£¨1£©ÏÂÁÐÎïÖÊÖУ¬ÊôÓÚÇ¿µç½âÖʵÄÊÇ                   £¨ÌîÐòºÅ£¬ÏÂͬ£©£¬ÊôÓÚÈõµç½âÖʵÄÊÇ                  £¬ÊôÓڷǵç½âÖʵÄÊÇ                  ¡£

A¡¢NH4Cl  B¡¢CaCO3  C¡¢SO2  D¡¢ÕáÌÇ  E¡¢NaClÈÜÒº 

F¡¢NaHCO3  G¡¢NH3¡¤H2O   H¡¢ÒºÂÈ   I¡¢Í­

     £¨2£©0.3 mol B2H6(Æø̬¸ßÄÜȼÁÏ)ÔÚÑõÆøÖÐȼÉÕ£¬Éú³É¹Ì̬ÈýÑõ»¯¶þÅðºÍҺ̬ˮ£¬·Å³ö649.5kJ µÄÈÈÁ¿£¬Ð´³ö¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ                                  

                                             ¡£

£¨3£©ÒÑÖªÏÂÁÐÈÈ»¯Ñ§·½³Ìʽ£º

¢ÙCH3COOH(l)£«2O2(g)===2CO2(g)£«2H2O(l)

¦¤H1£½£­870.3 kJ/mol

¢ÚC(s)£«O2(g)===CO2(g)¡¡¦¤H2£½£­393.5 kJ¡¤mol-1

¢ÛH2(g)£«O2(g)===H2O(l)¦¤H3£½£­285.8 kJ¡¤mol-1

д³öÓÉC(s)¡¢H2(g)ºÍO2(g)»¯ºÏÉú³ÉCH3COOH(l)µÄÈÈ»¯Ñ§·½³Ìʽ                        

£¨1£©ABF¡¢G¡¢CD£¨Ã¿¸öÑ¡Ïî1·Ö£©£¨3£©2C(s)£«2H2(g)£«O2(g)===CH3COOH(l) ¦¤H£½£­488.3 kJ¡¤mol-1£¨2·Ö£©

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨1£©ÏÂÁÐÎïÖÊ£º¢ÙCO2  ¢ÚÑÎËá   ¢ÛÉռÌå  ¢Ü¾Æ¾«  ¢ÝÌú  ¢Þ´¿´×ËᣨÓñàºÅ»Ø´ð£©ÆäÖÐÄܵ¼µçµÄÊÇ
¢Ú¢Ý
¢Ú¢Ý
£¬ÊôÓÚµç½âÖʵÄÓÐ
¢Û¢Þ
¢Û¢Þ
£»Êô·Çµç½âÖʵÄÊÇ
¢Ù¢Ü£®
¢Ù¢Ü£®
£®
£¨2£©0.270kgÖÊÁ¿·ÖÊýΪ10%µÄCuCl2ÈÜÒºÖÐCuCl2µÄÎïÖʵÄÁ¿
0.2 mol
0.2 mol
£¬£¨´øµ¥Î»£©ÈÜÒºÖÐCu2+ºÍCl-µÄÎïÖʵÄÁ¿·Ö±ðΪ
0.2 mol
0.2 mol
£¬
0.4 mol
0.4 mol
£®£¨´øµ¥Î»£©
£¨3£©±ê×¼×´¿öÏ£¬224mlijÆøÌåµÄÖÊÁ¿Îª0.32¿Ë£¬Ôò¸ÃÆøÌåµÄÏà¶Ô·Ö×ÓÖÊÁ¿
32
32
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÔÚÏÂÁÐÎïÖÊÖл¥ÎªÍ¬·ÖÒì¹¹ÌåµÄÓÐ
£¨9£©Ó루10£©
£¨9£©Ó루10£©
£»»¥ÎªÍ¬ËØÒìÐÎÌåµÄÓÐ
£¨3£©Ó루4£©
£¨3£©Ó루4£©
£»»¥ÎªÍ¬Î»ËصÄÓÐ
£¨5£©Ó루6£©
£¨5£©Ó루6£©
£»ÊôͬһÎïÖʵÄÊÇ
£¨1£©Ó루2£©
£¨1£©Ó루2£©
£»»¥ÎªÍ¬ÏµÎïµÄÊÇ
£¨7£©Ó루8£©
£¨7£©Ó루8£©
£®
£¨1£©ÒºÂÈ £¨2£©ÂÈÆø      £¨3£©°×Á× £¨4£©ºìÁ×       £¨5£©D £¨6£©T£¨7£©CH3-CH3 £¨8£©CH3-CH2-CH3£¨9£©CH3-CH2-CH2-CH2-CH3£®£¨10£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â

