ÌâÄ¿ÄÚÈÝ

£¨24·Ö£©ÏÂͼÊÇÔªËØÖÜÆÚ±íµÄÒ»²¿·Ö£¬Õë¶Ô±íÖеĢ١«¢âÖÐÔªËØ£¬ÌîдÏÂÁпոñ£º

 
 
 
 
 
 
 
¢Ù
¢â
¢Ú
¢Û
 
 
¢Ü
¢Ý
 
 
¢ß
¢à
¢á
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
¢Þ
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
¢ÅÔÚÕâ10ÖÖÔªËØÖУ¬Ô­×Ó°ë¾¶½ÏСµÄÊÇ________(ÌîÔªËØ·ûºÅ)£¬»¯Ñ§ÐÔÖÊ×î²»»îÆÃµÄÔªËØÊÇ_____(Ìî¡°ÔªËØ·ûºÅ¡±)£»ÆäÖÐÒ»ÖÖºËËØ¿É²â¶¨ÎÄÎïÄê´ú£¬ÕâÖÖºËËØµÄ·ûºÅÊÇ     ¡£
¢ÆÔªËآٵÄÔ­×ӽṹʾÒâͼΪ__________£»ÔªËآٵÄ×î¸ß¼ÛÑõ»¯Îï½á¹¹Ê½Îª£º________£¬ÔªËØ¢âµÄµ¥Öʵç×ÓʽΪ£º__________¡£
ÇëÓõç×Óʽ±íʾ»¯ºÏÎï¢ÜºÍ¢àµÄÐγɹý³Ì
________________________________________________¡£
¢ÇÔªËØ¢ÝµÄÑõ»¯ÎïÓëÑÎËá·´Ó¦µÄÀë×Ó·½³ÌʽΪ£º    _________________________________¡£
ÔªËØ¢ÝµÄµ¥ÖÊÓëÇâÑõ»¯ÄÆÈÜÒº·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º ________________________________¡£      
¢ÈÔªËØ¢ÝµÄµ¥ÖÊÓëFeºÍÏ¡ÁòËá¹¹³ÉÔ­µç³Ø£¬ÊÔÔÚÏÂÃæµÄ·½¿òÄÚ»­³öÔ­µç³Ø×°ÖÃͼ£¬±ê³öÔ­µç³ØµÄµç¼«²ÄÁϺ͵ç½âÖÊÈÜÒº£¬²¢Ð´³ö¸º¼«µÄµç¼«·´Ó¦Îª___________________________¡£

£¨5£©ÔªËآ൥ÖÊÄÜÈÜÓÚË®£¬Ë®Òº³Ê   É«£¬ÔÚÆäÖÐͨÈëÔªËØ¢ßµÄijÖÖÑõ»¯ÎÈÜÒºÑÕÉ«ÍÊÈ¥£¬Óû¯Ñ§·½³Ìʽ±íʾԭÒò                     ¡£
£¨6£©ÔªËآߵÄ×î¸ßÕý¼ÛºÍ×îµÍ¸º¼Û·Ö±ðΪ       ¡¢ ¡¡  ¡¡£¬ÔÚÒ»¶¨Ìõ¼þÏ£¬ÔªËØ¢ßÓëH2·´Ó¦ÓÐÒ»¶¨ÏÞ¶È(¿ÉÀí½âΪ·´Ó¦½øÐеij̶È)£¬ÇëÅжÏÔÚÏàͬÌõ¼þÏÂÔªËØ¢ÞÓëH2·´Ó¦µÄÏÞ¶È(Ñ¡Ìî¡°¸ü´ó¡±¡¢¡°¸üС¡±»ò¡°Ïàͬ¡±)  ¡¡¡¡¡¡¡£

£¨24·Ö£©
(1) F,  Ar £¬14C     £¨3·Ö£©£¬  (2)   £»O="C=O" £»£¨3·Ö£©£¬
£¨2·Ö£©£¬
(3) Al2O3 + 6H+="2" Al3++3H2O£»2Al+2NaOH+2H2O=2NaAlO2+3H2¡ü£¨4·Ö£©
(4)

Al£­3e£­£½Al3£«£¨2·Ö£©£¨2·Ö£©£¬
(5) dz»ÆÂÌÉ«£¬SO2 + Cl2 + 2 H2O = H2SO4 + 2 H Cl £¨2·Ö£©
(6) +6£¨1·Ö£©£» -2 £¨1·Ö£©£»¸üС£¨1·Ö£©

