ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿°´ÒªÇó»Ø´ðÏÂÁÐÎÊÌ⣺

¢ñ¡¢ ÏÂÁи÷×éÎïÖÊ£º¢ÙO2ºÍO3 ¢ÚH2¡¢D2¡¢T2 ¢Û12 CºÍ14 C ¢ÜCH3CH2CH2CH3ºÍ(CH3)2CHCH3 ¢Ý¹ïÍéºÍÊ®ÁùÍé ¢ÞCH3(CH2)5CH3ºÍCH3CH2CH2CH(CH3)C2H5¢ß(ÔÚºáÏßÉÏÌîÏàÓ¦µÄÐòºÅ)

A¡¢»¥ÎªÍ¬Î»ËصÄÊÇ_____________£»B¡¢»¥ÎªÍ¬·ÖÒì¹¹ÌåµÄÊÇ____________£»

C¡¢»¥ÎªÍ¬ËØÒìÐÎÌåµÄÊÇ____________£»D¡¢Í¬Ò»ÖÖÎïÖʵÄÊÇ_____________¡£

¢ò¡¢ д³öÏÂÁÐÍéÌþµÄ·Ö×Óʽ

£¨1£©¼ÙÈçijÍéÌþµÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª142£¬Ôò¸ÃÍéÌþµÄ·Ö×ÓʽΪ £»

£¨2£©ÍéÌþAÔÚͬÎÂͬѹÏÂÕôÆøµÄÃܶÈÊÇH2µÄ36±¶ £»

£¨3£©1LÍéÌþDµÄÕôÆøÍêȫȼÉÕʱ,Éú³ÉͬÎÂͬѹÏÂ15LË®ÕôÆø ¡£

¢ó¡¢£¨1£© °´ÏµÍ³ÃüÃû·¨ÃüÃû:

¢Ù ,д³öËüºÍÂÈÆø·¢Éúһȡ´ú·´Ó¦µÄ·½³Ìʽ

¢Ú £»ËüµÄÒ»ÂÈ´úÎï¾ßÓв»Í¬·ÐµãµÄ²úÎïÓÐ ÖÖ

£¨2£© д³öÏÂÁи÷ÓлúÎïµÄ½á¹¹¼òʽ£º

¢Ù2£¬3£­¶þ¼×»ù£­4£­ÒÒ»ùÒÑÍé £º_____________________£»

¢ÚÖ§Á´Ö»ÓÐÒ»¸öÒÒ»ùÇÒÏà¶Ô·Ö×ÓÖÊÁ¿×îСµÄÍéÌþ£º___________________£»

£¨3£©ôÇ»ùµÄµç×Óʽ £»

¡¾´ð°¸¡¿

¢ñ¡¢A.¢Û B.¢Ü¢Þ C.¢Ù D.¢ß

¢ò¡¢£¨1£©C10H22 £¨2£© C5H12 £¨3£© C14H30

¢ó£¨1£©¢Ù2£¬2¡ª¶þ¼×»ù±ûÍ飻

+Cl2(CH3)3CCH2Cl+HCl £»

¢Ú3,3,5,5¡ªËļ׻ù¸ýÍ飻4£»

£¨2£©¢ÙCH3CH(CH3)CH(CH3)CH(C2H5)CH2CH3£»

¢ÚCH3CH2CH(C2H5)CH2CH3 £»

£¨3£©£»

