¡¾ÌâÄ¿¡¿¢¡ÓÃ1¡«18ºÅÔªËØ·ûºÅ¼°ÆäÐγɵĻ¯ºÏÎïµÄ»¯Ñ§Ê½Ìî¿Õ¡£

£¨1£©Ô­×Ӱ뾶×îСµÄÔªËØÊÇ_____¡£

£¨2£©ÓëË®·´Ó¦×î¾çÁҵĽðÊôÊÇ____¡£

£¨3£©×î¸ß¼ÛÑõ»¯ÎïµÄË®»¯Îï¼îÐÔ×îÇ¿µÄÊÇ______¡£

£¨4£©×î¸ß¼ÛÑõ»¯ÎïµÄË®»¯ÎïΪÁ½ÐÔÇâÑõ»¯ÎïµÄÊÇ____________¡£

£¨5£©Æø̬Ç⻯ÎïµÄË®ÈÜÒº³Ê¼îÐÔµÄÔªËØÊÇ______¡£

¢¢£¨6£©Ð´³öCO2¡¢H2O2µÄµç×Óʽ£º

CO2__________¡¢H2O2__________¡£

£¨7£©Óõç×Óʽ±íʾHClºÍMgBr2µÄÐγɹý³Ì£º

HCl__________£»MgBr2__________¡£

¢£ÓÃX±íʾԭ×Ó£º

£¨8£©ÖÐÐÔÔ­×ÓµÄÖÐ×ÓÊýN£½________£»

£¨9£©AXn£­¹²ÓÐx¸öµç×Ó£¬Ôò¸ÃÒõÀë×ÓµÄÖÊ×ÓÊýZ£½______£»

£¨10£©12C16O2·Ö×ÓÖеÄÖÐ×ÓÊýN£½_______¡£

¡¾´ð°¸¡¿H Na Na£¨»òNaOH£© Al[»òAl(OH)3£©] N A-Z x-n 22

¡¾½âÎö¡¿

¢¡£¨1£©Ô­×ӵĵç×Ó²ãÊýÔ½ÉÙ£¬ÆäÔ­×Ӱ뾶ԽС£¬Í¬Ò»ÖÜÆÚÔªËØÖУ¬Ô­×Ӱ뾶Ëæ×ÅÔ­×ÓÐòÊýÔö´ó¶ø¼õС£¬Ô­×Ӱ뾶×îСµÄÊÇHÔªËØ£¬¹Ê´ð°¸Îª£ºH£»

£¨2£©Í¬Ö÷×åÔªËØ£¬Ô­×Ӱ뾶Ëæ×ÅÔ­×ÓÐòÊýÔö´ó¶øÔö´ó£¬Í¬ÖÜÆÚÔªËØ£¬Ô­×Ӱ뾶Ëæ×ÅÔ­×ÓÐòÊýÔö´ó¶ø¼õС£¬Ôò1¡«18ºÅÔªËØÖÐÔ­×Ӱ뾶×î´óµÄÊÇNaÔªËØ£¬¹Ê´ð°¸Îª£ºNa£»

£¨3£©ÔªËؽðÊôÐÔԽǿ£¬×î¸ß¼ÛÑõ»¯Îï¶ÔӦˮ»¯Îï¼îÐÔԽǿ£¬Í¬ÖÜÆÚÔªËØ´Ó×óÏòÓÒ£¬½ðÊôÐÔÒÀ´Î¼õÈõ£¬Í¬Ö÷×åÔªËØ´ÓÉϵ½Ï£¬½ðÊôÐÔÒÀ´ÎÔöÇ¿£¬Ôò1¡«18ºÅÔªËØÖнðÊôÐÔ×îÇ¿µÄÊÇNa£¬×î¸ß¼ÛÑõ»¯ÎïµÄË®»¯Îï¼îÐÔ×îÇ¿µÄÊÇÇâÑõ»¯ÄÆ£¬¹Ê´ð°¸Îª£ºNa£¨»òNaOH£©£»

