¡¾ÌâÄ¿¡¿ºÏ³ÉÆø(CO+H2) ¹ã·ºÓÃÓںϳÉÓлúÎ¹¤ÒµÉϳ£²ÉÓÃÌìÈ»ÆøÓëË®ÕôÆø·´Ó¦µÈ·½·¨À´ÖÆÈ¡ºÏ³ÉÆø£¬
(1)ÒÑÖª±ê¿öÏ£¬5.6LCH4ÓëË®ÕôÆøÍêÈ«·´Ó¦Ê±ÎüÊÕ51.5kJµÄÈÈÁ¿£¬Çëд³ö¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ________________£»
(2)ÔÚ150¡æʱ2L µÄÃܱÕÈÝÆ÷ÖУ¬½«2molCH4ºÍ2mol H2O(g)»ìºÏ£¬¾¹ý15min´ïµ½Æ½ºâ£¬´ËʱCH4µÄת»¯ÂÊΪ60%¡£»Ø´ðÏÂÁÐÎÊÌ⣺
¢Ù´Ó·´Ó¦¿ªÊ¼ÖÁƽºâ£¬ÓÃÇâÆøµÄ±ä»¯Á¿À´±íʾ¸Ã·´Ó¦ËÙÂÊv(H2)=________¡£
¢ÚÔÚ¸ÃζÈÏ£¬¼ÆËã¸Ã·´Ó¦µÄƽºâ³£ÊýK=________¡£
¢ÛÏÂÁÐÑ¡ÏîÖÐÄܱíʾ¸Ã·´Ó¦ÒѴﵽƽºâ״̬µÄÊÇ________¡£
A£®V(H2)Äæ=3v (CO)Õý B£®ÃܱÕÈÝÆ÷ÖлìºÏÆøÌåµÄÃܶȲ»±ä
C£®ÃܱÕÈÝÆ÷ÖÐ×Üѹǿ²»±ä D£®C(CH4)=C(CO)
(3)ºÏ³ÉÆøÖеÄÇâÆøÒ²ÓÃÓںϳɰ±Æø£ºN2+3H22NH3¡£±£³ÖζȺÍÌå»ý²»±ä£¬Ôڼס¢ÒÒ¡¢±ûÈý¸öÈÝÆ÷Öн¨Á¢Æ½ºâµÄÏà¹ØÐÅÏ¢ÈçÏÂ±í¡£ÔòÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ_______________£»
ÈÝÆ÷ | Ìå»ý | ÆðʼÎïÖÊ | ƽºâʱNH3µÄÎïÖʵÄÁ¿ | ƽºâʱN2 µÄÌå»ý·ÖÊý | ·´Ó¦¿ªÊ¼Ê±µÄËÙÂÊ | ƽºâʱÈÝÆ÷ÄÚѹǿ |
¼× | 1L | 1molN2+3molH2 | 1.6mol | ¦Õ¼× | v¼× | P¼× |
ÒÒ | 1L | 2molN2+6molH2 | n1 mol | ¦ÕÒÒ | vÒÒ | PÒÒ |
±û | 2L | 2molN2+6molH2 | n2 mol | ¦Õ±û | v±û | P±û |
A£®n1=n2=3.2 B£®¦Õ¼×=¦Õ±û>¦ÕÒÒ C£®vÒÒ >v>v¼× D£®PÒÒ>P¼×=P±û
(4)ºÏ³ÉÆø¿ÉÒÔÖÆÈ¡¼×ÃÑ£¬ÂÌÉ«µçÔ´¡°¶þ¼×ÃÑ-ÑõÆøȼÁϵç³Ø¡±¹¤×÷ÔÀíÈçÏÂͼËùʾ
¢Ù µç¼«Y ÉÏ·¢ÉúµÄ·´Ó¦Ê½Îª__________ £»
¢Úµç³ØÔڷŵç¹ý³ÌÖУ¬µç¼«XÖÜΧÈÜÒºµÄpH_______(Ìî¡°Ôö´ó¡¢¼õС¡¢²»±ä¡±)¡£