26£®²ÄÁÏÊÇ¿Æѧ¼¼Êõ½ø²½µÄ¹Ø¼ü£¬ÊÇ¿Æѧ¼¼ÊõºÍÉç»á·¢Õ¹µÄÎïÖÊ»ù´¡£®²ÄÁϵķ¢Õ¹²»½öÓ°ÏìÁËÈËÀàµÄ×òÌìºÍ½ñÌ죬¶øÇÒ»¹½«Ó°Ïìµ½ÈËÀàµÄÃ÷Ì죮Çë»Ø´ðÒÔÏÂÓë²ÄÁÏÓйصÄÎÊÌ⣮
£¨1£©ÎÞ»ú·Ç½ðÊô²ÄÁÏ£®µ¥¾§¹èÊÇÒ»ÖֱȽϻîÆõķǽðÊôÔªËØ£¬ÊǾ§Ìå²ÄÁϵÄÖØÒª×é³É²¿·Ö£¬´¦ÓÚвÄÁÏ·¢Õ¹µÄÇ°ÑØ£®ÆäÖ÷ÒªÓÃ;ÊÇÓÃ×ö°ëµ¼Ìå²ÄÁϺÍÀûÓÃÌ«ÑôÄܹâ·ü·¢µç¡¢¹©Èȵȣ®µ¥¾§¹èµÄÖƱ¸·½·¨ÈçÏ£º
SiO2
¢ÙC
¸ßÎÂ
Si£¨´Ö£©
¢ÚHCI
300¡æ
SiHCl3
¢Û¹ýÁ¿H2
1000¡«1100¡æ
Si£¨´¿£©
¢Ùд³ö²½Öè¢ÙµÄ»¯Ñ§·½³Ìʽ£º
 
£®
¢ÚÒÑÖªÒÔϼ¸ÖÖÎïÖʵķе㣺
ÎïÖÊ SiHCl3 SiCl4 HCl
·Ðµã 33.0¡æ 57.6¡æ -84.7¡æ
ÔÚ²½Öè¢ÚÖÐÌá´¿SiHCl3Ëù½øÐеÄÖ÷Òª²Ù×÷µÄÃû³ÆÊÇ
 
£®
£¨2£©½ðÊô²ÄÁÏ£®½ðÊô²ÄÁÏÊÇÖ¸½ðÊôÔªËØ»òÒÔ½ðÊôÔªËØΪÖ÷¹¹³ÉµÄ¾ßÓнðÊôÌØÐԵIJÄÁϵÄͳ³Æ£®°üÀ¨´¿½ðÊô¡¢ºÏ½ðºÍÌØÖÖ½ðÊô²ÄÁϵȣ®
¢ÙÏÂÁÐÎïÖÊÖв»ÊôÓںϽðµÄÊÇ
 
£®A£®¸ÖÌú  B£®ÇàÍ­  C£®Ó²ÂÁ  D£®Ë®Òø
¢ÚÍ­Æ÷ÖÆÆ·³£Òò½Ó´¥¿ÕÆøÖеÄO2¡¢CO2ºÍH2O¶øÒ×Éú³ÉÍ­Ð⣮ÊÔд³ö±£»¤Í­ÖÆÆ·µÄ·½·¨£º
 
£®
£¨3£©ÄÉÃײÄÁÏ£¬½ºÌåÁ£×ÓµÄÖ±¾¶´óÔ¼ÊÇ
 
£¬ÓëÄÉÃײÄÁϵijߴçÏ൱£®ÊµÑéÊÒÖÆÈ¡Fe£¨OH£©3½ºÌåÈÜÒºµÄ·½·¨ÊÇ
 
£¬ÓÃ
 
·½·¨Ïû³ý½ºÌåÖеĻë×Ç£¬¸ù¾Ý
 
ÏÖÏóÖ¤Ã÷½ºÌåÒѾ­ÖƳɣ®ÊµÑéÖбØÐëÒªÓÃÕôÁóË®£¬¶ø²»ÄÜÓÃ×ÔÀ´Ë®£¬ÆäÔ­ÒòÊÇ
 
£®
£¨4£©´ÅÐÔ²ÄÁÏ£®Ä³´ÅÐÔ·ÛÄ©²ÄÁÏÊÇÒ»ÖÖ¸´ºÏÐÍÑõ»¯ÎΪ²â¶¨Æä×é³É£¬ÏÖ³ÆÈ¡6.26gÑùÆ·£¬½«ÆäÈ«²¿ÈÜÓÚ¹ýÁ¿Ï¡HNO3£¬¼ÓÈë¹ýÁ¿Na2SO4ÈÜÒº£¬Éú³É4.66g°×É«³Áµí¡¢¹ýÂË¡¢ÔÚÂËÒºÖмÓÈë¹ýÁ¿NaOHÈÜÒº£¬Éú³ÉºìºÖÉ«³Áµí£¬½«³Áµí¹ýÂË¡¢Ï´µÓ¡¢×ÆÉÕºóµÃ3.20g¹ÌÌ壮
¢Ù¸Ã´ÅÐÔ·ÛÄ©ÖÐÑõÔªËصÄÖÊÁ¿·ÖÊýΪ
 
£»
¢Ú¸Ã²ÄÁϵĻ¯Ñ§Ê½Îª
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2011-2012ѧÄê¹óÖÝÊ¡×ñÒåËÄÖиßÒ»ÉÏѧÆÚÆÚÖп¼ÊÔ»¯Ñ§ÊÔ¾í ÌâÐÍ£ºÌî¿ÕÌâ

£¨14·Ö£©£¨1£©ÏÂÁÐÎïÖÊ£º¢ÙCO2 ¢ÚÑÎËá  ¢ÛÉռÌ堢ܾƾ« ¢ÝÌú ¢Þ´¿´×ËᣨÓñàºÅ»Ø´ð£©ÆäÖÐÄܵ¼µçµÄÊÇ      £¬ÊôÓÚµç½âÖʵÄÓР   £»Êô·Çµç½âÖʵÄÊÇ       ¡£
£¨2£©0.270kgÖÊÁ¿·ÖÊýΪ10%µÄCuCl2ÈÜÒºÖÐCuCl2µÄÎïÖʵÄÁ¿           £¬£¨´øµ¥Î»£©
ÈÜÒºÖÐCu2+ºÍCl-µÄÎïÖʵÄÁ¿·Ö±ðΪ          £¬           ¡££¨´øµ¥Î»£©
£¨3£©±ê×¼×´¿öÏ£¬224mlijÆøÌåµÄÖÊÁ¿Îª0.32¿Ë£¬Ôò¸ÃÆøÌåµÄÏà¶Ô·Ö×ÓÖÊÁ¿          ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2015½ì¼ªÁÖÊ¡¸ß¶þÉÏѧÆÚÆÚÄ©¿¼ÊÔ»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÌî¿ÕÌâ

²ÄÁÏÊÇ¿Æѧ¼¼Êõ½ø²½µÄ¹Ø¼ü£¬ÊÇ¿Æѧ¼¼ÊõºÍÉç»á·¢Õ¹µÄÎïÖÊ»ù´¡¡£²ÄÁϵķ¢Õ¹²»½öÓ°ÏìÁËÈËÀàµÄ×òÌìºÍ½ñÌ죬¶øÇÒ»¹½«Ó°Ïìµ½ÈËÀàµÄÃ÷Ìì¡£Çë»Ø´ðÒÔÏÂÓë²ÄÁÏÓйصÄÎÊÌâ¡£

