ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿(1)ÔÚ¢ÙLi¡¡¢Úʯī¡¡¢ÛC60¡¡¢ÜMg ¢Ý CH3CH2OH ¢ÞC ¢ßLi ¢à CH3OCH3 ÖУº____»¥ÎªÍ¬Î»ËØ£» ____»¥ÎªÍ¬·ÖÒì¹¹Ì壻___»¥ÎªÍ¬ËØÒìÐÎÌå(ÌîÐòºÅ)

(2)ÏÖÓТÙCaCl2 ¢Ú½ð¸Õʯ ¢ÛNH4Cl ¢ÜNa2SO4 ¢Ý±ù µÈÎåÖÖÎïÖÊ£¬°´ÏÂÁÐÒªÇó»Ø´ð£º

¢ÙÈÛ»¯Ê±²»ÐèÒªÆÆ»µ»¯Ñ§¼üµÄÊÇ___________£¬ÈÛµã×î¸ßµÄÊÇ_______¡£(ÌîÐòºÅ)

¢ÚÖ»º¬ÓÐÀë×Ó¼üµÄÎïÖÊÊÇ______£¬¾§ÌåÒÔ·Ö×Ó¼ä×÷ÓÃÁ¦½áºÏµÄÊÇ______¡£(ÌîÐòºÅ)

(3)д³öÏÂÁÐÎïÖʵĵç×Óʽ

¢ÙH2O______

¢ÚNaOH______

¢ÛNH3______

¡¾´ð°¸¡¿¢Ù¢ß ¢Ý¢à ¢Ú¢Û ¢Ý ¢Ú ¢Ù ¢Ý

¡¾½âÎö¡¿

(1)ÖÊ×ÓÊýÏàͬ¡¢ÖÐ×ÓÊý²»Í¬µÄÔ­×Ó»¥ÎªÍ¬Î»ËØ£»·Ö×ÓʽÏàͬ£¬½á¹¹²»Í¬µÄÓлúÎﻥΪͬ·ÖÒì¹¹Ì壻ͬÖÖÔªËØ×é³ÉµÄ²»Í¬µ¥ÖÊ»¥ÎªÍ¬ËØÒìÐÎÌ壻

(2)¢ÙCaCl2ÊÇÀë×Ó¾§Ì壻¢Ú½ð¸ÕʯÊÇÔ­×Ó¾§Ì壻 ¢ÛNH4ClÊÇÀë×Ó¾§Ì壻¢ÜNa2SO4ÊÇÀë×Ó¾§Ì壻¢Ý±ùÊÇ·Ö×Ó¾§Ì壻

(3)¢ÙH2OÊǹ²¼Û»¯ºÏÎ¢ÚNaOHÊÇÀë×Ó»¯ºÏÎ¢ÛNH3Êǹ²¼Û»¯ºÏÎï¡£

(1)¢ÙLi¡¢ ¢ßLiÊÇÖÊ×ÓÊýÏàͬ¡¢ÖÐ×ÓÊý²»Í¬µÄÔ­×Ó£¬¢Ù¡¢¢ß»¥ÎªÍ¬Î»ËØ£»CH3CH2OH¡¢CH3OCH3µÄ·Ö×Óʽ¶¼ÊÇC2H6O£¬½á¹¹²»Í¬£¬¢Ý¡¢¢à»¥ÎªÍ¬·ÖÒì¹¹Ì壻ʯī¡¢C60¶¼ÊÇÓÉCÔªËØ×é³ÉµÄµ¥ÖÊ£¬¢Ú¡¢¢Û»¥ÎªÍ¬ËØÒìÐÎÌ壻

(2)¢Ù·Ö×Ó¾§ÌåÈÛ»¯Ê±ÆÆ»µ·Ö×Ó¼ä×÷ÓÃÁ¦£¬ÈÛ»¯Ê±²»ÐèÒªÆÆ»µ»¯Ñ§¼üµÄÊDZù£¬Ñ¡¢Ý£»Ô­×Ó¾§ÌåÈÛ»¯ÐèÆÆ»µ¹²¼Û¼ü£¬Ô­×Ó¾§ÌåµÄÈÛµã¸ß£¬ÈÛµã×î¸ßµÄÊǽð¸Õʯ£¬Ñ¡¢Ú£»

¢ÚCaCl2ÊÇÀë×Ó¾§Ì壬ֻº¬ÓÐÀë×Ó¼ü£¬Ñ¡¢Ù£»·Ö×Ó¾§ÌåÒÔ·Ö×Ó¼ä×÷ÓÃÁ¦½áºÏ£¬ÒÔ·Ö×Ó¼ä×÷ÓÃÁ¦½áºÏµÄ¾§ÌåÊDZù£¬Ñ¡¢Ý£»

