ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿£¨17·Ö£©Ï±íÊÇÔªËØÖÜÆÚ±íµÄÒ»²¿·Ö, Õë¶Ô±íÖеĢ١«¢áÖÖÔªËØ,ÌîдÏÂÁпհ×(ÌîдÐòºÅ²»µÃ·Ö):

×å¡¡

ÖÜÆÚ

¢ñA

¢òA

¢óA

¢ôA

¢õA

¢öA

¢÷A

0×å

2

¢Ù

¢Ú

¢Û

3

¢Ü

¢Ý

¢Þ

¢ß

¢à

4

¢á

¢â

£¨1£©ÔÚÕâÐ©ÔªËØÖÐ,»¯Ñ§ÐÔÖÊ×î²»»îÆÃµÄÊÇ:_____¡££¨ÌîÔªËØ·ûºÅ£©

£¨2£©ÔÚ×î¸ß¼ÛÑõ»¯ÎïµÄË®»¯ÎïÖУ¬ËáÐÔ×îÇ¿µÄ»¯ºÏÎïµÄ»¯Ñ§Ê½ÊÇ/span>______£¬¼îÐÔ×îÇ¿µÄ»¯ºÏÎïµÄ»¯Ñ§Ê½ÊÇ:__________¡£

£¨3£©±È½Ï¢ÙÓë¢ÝµÄ×î¸ß¼ÛÑõ»¯Îï¶ÔÓ¦µÄË®»¯Î_________µÄËáÐÔÇ¿£¨Ìѧʽ£©£»ÄÜͨ¹ý________________________________˵Ã÷£¨Ð´·´Ó¦µÄ»¯Ñ§·½³Ìʽ£©¡£

£¨4£©ÊµÑéÊÒÖÆÈ¡¢ÚµÄÇ⻯ÎïµÄ»¯Ñ§·½³Ìʽ_______________________________ £¬¢ÚµÄÇ⻯ÎïÓë¢ÚµÄ×î¸ß¼ÛÑõ»¯ÎïµÄË®»¯Îï·´Ó¦ËùµÃµÄ²úÎﻯѧʽΪ______

£¨5£©¢Ú¿ÉÒÔÐγɶàÖÖÑõ»¯ÎÆäÖÐÒ»ÖÖÊǺì×ØÉ«ÆøÌ壬ÊÔÓ÷½³Ìʽ˵Ã÷¸ÃÆøÌå²»Ò˲ÉÓÃÅÅË®·¨ÊÕ¼¯µÄÔ­Òò________________£¨Óû¯Ñ§·½³Ìʽ±íʾ£©

£¨6£©±È½Ï¢ÛÓë¢ÞµÄÇ⻯Î_________ ¸üÎȶ¨£¨Ìѧʽ£©

£¨7£©Ð´³ö¢ÜµÄµ¥ÖÊÓëË®·´Ó¦µÄÀë×Ó·½³Ìʽ_______________________¡£

£¨8£©Ð´³ö¢ßÔªËØµÄÀë×ӽṹʾÒâͼ______,¸ÃÀë×Ó°ë¾¶_________ S2-£¨Ìî¡°©ƒ¡±»ò¡°©‚¡±£©Ð´³ö¢âÔªËØÔÚÖÜÆÚ±íµÄλÖÃ_______________________

¡¾´ð°¸¡¿ Ar HClO4 KOH H2CO3 Na2SiO3+CO2+H2O=Na2CO3+H2SiO3¡ý Ca(OH)2 +2NH4Cl CaCl2+2NH3¡ü+2H2O NH4NO3 3NO2+H2O=2HNO3+NO H2O ¡¾´ðÌâ¿Õ10¡¿2Na+2H2O==2Na++2OH-+H2¡ü £¼ µÚËÄÖÜÆÚµÚ¢÷A×å