£¨4£©×î¸ß¼ÛÑõ»¯ÎïµÄË®»¯ÎïΪÁ½ÐÔÇâÑõ»¯ÎïµÄÊÇÇâÑõ»¯ÂÁ£»¹Ê´ð°¸Îª£ºAl[»òAl(OH)3£©]£»

£¨5£©Æø̬Ç⻯ÎïµÄË®ÈÜÒº³Ê¼îÐÔµÄÆøÌåÊÇNH3£¬¹Ê´ð°¸Îª£ºN£»

¢¢£¨6£©¶þÑõ»¯Ì¼Îª¹²¼Û»¯ºÏÎ·Ö×ÓÖдæÔÚÁ½¶Ô̼Ñõ¹²Óõç×Ó¶Ô£¬µç×ÓʽΪ£º£»Ë«ÑõˮΪ¹²¼Û»¯ºÏÎ·Ö×ÓÖдæÔÚÁ½¸öÑõÇâ¼üºÍÒ»¸öO-O¼ü£¬µç×ÓʽΪ£º£¬¹Ê´ð°¸Îª£º£»£»

£¨7£©ÂÈ»¯ÇâΪ¹²¼Û»¯ºÏÎ·Ö×ÓÖÐÇâÔ­×ÓÓëÂÈÔ­×ÓÖ®¼äÐγÉ1¶Ô¹²Óõç×Ó¶Ô£¬Óõç×Óʽ±íʾÐγɹý³ÌΪ£»ä廯þÊÇÀë×Ó»¯ºÏÎÓÉäåÀë×ÓÓëþÀë×Ó¹¹³É£¬Óõç×Óʽ±íʾµÄÐγɹý³ÌΪ£¬¹Ê´ð°¸Îª£º£»£»

¢££¨7£©XÔ­×ÓµÄÖÊ×ÓÊýΪZ£¬ÖÊÁ¿ÊýΪA£¬ÓÉÖÊÁ¿Êý£½ÖÊ×ÓÊý£«ÖÐ×ÓÊý¿ÉÖª£¬XÔ­×ÓµÄÖÐ×ÓÊýN£½A£­Z£¬¹Ê´ð°¸Îª£ºA-Z£»

£¨8£©AXn£­¹²ÓÐx¸öµç×Ó£¬Ô­×ÓAXµÄµç×ÓÊýΪx£­n£¬Ô­×ÓÖÐÖÊ×ÓÊýµÈÓÚºËÍâµç×ÓÊý£¬ÔòAXn£­µÄÖÊ×ÓÊýΪx-n£¬¹Ê´ð°¸Îª£ºx-n£»

£¨9£©ÓÉÖÊÁ¿Êý£½ÖÊ×ÓÊý£«ÖÐ×ÓÊý¿ÉÖª£¬12C16O2·Ö×ÓÖеÄÖÐ×ÓÊýΪ£¨12¡ª6£©£«£¨16¡ª8£©¡Á2£½22£¬¹Ê´ð°¸Îª£º22¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÏÂÁи÷×éÀë×ÓÒ»¶¨¿ÉÒÔ´óÁ¿¹²´æµÄÊÇ

A. ³ÎÇå͸Ã÷ÈÜÒºÖÐ:K+¡¢Cu2+¡¢ClO£­¡¢S2£­

B. ÄÜÓëÂÁ·Û·´Ó¦Éú³ÉÇâÆøµÄÈÜÒºÖУºNa+¡¢Fe3+¡¢NO3-¡¢SO42-

C. c(Al3+)=0.1mol/LµÄÈÜÒºÖÐ:K+¡¢Mg2+¡¢SO42-¡¢AlO2-

D. c(Fe3+)=1 mol¡¤L-1µÄÈÜÒºÖУºMg2+¡¢H+¡¢MnO4-¡¢SO42-

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿(Ë«Ñ¡)ÏÂÁйØÓÚÏ©Ìþ¡¢È²ÌþµÄÐðÊöÖУ¬ÕýÈ·µÄÊÇ(¡¡¡¡)

