ÌâÄ¿ÄÚÈÝ

Ö¸³öÔÚʹÓÃÏÂÁÐÒÇÆ÷£¨ÒѾ­Ï´µÓ¸É¾»£©»òÓÃƷʱµÄµÚÒ»²½²Ù×÷£º

¢ÙʯÈïÊÔÖ½£¨¼ìÑéÆøÌ壩£º___________________________________________¡£

¢ÚÈÝÁ¿Æ¿£º___________________________________________¡£

¢ÛËáʽµÎ¶¨¹Ü£º___________________________________________¡£

¢Ü¼¯ÆøÆ¿£¨ÊÕ¼¯ÂÈ»¯Ç⣩£º___________________________________________¡£

£¨2£©ÏÂÁÐʵÑé²Ù×÷»ò¶ÔʵÑéʹʵÄÐðÊöÕýÈ·µÄÊÇ_________£¨ÌîÐòºÅ£©¡£

¢ÙÓÃÏ¡HNO3ÇåÏ´×ö¹ýÒø¾µ·´Ó¦ÊµÑéµÄÊԹܣ»

¢ÚÅäÖÆŨÁòËáºÍŨÏõËáµÄ»ìºÏËáʱ£¬½«Å¨ÁòËáÑØÆ÷±ÚÂýÂý¼ÓÈ뵽ŨÏõËáÖУ¬²¢²»¶Ï½Á°è£»

¢ÛÓüîʽµÎ¶¨¹ÜÁ¿È¡20.00 mL 0.1000 mol/L KMnO4ÈÜÒº£»

¢ÜÓÃÍÐÅÌÌìƽ³ÆÈ¡10.50 g¸ÉÔïµÄNaCl¹ÌÌ壻

¢Ý²»É÷½«ÒºäåÕ´µ½Æ¤·ôÉÏ£¬Á¢¼´Óþƾ«ÇåÏ´£»

¢ÞÓôÉÛáÛö¸ßÎÂÈÛÈÚNaOHºÍNa2CO3¹ÌÌ壻

¢ßÏò·ÐÌÚµÄNaOHÏ¡ÈÜÒºÖеμÓFeCl3±¥ºÍÈÜÒº£¬ÒÔÖƱ¸Fe(OH)3½ºÌ壻

¢àÅäÖÆAl2(SO4)3ÈÜҺʱ£¬¼ÓÈëÉÙÁ¿µÄÏ¡ÁòË᣻

¢áŨH2SO4Õ´µ½Æ¤·ôÉÏ£¬Á¢¼´ÓÃNaOHÈÜÒºÇåÏ´¡£

½âÎö£ºÀûÓÃÏ¡HNO3ÓëAg·¢Éú»¯Ñ§·´Ó¦£¬Í¨³£ÓÃÏ¡HNO3ÇåÏ´×ö¹ýÒø¾µ·´Ó¦µÄÊԹܣ¬¢ÙÕýÈ·£»ÈÜÒº»ìºÏʱ×ñÑ­ÃܶȴóµÄÒºÌå¼Óµ½ÃܶÈСµÄÒºÌåÖУ¬¢ÚÕýÈ·£»KMnO4¾ßÓÐÇ¿Ñõ»¯ÐÔ£¬¿ÉÑõ»¯Ï𽺣¬¹Ê²»ÄÜÓüîʽµÎ¶¨¹ÜÁ¿È¡KMnO4ÈÜÒº£¬¢Û´íÎó£»ÍÐÅÌÌìƽµÄ×îС·Ö¶ÈֵΪ0.1 g£¬²»ÄܳÆÁ¿×¼È·ÖÁ0.01 gµÄÑùÆ·£¬¢Ü´íÎó£»äåµ¥ÖÊÓи¯Ê´ÐÔ£¬²»É÷Õ´µ½Æ¤·ôÉÏ£¬Ó¦Óþƾ«½«ÆäÈܽâµô£¬¢ÝÕýÈ·£»´ÉÛáÛöÖÐÓÐSiO2,SiO2¿ÉÓëNaOH¡¢Na2CO3µÈ¼îÐÔÎïÖÊÔÚ¸ßÎÂÏ·´Ó¦£¬¢Þ´íÎó£»ÏòNaOHÈÜÒºÖмÓÈëFeCl3»áÉú³ÉFe(OH)3³Áµí£¬ÖƱ¸Fe(OH)3½ºÌåÓ¦Ïò·ÐË®ÖмÓÈë±¥ºÍFeCl3ÈÜÒº£¬¢ß´íÎó£»ÎªÒÖÖÆAl2(SO4)3µÄË®½â£¬¿É¼ÓÉÙÁ¿Ï¡H2SO4£¬¢àÕýÈ·£»Å¨H2SO4Õ´µ½Æ¤·ôÉÏ£¬Ó¦ÏÈÓò¼²ÁÈ¥£¬ÔÙÓôóÁ¿Ë®³åÏ´£¬¢á´íÎó¡£

´ð°¸£º£¨1£©¢ÙÓÃÕôÁóË®Èóʪ   ¢Ú¼ìÑéÆøÃÜÐÔ£¨»òÊÇ·ñ©ˮ£©   ¢Û¼ìÑéÊÇ·ñ©ˮ   ¢Ü¼ìÑé¸ÉÔïÐÔ   £¨2£©¢Ù¢Ú¢Ý¢à

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
(¢ñ)Ö¸³öÔÚʹÓÃÏÂÁÐÒÇÆ÷(ÒѾ­Ï´µÓ¸É¾»)»òÓÃƷʱµÄµÚÒ»²½²Ù×÷£º

¢ÙʯÈïÊÔÖ½(¼ìÑéÆøÌå)£º¡£?