¡¾½âÎö¡¿ÓÉÔªËØÔÚÖÜÆÚ±íÖеÄλÖÿÉÖª£¬¢ÙΪC¡¢¢ÚΪN¡¢¢ÛΪO¡¢¢ÜΪNa¡¢¢ÝΪSi¡¢¢ÞΪS¡¢¢ßΪCl¡¢¢àΪAr¡¢¢áΪK¡¢¢âΪBr¡£

£¨1£©Ï¡ÓÐÆøÌåAr×îÍâ²ãΪ8µç×ÓÎȶ¨½á¹¹£¬»¯Ñ§ÐÔÖÊ×î²»»îÆÃ£¬¹ÊÌîAr£»£¨2£©ÉÏÊöÔªËØÖÐClµÄ×î¸ß¼Ûº¬ÑõËá¸ßÂÈËáËáÐÔ×îÇ¿£¬¸ßÂÈËữѧʽΪHClO4£¬KµÄ½ðÊôÐÔ×îÇ¿£¬¹ÊKOH¼îÐÔ×îÇ¿£»£¨3£©Í¬Ö÷×å×ÔÉ϶øÏ£¬·Ç½ðÊôÐÔ¼õÈõ£¬×î¸ß¼ÛÑõ»¯Îï¶ÔӦˮ»¯ÎïµÄËáÐÔ¼õÈõ£¬¹ÊËáÐÔH2CO3£¾H2SiO3£»¿ÉÒÔÀûÓÃÇ¿ËáÖÆ±¸ÈõËáÔ­Àí½øÐÐÑéÖ¤£¬·½³ÌʽΪ£ºNa2SiO3+CO2+H2O=H2SiO3¡ý+Na2CO3£»£¨4£©ÊµÑéÊÒÓÃÂÈ»¯ï§ÓëÇâÑõ»¯¸Æ·´Ó¦ÖƱ¸°±Æø£¬·´Ó¦Éú³ÉÂÈ»¯¸Æ¡¢°±ÆøÓëË®£¬·´Ó¦·½³ÌʽΪ£º Ca(OH)2 +2NH4Cl CaCl2+2NH3¡ü+2H2O£»¢ÚµÄÇ⻯ÎïÓë¢ÚµÄ×î¸ß¼ÛÑõ»¯ÎïµÄË®»¯Îï·´Ó¦ËùµÃµÄ²úÎﻯѧʽΪNH4NO3£»£¨5£©¸Ãºì×ØÉ«ÆøÌåΪ¶þÑõ»¯µª£¬¶þÑõ»¯µªÓëË®·´Ó¦Éú³ÉÏõËáÓëNO£¬·´Ó¦·½³ÌʽΪ£º3NO2+H2O=2HNO3+NO£¬¹Ê¶þÑõ»¯µª²»ÄÜÓÃÅÅË®·¨ÊÕ¼¯£»£¨6£©Í¬Ö÷×å×ÔÉ϶øÏ·ǽðÊôÐÔ¼õÈõ£¬ÆøÌ¬Ç⻯ÎïÎȶ¨ÐÔ¼õÈõ£¬¹ÊH2O¸üÎȶ¨±È½Ï¢ÛÓë¢ÞµÄÇ⻯ÎH2O±ÈH2S¸üÎȶ¨£»£¨7£©ÄÆÓëË®·´Ó¦Éú³ÉÇâÑõ»¯ÄÆÓëÇâÆø£¬·´Ó¦Àë×Ó·½³ÌʽΪ£º2Na+2H2O=2Na£«+2OH£­+H2¡ü£»£¨8£©ÂÈÀë×ÓºËÍâÓÐ18¸öµç×Ó£¬ÓÐ3¸öµç×Ӳ㣬¸÷²ãµç×ÓÊýΪ2¡¢8¡¢8£¬Àë×ӽṹʾÒâͼΪ£»µç×Ó²ã½á¹¹Ïàͬ£¬ºËµçºÉÊýÔ½´óÀë×Ӱ뾶ԽС£¬¹ÊÀë×Ó°ë¾¶Cl£­£¼S2£­£»äå´¦ÓÚµÚËÄÖÜÆÚ¢÷A×å¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿Ä³¿ÎÍâÐËȤС×éÓû²â¶¨Ä³NaOHÈÜÒºµÄŨ¶È£¬Æä²Ù×÷²½ÖèÈçÏ£º