¡¾½âÎö¡¿

ÊÔÌâ·ÖÎö£º¢ñ¡¢¢ÙO2ºÍO3 ÊÇͬÖÖÔªËØËùÐγɵÄÐÔÖʲ»Í¬µÄµ¥ÖÊ£¬»¥ÎªÍ¬ËØÒìÐÎÌ壻¢ÚH2¡¢D2¡¢T2 ¶¼ÊÇÇâÔªËØ×é³ÉµÄ£¬ÊÇÊôÓÚͬÖÖÎïÖʵIJ»Í¬·Ö×Ó£»¢Û12 CºÍ14 C ÖÐ×ÓÊý²»Í¬£¬ÊÇ̼ԪËصIJ»Í¬Ô­×Ó£¬»¥ÎªÍ¬Î»ËØ£»¢ÜCH3CH2CH2CH3ºÍ(CH3)2CHCH3 ·Ö×ÓʽÏàͬ£¬½á¹¹²»Í¬£¬ÎªÌ¼Á´Òì¹¹£¬»¥ÎªÍ¬·ÖÒì¹¹Ì壻¢Ý¹ïÍéºÍÊ®ÁùÍé ½á¹¹ÏàËÆ£¬¶¼ÊôÓÚÍéÌþ£¬·Ö×Ó×é³ÉÉÏÏà²î6¸öCH2Ô­×ÓÍÅ£»¢ÞCH3(CH2)5CH3ºÍCH3CH2CH2CH(CH3)C2H5 ·Ö×ÓʽÏàͬ£¬½á¹¹²»Í¬£¬ÎªÌ¼Á´Òì¹¹£¬»¥ÎªÍ¬·ÖÒì¹¹Ì壻¢ß¶¼ÊǼ×ÍéµÄ¶þäå´úÎÊôÓÚͬÖÖÎïÖÊ£»A¡¢»¥ÎªÍ¬Î»ËعÊÑ¡¢Û£» B¡¢»¥ÎªÍ¬·ÖÒì¹¹Ìå¹ÊÑ¡¢Ü¢Þ£»C¡¢»¥ÎªÍ¬ËØÒìÐÎÌå¹ÊÑ¡¢Ù£»D¡¢Í¬Ò»ÖÖÎïÖʹÊÑ¡¢ß£»¹Ê´ð°¸Îª£ºA£®¢Û£»B£®¢Ü¢Þ£»C£®¢Ù£»D£®¢ß£»

¢ò¡¢£¨1£©Ä³ÍéÌþµÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª142£¬12n+2n+2=142£¬n=10£¬·Ö×ÓʽΪC10H22£¬Ôò¹Ê´ð°¸Îª£ºC10H22£»

£¨2£©ÍéÌþAÔÚͬÎÂͬѹÏÂÕôÆøµÄÃܶÈÊÇH2µÄ36±¶£¬ÔòÍéÌþµÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª72£¬Ôò12n+2n+2=72£¬n=5£¬·Ö×ÓʽΪC5H12£¬¹Ê´ð°¸Îª£ºC5H12£»

£¨3£©1LÍéÌþDµÄÕôÆøÍêȫȼÉÕʱ£¬Éú³ÉͬÎÂͬѹÏÂ15LË®ÕôÆø£¬ËµÃ÷1molÌþÉú³É15molË®£¬Ôò1molÌþº¬ÓÐ30molHÔ­×Ó£¬Ôò2n+2=30£¬n=14£¬·Ö×ÓʽΪC14H30£¬¹Ê´ð°¸Îª£ºC14H30£»

¢ó¡¢£¨1£©¢Ù¸ù¾ÝÍéÌþµÄÃüÃû¹æÔò£¬µÄÃû³ÆΪ2£¬2¡ª¶þ¼×»ù±ûÍ飬ËüºÍÂÈÆø·¢Éúһȡ´ú·´Ó¦µÄ·½³ÌʽΪ+Cl2(CH3)3CCH2Cl+HCl£¬¹Ê´ð°¸Îª£º2£¬2¡ª¶þ¼×»ù±ûÍ飻+Cl2(CH3)3CCH2Cl+HCl£»

¢Ú¸ù¾ÝÍéÌþµÄÃüÃû¹æÔò£¬µÄÃû³ÆΪ3,3,5,5¡ªËļ׻ù¸ýÍ飬ËüµÄÒ»ÂÈ´úÎïµÄͬ·ÖÒì¹¹ÌåÓÐ(¡ñ²»ÊÇÂÈÔ­×Ó¿ÉÄܵÄλÖÃ)£¬¹²4ÖÖ£¬¹Ê´ð°¸Îª£º3,3,5,5¡ªËļ׻ù¸ýÍ飻4£»

£¨2£©¢Ù 2£¬3£­¶þ¼×»ù£­4£­ÒÒ»ùÒÑÍéµÄ½á¹¹¼òʽΪCH3CH(CH3)CH(CH3)CH(C2H5)CH2CH3£¬¹Ê´ð°¸Îª£ºCH3CH(CH3)CH(CH3)CH(C2H5)CH2CH3£»

¢ÚÖ§Á´Ö»ÓÐÒ»¸öÒÒ»ùÇÒÏà¶Ô·Ö×ÓÖÊÁ¿×îСµÄÍéÌþÊÇ3-ÒÒ»ùÎìÍ飬½á¹¹¼òʽΪCH3CH2CH(C2H5)CH2CH3£¬¹Ê´ð°¸Îª£ºCH3CH2CH(C2H5)CH2CH3£»

£¨3£©ôÇ»ùµÄµç×ÓʽΪ£¬¹Ê´ð°¸Îª£º¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø