¡¾ÌâÄ¿¡¿ÒÑÖª²ð¿ª1molÇâÆøÖеĻ¯Ñ§¼üÐèÒªÏûºÄ436kJÄÜÁ¿£¬²ð¿ª1molÑõÆøÖеĻ¯Ñ§¼üÐèÒªÏûºÄ498kJÄÜÁ¿£¬¸ù¾ÝͼÖеÄÄÜÁ¿Í¼£¬»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©·Ö±ðд³ö¢Ù¢ÚµÄÊýÖµ£º

¢Ù____________£» ¢Ú ____________ £»

£¨2£©Éú³ÉH2O£¨g£©ÖеÄ1mol H-O¼ü·Å³ö____________kJµÄÄÜÁ¿£»

£¨3£©ÒÑÖª£ºH2O£¨l£©= H2O£¨g£© ¡÷H = +44 kJ¡¤mol£­1 £¬ÊÔд³öÇâÆøÔÚÑõÆøÖÐÍêȫȼÉÕÉú³ÉҺ̬ˮµÄÈÈ»¯Ñ§·½³Ìʽ£º_____________________________________¡£

¡¾´ð°¸¡¿1370 1852 463 2H2£¨g£©+O2(g) == 2H2O£¨l£©¡÷H= ¡ª570kJ/mol

¡¾½âÎö¡¿

ÓÉͼÏó¿ÉÖª2H2£¨g£©+O2£¨g£©=2H2O£¨g£©¡÷H=-482kJmol-1£¬»¯Ñ§·´Ó¦µÄʵÖÊÊǾɼüµÄ¶ÏÁѺÍмüµÄÐγɣ¬»¯Ñ§¼ü¶ÏÁÑÒªÎüÊÕÄÜÁ¿£¬ÐγÉмüÒª·Å³öÄÜÁ¿£¬·´Ó¦ÈÈΪ·´Ó¦ÎïµÄ×ܼüÄܼõÈ¥Éú³ÉÎïµÄ×ܼüÄÜ£¬ÒԴ˽â´ð¸ÃÌâ¡£

£¨1£©¢ÙÖУ¬ÒÑÖª²ð¿ª1 molÇâÆøÖеĻ¯Ñ§¼üÐèÒªÏûºÄ436 kJÄÜÁ¿£¬²ð¿ª1 molÑõÆøÖеĻ¯Ñ§¼üÐèÒªÏûºÄ498 kJ ÄÜÁ¿£¬ËùÒÔ²ð¿ª2Ħ¶ûÇâÆøºÍ1Ħ¶ûÑõÆø£¬ÐèÒªµÄÄÜÁ¿Îª2¡Á436+498=1370kJ£¬¢ÚÖиù¾ÝÄÜÁ¿Êغ㣬¿ÉÒÔÖªµÀ·Å³öµÄÄÜÁ¿Îª1370+482=1852kJ£»

£¨2£©Éú³ÉH2O£¨g£©ÖÐ1 mol H¡ªO¼ü·Å³öµÄÄÜÁ¿Îª 1852 /4=463kJ£»

£¨3£©ÓÉͼÖÐÐÅÏ¢¿ÉÒÔÖªµÀ£¬2H2£¨g£©+O2£¨g£©=2H2O£¨g£©¦¤H =-482kJ¡¤mol-1£¬

ÓÖÒòΪH2O£¨l£©=H2O£¨g£©¦¤H = +44 kJ¡¤mol-1£¬ËùÒÔ2 molÇâÆøÔÚ×ãÁ¿ÑõÆøÖÐÍêȫȼÉÕÉú³ÉҺ̬ˮµÄÈÈ»¯Ñ§·½³ÌΪ2H2£¨g£©+O2£¨g£©=2H2O£¨l£©¦¤H =-570 kJ¡¤mol-1¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿¢ôA×åÔªËØ(C¡¢Si¡¢Ge¡¢Sn¡¢Pb)¼°Æ仯ºÏÎïÔÚ²ÄÁϵȷ½ÃæÓÐÖØÒªÓÃ;¡£»Ø´ðÏÂÁÐÎÊÌ⣺