A. ijÎïÖʵÄÃû³ÆΪ2¡ªÒÒ»ù¡ª1¡ª¶¡Ï©£¬ËüµÄ½á¹¹¼òʽΪ

B. CH3¡ªCH===CH¡ªC¡ÔC¡ªCF3·Ö×ӽṹÖÐ×î¶àÓÐ4¸ö̼ԭ×ÓÔÚÒ»ÌõÖ±ÏßÉÏ

C. ÏàͬÎïÖʵÄÁ¿µÄÒÒȲÓë±½·Ö±ðÔÚ×ãÁ¿µÄÑõÆøÖÐÍêȫȼÉÕ£¬ÏûºÄÑõÆøµÄÁ¿Ïàͬ

D. ¦ÂÔ¹ðÏ©µÄ½á¹¹ÈçͼËùʾ£¬¸ÃÎïÖÊÓëµÈÎïÖʵÄÁ¿äå·¢Éú¼Ó³É·´Ó¦µÄ²úÎï(²»¿¼ÂÇÁ¢ÌåÒì¹¹)ÀíÂÛÉÏ×î¶àÓÐ5ÖÖ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÏÂÁÐÊôÓÚÓлúÎÓÖÊǷǵç½âÖʵÄÊÇ

A.´¿¼îB.ÒÒËáC.ÆÏÌÑÌÇD.¸Ê°±Ëá

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÒÑÖªÏÂÁÐÈÈ»¯Ñ§·½³Ìʽ£º¢ÙH2£¨g£©+1/2O2 £¨g£©£½H2O£¨l£©¦¤H=-285kJ¡¤mol-1£¬¢ÚH2£¨g£©+1/2O2£¨g£©£½H2O£¨g£© ¦¤H=-241.8 kJ¡¤mol-1 £¬¢ÛC£¨s£©+1/2O2£¨g£©£½CO£¨g£© ¦¤H=-110.5 kJ¡¤mol-1£¬¢ÜC£¨s£©+O2£¨g£©£½CO2£¨g£© ¦¤H=-393.5 kJ¡¤mol-1£¬»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©H2ȼÉÕÈȵÄÈÈ»¯Ñ§·½³ÌʽΪ___________ £»CȼÉÕÈȵÄÈÈ»¯Ñ§·½³ÌʽΪ ___________¡££¨Ñ¡Êý×Ö£©

£¨2£©È¼ÉÕ1gH2Éú³ÉҺ̬ˮ£¬·Å³öµÄÈÈÁ¿Îª_________________¡£

£¨3£©ÒºÌ¬Ë®µÄÎȶ¨ÐÔ_______Æø̬ˮµÄÎȶ¨ÐÔ£¨Ìî¡°´óÓÚ¡±¡¢¡°Ð¡ÓÚ¡±¡¢¡°µÈÓÚ¡±£©¡£ ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÔÚÒ»¶¨Ìå»ýµÄÃܱÕÈÝÆ÷ÖУ¬½øÐÐÈçÏ»¯Ñ§·´Ó¦£ºÆ仯ѧƽºâ³£ÊýKÓëζÈtµÄ¹ØϵÈçÏ£ºCO2(g)+H2(g) CO(g)+H2O(g),Æ仯ѧƽºâ³£ÊýKºÍζÈtµÄ¹ØϵÈçÏÂ±í£º

t¡æ

700

800

830

1000

1200

K

0.6

0.9

1.0

1.7

2.6

Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©¸Ã·´Ó¦µÄ»¯Ñ§Æ½ºâ³£Êý±í´ïʽΪK =__________________¡£

£¨2£©¸Ã·´Ó¦Îª_________·´Ó¦¡££¨Ìî¡°ÎüÈÈ¡±»ò¡°·ÅÈÈ¡±£©

£¨3£©Èô·´Ó¦Î¶ÈÉý¸ß£¬CO2µÄת»¯ÂÊ________(Ìî¡°Ôö´ó¡±¡°¼õС¡±»ò¡°²»±ä¡±)£»