£¨1£©ÎÞ»ú·Ç½ðÊô²ÄÁÏ¡£µ¥¾§¹èÊÇÒ»ÖֱȽϻîÆõķǽðÊôÔªËØ£¬ÊǾ§Ìå²ÄÁϵÄÖØÒª×é³É²¿·Ö£¬´¦ÓÚвÄÁÏ·¢Õ¹µÄÇ°ÑØ¡£ÆäÖ÷ÒªÓÃ;ÊÇÓÃ×ö°ëµ¼Ìå²ÄÁϺÍÀûÓÃÌ«ÑôÄܹâ·ü·¢µç¡¢¹©Èȵȡ£µ¥¾§¹èµÄÖƱ¸·½·¨ÈçÏ£º

¢Ùд³ö²½Öè¢ÙµÄ»¯Ñ§·½³Ìʽ£º______________________¡£

¢ÚÒÑÖªÒÔϼ¸ÖÖÎïÖʵķе㣺

ÎïÖÊ

SiHCl3

SiCl4

HCl

·Ðµã

33.0 ¡æ

57.6 ¡æ

£­84.7 ¡æ

ÔÚ²½Öè¢ÚÖÐÌá´¿SiHCl3Ëù½øÐеÄÖ÷Òª²Ù×÷µÄÃû³ÆÊÇ________¡£

£¨2£©½ðÊô²ÄÁÏ¡£½ðÊô²ÄÁÏÊÇÖ¸½ðÊôÔªËØ»òÒÔ½ðÊôÔªËØΪÖ÷¹¹³ÉµÄ¾ßÓнðÊôÌØÐԵIJÄÁϵÄͳ³Æ¡£°üÀ¨´¿½ðÊô¡¢ºÏ½ðºÍÌØÖÖ½ðÊô²ÄÁϵȡ£

¢ÙÏÂÁÐÎïÖÊÖв»ÊôÓںϽðµÄÊÇ(¡¡¡¡)¡£

A£®¸ÖÌú       B£®ÇàÍ­       C£®Ó²ÂÁ        D£®Ë®Òø

¢ÚÍ­Æ÷ÖÆÆ·³£Òò½Ó´¥¿ÕÆøÖеÄO2¡¢CO2ºÍH2O¶øÒ×Éú³ÉÍ­Ðâ¡£ÊÔд³ö±£»¤Í­ÖÆÆ·µÄ·½·¨£º___________________________________________________¡£

£¨3£©ÄÉÃײÄÁÏ£¬½ºÌåÁ£×ÓµÄÖ±¾¶´óÔ¼ÊÇ________£¬ÓëÄÉÃײÄÁϵijߴçÏ൱¡£ÊµÑéÊÒÖÆÈ¡Fe(OH)3½ºÌåÈÜÒºµÄ·½·¨ÊÇ_________________________________£¬

ÓÃ________·½·¨Ïû³ý½ºÌåÖеĻë×Ç£¬¸ù¾Ý________ÏÖÏóÖ¤Ã÷½ºÌåÒѾ­ÖƳɡ£ÊµÑéÖбØÐëÒªÓÃÕôÁóË®£¬¶ø²»ÄÜÓÃ×ÔÀ´Ë®£¬ÆäÔ­ÒòÊÇ____________________¡£

£¨4£©´ÅÐÔ²ÄÁÏ¡£Ä³´ÅÐÔ·ÛÄ©²ÄÁÏÊÇÒ»ÖÖ¸´ºÏÐÍÑõ»¯ÎΪ²â¶¨Æä×é³É£¬ÏÖ³ÆÈ¡6.26 gÑùÆ·£¬½«ÆäÈ«²¿ÈÜÓÚ¹ýÁ¿Ï¡HNO3£¬¼ÓÈë¹ýÁ¿Na2SO4ÈÜÒº£¬Éú³É4.66 g°×É«³Áµí¡¢¹ýÂË¡¢ÔÚÂËÒºÖмÓÈë¹ýÁ¿NaOHÈÜÒº£¬Éú³ÉºìºÖÉ«³Áµí£¬½«³Áµí¹ýÂË¡¢Ï´µÓ¡¢×ÆÉÕºóµÃ3.20 g¹ÌÌå¡£¢Ù¸Ã´ÅÐÔ·ÛÄ©ÖÐÑõÔªËصÄÖÊÁ¿·ÖÊýΪ________£»¢Ú¸Ã²ÄÁϵĻ¯Ñ§Ê½Îª________¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