(3)¢ÙH2OÊǹ²¼Û»¯ºÏÎµç×ÓʽÊÇ£»

¢ÚNaOHÊÇÀë×Ó»¯ºÏÎµç×ÓʽÊÇ£»

¢ÛNH3Êǹ²¼Û»¯ºÏÎµç×ÓʽÊÇ¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿£¨1£©»ý¼«±£»¤Éú̬»·¾³¿ÉʵÏÖÈËÓë×ÔÈ»µÄºÍг¹²´¦¡£

¢ÙÏÂÁÐ×ö·¨»á¼Ó¾çÎÂÊÒЧӦµÄÊÇ__________£¨Ìî×Öĸ£©¡£

a.Ö²Ê÷ÔìÁÖ b.ȼú¹©Å¯ c.·çÁ¦·¢µç

¢ÚÏÂÁзÀÖΡ°°×É«ÎÛȾ¡±µÄÕýÈ··½·¨ÊÇ_____________£¨Ìî×Öĸ£©¡£

a.ʹÓÿɽµ½âËÜÁÏ b.¶Ìì·ÙÉշϾÉËÜÁÏ c.Ö±½ÓÌîÂñ·Ï¾ÉËÜÁÏ

¢ÛΪ¼õÇá´óÆøÎÛȾ£¬¶à¸ö³ÇÊÐÒѽûֹȼ·ÅÑÌ»¨±¬Öñ¡£¡°½ûֹȼ·ÅÑÌ»¨±¬Öñ¡±µÄ±êʶÊÇ_____£¨Ìî×Öĸ£©¡£

£¨2£©ºÏÀíʹÓû¯Ñ§ÖªÊ¶¿ÉÌá¸ßÈËÃǵÄÉú»îÖÊÁ¿¡£

ijƷÅÆÑÀ¸àµÄ³É·ÖÓиÊÓÍ¡¢É½ÀæËá¼Ø¡¢·ú»¯ÄƵȡ£

¢ÙÔÚÉÏÊöÑÀ¸à³É·ÖÖУ¬ÊôÓÚ·À¸¯¼ÁµÄÊÇ_______________¡£

¢Ú¸ÊÓ͵Ľṹ¼òʽΪ____________£»ÓÍ֬ˮ½â¿ÉÉú³É¸ÊÓͺÍ_____________¡£

¢Û·ú»¯ÄÆ(NaF)¿ÉÓëÑÀ³ÝÖеÄôÇ»ùÁ×Ëá¸Æ[Ca5(PO4)3OH]·´Ó¦£¬Éú³É¸üÄÑÈܵķúÁ×Ëá¸Æ[Ca5(PO4)3F]£¬´Ó¶ø´ïµ½·ÀÖÎÈ£³ÝµÄÄ¿µÄ¡£Ð´³ö¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º____________________¡£

£¨3£©´´Ð·¢Õ¹²ÄÁϼ¼Êõ¿ÆÍƶ¯ÈËÀàÉç»áµÄ½ø²½¡£

¢Ùʯīϩ£¨¼ûÏÂͼ£©¿ÉÓÃ×÷Ì«ÑôÄܵç³ØµÄµç¼«£¬ÕâÀïÖ÷ÒªÀûÓÃÁËʯīϩµÄ______________ÐÔ¡£

¢Ú»ù´¡¹¤³Ì½¨ÉèÖг£Óõ½Ë®Äà¡¢²£Á§¡¢¸Ö²ÄµÈ¡£Éú³ÉË®ÄàºÍ²£Á§¶¼Óõ½µÄÔ­ÁÏÊÇ__________£»ÔÚ¸Ö²ÄÖÐÌí¼Ó¸õ¡¢ÄøµÈÔªËصÄÄ¿µÄÊÇ___________¡£

¢ÛÐÂÐÍÕ½¶·»ú³£ÓÃÄÉÃ×SiC·ÛÌå×÷ΪÎü²¨²ÄÁÏ¡£¸ßÎÂϽ¹Ì¿ºÍʯӢ·´Ó¦¿ÉÖƵÃSiC£¬Ê¯Ó¢µÄ»¯Ñ§Ê½Îª________________£»¸ßηֽâSi(CH3)2Cl2Ò²¿ÉÖƵÃSiC£¬Í¬Ê±»¹Éú³ÉCH4ºÍÒ»ÖÖ³£¼ûËáÐÔÆøÌ壬д³ö¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º____________________________________¡£