£¨4£©ÄÜÅжϸ÷´Ó¦ÊÇ·ñ´ïµ½»¯Ñ§Æ½ºâ״̬µÄÒÀ¾ÝÊÇ______¡£

A.ÈÝÆ÷ÖÐѹǿ²»±ä B.»ìºÏÆøÌåÖÐ c£¨CO£©²»±ä

C.vÕý£¨H2£©£½vÄ棨H2O£© D.c£¨CO2£©£½c£¨CO£©

£¨5£©Ä³Î¶ÈÏ£¬Æ½ºâŨ¶È·ûºÏÏÂʽ£ºc£¨CO2£©¡¤c£¨H2£©£½c£¨CO£©¡¤c£¨H2O£©£¬ÊÔÅжϴËʱµÄζÈΪ_____¡æ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÒÑÖªW¡¢X¡¢Y¡¢ZΪ¶ÌÖÜÆÚÔªËØ£¬W¡¢ZͬÖ÷×壬X¡¢Y¡¢ZͬÖÜÆÚ£¬WµÄÆø̬Ç⻯ÎïµÄÎȶ¨ÐÔ´óÓÚZµÄÆø̬Ç⻯ÎïÎȶ¨ÐÔ£¬X¡¢YΪ½ðÊôÔªËØ£¬XµÄÑôÀë×ÓµÄÑõ»¯ÐÔСÓÚYµÄÑôÀë×ÓµÄÑõ»¯ÐÔ£¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ

A. X¡¢Y¡¢Z¡¢WµÄÔ­×Ӱ뾶ÒÀ´Î¼õС

B. WÓëXÐγɵĻ¯ºÏÎïÖÐÖ»º¬Àë×Ó¼ü

C. WµÄÆø̬Ç⻯ÎïµÄ·ÐµãÒ»¶¨¸ßÓÚZµÄÆø̬Ç⻯ÎïµÄ·Ðµã

D. ÈôWÓëYµÄÔ­×ÓÐòÊýÏà²î5£¬Ôò¶þÕßÐγɻ¯ºÏÎïµÄ»¯Ñ§Ê½Ò»¶¨ÎªY2W3

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿Çø±ðŨÁòËáºÍÏ¡ÁòËᣬ¼È¼òµ¥ÓÖ·½±ãµÄ·½·¨ÊÇ(¡¡¡¡)

A.³£ÎÂÏÂÓëͭƬ·´Ó¦B.ÓëʯÈïÈÜÒº·´Ó¦

C.Óò£Á§°ô¸÷ÕºÉÙÐíÈÜҺͿÔÚÖ½ÉÏD.¼ÓÈëпƬ¿´ÊÇ·ñÓÐÆøÌåÉú³É

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÔÚÒ»ÃܱÕÈÝÆ÷ÖнøÐз´Ó¦£º2SO2(g)£«O2(g)2SO3(g)¡£ÒÑÖª·´Ó¦¹ý³ÌÖÐijһʱ¿ÌSO2¡¢O2¡¢SO3µÄŨ¶È·Ö±ðΪ0.2 mol¡¤L£­1¡¢0.1 mol¡¤L£­1¡¢0.2 mol¡¤L£­1¡£µ±·´Ó¦´ïµ½Æ½ºâʱ£¬¿ÉÄÜ´æÔÚµÄÊý¾ÝÊÇ(¡¡¡¡)

A. SO2µÄŨ¶ÈΪ0.4 mol¡¤L£­1£¬O2µÄŨ¶ÈΪ0.2 mol¡¤L£­1

B. SO2µÄŨ¶ÈΪ0.25 mol¡¤L£­1

C. SO3µÄŨ¶ÈΪ0.4 mol¡¤L£­1

D. SO2¡¢SO3µÄŨ¶È¾ùΪ0.15 mol¡¤L£­1

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