¡¾´ð°¸¡¿ CH4(g)+H2O(g)=CO(g)+3H2(g) ¡÷H= +206kJ/mol 0.12mol¡¤L-1¡¤min-1 21.87 AC BD O2+4e-+4H+=2H2O ¼õС
¡¾½âÎö¡¿(1). ±ê¿öÏÂ5.6LCH4Ϊ0.25mol£¬Ôò·½³ÌʽΪ: CH4(g)+H2O(g)=CO(g)+3H2(g)£¬ÎüÈÈ·´Ó¦£¬¡÷HΪÕý£¬¡÷H=+51.5¡Á4=+206kJ/mol£»
(2). ¢Ù¼×Íéת»¯1.2mol£¬Éú³ÉÇâÆø3.6mol£¬ÔòÇâÆøµÄ±ä»¯Á¿Îª3.6mol/2L/15min=0.12 mol¡¤L-1¡¤min-1£»¢Ú£»¢ÛA. ´Ó·´Ó¦¿ªÊ¼ÖÁƽºâÇâÆøµÄ±ä»¯Á¿ÎªCO(g)µÄ3±¶£¬AÕýÈ·£»B. ·´Ó¦ÖÊÁ¿Êغ㣬ÎÞÂÛ·´Ó¦ÈçºÎ£¬ÖÊÁ¿¾ù²»±ä£¬ÈÝÆ÷Ìå»ý²»±ä£¬ÔòÃܶȲ»±ä£»C. ·½³ÌÁ½¶ËÆøÌåÌå»ý²»Í¬£¬ÈôÈÝÆ÷ÄÚѹǿ²»±ä£¬ÔòÆøÌåÌå»ý²»±ä»¯£¬¼´·´Ó¦´ïµ½Æ½ºâ£¬CÕýÈ·£»D. ·´Ó¦Æ½ºâʱCH4µÄת»¯ÂÊΪ²»Ò»¶¨Îª50%£¬¼´C(CH4)= C(CO)²»Äܱ£Ö¤·´Ó¦Æ½ºâ£¬D´íÎó¡£ËùÒÔÑ¡ÔñAC¡£
(3).ÒÒÈÝÆ÷ÖеÄÆðʼѹǿ´óÓÚ±ûÈÝÆ÷ÖеÄÆðʼѹǿ£¬¹Ên1n2²»ÏàµÈ£¬A´íÎó£»B. ÈÝÆ÷¼×ºÍ±ûµÄÆðʼѹǿÏàµÈÇÒСÓÚÒÒÈÝÆ÷£¬ËùÒÔÕýÏò·´Ó¦³Ì¶È¦ÕÒÒ>¦Õ¼×=¦Õ±û£¬ËùÒÔN2 µÄÌå»ý·ÖÊý¦Õ¼×=¦Õ±û>¦ÕÒÒ£¬BÕýÈ·£»C. ͬB£¬vÒÒ >v±û=v¼×£¬C´íÎó£»D. ÓÉÓÚ·½³ÌÁ½¶ËÆøÌåµÄÎïÖʵÄÁ¿²»·¢Éú±ä»¯£¬ËùÒÔƽºâʱÈÝÆ÷ѹǿµÈÓÚÆðʼѹǿ£¬¹ÊPÒÒ>P¼×=P±û£¬DÕýÈ·¡£ËùÒÔÑ¡ÔñBD¡£
(4). ¢ÙÒòΪÇâÀë×ÓÏòY¼«Òƶ¯£¬ÔòY¼«µÃµ½µç×ÓΪÕý¼«£¬ÓÖȼÁϵç³ØÒÔÑõÆøΪ·´Ó¦ÎÆ䷴ӦΪ£ºO2+4e-+4H+=2H2O£»¢Úµç³ØÔڷŵç¹ý³ÌÖУ¬X¼«´¦Éú³ÉÇâÀë×Ó£¬ËùÒÔpH¼õС¡£
Ä꼶 | ¸ßÖÐ¿Î³Ì | Ä꼶 | ³õÖÐ¿Î³Ì |
¸ßÒ» | ¸ßÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ | ³õÒ» | ³õÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ |
¸ß¶þ | ¸ß¶þÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õ¶þ | ³õ¶þÃâ·Ñ¿Î³ÌÍƼö£¡ |