¢ÚµÎ¶¨¹Ü£º                ¡£?

¢Û¼¯ÆøÆ¿(ÊÕ¼¯°±Æø)£º                 ¡£?

¢ÜÍÐÅÌÌìƽ                           ¡£?

(¢ò)Ä¿Ç°¹ú¼ÊÓͼ۾Ӹ߲»Ï¡£ÒÒ´¼ÆûÓÍÊÇÖ¸ÔÚÆûÓÍÖмÓÈë10%µÄÒÒ´¼(Ìå»ý±È)¡£ÒÒ´¼ÆûÓ;ßÓÐÐÁÍéÖµ¸ß¡¢¿¹±¬ÐԺõÄÌص㣬ÍƹãʹÓÿɻº½âÄÜÔ´½ôȱ¡¢´Ù½ø¹úÃñ¾­¼Ã·¢Õ¹¡£ÆûÓÍÖÐÌí¼ÓµÄÒÒ´¼²»Äܺ¬Ë®£¬·ñÔò»áÓ°Ïì·¢¶¯»úÕý³£ÔËתºÍʹÓÃÊÙÃü¡£ÖƱ¸ÎÞË®ÒÒ´¼¿É²ÉÈ¡ÈçÏ·½·¨£º??

¢ÙÔÚ250 mLÔ²µ×ÉÕÆ¿ÖмÓÈë95%µÄÒÒ´¼100 mLºÍÐÂÖƵÄÉúʯ»Ò30 g£¬ÔÚˮԡÖмÓÈÈ»ØÁ÷1ÖÁ2Сʱ¡£(Èçͼ1Ëùʾ)?

¢ÚÈ¡ÏÂÀäÄý¹Ü£¬¸Ä³ÉÈçͼ2ËùʾµÄ×°Öã¬ÔÙ½«AÖÃÓÚˮԡÖÐÕôÁó¡£?

¢Û°Ñ×î³õÕô³öµÄ5 mLÁó³öÒºÁíÍâ»ØÊÕ¡£?

¢ÜÓúæ¸ÉµÄÎüÂËÆ¿×÷Ϊ½ÓÊÜÆ÷£¬Æä²à¹Ü½Óһ֧װÓÐCaCl2µÄ¸ÉÔï¹ÜC£¬Ê¹ÆäÓë´óÆøÏàͨ£¬ÕôÖÁÎÞÒºµÎ³öÀ´ÎªÖ¹£¬¼´µÃ99.5%µÄ¾Æ¾«¡£ÊԻشð£º?

(1)ÒÑÖªÔÚ101 kPa¡¢25 ¡æʱ£¬?

C2H5OH(l)+3O2(g)=2CO2(g)+3H2O(l);¦¤H =-1 367 kJ¡¤mol-1??

1 gÒÒ´¼ÍêȫȼÉÕÉú³ÉҺ̬ˮʱ·Å³ö            ÈÈÁ¿¡£?

(2)ͼ2ÖеĸÉÔï¹ÜCµÄ×÷ÓÃÊÇ            ¡£ÒÇÆ÷BµÄÃû³ÆÊÇ            £¬ÀäÄýË®ÊÇ´Ó¿Ú            (Ìî¡°a¡±»ò¡°b¡±)½øÈëÀäÄý¹Ü¡£?

(3)ÎÞË®CaCl2³£ÓÃ×÷ÎüË®¼Á£¬ÔÚÉÕÆ¿AÖÐÄÜ·ñÓÃÎÞË®CaCl2´úÌæÉúʯ»Ò£¿            (Ìîд¡°ÄÜ¡±¡°²»ÄÜ¡±)£¬Ô­ÒòÊÇ                           ¡£?

(4)¼ìÑéËùµÃ²úÆ·ÖÐÊÇ·ñº¬Ë®µÄ²Ù×÷·½·¨ÊÇ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡£

(5)ijͬѧÏëÒª¼ìÑéͼ2×°ÖõÄÆøÃÜÐÔ£¬Ëû¿É´Óͼ3ÖÐÑ¡ÔñÄļ¸ÖÖÒÇÆ÷(ÌîÒÇÆ÷´úºÅ)            ¡£Çë¼òÊö¼ìÑéÆøÃÜÐԵIJÙ×÷¹ý³Ì            ¡£?

(6)д³öÓÉÆÏÌÑÌÇת»¯Îª¾Æ¾«µÄ»¯Ñ§·½³Ìʽ£º                                 ¡£?

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

¾«Ó¢¼Ò½ÌÍø