½âÎö

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
ÏÂͼÊÇÔªËØÖÜÆÚ±íµÄÒ»²¿·Ö£¬¸ù¾Ý¢Ù¡«¢àÔÚÖÜÆÚ±íÖеÄλÖð´ÌâĿҪÇ󻨴ð£º
    ×å
ÖÜÆÚ
¢ñA ¢òA ¢óA ¢ôA ¢õA ¢öA ¢õ¢òA 0
¶þ ¢Ù ¢Ú ¢Û
Èý ¢Ü ¢Ý ¢Þ ¢ß ¢à
£¨1£©ÔªËآ١«¢àÖУ¬»¯Ñ§ÐÔÖÊ×î²»»îÆÃÔªËØµÄÔ­×ӽṹʾÒâͼÊÇ
£®³ý¢àÍ⣬ԭ×Ó°ë¾¶×î´óµÄÊÇ
Na
Na
£¨ÌîÔªËØ·ûºÅ£©£¬ÔªËآߵÄÇ⻯ÎïµÄµç×ÓʽÊÇ
£®
£¨2£©¢Ù¢Ú¢ÛÈýÖÖÔªËØµÄÆøÌ¬Ç⻯ÎïµÄÎȶ¨ÐÔÓÉÇ¿µ½ÈõµÄ˳ÐòÊÇ
HF£¾NH3£¾CH4
HF£¾NH3£¾CH4
£¨ÓÃÏàÓ¦Ç⻯ÎïµÄ»¯Ñ§Ê½×÷´ð£©£®ÔªËآ١«¢àÖеÄ×î¸ß¼ÛÑõ»¯Îï¶ÔÓ¦µÄË®»¯ÎïÖгÊÁ½ÐÔµÄÇâÑõ»¯ÎïÊÇ
Al£¨OH£©3
Al£¨OH£©3
£¨Ìѧʽ£©£¬ËüÓëÔªËØ¢ÜµÄ×î¸ß¼ÛÑõ»¯Îï¶ÔÓ¦µÄË®»¯ÎïµÄÈÜÒº·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ
Al£¨OH£©3+OH-=AlO2-+2H2O
Al£¨OH£©3+OH-=AlO2-+2H2O
£®
£¨3£©Ä³Í¬Ñ§Éè¼Æ²¢½øÐÐʵÑ飬̽¾¿ÔªËآܢݢ޽ðÊôÐÔµÝ±ä¹æÂÉ£¬Ç뽫ʵÑ鱨¸æ²¹³äÍêÈ«£®
      ʵÑé²½Öè ʵÑéÏÖÏó
1£®½«Ò»Ð¡¿é¢ÜµÄµ¥ÖÊ·ÅÈëµÎÓзÓ̪ÈÜÒºµÄÀäË®ÖÐ
ÄÆ¸¡ÔÚË®ÃæÉÏ£¬ÈÛ³ÉСÇò£¬ËÄ´¦Óζ¯
ÄÆ¸¡ÔÚË®ÃæÉÏ£¬ÈÛ³ÉСÇò£¬ËÄ´¦Óζ¯
£¬·¢³ö˻˻ÏìÉù£¬Öð½¥Ïûʧ£¬ÈÜÒº±ä³ÉºìÉ«
2£®½«ÉÙÁ¿ÀäË®×¢ÈëÊ¢ÓдòÄ¥¹ýµÄ¢ÝµÄµ¥ÖʵÄÊÔ¹ÜÖУ¬ÔٵμӷÓ̪£¬Ò»¶Îʱ¼äºó¼ÓÈÈÖÁ·ÐÌÚ
¼ÓÈȺó
þÌõ±íÃæ²úÉúÎÞÉ«ÆøÅÝ£¬ÈÜÒº³ÊºìÉ«
þÌõ±íÃæ²úÉúÎÞÉ«ÆøÅÝ£¬ÈÜÒº³ÊºìÉ«
3£®½«2mL 1mol/LÑÎËá¼ÓÈëÊ¢ÓдòÄ¥¹ýµÄ¢ÝµÄµ¥ÖʵÄÊÔ¹ÜÖÐ ¾çÁÒ·´Ó¦£¬Ñ¸ËÙ²úÉú´óÁ¿ÎÞÉ«ÆøÅÝ
4£®½«2mL 1mol/LÑÎËá¼ÓÈëÊ¢ÓдòÄ¥¹ýµÄ
Al
Al
µÄµ¥ÖʵÄÊÔ¹ÜÖÐ
·´Ó¦»ºÂý£¬Ò»¶Îʱ¼äºó£¬²úÉúÎÞÉ«ÆøÅÝ
½áÂÛ£º
Na¡¢Mg¡¢AlÈýÖÖÔªËØµÄ½ðÊôÐÔÒÀ´Î¼õÈõ
Na¡¢Mg¡¢AlÈýÖÖÔªËØµÄ½ðÊôÐÔÒÀ´Î¼õÈõ
£¨ÓÃÔªËØ·ûºÅ½áºÏÎÄ×Ö˵Ã÷£©
ÏÂͼÊÇÔªËØÖÜÆÚ±íµÄ²¿·Ö¿ò¼Ü£¬Çë»Ø´ð£º