¢Ù½«¼îʽµÎ¶¨¹ÜÓÃÕôÁóˮϴ¾»£¬ÔÙÓôý²âÈÜÒºÈóÏ´ºó£¬×¢Èë´ý²âÈÜÒº£¬È»ºóµ÷½ÚµÎ¶¨¹ÜµÄ¼â×첿·Ö³äÂúÈÜÒº£¬²¢Ê¹ÒºÃæ´¦ÓÚ"0"¿Ì¶ÈÒÔϵÄλÖÃ,¼Ç϶ÁÊý£»

¢Ú½«ËáʽµÎ¶¨¹ÜÓÃÕôÁóˮϴ¾»£¬ÔÙÓñê×¼ËáÒºÈóÏ´ºó£¬ÏòÆäÖÐ×¢Èë0.1000mol/L±ê×¼ÑÎËᣬµ÷½ÚµÎ¶¨¹ÜµÄ¼â×첿·Ö³äÂúÈÜÒº£¬²¢Ê¹ÒºÃæ´¦ÓÚ"0"¿Ì¶ÈÒÔϵÄλÖ㬼Ç϶ÁÊý£»

¢ÛÓÃÕôÁóË®½«×¶ÐÎÆ¿Ï´¾»ºó£¬´Ó¼îʽµÎ¶¨¹ÜÖзÅÈë20.00mL´ý²âÈÜÒº£¬µÎÈë¼×»ù³È×÷ָʾ¼Á£¬È»ºóÓñê×¼ÑÎËá½øÐеζ¨¡£µÎ¶¨ÖÁָʾ¼Á¸ÕºÃ±äÉ«£¬ÇÒ°ë·ÖÖÓÄÚÑÕÉ«²»ÔٸıäΪֹ£¬²âµÃËùºÄÑÎËáµÄÌå»ýΪV1mL£»

¢ÜÖØ¸´ÒÔÉϹý³Ì£¬²âµÃËùºÄÑÎËáµÄÌå»ýΪV2mL¡£

ÊԻشðÏÂÁÐÎÊÌ⣺

£¨1£©×¶ÐÎÆ¿ÖÐÈÜÒºµÄÑÕÉ«´Ó_________É«±äΪ___________ɫʱ£¬Í£Ö¹µÎ¶¨¡£

£¨2£©ÏÂͼÖУ¬µÚ¢Ú²½¡°µ÷½ÚµÎ¶¨¹ÜµÄ¼â×첿·Ö³äÂúÈÜÒº¡±·½·¨ÕýÈ·µÄÊÇ_________£¬Èç¹ûµÎ¶¨Ç°ÓÐÆøÅÝ£¬µÎ¶¨ºóÆøÅÝÏûʧ£¬Óɴ˶Բⶨ½á¹ûÐγɵÄÓ°ÏìÊÇ___________£¨Ìî¡°Æ«¸ß¡±¡¢¡°Æ«µÍ¡±»ò¡°ÎÞÓ°Ï족£©

£¨3£©Í¼ÖÐÊÇij´ÎµÎ¶¨Ê±µÄµÎ¶¨¹ÜÖеÄÒºÃæ£¬Æä¶ÁÊýΪ__________mL¡£

£¨4£©¸ù¾ÝÏÂÁÐÊý¾Ý£º

µÎ¶¨´ÎÊý

´ý²âÒºÌå»ý

±ê×¼ÑÎËáÌå»ý(mL)

µÎ¶¨Ç°¶ÁÊý(mL)

µÎ¶¨ºó¶ÁÊý(mL)

