ÌâÄ¿ÄÚÈÝ

ÏÂÁÐÎªÔªËØÖÜÆÚ±íÖеÄÒ»²¿·Ö£¬Óû¯Ñ§Ê½»òÔªËØ·ûºÅ»Ø´ðÏÂÁÐÎÊÌ⣺

(1)11ÖÖÔªËØÖУ¬»¯Ñ§ÐÔÖÊ×î²»»îÆÃµÄÊÇ________£®

(2)¢Ù¢Ú¢ÝÖУ¬×î¸ß¼ÛÑõ»¯ÎïµÄË®»¯Î¼îÐÔ×îÇ¿µÄÊÇ________£¬

(3)¢Ú¢Û¢ÜÖÐÐγɵļòµ¥Àë×Ó°ë¾¶ÓÉ´óµ½Ð¡µÄ˳ÐòÊÇ________£®

(4)ÔªËØ¢ßµÄÇ⻯Îï·Ö×ÓʽΪ________£¬¸ÃÇ⻯Îï³£ÎÂϺÍÔªËØ¢ÚµÄµ¥ÖÊ·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ________£¬¸ÃÇ⻯ÎïÓëÔªËØ¢àµÄµ¥ÖÊ·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ________£®

(5)¢ÙºÍ¢áµÄ×î¸ß¼ÛÑõ»¯Îï¶ÔÓ¦µÄË®»¯ÎﻯѧʽΪ________ºÍ________£®¢ÙºÍ¢áÁ½ÔªËØÐγɵϝºÏÎïµÄÈÜÒºÓëÔªËØ¢àµÄµ¥ÖÊ·´Ó¦µÄÀë×Ó·½³ÌʽΪ_________£®

(6)¢ÙºÍ¢Ý×î¸ß¼ÛÑõ»¯Îï¶ÔÓ¦µÄË®»¯ÎïÏ໥·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ________£®

(7)¢ÞºÍ¢àÐγɵϝºÏÎïµÄ»¯Ñ§Ê½Îª________£¬¸Ã»¯ºÏÎïÈܽâ¢áµÄµ¥ÖÊËùµÃÈÜҺΪ________£®

(8)¢à¢áÈýÖÖÔªËØÐÎ³ÉµÄÆøÌ¬Ç⻯Îï×îÎȶ¨µÄÊÇ________£®

´ð°¸£ºÂÔ
½âÎö£º

(1)Ar

(2)KOH

(3)

(4)

(5)£»£»

(6)

(7)£»µÄÈÜÒº

(8)HF


Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
ÏÂÁÐÎªÔªËØÖÜÆÚ±íÖеÄÒ»²¿·Ö£¬°´ÒªÇ󻨴ðÏÂÁÐÎÊÌ⣮
¢ñA 0
1 ¢Ù ¢òA ¢óA ¢ôA ¢õA ¢öA ¢÷A
2 ¢Ú ¢Ü ¢Ý
3 ¢Þ ¢ß ¢Û ¢à ¢á
4 ¢â
£¨1£©ÕâÐ©ÔªËØÖÐ×î²»»îÆÃµÄÔªËØÃû³ÆÎª£º
ÄÊ
ÄÊ
£»
£¨2£©ÔªËآޢߢâµÄ×î¸ß¼ÛÑõ»¯Îï¶ÔÓ¦µÄË®»¯ÎïÖмîÐÔ×îÇ¿µÄÊÇ£¨Ìѧʽ£©£º
KOH
KOH
£»
£¨3£©ÔªËØ¢áµÄµ¥ÖÊÓëH2O·´Ó¦µÄÀë×Ó·½³ÌʽΪ£º
Cl2+H2O¨TH++Cl-+HClO
Cl2+H2O¨TH++Cl-+HClO
£»
£¨4£©ÔªËØ¢ÛµÄÑõ»¯ÎïÓëÇâÑõ»¯¼ØÈÜÒº·´Ó¦µÄÀë×Ó·½³ÌʽΪ
Al2O3+2OH-+3H2O=2[Al£¨OH£©4]-
Al2O3+2OH-+3H2O=2[Al£¨OH£©4]-
£»
£¨5£©ÔªËآ٢ܢâÐγɵϝºÏÎïÖеĻ¯Ñ§¼üÀàÐÍΪ£º
Àë×Ó¼ü£¬¹²¼Û¼ü
Àë×Ó¼ü£¬¹²¼Û¼ü
£®
£¨6£©±È½ÏÔªËØ¢Þ¢ß¢à¢áÀë×ӵİ뾶ÓÉ´óµ½Ð¡µÄ˳Ðò£¨Ìѧ·ûºÅ£©Îª£º
S2-£¾Cl-£¾Na+£¾Mg2+
S2-£¾Cl-£¾Na+£¾Mg2+
£®
£¨7£©ÔªËØÖÜÆÚ±íÌåÏÖÁËÔªËØÖÜÆÚÂÉ£¬ÔªËØÖÜÆÚÂɵı¾ÖÊÊÇÔ­×ÓºËÍâµç×ÓÅŲ¼µÄ
ÖÜÆÚÐԱ仯
ÖÜÆÚÐԱ仯
£¬Çëд³öÖ÷×åÔªËØÔÚÔªËØÖÜÆÚ±íÖеÄλÖÃÓëÔªËØÔ­×ӽṹµÄ¹ØÏµ£º
ÔªËØµÄÖÜÆÚÊý¼´ÎªÔ­×ÓºËÍâµç×Ó²ãÊý£¬ÔªËصÄÖ÷×åÐòÊý¼´ÎªÔªËØÔ­×ÓµÄ×îÍâ²ãµç×ÓÊý
ÔªËØµÄÖÜÆÚÊý¼´ÎªÔ­×ÓºËÍâµç×Ó²ãÊý£¬ÔªËصÄÖ÷×åÐòÊý¼´ÎªÔªËØÔ­×ÓµÄ×îÍâ²ãµç×ÓÊý
£®

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

¾«Ó¢¼Ò½ÌÍø