(1) PbµÄ¼Û²ãµç×ÓÅŲ¼Í¼Îª£º____________£»

(2)GeC14µÄÖÐÐÄÔ­×ӵļ۲ãµç×Ó¶ÔÊýΪ____________£¬·Ö×ÓµÄÁ¢Ìå¹¹ÐÍΪ___________¡£GeC14¿ÉË®½âÉú³ÉÒ»ÖÖÑõ»¯ÎïºÍÒ»ÖÖÎÞÑõËᣬÆ仯ѧ·´Ó¦·½³ÌʽΪ£º________________________________¡£

(3)Ëı»¯¹èµÄ·ÐµãºÍ¶þ±»¯Ç¦µÄÈÛµãÈçÏÂͼËùʾ¡£

¢ÙµÄ·ÐµãÒÀF¡¢CI¡¢Br¡¢I´ÎÐòÉý¸ßµÄÔ­ÒòÊÇ____________¡£

¢Ú½áºÏµÄ·ÐµãºÍµÄÈÛµãµÄ±ä»¯¹æÂÉ£¬¿ÉÍƶϣºÒÀF¡¢Cl¡¢Br¡¢I´ÎÐò£¬Öл¯Ñ§¼üµÄÀë×ÓÐÔ__________(Ìî¡°ÔöÇ¿¡±¡°²»±ä¡±»ò¡°¼õÈõ¡±£¬ÏÂͬ)¡¢¹²¼ÛÐÔ__________¡£

(4)̼µÄÁíÒ»ÖÖµ¥ÖÊ¿ÉÒÔÓë¼ØÐγɵÍ㬵¼»¯ºÏÎ¾§°û½á¹¹ÈçͼËùʾ¡£´Ë»¯ºÏÎï¿É¿´³ÉÊÇKÌî³äÔÚÐγɵÄËùÓÐËÄÃæÌå¼ä϶ºÍ°ËÃæÌå¼ä϶֮ÖУ¬ÆäËÄÃæÌå¼ä϶ÊýĿΪ_________¡£ÁíÓÐÒ»ÖÖ¼î½ðÊôX(Ïà¶ÔÔ­×ÓÖÊÁ¿ÎªM)Óë¿ÉÐγÉÀàËÆ»¯ºÏÎµ«XÖ»Ìî³äÐγɵİËÃæÌå¼ä϶µÄÒ»°ë£¬´Ë»¯ºÏÎïµÄ»¯Ñ§Ê½Îª£º__________£¬Æ侧°û²ÎÊýΪ1.4nm£¬¾§ÌåÃܶÈΪ_________(Óú¬MºÍ°¢·üÙ¤µÂÂÞ³£ÊýµÄÖµµÄʽ×Ó±íʾ)

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÄαØÂå¶ûÊÇÒ»ÖÖÓÃÓÚѪ¹ÜÀ©ÕŵĽµÑªÑ¹Ò©ÎһÖֺϳÉÄαØÂå¶ûÖмäÌåGµÄ²¿·ÖÁ÷³ÌÈçÏ£º

ÒÑÖª£ºÒÒËáôûµÄ½á¹¹¼òʽΪ¡£

Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©AµÄÃû³ÆÊÇ______£»BÖÐËùº¬¹ÙÄÜÍŵÄÃû³ÆÊÇ______¡£