µÚÒ»´Î

20.00

0.20

24.10

µÚ¶þ´Î

20.00

3.00

27.10

Çë¼ÆËã´ý²âÉÕ¼îÈÜÒºµÄŨ¶ÈΪ____________mol/L¡£

¡¾ÌâÄ¿¡¿¹ýÑõ»¯ÇâÊÇÖØÒªµÄÑõ»¯¼Á¡¢»¹Ô­¼Á£¬ËüµÄË®ÈÜÒºÓÖ³ÆÎªË«ÑõË®£¬³£ÓÃ×÷Ïû¶¾¡¢É±¾ú¡¢Æ¯°×µÈ¡£Ä³»¯Ñ§ÐËȤС×éȡһ¶¨Á¿µÄ¹ýÑõ»¯ÇâÈÜÒº£¬×¼È·²â¶¨Á˹ýÑõ»¯ÇâµÄº¬Á¿¡£ÇëÌîдÏÂÁпհףº

ÒÆÈ¡10.00 mLÃܶÈΪ¦Ñ g/mLµÄ¹ýÑõ»¯ÇâÔ­ÈÜҺϡÊͳÉ250 mL¡£Á¿È¡Ï¡¹ýÑõ»¯ÇâÈÜÒº25.00 mLÖÁ×¶ÐÎÆ¿ÖУ¬¼ÓÈëÏ¡ÁòËáËữ£¬ÓÃÕôÁóˮϡÊÍ£¬×÷±»²âÊÔÑù¡£

£¨1£©ÓøßÃÌËá¼Ø·¨(Ò»ÖÖÑõ»¯»¹Ô­µÎ¶¨·¨)¿É²â¶¨´ý²âÒºÖеÄH2O2µÄº¬Á¿¡£ÈôÐèÅä֯ō¶ÈΪ0.10 mol¡¤L1µÄKMnO4±ê×¼ÈÜÒº500 mL£¬Ó¦×¼È·³ÆÈ¡_________g KMnO4[ÒÑÖªM(KMnO4)=158.0 g¡¤mol1]¡£

a.ÅäÖÆ¸Ã±ê×¼ÈÜҺʱ£¬ÏÂÁÐÒÇÆ÷Öв»±ØÒªÓõ½µÄÓÐ_________¡£(ÓñàºÅ±íʾ)¡£

¢ÙÍÐÅÌÌìÆ½ ¢ÚÉÕ±­ ¢ÛÁ¿Í² ¢Ü²£Á§°ô ¢ÝÈÝÁ¿Æ¿ ¢Þ½ºÍ·µÎ¹Ü ¢ßÒÆÒº¹Ü

b.¶¨ÈÝʱÑöÊÓ¶ÁÊý£¬µ¼ÖÂ×îÖÕ½á¹û__________(Ìî¡°Æ«´ó¡±¡°Æ«Ð¡¡±»ò¡°²»±ä¡±)¡£

£¨2£©Íê³É²¢Å䯽Àë×Ó·½³Ìʽ£º

£«H2O2£«Mn2+£«O2¡ü£«

£¨3£©µÎ¶¨Ê±£¬½«¸ßÃÌËá¼Ø±ê×¼ÈÜҺעÈë____________(Ìî¡°Ëáʽ¡±»ò¡°¼îʽ¡±)µÎ¶¨¹ÜÖС£µÎ¶¨µ½´ïÖÕµãµÄÏÖÏóÊÇ____________________________________________¡£

£¨4£©Öظ´µÎ¶¨Èý´Î£¬Æ½¾ùºÄÓÃKMnO4±ê×¼ÈÜÒºV mL£¬ÔòÔ­¹ýÑõ»¯ÇâÈÜÒºÖйýÑõ»¯ÇâµÄÖÊÁ¿·ÖÊýΪ_____________¡£

£¨5£©ÈôµÎ¶¨Ç°µÎ¶¨¹Ü¼â×ìÖÐÓÐÆøÅÝ£¬µÎ¶¨ºóÆøÅÝÏûʧ£¬Ôò²â¶¨½á¹û______________ (Ìî¡°Æ«¸ß¡±»ò¡°Æ«µÍ¡±»ò¡°²»±ä¡±)¡£

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

¾«Ó¢¼Ò½ÌÍø