ÌâÄ¿ÄÚÈÝ

º£Ë®ÖÐÓзdz£·á¸»µÄ»¯Ñ§×ÊÔ´£¬´Óº£Ë®ÖпÉÌáÈ¡¶àÖÖ»¯¹¤Ô­ÁÏ£®ÊԻشðÏÂÁÐÎÊÌ⣺
£¨1£©´ÖÑÎÖк¬ÓÐCa2+¡¢Mg2+¡¢SO42-µÈÔÓÖÊ£¬¾«Öƺó¿ÉµÃ±¥ºÍNaClÈÜÒº£¬¾«ÖÆʱͨ³£ÔÚÈÜÒºÖмÓÈëÒÔÏÂÎïÖÊ£¬²¢×îºóʹÈÜÒº³ÊÖÐÐÔ£®
¢Ù¹ýÁ¿µÄNa2CO3ÈÜÒº¡¢¢ÚÑÎËá¡¢¢Û¹ýÁ¿µÄNaOHÈÜÒº¡¢¢Ü¹ýÁ¿µÄBaCl2ÈÜÒºÔò£¬¼ÓÈëµÄ˳ÐòÊÇ______
£¨2£©Ð´³ö¼ÓÈë×îºóÒ»ÖÖÊÔ¼Áºó¿ÉÄÜ·¢ÉúµÄ»¯Ñ§·´Ó¦µÄÀë×Ó·½³Ìʽ______£®
£¨1£©SO42-¡¢Ca2+¡¢Mg2+µÈ·Ö±ðÓëBaCl2ÈÜÒº¡¢Na2CO3ÈÜÒº¡¢NaOHÈÜÒº·´Ó¦Éú³É³Áµí£¬¿ÉÔÙͨ¹ý¹ýÂ˳ýÈ¥£¬Na2CO3ÈÜÒºÄܳýÈ¥¹ýÁ¿µÄBaCl2ÈÜÒº£¬ÑÎËáÄܳýÈ¥¹ýÁ¿µÄNa2CO3ÈÜÒººÍNaOHÈÜÒº£¬ËùÒÔÓ¦ÏȼÓBaCl2ÈÜÒºÔÙ¼ÓNa2CO3ÈÜÒº£¬×îºó¼ÓÈëÑÎËᣬËùÒÔÕýȷ˳ÐòΪ£º¢Ü¢Û¢Ù¢Ú»ò¢Û¢Ü¢Ù¢Ú£¬
¹Ê´ð°¸Îª£º¢Ü¢Û¢Ù¢Ú»ò¢Û¢Ü¢Ù¢Ú£»
£¨2£©ÒòNaOHºÍNa2CO3¹ýÁ¿£¬ÄÜÓëÑÎËá·´Ó¦£¬ËùÒÔÀë×Ó·½³ÌʽΪ£ºH++OH-¨TH2O¡¢2H++CO32-¨TH2O+CO2¡ü£¬
¹Ê´ð°¸Îª£ºH++OH-¨TH2O¡¢2H++CO32-¨TH2O+CO2¡ü£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
£¨10·Ö£©¶þÑõ»¯ÂÈ£¨ClO2£©ÊÇÒ»ÖÖÔÚË®´¦ÀíµÈ·½ÃæÓй㷺ӦÓõĸßЧ°²È«Ïû¶¾¼Á¡£ÓëCl2±È£¬ClO2²»µ«¾ßÓиüÏÔÖøµÄɱ¾úÄÜÁ¦£¬¶øÇÒ²»»á²úÉú¶ÔÈËÌåÓÐDZÔÚΣº¦µÄÓлúÂÈ´úÎï¡£
£¨1£©ÔÚClO2µÄÖƱ¸·½·¨ÖУ¬ÓÐÏÂÁÐÁ½ÖÖÖƱ¸·½·¨£º
·½·¨Ò»£º2NaClO3£«4HCl£½2ClO2¡ü£«Cl2¡ü£«2NaCl£«2H2O
·½·¨¶þ2NaClO3£«H2O2£«H2SO4£½2ClO2¡ü£«Na2SO4£«O2¡ü£«2H2O£º
Ó÷½·¨¶þÖƱ¸µÄClO2¸üÊʺÏÓÃÓÚÒûÓÃË®µÄÏû¶¾£¬ÆäÖ÷ÒªÔ­ÒòÊÇ               ¡£
£¨2£©ÓÃClO2´¦Àí¹ýµÄÒûÓÃË®£¨pHΪ5.5~6.5£©³£º¬ÓÐÒ»¶¨Á¿¶ÔÈËÌå²»ÀûµÄÑÇÂÈËá¸ùÀë×Ó
£¨ClO2£­£©¡£2001ÄêÎÒ¹úÎÀÉú²¿¹æ¶¨£¬ÒûÓÃË®ÖÐClO2£­µÄº¬Á¿Ó¦²»³¬¹ý0.2mg¡¤L£­1¡£
ÒûÓÃË®ÖÐClO2¡¢ClO2£­µÄº¬Á¿¿ÉÓÃÁ¬ÐøµâÁ¿·¨½øÐвⶨ¡£ClO2±»I£­»¹Ô­ÎªClO2£­¡¢Cl£­µÄת»¯ÂÊÓëÈÜÒºpHµÄ¹ØϵÈçͼËùʾ¡£