£¨2£©·´Ó¦¢ÝµÄ»¯Ñ§·½³ÌʽΪ______£¬¸Ã·´Ó¦µÄ·´Ó¦ÀàÐÍÊÇ______¡£

£¨3£©GµÄ·Ö×ÓʽΪ______¡£

£¨4£©Ð´³öÂú×ãÏÂÁÐÌõ¼þµÄEµÄͬ·ÖÒì¹¹ÌåµÄ½á¹¹¼òʽ£º______¡¢______¡£

¢ñ£®±½»·ÉÏÖ»ÓÐÈý¸öÈ¡´ú»ù

¢òºË´Å¹²ÕñÇâÆ×ͼÖÐÖ»ÓÐ4×éÎüÊÕ·å

¢ó.1mol¸ÃÎïÖÊÓë×ãÁ¿NaHCO3ÈÜÒº·´Ó¦Éú³É2molCO2

£¨5£©¸ù¾ÝÒÑÓÐ֪ʶ²¢½áºÏÏà¹ØÐÅÏ¢£¬Ð´³öÒÔΪԭÁÏÖƱ¸µÄºÏ³É·ÏßÁ÷³Ìͼ£¨ÎÞ»úÊÔ¼ÁÈÎÑ¡£©______£¬ºÏ³É·ÏßÁ÷³ÌͼʾÀýÈçÏ£ºCH3CH2BrCH3CH2OHCH3COOCH2CH3

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿¿Æѧ¼Ò·¢ÏÖ¶ÔÒ±½ð¹è½øÐеç½â¾«Á¶Ìá´¿¿É½µµÍ¸ß´¿¹èÖƱ¸³É±¾¡£Ïà¹Øµç½â²Û×°ÖÃÈç×óͼ£¬ÓÃCu£­SiºÏ½ð×÷¹èÔ´£¬ÔÚ950¡æÀûÓÃÈý²ãÒºÈÛÑνøÐеç½â¾«Á¶£¬²¢ÀûÓÃijCH4ȼÁϵç³Ø£¨ÓÒͼ£©×÷ΪµçÔ´¡£ÓйØ˵·¨²»ÕýÈ·µÄÊÇ

A. µç¼«cÓëaÏàÁ¬£¬dÓëbÏàÁ¬

B. ×ó²àµç½â²ÛÖУºSi ÓÅÏÈÓÚCu±»»¹Ô­£¬CuÓÅÏÈÓÚSi±»Ñõ»¯

C. Èý²ãÒºÈÛÑεÄ×÷ÓÃÊÇÔö´óµç½â·´Ó¦Ãæ»ý£¬Ìá¸ß¹è³Á»ýЧÂÊ

D. Ïàͬʱ¼äÏ£¬Í¨ÈëCH4¡¢O2µÄÌå»ý²»Í¬£¬»áÓ°Ïì¹èÌá´¿ËÙÂÊ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿Ì¼ÊǵØÇòÉÏ×é³ÉÉúÃüµÄ×î»ù±¾µÄÔªËØÖ®Ò»¡£°´ÒªÇó»Ø´ð£º

£¨1£©Ì¼Ô­×ÓºËÍâÓÐ________ÖÖ²»Í¬¿Õ¼äÔ˶¯×´Ì¬µÄµç×Ó£¬µÚÒ»µçÀëÄܽéÓÚBºÍCÖ®¼äµÄÔªËصÄÃû³ÆΪ_________¡£

£¨2£©Ì¼ÔªËØÄÜÐγɶàÖÖÎÞ»úÎï¡£

¢ÙCO32-µÄÁ¢Ìå¹¹ÐÍÊÇ_______________¡£

¢ÚMgCO3·Ö½âζȱÈCaCO3µÍµÄÔ­ÒòÊÇ_________________________¡£

¢ÛʯīÓë¼Ø¿ÉÐγÉʯī¼Ð²ãÀë×Ó¾§ÌåC8K£¨Èçͼ£©£¬Æä½á¹¹ÎªÃ¿¸ôÒ»²ã̼ԭ×Ó²åÈëÒ»²ã¼ØÔ­×Ó£¬Óë¼ØÔ­×Ó²ãÏàÁÚµÄÉÏÏÂÁ½²ã̼ԭ×ÓÅÅÁз½Ê½Ïàͬ£¬ÔòÓë¼Ø×î½üµÈ¾àµÄÅäλ̼ԭ×ÓÓÐ_________¸ö¡£

£¨3£©Ì¼Ò²¿ÉÐγɶàÖÖÓлú»¯ºÏÎÏÂͼËùʾÊÇÒ»ÖÖàÑßʺÍÒ»ÖÖßÁवĽṹ£¬Á½ÖÖ·Ö×ÓÖÐËùÓÐÔ­×Ó¶¼ÔÚÒ»¸öƽÃæÉÏ¡£

