖмÓÈë×ãÁ¿ËáÐÔKMnO4ÈÜÒº£º2MnO4-+3H2O2+6H+=2Mn2++6H2O+4O2¡ü
C£®Ca(HCO3)2ÈÜÒºÖмÓÈë×ãÁ¿³ÎÇåʯ»ÒË®£ºCa2++HCO3-+OH-=CaCO3¡ý+H2O
D£®NH4HSO4µÄÏ¡ÈÜÒºÖÐÖðµÎ¼ÓÈëBa(OH)2ÈÜÒºÖÁSO42¡ªÇ¡ºÃ³ÁµíÍêÈ«£ºNH4++H++SO42-+Ba2++2OH-=NH3?H2O+BaSO4¡ý+H2O

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