¸ßÈý | ¸ßÈýÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õÈý | ³õÈýÃâ·Ñ¿Î³ÌÍƼö£¡ |
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÒÔÏÂÊÇA¡«FÎïÖʼäµÄÏ໥ת»¯¹Øϵͼ¡£ÒÑÖªA¡«FÖоùº¬ÓÐÒ»ÖÖÏàͬµÄÔªËØ,BÊÇÒ»ÖÖ¹¤Òµ²úÆ·,Ó¦Óù㷺,DÔÚ³£ÎÂÏÂÊÇÆøÌå¡£CÊǵ¥ÖÊ,ºÜ´àµÄµ»ÆÉ«¾§Ìå¡£CÓëFÎïÖʵÄÁ¿Ö®±È1¡Ã1·´Ó¦Éú³ÉA¡£
Çë»Ø´ð£º
£¨1£©AµÄ»¯Ñ§Ê½________¡£
£¨2£©·´Ó¦¢ÙµÄÀë×Ó·½³Ìʽ________¡£
£¨3£©½«CÓë¹ýÁ¿Å¨NaOHÈÜÒº»ìºÏ¼ÓÈÈ,ÓÐFÉú³É¡£Ð´³ö·´Ó¦µÄ»¯Ñ§·½³Ìʽ________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÏÂÁÐÓйØÎïÖÊÓÃ;µÄ˵·¨ÖдíÎóµÄÊÇ£¨ £©
A.ÂÈÆø¿ÉÓÃÓÚ×ÔÀ´Ë®µÄÏû¶¾
B.ËÄÑõ»¯ÈýÌú³£ÓÃ×÷ºìÉ«ÓÍÆáºÍÍ¿ÁÏ
C.Ã÷·¯¿ÉÓÃÓÚ¾»Ë®
D.¹ýÑõ»¯ÄÆÓÃÓÚºôÎüÃæ¾ßºÍDZˮͧÀïÑõÆøµÄÀ´Ô´
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿C60¡¢½ð¸Õʯ¡¢Ê¯Ä«¡¢¶þÑõ»¯Ì¼ºÍÂÈ»¯ï¤µÄ½á¹¹Ä£ÐÍÈçͼËùʾ(ʯī½ö±íʾ³öÆäÖеÄÒ»²ã½á¹¹):
£¨1£©C60¡¢½ð¸ÕʯºÍʯīÈýÕߵĹØϵÊÇ»¥Îª____¡£
A.ͬ·ÖÒì¹¹Ìå B.ͬËØÒìÐÎÌå C.ͬϵÎï D.ͬλËØ
£¨2£©¹Ì̬ʱ,C60ÊôÓÚ____(Ìî¡°Ô×Ó¡±»ò¡°·Ö×Ó¡±)¾§Ìå,C60·Ö×ÓÖк¬ÓÐË«¼üµÄÊýÄ¿ÊÇ____¡£
£¨3£©¾§Ìå¹èµÄ½á¹¹¸ú½ð¸ÕʯÏàËÆ,1 mol¾§Ìå¹èÖк¬Óй衪¹èµ¥¼üµÄÊýÄ¿Ô¼ÊÇ____NA¡£
£¨4£©Ê¯Ä«²ã×´½á¹¹ÖÐ,ƽ¾ùÿ¸öÕýÁù±ßÐÎÕ¼ÓеÄ̼Ô×ÓÊýÊÇ____¡£
£¨5£©¹Û²ìCO2·Ö×Ó¾§Ìå½á¹¹µÄÒ»²¿·Ö,ÊÔ˵Ã÷ÿ¸öCO2·Ö×ÓÖÜΧÓÐ____¸öÓëÖ®½ôÁÚÇҵȾàµÄCO2·Ö×Ó;¸Ã½á¹¹µ¥ÔªÆ½¾ùÕ¼ÓÐ____¸öCO2·Ö×Ó¡£