£¨1£©¸ù¾ÝÔªËØÖÜÆÚ±í»Ø´ðÏÂÃæÎÊÌ⣺
a£©ÔªËØÖÜÆÚ±íÖеÄÔªËØ¢ÝºÍÔªËØ¢ÞµÄ×î¸ß¼ÛÑõ»¯ÎïµÄË®»¯ÎïÖÐËáÐÔ½ÏÇ¿ÊÇ
H3PO4
H3PO4
£¨Óû¯Ñ§Ê½±íʾ£©£®Óõç×Óʽ±íÊ¾ÔªËØ¢ÙÓë¢ÜµÄ»¯ºÏÎïµÄÐγɹý³Ì£º
£¬¸Ã»¯ºÏÎïÊôÓÚ
¹²¼Û
¹²¼Û
£¨Ìî¡°¹²¼Û¡±»ò¡°Àë×Ó¡±£©»¯ºÏÎ
b£©ÔªËØ¢Ù-¢ßÖеÄijµ¥ÖÊÔÚ³£ÎÂÏ»¯Ñ§ÐÔÖÊÎȶ¨£¬Í¨³£¿ÉÒÔ×ö±£»¤ÆøµÄÊÇ
£¨Óõç×Óʽ±íʾ£©£®
c£©Ì¼¡¢ÇâÁ½ÖÖÔªËØÄÜÐγÉÐí¶àÖÖ»¯ºÏÎÓÐЩ·Ö×Ó»¹»áÐγÉͬ·ÖÒì¹¹Ì壮д³öÓ뼺Í黥Ϊͬ·ÖÒì¹¹Ì壬ÇÒÒ»ÂÈ´úÎïÓÐÁ½ÖֵϝºÏÎïµÄ½á¹¹¼òʽ²¢ÃüÃû
CH3CH£¨CH3£©CH£¨CH3£©CH3
CH3CH£¨CH3£©CH£¨CH3£©CH3
£¬
2£¬3-¶þ¼×»ù¶¡Íé
2£¬3-¶þ¼×»ù¶¡Íé
£®
£¨2£©ÔªËØÖÜÆÚ±íÖÐλÓÚб¶Ô½ÇÏßµÄÁ½ÖÖÔªËØ¼°ÆäÐÔÖÊÊ®·ÖÏàËÆ£¬³ÆÖ®Îª¶Ô½ÇÏßÔ­Ôò£®¸ù¾ÝÔªËØÖÜÆÚ±í¶Ô½ÇÏßÔ­Ôò£¬½ðÊôBeÓëÂÁµ¥Öʼ°Æä»¯ºÏÎïÐÔÖÊÏàËÆ£®
a£©Ð´³öÖ¤Ã÷Al2O3ÊÇÁ½ÐÔÎïÖʵĻ¯Ñ§·´Ó¦µÄÀë×Ó·½³Ìʽ£º
Al2O3+6H+=2Al3++3H2O
Al2O3+6H+=2Al3++3H2O
£¬
Al2O3+2OH-=2AlO2-+H2O
Al2O3+2OH-=2AlO2-+H2O
£®
b£©Be£¨OH£©2ºÍMg£¨OH£©2¿ÉÓÃÊÔ¼Á
ÇâÑõ»¯ÄÆÈÜÒº
ÇâÑõ»¯ÄÆÈÜÒº
¼ø±ð£®

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

¾«Ó¢¼Ò½ÌÍø