ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿×î½üÃÀ¹úÓ¾Ö£¨NASA£©ÂíÀïŵÍÞ²©Ê¿ÕÒµ½ÁËÒ»ÖֱȶþÑõ»¯Ì¼ÓÐЧ104±¶µÄ¡°³¬¼¶ÎÂÊÒÆøÌ塱ȫ·ú±ûÍ飨C3F8£©£¬²¢Ìá³öÓÃÆä¡°ÎÂÊÒ»¯»ðÐÇ¡±Ê¹Æä³ÉΪµÚ¶þ¸öµØÇòµÄ¼Æ»®¡£ÓйØÈ«·ú±ûÍéµÄ˵·¨ÕýÈ·µÄÊÇ£¨ £©

A.È«·ú±ûÍ飨C3F8£©·Ö×ÓÖмÈÓм«ÐÔ¼üÓÖÓзǼ«ÐÔ¼ü

B.È«·ú±ûÍ飨C3F8£©µÄµç×ÓʽΪ£º

C.ÏàͬѹǿÏ£¬·Ðµã£ºC3F8£¼C3H8

D.·Ö×ÓÖÐÈý¸ö̼ԭ×Ó¿ÉÄÜ´¦ÓÚͬһֱÏßÉÏ

¡¾´ð°¸¡¿A

¡¾½âÎö¡¿

A£®È«·ú±ûÍ飨C3F8£©µÄ½á¹¹¼òʽΪCF3CF2CF3£¬CºÍCÖ®¼äΪ·Ç¼«ÐÔ¹²¼Û¼ü£¬CºÍFÖ®¼äΪ¼«ÐÔ¹²¼Û¼ü£¬AÕýÈ·£»

B£®È«·ú±ûÍéÖÐFÔ­×ÓÓ¦Âú×ã8µç×ÓÎȶ¨½á¹¹£¬B´íÎó£»

C£®C3F8ºÍC3H8½á¹¹ÏàËÆ£¬C3F8µÄÏà¶Ô·Ö×ÓÖÊÁ¿±ÈC3H8µÄÏà¶Ô·Ö×ÓÖÊÁ¿´ó£¬¹ÊC3F8ÖзÖ×Ó¼ä×÷ÓÃÁ¦±ÈC3H8Ç¿£¬È۷еãC3F8±ÈC3H8¸ß£¬C´íÎó£»

D£®C3F8ΪC3H8µÄÈ«·úÈ¡´úÎÈý¸öCÔ­×ÓÓ¦³Ê¾â³Ý×´£¬²»¿ÉÄÜÔÚͬһֱÏßÉÏ£¬D´íÎó¡£

´ð°¸Ñ¡A¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿°±ÊÇÒ»ÖÖÖØÒªµÄ»¯¹¤Ô­ÁÏ£¬¿ÉÒÔÓÃÀ´ÖƱ¸µª»¯¹è(Si3N4)ëÂ(N2H4)¡¢ÇâÇèËá(HCN)¡£

(1)ÒÑÖª£ºSi(s)+2Cl2(g)====SiCl4(g) ¡÷H1=akJ¡¤mol£­1

N2(g)+3H2(g) 2NH3(g) ¡÷H2=bkJ¡¤mol£­1

3Si(s)+2N2(g)====Si3N4(s) ¡÷H3=ckJ¡¤mol£­1

H2(g)+Cl2(g)====2HCl(g) ¡÷H4=dkJ¡¤mol£­1

Ôò·´Ó¦3SiCl4(g)+4NH3(g)====Si3N4(s)+12HCl(g)µÄ¡÷H=________________kJ¡¤mol£­1(ÓÃa¡¢b¡¢c¡¢d±íʾ)¡£

(2)ëµÄÖƱ¸·½·¨ÊÇÓôÎÂÈËáÄÆÑõ»¯¹ýÁ¿µÄ°±¡£

ÒÑÖªClO£­Ë®½âµÄ·½³ÌʽΪ£ºClO£­+H2 O=HClO+OH£­¡£³£ÎÂÏ£¬¸ÃË®½â·´Ó¦µÄƽºâ³£ÊýΪK=1.0¡Á10£­6mol¡¤L£­1£¬Ôò1.0mol¡¤ L £­1NaCIOÈÜÒºµÄpH=________¡£

(3)¹¤ÒµÉÏÀûÓð±ÆøÉú²úÇâÇèËá(HCN)µÄ·´Ó¦Îª£ºCH4(g)+NH3(g) HCN(g)+3H2 (g) ¡÷H>O

¢ÙÆäËûÌõ¼þÒ»¶¨£¬´ïµ½Æ½ºâʱNH3ת»¯ÂÊËæÍâ½çÌõ¼þX±ä»¯µÄ¹ØϵÈçͼËùʾ¡£X´ú±íµÄÊÇ________(Ìζȡ±»ò¡°Ñ¹Ç¿¡±)¡£

¢ÚÆäËûÌõ¼þÒ»¶¨£¬Ïò2LÃܱÕÈÝÆ÷ÖмÓÈË n mol CH4ºÍ2 mol NH3£¬Æ½ºâʱNH3Ìå»ý·ÖÊýËæn±ä»¯µÄ¹ØϵÈçͼËùʾ¡£Èô·´Ó¦´Ó¿ªÊ¼µ½aµãËùÓÃʱ¼äΪ10min£¬¸Ãʱ¼ä¶ÎÄÚÓÃCH4µÄŨ¶È±ä»¯±íʾµÄ·´Ó¦ËÙÂÊΪ________mol¡¤L£­1¡¤min£­1£»Æ½ºâ³£Êý£ºK(a) ________K(b)(Ìî¡°>¡±¡°=¡±»ò¡°<¡±)

¢Û¹¤ÒµÉÏÓõç½â·¨´¦Àíº¬Çèµç¶Æ·ÏË®(pH=10)µÄ×°ÖÃÈçͼËùʾ¡£

Ñô¼«²úÉúµÄÂÈÆøÓë¼îÐÔÈÜÒº·´Ó¦Éú³ÉClO-£¬ClO£­½«CN£­Ñõ»¯µÄÀë×Ó·½³ÌʽΪ£º_____CN£­+ _____ClO£­+ ________====_____CO32£­+_____N2¡ü+________+________Èôµç½â´¦Àí2 mol CN£­£¬ÔòÒõ¼«²úÉúÆøÌåµÄÌå»ý(±ê×¼×´¿öÏÂ)Ϊ________L¡£

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

¾«Ó¢¼Ò½ÌÍø