·ÖÎö £¨1£©¿ª·¢½Ï¾¼ÃÇÒ×ÊÔ´¿É³ÖÐøÀûÓõÄÖÆÇâÆø·½·¨£¬ÏûºÄÄÜÔ´Ô½ÉÙÔ½ºÃ£»
£¨2£©´ß»¯¼ÁÄܽµµÍ·´Ó¦»î»¯ÄÜ£¬µ±·´Ó¦Îï×ÜÄÜÁ¿´óÓÚÉú³ÉÎï×ÜÄÜÁ¿Ê±£¬¸Ã·´Ó¦ÊÇ·ÅÈÈ·´Ó¦£¬µ±·´Ó¦Îï×ÜÄÜÁ¿Ð¡ÓÚÉú³ÉÎï×ÜÄÜÁ¿Ê±£¬¸Ã·´Ó¦ÊÇÎüÈÈ·´Ó¦£»
£¨3£©1gÇâÆøµÄÎïÖʵÄÁ¿Îª0.5mol£¬1gÇâÆøÍêȫȼÉÕÉú³ÉҺ̬ˮÊͷųö142.9kJµÄÈÈÁ¿£¬Ôò2molÇâÆøÍêȫȼÉÕÉú³ÉҺ̬ˮ·Å³öÈÈÁ¿=$\frac{142.9kJ}{0.5mol}$=571.6 kJ£¬¾Ý´ËÊéдÆäÈÈ»¯Ñ§·´Ó¦·½³Ìʽ£»
£¨4£©¾Ý¸Ç˹¶¨ÂÉÓÉÒÑÖªµÄÈÈ»¯Ñ§·½³Ìʽ³ËÒÔÏàÓ¦µÄÊýÖµ½øÐмӼõ£¬À´¹¹ÔìÄ¿±êÈÈ»¯Ñ§·½³Ìʽ£¬·´Ó¦ÈÈÒ²³ËÒÔÏàÓ¦µÄÊýÖµ½øÐмӼõ£¬ÓɸÇ˹¶¨ÂÉ¿ÉÖª£¬¢Ù¡Á2+¢Ú+¢ÛµÃµ½£»
£¨5£©¢ÙÇâÑõȼÁÏµç³ØÖУ¬ÑõÆøÔÚÕý¼«µÃµç×ÓÉú³ÉÇâÑõ¸ùÀë×Ó£»
¢Ú͵缫ÖÊÁ¿¼õÇá6.4g£¬Ôò×ªÒÆµç×ÓµÄÎïÖʵÄÁ¿=$\frac{6.4g}{64g/mol}$¡Á2=0.2mol£¬¸ù¾Ý×ªÒÆµç×ÓÊØºã¼ÆËãÏûºÄÑõÆøÌå»ý£»
£¨6£©Ôµç³ØµÄÕý¼«ÉÏ·¢ÉúµÃµç×ӵĻ¹Ô·´Ó¦£¬¾Ý´ËÊéдµç¼«·´Ó¦£®
½â´ð ½â£º£¨1£©¿ª·¢½Ï¾¼ÃÇÒ×ÊÔ´¿É³ÖÐøÀûÓõÄÖÆÇâÆø·½·¨£¬ÏûºÄÄÜÔ´Ô½ÉÙÔ½ºÃ£¬
A£®µç½âË®ÀË·ÑÄÜÔ´£¬ËùÒԸ÷½·¨²»ºÃ£¬¹Ê²»Ñ¡£»
B£®Ð¿¡¢Ï¡ÁòËá¼Û¸ñ½Ï¸ß£¬²»¾¼Ã£¬ËùÒԸ÷½·¨²»ºÃ£¬¹Ê²»Ñ¡£»
C£®¹â½âº£Ë®ÄÜÔ´ÏûºÄµÍ£¬¾¼Ã£¬¸Ã·½·¨ºÃ£¬¹ÊÑ¡£»
D£®·Ö½âÌìÈ»ÆøÀË·ÑÄÜÔ´£¬ËùÒԸ÷½·¨²»ºÃ£¬¹Ê²»Ñ¡£»
¹ÊÑ¡C£»
£¨2£©´ß»¯¼ÁÄܽµµÍ·´Ó¦»î»¯ÄÜ£¬ËùÒÔbÇúÏßʹÓô߻¯¼Á£¬¸ù¾Ý·´Ó¦ÎïºÍÉú³ÉÎï×ÜÄÜÁ¿Ïà¶Ô´óС֪£¬¸Ã·´Ó¦ÊÇÎüÈÈ·´Ó¦£¬
¹Ê´ð°¸Îª£ºb£»ÎüÈÈ£»
£¨3£©1gÇâÆøµÄÎïÖʵÄÁ¿Îª0.5mol£¬1gÇâÆøÍêȫȼÉÕÉú³ÉҺ̬ˮÊͷųö142.9kJµÄÈÈÁ¿£¬Ôò2molÇâÆøÍêȫȼÉÕÉú³ÉҺ̬ˮ·Å³öÈÈÁ¿=$\frac{142.9kJ}{0.5mol}$=571.6 kJ£¬¸ÃÈÈ»¯Ñ§·´Ó¦·½³ÌʽΪ£º2H2£¨g£©+O2£¨g£©=2H2O£¨l£©¡÷H=-571.6 kJ•mol-1 £¨ÆäËüºÏÀí´ð°¸¾ù¿É£©£¬