¢Ù1 mol ßÁषÖ×ÓÖк¬ÓЦҼüÊýÄ¿ÊÇ__________¡£

¢ÚàÑßʽṹÖÐNÔ­×ÓµÄÔÓ»¯·½Ê½Îª________¡£

¢ÛàÑßÊÖйìµÀÖ®¼äµÄ¼Ð½Ç¡Ï1±È¡Ï2´ó£¬½âÊÍÔ­Òò________________________________¡£

£¨4£©½«Á¢·½½ð¸ÕʯÖеÄÿ¸ö̼ԭ×ÓÓÃÒ»¸öÓÉ4¸ö̼ԭ×Ó×é³ÉµÄÕýËÄÃæÌå½á¹¹µ¥ÔªÈ¡´ú¿ÉÐγÉ̼µÄÒ»ÖÖÐÂÐÍÈýάÁ¢·½¾§Ìå½á¹¹¡ª¡ªT-̼¡£ÒÑÖªT-̼ÃܶÈΪ¦Ñ g/cm£¬°¢·ü¼ÓµÂÂÞ³£ÊýΪNA£¬ÔòT-̼µÄ¾§°û²ÎÊýa=________ pm (д³ö±í´ïʽ¼´¿É)¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿µªºÍµªµÄ»¯ºÏÎïÔÚ¹ú·À½¨Éè¡¢¹¤Å©ÒµÉú²úºÍÉú»îÖж¼Óм«Æä¹ã·ºµÄÓÃ;¡£Çë»Ø´ðÏÂÁÐÓ뵪ԪËØÓйصÄÎÊÌ⣺

(1)NaCN¹ã·ºÓÃÓÚµç¶Æ¹¤ÒµÉÏ£¬ÊµÑé²âµÃŨ¶ÈÏàͬµÄNaCNÈÜÒººÍNaHCO3ÈÜÒº£¬Ç°ÕßµÄpH´ó£¬ÔòËáÐÔ£ºHCN___________H2CO3(Ìî¡°Ç¿ÓÚ¡±»ò¡°ÈõÓÚ¡±)¡£

(2)ÑÇÏõõ£ÂÈ(½á¹¹Ê½ÎªCl£­N=O)ÊÇÓлúºÏ³ÉÖеÄÖØÒªÊÔ¼Á¡£Ëü¿ÉÓÉC12ºÍNOÔÚͨ³£Ìõ¼þÏ·´Ó¦ÖƵ㬷´Ó¦·½³ÌʽΪ£º2NO(g)+Cl2(g)2ClNO(g)¡£ÒÑÖª¼¸ÖÖ»¯Ñ§¼üµÄ¼üÄÜÊý¾ÝÈçÏÂ±í£º

µ±Cl2ÓëNO·´Ó¦Éú³ÉClNOµÄ¹ý³ÌÖÐתÒÆÁË5molµç×Ó£¬ÀíÂÛÉÏÈÈÁ¿±ä»¯Îª___________kJ¡£

(3)ÔÚÒ»¸öºãÈÝÃܱÕÈÝÆ÷ÖгäÈë2 mol NO(g)ºÍ1molCl2(g)·¢Éú(2)Öз´Ó¦£¬ÔÚζȷֱðΪT1¡¢T2ʱ²âµÃNOµÄÎïÖʵÄÁ¿(µ¥Î»£ºmol)Óëʱ¼äµÄ¹ØϵÈçϱíËùʾ£º

¢ÙT1___________T2(Ìî¡°>¡±¡°¡Ü¡±»ò¡°=¡±)£¬ÀíÓÉÊÇ___________¡£

¢ÚÈôÈÝÆ÷ÈÝ»ýΪ1L£¬Î¶ÈΪT1¡æʱ£¬·´Ó¦¿ªÊ¼µ½5minʱ£¬C12µÄƽ¾ù·´Ó¦ËÙÂÊΪ_______¡£

¢ÛζÈΪT2¡æʱ£¬ÔÚÏàͬÈÝÆ÷ÖУ¬³äÈë1 molNO(g)ºÍ0.5mo1Cl2(g)£¬ÔòNOµÄƽºâת»¯ÂÊ___________50%(Ìî¡°´óÓÚ¡±¡¢¡°µÈÓÚ¡±»ò¡°Ð¡ÓÚ¡±)