µ±pH¡Ü2.0ʱ£¬ClO2£­Ò²Äܱ»I£­»¹Ô­³ÉCl£­¡£
·´Ó¦Éú³ÉµÄI2Óñê×¼Na2S2O3ÈÜÒºµÎ¶¨£º
Na2S2O3£«I2£½Na2S4O6£«2NaI
¢ÙÇëд³öpH¡Ü2.0ʱ£¬ClO2£­ÓëI£­·´Ó¦µÄÀë×Ó·½³Ìʽ                       ¡£
¢ÚÇëÍê³ÉÏàÓ¦µÄʵÑé²½Ö裺
²½Öè1£º×¼È·Á¿È¡VmLË®Ñù¼ÓÈ뵽׶ÐÎÆ¿ÖС£
²½Öè2£ºµ÷½ÚË®ÑùµÄpHΪ7.0~8.0¡£
²½Öè3£º¼ÓÈë×ãÁ¿µÄKI¾§Ìå¡£
²½Öè4£º¼ÓÈëÉÙÁ¿µí·ÛÈÜÒº£¬ÓÃcmol¡¤L£­1Na2S2O3ÈÜÒºµÎ¶¨ÖÁÖյ㣬ÏûºÄNa2S2O3
ÈÜÒºV1mL¡£
²½Öè5£º                       ¡£
²½Öè6£ºÔÙÓÃcmol¡¤L£­1Na2S2O3ÈÜÒºµÎ¶¨ÖÁÖյ㣬ÏûºÄNa2S2O3ÈÜÒºV2mL¡£
¢Û¸ù¾ÝÉÏÊö·ÖÎöÊý¾Ý£¬²âµÃ¸ÃÒûÓÃË®ÑùÖÐClO2£­Å¨¶ÈΪ             mol¡¤L£­1Óú¬×ÖĸµÄ´úÊýʽ±íʾ£©
¢ÜÈôÒûÓÃË®ÖÐClO2£­µÄº¬Á¿³¬±ê£¬¿ÉÏòÆäÖмÓÈëÊÊÁ¿µÄFe2£«½«ClO2£­»¹Ô­³ÉCl£­£¬
¸Ã·´Ó¦µÄÑõ»¯²úÎïÊÇ            £¨Ìѧʽ£©¡£
äåÒÒÍéÊÇÒ»ÖÖÖØÒªµÄÓлú»¯¹¤Ô­ÁÏ£¬ÖƱ¸äåÒÒÍéµÄÔ­ÁÏÓÐ95%ÒÒ´¼¡¢80%ÁòËᣨÓÃÕôÁóˮϡÊÍŨÁòËᣩ¡¢ÑÐϸµÄä廯ÄÆ·ÛÄ©ºÍ¼¸Á£Ëé´ÉƬ£¬¸Ã·´Ó¦µÄÔ­ÀíÈçÏ£º
NaBr+H2SO4¡úNaHSO4+HBr
CH3CH2OH+HBr
ÁòËá
CH3CH2Br+H2O
ij¿ÎÍâС×éÓûÔÚʵÑéÊÒÖƱ¸äåÒÒÍéµÄ×°ÖÃÈçͼ£®Êý¾ÝÈç±í£®
ÎïÖÊ
Êý¾Ý
ÒÒ´¼äåÒÒÍé1£¬2-¶þäåÒÒÍéÒÒÃÑŨÁòËá
ÃܶÈ/g?cm-30.791.462.20.711.84
È۵㣨¡æ£©-130-1199-11610
·Ðµã£¨¡æ£©78.538.413234.6338
ÔÚË®ÖеÄÈܽâ¶È£¨g/100gË®£©»¥ÈÜ0.91417.5»¥ÈÜ
Çë»Ø´ðÏÂÁÐÎÊÌ⣮
£¨1£©¼ÓÈëҩƷ֮ǰÐë×öµÄ²Ù×÷ÊÇ£º______£¬ÊµÑé½øÐеÄ;ÖÐÈô·¢ÏÖδ¼ÓÈëËé´ÉƬ£¬Æä´¦ÀíµÄ·½·¨ÊÇ______£®
£¨2£©×°ÖÃBµÄ×÷ÓÃÊdzýÁËʹäåÒÒÍéÁó³ö£¬»¹ÓÐÒ»¸öÄ¿µÄÊÇ______£®Î¶ȼƵÄζÈÓ¦¿ØÖÆÔÚ______Ö®¼ä£®
£¨3£©·´Ó¦Ê±ÓпÉÄÜÉú³ÉSO2ºÍÒ»ÖÖºì×ØÉ«ÆøÌ壬¿ÉÑ¡ÔñÇâÑõ»¯ÄÆÈÜÒº³ýÈ¥¸ÃÆøÌ壬ÓйصÄÀë×Ó·½³ÌʽÊÇ______£¬______£¬´Ë²Ù×÷¿ÉÔÚ______£¨Ìîд²£Á§ÒÇÆ÷Ãû³Æ£©ÖнøÐУ¬Í¬Ê±½øÐзÖÀ룮
£¨4£©ÊµÑéÖвÉÓÃ80%ÁòËᣬ¶ø²»ÄÜÓÃ98%ŨÁòËᣬһ·½ÃæÊÇΪÁ˼õÉÙ¸±·´Ó¦£¬ÁíÒ»·½ÃæÊÇΪÁË______£®
£¨5£©´Ö²úÆ·Öк¬ÓеÄÖ÷ÒªÓлúÒºÌåÔÓÖÊÊÇ______£¬Îª½øÒ»²½ÖƵô¿¾»µÄäåÒÒÍ飬¶Ô´Ö²úÆ·½øÐÐˮϴµÓ¡¢·ÖÒº£¬ÔÙ¼ÓÈëÎÞË®CaCl2£¬½øÐÐ______²Ù×÷£®

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

¾«Ó¢¼Ò½ÌÍø