£¨6£©¹Û²ìͼÐÎÍƲâ,CsCl¾§ÌåÖÐÁ½¾àÀë×î½üµÄCs+¼ä¾àÀëΪa,Ôòÿ¸öCs+ÖÜΧÓëÆä¾àÀëΪaµÄCs+ÊýĿΪ___,ÿ¸öCs+ÖÜΧ¾àÀëÏàµÈÇҴνüµÄCs+ÊýĿΪ___,¾àÀëΪ___,ÿ¸öCs+ÖÜΧ¾àÀëÏàµÈÇÒµÚÈý½üµÄCs+ÊýĿΪ____,¾àÀëΪ____,ÿ¸öCs+ÖÜΧ½ôÁÚÇҵȾàµÄCl-ÊýĿΪ___¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿PETÊÇÊÀ½çÉϲúÁ¿×î´óµÄºÏ³ÉÏËά£¬Æä½á¹¹¼òʽΪ£º
ÏÖÒÔúµÄ¸ÉÁó²úÆ·A ÓëF ΪÔÁÏÖƱ¸PET£¬Éú²úµÄ¹¤ÒÕÁ÷³ÌÈçͼËùʾ¡£
¢ÙAµÄ·Ö×ÓʽΪC8H10£¬ÇÒÄÜʹËáÐÔ¸ßÃÌËá¼ØÈÜÒºÍÊÉ«£¬µ«²»ÄÜʹäåË®ÍÊÉ«¡£
¢ÚM·Ö×ÓÀïËùÓÐÔ×Ó¹²Æ½Ãæ¡£
¢ÛG ΪCaC2£¬ÆäÓëH2O ·´Ó¦Ê±ÔªËØ»¯ºÏ¼Û²»±ä¡£
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
(1)AµÄÃû³ÆΪ__________¡£M¡úN µÄ·´Ó¦ÀàÐÍΪ__________£»
(2)·´Ó¦¢ÙµÄ·´Ó¦Ìõ¼þΪ£º__________£»
(3)д³öÓлúÎïA ËùÓÐÒ»ÂÈ´úÎïµÄ½á¹¹¼òʽ£º__________¡£
(4)д³öÏÂÁз´Ó¦µÄ»¯Ñ§·½³Ìʽ£º
¢ÙDÓë×ãÁ¿µÄÇâÑõ»¯ÍÐü×ÇÒºÖó·ð£º__________£»
¢Ú·´Ó¦¢Ý£º__________¡£
(5)PµÄÒ»ÖÖͬϵÎïXµÄ·Ö×ÓʽΪC3H8O2 £¬Ôں˴Ź²ÕñÇâÆ×ͼÖгöÏÖÈýÖÖÐźŷ壬Æä·åµÄÇ¿¶ÈÖ®±ÈΪ2¡Ã1¡Ã1¡£ÔòX µÄ½á¹¹¼òʽΪ__________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿µª»¯Ã¾(Mg3N2) ÔÚ¹¤ÒµÉϾßÓзdz£¹ã·ºµÄÓ¦Óá£Ä³»¯Ñ§ÐËȤС×éÓÃþÓ뵪Æø·´Ó¦ÖƱ¸Mg3N2²¢½øÐÐÓйØʵÑ顣ʵÑé×°ÖÃÈçÏÂËùʾ£º
ÒÑÖª£º¢Ùµª»¯Ã¾³£ÎÂÏÂΪdz»ÆÉ«·ÛÄ©£¬¼«Ò×ÓëË®·´Ó¦¡£
¢ÚÑÇÏõËáÄƺÍÂÈ»¯ï§ÖÆÈ¡µªÆøµÄ·´Ó¦¾çÁÒ·ÅÈÈ£¬²úÉúµªÆøµÄËٶȽϿ졣
¢ÛζȽϸßʱ£¬ÑÇÏõËáÄÆ»á·Ö½â²úÉúO2µÈ¡£
»Ø´ðÏÂÁÐÎÊÌ⣺
(1)ÒÇÆ÷a¡¢bµÄÃû³Æ·Ö±ðÊÇ____________£¬____________£»Ð´³ö×°ÖÃA Öз¢Éú·´Ó¦µÄ»¯Ñ§·½³Ìʽ____________¡£