¢ÜζÈΪT2¡æʱ£¬ÆðʼʱÈÝÆ÷ÄÚµÄѹǿΪp0£¬Ôò¸Ã·´Ó¦µÄƽºâ³£ÊýKp=___________(ÓÃƽºâ·Öѹ´úÌæƽºâŨ¶È¼ÆË㣬·Öѹ=×Üѹ¡ÁÎïÖʵÄÁ¿·ÖÊý)¡£

(4)½üÄêÀ´£¬µØÏÂË®ÖеĵªÎÛȾÒѳÉΪһ¸öÊÀ½çÐԵĻ·¾³ÎÊÌâ¡£ÔÚ½ðÊôPt¡¢CuºÍÒ¿(Ir)µÄ´ß»¯×÷ÓÃÏ£¬ÃܱÕÈÝÆ÷ÖеÄH2¿É¸ßЧת»¯ËáÐÔÈÜÒºÖеÄÏõ̬µª(NO3£­)£¬Æ乤×÷Ô­ÀíÈçͼËùʾ£º

¢ÙIr±íÃæ·¢Éú·´Ó¦µÄ·½³ÌʽΪ___________¡£

¢ÚÈôµ¼µç»ùÌåÉϵÄPt¿ÅÁ£Ôö¶à£¬Ôì³ÉµÄ½á¹ûÊÇ___________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿±ê×¼×´¿öÏÂ11.2LCH4ÔÚ¿ÕÆøÖÐÍêȫȼÉÕ£¬ÇëÍê³ÉÏÂÁÐÎÊÌ⣺

(1)д³ö¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ_______£»

(2)¼ÆËã±ê×¼×´¿öÏÂ11.2LCH4µÄÖÊÁ¿_______£»

(3)¼ÆËãÉú³ÉCO2µÄÖÊÁ¿_______¡£[(2)(3)д³ö¼ÆËã¹ý³Ì]

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿NAΪ°¢·ü¼ÓµÂÂÞ³£ÊýµÄÖµ¡£ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ

A. 1molH2ºÍ1molI2ÔÚ¼ÓÈÈÌõ¼þϳä·Ö·´Ó¦£¬Éú³ÉHIµÄ·Ö×ÓÊýΪ2NA

B. 10gÖÊÁ¿·ÖÊýΪ46%µÄÒÒ´¼ÈÜÒºº¬ÓеÄÇâÔ­×ÓÊýĿΪ0.6NA

C. 20mL0.1 mol/LAlCl3ÈÜÒºÖУ¬Ë®½âÐγÉAl(OH)3½ºÌåÁ£×ÓÊýΪ0.002NA

D. 0.1molNa2O2ºÍNa2OµÄ»ìºÏÎïÖк¬ÓеÄÀë×Ó×ÜÊýµÈÓÚ0.3NA

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÏÂÁйØÓÚʵÑé²Ù×÷µÄ˵·¨ÕýÈ·µÄÊÇ£¨ £©

A. Óñ½ÝÍÈ¡äåË®ÖеÄäåʱ£¬½«äåµÄ±½ÈÜÒº´Ó·ÖҺ©¶·Ï¿ڷųö

B. Öк͵ζ¨ÊµÑéÖУ¬×¶ÐÎÆ¿ÓÃÕôÁóˮϴ¾»Ðë¾­ºæÏäºæ¸Éºó²Å¿ÉʹÓÃ

C. ÅäÖÆ0.5mol¡¤L£­1480mLµÄNaOHÈÜÒºÐè³ÆÁ¿9.6 g NaOH¹ÌÌå

D. ijÈÜÒºÖеÎÈë2µÎK3[Fe(CN)6]ÈÜÒºÉú³É¾ßÓÐÌØÕ÷À¶É«µÄ³Áµí£¬ËµÃ÷Ô­ÈÜÒºÖк¬ÓÐFe2+

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