¹Ê´ð°¸Îª£º2H2£¨g£©+O2£¨g£©=2H2O£¨l£©¡÷H=-571.6 kJ•mol-1 £¨ÆäËüºÏÀí´ð°¸¾ù¿É£©£»
£¨4£©ÒÑÖª£º¢ÙCO£¨g£©+2H2£¨g£©?CH3OH£¨g£©¡÷H=-90.8kJ/mol£¬
¢Ú2CH3OH£¨g£©?CH3OCH3£¨g£©+H2O£¨g£©¡÷H=-23.5kJ/mol£¬
¢ÛCO£¨g£©+H2O£¨g£©?CO2£¨g£©+H2£¨g£©¡÷H=-41.3kJ/mol£¬
ÓɸÇ˹¶¨ÂÉ¿ÉÖª£¬¢Ù¡Á2+¢Ú+¢ÛµÃ3CO£¨g£©+3H2£¨g£©?CH3OCH3£¨g£©+CO2£¨g£©¡÷H=-246.4kJ/mol£¬
¹Ê´ð°¸Îª£º-246.4kJ/mol£»
£¨5£©¢ÙÇâÑõȼÁÏµç³ØÖУ¬ÑõÆøÔÚÕý¼«µÃµç×ÓÉú³ÉÇâÑõ¸ùÀë×Ó£¬µç¼«·´Ó¦Ê½ÎªO2+4e-+2H2O=4OH-£¬
¹Ê´ð°¸Îª£ºO2+4e-+2H2O=4OH-£»
¢Ú͵缫ÖÊÁ¿¼õÇá6.4g£¬Ôò×ªÒÆµç×ÓµÄÎïÖʵÄÁ¿=$\frac{6.4g}{64g/mol}$¡Á2=0.2mol£¬Í¨ÈëÑõÆøµÄµç¼«·´Ó¦Ê½ÎªO2+4e-+2H2O=4OH-£¬¸ù¾Ý×ªÒÆµç×ÓÏàµÈµÃ£¬
ÉèͨÈëÑõÆøÌå»ýΪV£¬
O2+4e-+2H2O=4OH-
22.4L 4mol
V 0.2mol
$\frac{22.4L}{V}=\frac{4mol}{0.2mol}$£¬
½âµÃV=1.12L£¬
¹Ê´ð°¸Îª£º1.12L£»
£¨6£©Õý¼«ÉϵªÆøµÃµç×ÓºÍÇâÀë×Ó·´Ó¦Éú³É笠ùÀë×Ó£¬¼´µç¼«·´Ó¦ÎªN2+8H++6e-¨T2NH4+£¬
¹Ê´ð°¸Îª£ºN2+8H++6e-¨T2NH4+£®
µãÆÀ ±¾Ì⿼²é½Ï×ۺϣ¬Éæ¼°Ôµç³ØºÍµç½â³ØÔÀí¡¢È¼ÉÕÈÈ»¯Ñ§·´Ó¦·½³ÌʽµÄÊéд¡¢´ß»¯¼ÁµÄ×÷ÓõÈ֪ʶµã£¬¸ù¾Ý×ªÒÆµç×ÓÏàµÈ½øÐеç½â³Ø¡¢Ôµç³Ø·´Ó¦µÄ¼ÆË㣬֪µÀ´ß»¯¼ÁÄܸıä»î»¯ÄÜ£¬µ«²»¸Ä±ä·´Ó¦ÈÈ£¬ÌâÄ¿ÄѶȲ»´ó£®
| Äê¼¶ | ¸ßÖÐ¿Î³Ì | Äê¼¶ | ³õÖÐ¿Î³Ì |
| ¸ßÒ» | ¸ßÒ»Ãâ·Ñ¿Î³ÌÍÆ¼ö£¡ | ³õÒ» | ³õÒ»Ãâ·Ñ¿Î³ÌÍÆ¼ö£¡ |
| ¸ß¶þ | ¸ß¶þÃâ·Ñ¿Î³ÌÍÆ¼ö£¡ | ³õ¶þ | ³õ¶þÃâ·Ñ¿Î³ÌÍÆ¼ö£¡ |
| ¸ßÈý | ¸ßÈýÃâ·Ñ¿Î³ÌÍÆ¼ö£¡ | ³õÈý | ³õÈýÃâ·Ñ¿Î³ÌÍÆ¼ö£¡ |
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ
| A£® | CO£¨g£© ÓëNa2O2£¨s£©·´Ó¦·Å³ö509kJÈÈÁ¿Ê±£¬µç×Ó×ªÒÆÊýΪ6.02¡Á1023 | |
| B£® | ͼ¿É±íʾÓÉCOÉú³ÉCO2µÄ·´Ó¦¹ý³ÌºÍÄÜÁ¿¹ØÏµ | |
| C£® | 2Na2O2£¨s£©+2CO2£¨s£©¨T2Na2CO3£¨s£©+O2£¨g£©¡÷H£¾-452kJ/mol | |
| D£® | COµÄȼÉÕÈÈΪ283kJ |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ
| A£® | ÍéÌþ | |
| B£® | ÓÉʯÓÍ·ÖÁó¿ÉÒÔ»ñµÃʯÀ¯£¬ÓÉʯÀ¯ÁÑ»¯¿É»ñµÃÒÒÏ© | |
| C£® | ͼ | |
| D£® | ÔÚÏ¡ÁòËáÈÜÒºÖУ¬CH3CO18OC2H5µÄË®½â²úÎïÊÇCH3CO18OHºÍC2H5OH |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ
| A£® | H+H¡úH-H | B£® | H-C1¡úH+C1 | ||
| C£® | Mg+2HCl¨TMgCl2+H2¡ü | D£® | H2SO4+2NaOH¨TNa2SO4+2H2O |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ
| A£® | H 2O£¨g£©¨TH 2£¨g£©+O 2£¨g£©¡÷H=-485 kJ•mol - 1 | |
| B£® | H 2O£¨g£©¨TH 2£¨g£©+O 2£¨g£©¡÷H=+485 kJ•mol - 1 | |
| C£® | 2H 2£¨g£©+O 2£¨g£©¨T2H 2O£¨g£©¡÷H=+485 kJ•mol - 1 | |
| D£® | 2H 2£¨g£©+O 2£¨g£©¨T2H 2O£¨g£©¡÷H=-485 kJ•mol - 1 |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ
| A£® | Éý¸ßζÈÄÜʹ»¯Ñ§·´Ó¦ËÙÂÊÔö´ó£¬Ö÷ÒªÔÒòÊǽµµÍÁË·´Ó¦»î»¯ÄÜ | |
| B£® | ÓÐÆøÌå²Î¼ÓµÄ»¯Ñ§·´Ó¦£¬Ôö´óѹǿһ¶¨¿ÉÒÔÔö´ó»¯Ñ§·´Ó¦ËÙÂÊ | |
| C£® | Ôö´ó·´Ó¦ÎïŨ¶È£¬¿ÉÌá¸ßµ¥Î»Ìå»ýÄڻ·Ö×ӵİٷÖÊý£¬´Ó¶øÊ¹ÓÐЧÅöײ´ÎÊýÔö´ó | |
| D£® | ´ß»¯¼ÁµÄ¼ÓÈëÄÜÌá¸ßµ¥Î»Ìå»ýÄڻ·Ö×Ó°Ù·ÖÊý£¬´Ó¶øÔö´ó»¯Ñ§·´Ó¦ËÙÂÊ |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
²é¿´´ð°¸ºÍ½âÎö>>
¹ú¼ÊѧУÓÅÑ¡ - 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