(2)×°ÖÃCÖÐΪ±¥ºÍÁòËáÑÇÌúÈÜÒº£¬×÷ÓÃÊÇ_________£¬¸Ã×°ÖÃÖз¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ____________£»×°ÖÃD ÖеÄÊÔ¼ÁÊÇ____________£¬F ×°ÖõÄ×÷ÓÃÊÇ____________¡£
(3)¼ÓÈÈÖÁ·´Ó¦¿ªÊ¼·¢Éú£¬ÐèÒÆ×ßA ´¦¾Æ¾«µÆ£¬ÔÒòÊÇ____________¡£
(4)ʵÑé½áÊøºó£¬È¡×°ÖÃEµÄÓ²Öʲ£Á§¹ÜÖеÄÉÙÁ¿¹ÌÌåÓÚÊÔ¹ÜÖУ¬¼ÓÉÙÁ¿ÕôÁóË®£¬°ÑÈóʪµÄºìɫʯÈïÊÔÖ½·ÅÔڹܿڣ¬¹Û²ìʵÑéÏÖÏ󣬸òÙ×÷µÄÄ¿µÄÊÇ______________¡£·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ²Ù×÷µÄÄ¿µÄÊÇ__________£»ÔÙÆúÈ¥ÉϲãÇåÒº£¬¼ÓÈëÑÎËᣬ¹Û²ìÊÇ·ñÓÐÆøÅݲúÉú£¬¸Ã²Ù×÷µÄÄ¿µÄÊÇ__________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿Ä³ÊÐÖп¼»¯Ñ§ÊµÑé²Ù×÷¿¼ÊÔÓÐËĸö¿¼Ì⣺¢ÙÕôÁ󣻢ÚH2µÄ»¯Ñ§ÐÔÖÊ£»¢Û¶þÑõ»¯Ì¼µÄÖÆÈ¡¡¢ÊÕ¼¯ºÍÑéÂú£»¢ÜÑõÆøµÄÖÆÈ¡¡¢ÊÕ¼¯ºÍÑéÂú¡£¿¼ÊԵķ½·¨ÊÇÓÉ¿¼Éú³éÇ©È·¶¨¿¼Ì⣬С¿Í¬Ñ§³éÇ©ºó±»¼à¿¼ÀÏʦÒýµ¼ÖÁ×¼±¸ÁËÏÂÁÐÒÇÆ÷ºÍÒ©Æ·µÄʵÑę́ǰ£º
Çë»Ø´ð£º
£¨1£©Ö¸³öÉÏͼÖÐÒÇÆ÷aµÄÃû³Æ£º_______£»
£¨2£©ÓÉʵÑę́ÉÏÌṩµÄÒÇÆ÷ºÍÒ©Æ·£¬ÄãÈÏΪС¿³éµ½µÄÊǵÚ____¸ö¿¼Ì⣻
£¨3£©ÒÔÏÂÊÇС¿Íê³É¸ÃʵÑéÖ÷Òª²Ù×÷¹ý³ÌµÄʾÒâͼ¡£°´ÆÀ·Ö±ê×¼£¬Ã¿Ïî²Ù×÷ÕýÈ·µÃ1·Ö£¬Âú·Ö5·Ö£¬ÊµÑéÍê±ÏºóС¿µÃÁË3·Ö¡£ÇëÕÒ³öËûʧ·ÖµÄ²Ù×÷²¢ËµÃ÷ÔÒò£º______________________¡¢______________________£»
£¨4£©½öÓÃÉÏÊöÒÇÆ÷£¨Ò©Æ·ÁíÑ¡£©£¬Ò²ÄÜÍê³ÉÁíÒ»ÖÖ³£¼ûÆøÌåµÄʵÑéÊÒÖÆÈ¡£¬»¯Ñ§·½³ÌʽΪ£º______________£»ÈôÔö¼Ó________ £¨ÌîÒ»ÖÖ²£Á§ÒÇÆ÷Ãû³Æ£©»¹ÄÜ×é×°³É¸ßÃÌËá¼ØÖÆÑõÆøµÄ·¢Éú×°Öá£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿¹¤ÒµÉÏÓúϳÉÆø(COºÍH2)ÖÆÈ¡ÒÒ´¼µÄ·´Ó¦Îª2CO+4H2CH3CH2OH+H2O¡£Ñо¿·¢ÏÖ£¬Ê¹ÓÃTiO2×÷ΪÔØÌ帺ÔØîî»ù´ß»¯¼Á¾ßÓнϸߵÄÒÒ´¼²úÁ¿¡£»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Ti»ù̬Ô×ÓºËÍâµç×ÓÅŲ¼Ê½Îª_________¡£ºÍOͬһÖÜÆÚÇÒÔªËصĵÚÒ»µçÀëÄܱÈO´óµÄÓÐ______£¨ÌîÔªËØ·ûºÅ£©£¬ºÍOͬһÖÜÆÚÇÒ»ù̬Ô×ÓºËÍâδ³É¶Ôµç×ÓÊý±ÈO¶àµÄÓÐ____£¨ÌîÔªËØ·ûºÅ£©¡£
£¨2£©H2O·Ö×ÓÖÐOÔ×ӵļ۲ãµç×Ó¶ÔÊýÊÇ________£¬CH3CH2OH·Ö×ÓÖÐÑǼ׻ù(-CH2-)ÉϵÄCÔ×ÓµÄÔÓ»¯ÐÎʽΪ_______¡£
£¨3£©ÔÚÓúϳÉÆøÖÆÈ¡ÒÒ´¼·´Ó¦ËùÉæ¼°µÄ4ÖÖÎïÖÊÖУ¬·Ðµã´ÓµÍµ½¸ßµÄ˳ÐòΪ_________£¬ÔÒòÊÇ__________¡£
£¨4£©¹¤ÒµÉÏÒÔCO¡¢O2¡¢NH3ΪÔÁÏ£¬¿ÉºÏ³Éµª·ÊÄòËØ[CO(NH2)2]£¬CO(NH2)2·Ö×ÓÖк¬ÓеĦҼüÓë¦Ð¼üµÄÊýÄ¿Ö®±ÈΪ_________¡£
£¨5£©CÔªËØÓëNÔªËØÐγɵÄijÖÖ¾§ÌåµÄ¾§°ûÈçͼËùʾ£¨8¸ö̼Ô×ÓλÓÚÁ¢·½ÌåµÄ¶¥µã£¬4¸ö̼Ô×ÓλÓÚÁ¢·½ÌåµÄÃæÐÄ£¬4¸öµªÔ×ÓÔÚÁ¢·½ÌåÄÚ£©£¬¸Ã¾§ÌåÓ²¶È³¬¹ý½ð¸Õʯ£¬³ÉΪÊ×ÇüÒ»Ö¸µÄ³¬Ó²Ð²ÄÁÏ¡£
¢Ù¸Ã¾§ÌåÓ²¶È³¬¹ý½ð¸ÕʯµÄÔÒòÊÇ____________¡£
¢ÚÒÑÖª¸Ã¾§°ûµÄÃܶÈΪd g/cm3£¬NÔ×ӵİ뾶Ϊr1cm£¬CÔ×ӵİ뾶Ϊr2cm£¬ÉèNAΪ°¢·ü¼ÓµÂÂÞ³£Êý£¬Ôò¸Ã¾§°ûµÄ¿Õ¼äÀûÓÃÂÊΪ________(Óú¬d¡¢r1¡¢r2¡¢NAµÄ´úÊýʽ±íʾ£©¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿Ì¼µÄÑõ»¯Îï¶Ô»·¾³µÄÓ°Ïì½Ï´ó£¬COÊÇȼú¹¤ÒµÉú²úÖеĴóÆøÎÛȾÎCO2Ôò´Ù½øÁ˵ØÇòµÄÎÂÊÒЧӦ¡£¸øµØÇòÉúÃü´øÀ´Á˼«´óµÄÍþв¡£
(1)ÒÑÖª£º¢Ù¼×´¼µÄȼÉÕÈÈ¡÷H=-726.4kJ¡¤mol-1
¢ÚH2(g)+ O2(g)=H2O(l) ¡÷H=-285.8kJ¡¤mol-1¡£
Ôò¶þÑõ»¯Ì¼ºÍÇâÆøºÏ³ÉҺ̬¼×´¼£¬Éú³ÉҺ̬ˮµÄÈÈ»¯Ñ§·½³ÌʽΪ_____________________________¡£