¡¾ÌâÄ¿¡¿¡°84Ïû¶¾Òº¡±Òò1984Äê±±¾©Ä³Ò½ÔºÑÐÖÆʹÓöøµÃÃû£¬ÔÚÈÕ³£Éú»îÖÐʹÓù㷺£¬ÆäÓÐЧ³É·ÖÊÇNaClO¡£Ä³»¯Ñ§Ñо¿ÐÔѧϰС×éÔÚʵÑéÊÒÖƱ¸NaClOÈÜÒº£¬²¢½øÐÐÐÔÖÊ̽¾¿ºÍ³É·Ö²â¶¨

£¨1£©¸ÃѧϰС×é°´ÉÏͼװÖýøÐÐʵÑé(²¿·Ö¼Ð³Ö×°ÖÃÊ¡È¥)£¬·´Ó¦Ò»¶Îʱ¼äºó£¬·Ö±ðÈ¡B¡¢CÆ¿ÖеÄÈÜÒº½øÐÐʵÑ飬ʵÑéÏÖÏóÈçÏÂ±í¡£

ÒÑÖª£º1£®±¥ºÍNaClOÈÜÒºpHΪ11£»

2£®25¡ãCʱ£¬ÈõËáµçÀë³£ÊýΪ£ºH2CO3£ºK1=4.4¡Á10£­7£¬K2=4.7¡Á10£­11£»HClO£ºK=3¡Á10£­8

BÆ¿

CÆ¿

ʵÑé1£ºÈ¡Ñù£¬µÎ¼Ó×ÏɫʯÈïÊÔÒº

±äºì£¬²»ÍÊÉ«

±äÀ¶£¬²»ÍÊÉ«

ʵÑé2£º²â¶¨ÈÜÒºµÄpH

3

12

»Ø´ðÏÂÁÐÎÊÌ⣺

¢ÙÒÇÆ÷aµÄÃû³Æ_____£¬×°ÖÃAÖз¢Éú·´Ó¦µÄÀë×Ó·½³Ìʽ________¡£

¢ÚʵÑé1ÖÐBÆ¿ÈÜÒºÖвúÉúÏÖÏóµÄÔ­ÒòÊÇ_________¡£

¢ÛÈô½«CÆ¿ÈÜÒº»»³É NaHCO3ÈÜÒº£¬°´ÉÏÊö²Ù×÷²½Öè½øÐÐʵÑ飬CÆ¿ÏÖÏóΪ£ºÊµÑé1ÖÐ×ÏɫʯÈïÊÔÒºÁ¢¼´ÍÊÉ«£»ÊµÑé2ÖÐÈÜÒºµÄpH=7¡£½áºÏƽºâÒƶ¯Ô­Àí½âÊÍ×ÏɫʯÈïÊÔÒºÁ¢¼´ÍÊÉ«µÄÔ­Òò______¡£

£¨2£©²â¶¨CÆ¿ÈÜÒºÖÐNaClOº¬Á¿(µ¥Î»£ºg/L)µÄʵÑé²½ÖèÈçÏ£º

¢ñ£®È¡CÆ¿ÈÜÒº20mlÓÚ׶ÐÎÆ¿ÖУ¬¼ÓÈëÁòËáËữ£¬¼ÓÈë¹ýÁ¿KIÈÜÒº£¬¸Ç½ôÆ¿Èû²¢ÔÚ°µ´¦³ä·Ö·´Ó¦¡£

¢ò£®ÓÃ0.1000mol/LNa2S2O3±ê×¼ÈÜÒºµÎ¶¨×¶ÐÎÆ¿ÖеÄÈÜÒº£¬Ö¸Ê¾¼ÁÏÔʾÖÕµãºó£¬Öظ´²Ù×÷2~3´Î£¬Na2S2O3ÈÜÒºµÄƽ¾ùÓÃÁ¿Îª24.00ml¡£(ÒÑÖª£ºI2+2S2O32£­=2I£­+S4O62£­)

¢Ù²½Öè¢ñµÄCÆ¿Öз¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ_______¡£

¢Ú²½Öè¢òͨ³£Ñ¡ÓÃ___×÷ָʾ¼Á£¬µÎ¶¨ÖÁÖÕµãµÄÏÖÏó______¡£

¢ÛCÆ¿ÈÜÒºÖÐNaClOº¬Á¿Îª_____g/L(±£Áô2λСÊý)

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

¾«Ó¢¼Ò½ÌÍø