(2)¶þÑõ»¯Ì¼ºÏ³ÉCH3OH µÄÈÈ»¯Ñ§·½³ÌʽΪCO2(g)+3H2(g)CH3OH(g)+H2O(g)¡÷H1£¬¹ý³ÌÖлá²úÉú¸±·´Ó¦£ºCO2(g)+H2(g)CO(g)+H2O(g)¡÷H2¡£Í¼1ÊǺϳɼ״¼·´Ó¦ÖÐζȶÔCH3OH¡¢COµÄ²úÂÊÓ°ÏìÇúÏßͼ£¬¡÷H2________0(Ìî¡°£¾¡±»ò¡°£¼¡±)¡£Ôö´ó·´Ó¦ÌåϵµÄѹǿ£¬ºÏ³É¼×´¼µÄ·´Ó¦ËÙÂÊ___________(Ìî¡°Ôö´ó¡±¡°¼õС¡±»ò¡° ²»±ä¡±)£¬¸±·´Ó¦µÄ»¯Ñ§Æ½ºâ________(Ìî¡°ÏòÕý·´Ó¦·½Ïò¡±¡°ÏòÄæ·´Ó¦·½Ïò¡±»ò¡°²»¡±)Òƶ¯¡£
(3)ÒÔijЩ¹ý¶É½ðÊôÑõ»¯Îï×÷´ß»¯¼Á£¬¶þÑõ»¯Ì¼Óë¼×Íé¿Éת»¯ÎªÒÒË᣺CO2(g) +CH4(g)CH3COOH(g) ¡÷H=+36.0kJ¡¤mol-1¡£²»Í¬Î¶ÈÏ£¬ÒÒËáµÄÉú³ÉËÙÂʱ仯ÇúÏßÈçͼ2¡£½áºÏ·´Ó¦ËÙÂÊ£¬Ê¹Óô߻¯¼ÁµÄ×î¼ÑζÈÊÇ________¡æ£¬ÓûÌá¸ßCH4µÄת»¯ÂÊ£¬ÇëÌṩһÖÖ¿ÉÐеĴëÊ©£º____________________________¡£
(4)Ò»¶¨Ìõ¼þÏ£¬CO2 ÓëNH3 ¿ÉºÏ³ÉÄòËØ[CO(NH2)2]£ºCO2(g)+2NH3(g)CO(NH2)2(g)+H2O(g)¡÷H¡£Ä³Î¶ÈÏ¡£ÔÚÈÝ»ýΪ1LµÄºãÈÝÃܱÕÈÝÆ÷ÖÐ,¼ÓÈëÒ»¶¨°±Ì¼±ÈµÄ3molCO2 ºÍNH3µÄ»ìºÏÆøÌ塣ͼ3ÊÇÓйØÁ¿µÄ±ä»¯ÇúÏߣ¬ÆäÖбíʾNH3ת»¯ÂʵÄÊÇÇúÏß________(Ìî¡°a¡±»ò¡°b¡±),ÇúÏßc ±íʾÄòËØÔÚƽºâÌåϵÖеÄÌå»ý·ÖÊý±ä»¯ÇúÏß,ÔòMµãµÄƽºâ³£ÊýK=________£¬y=________¡£
²é¿´´ð°¸ºÍ½âÎö>>
°Ù¶ÈÖÂÐÅ - Á·Ï°²áÁбí - ÊÔÌâÁбí
ºþ±±Ê¡»¥ÁªÍøÎ¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨Æ½Ì¨ | ÍøÉÏÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | µçÐÅթƾٱ¨×¨Çø | ÉæÀúÊ·ÐéÎÞÖ÷ÒåÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | ÉæÆóÇÖȨ¾Ù±¨×¨Çø
Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com