³ÇÊÐʹÓõÄȼÁÏ£¬ÏÖ´ó¶àΪúÆø¡¢Òº»¯Ê¯ÓÍÆø£®ÃºÆøµÄÖ÷Òª³É·ÖΪCOºÍµÄ»ìºÏÆøÌ壬ËüÓÉú̿ÓëË®ÕôÆøÔÚ¸ßÎÂÏ·´Ó¦ÖƵ㬹ÊÓÖ³ÆˮúÆø£®ÊÔ½â´ðÏÂÁÐÎÊÌ⣺
(1)д³öÖÆȡˮúÆøµÄÖ÷Òª»¯Ñ§·½³Ìʽ________£¬¸Ã·´Ó¦ÊÇ________·´Ó¦£®(Ìî¡°ÎüÈÈ¡±»ò¡°·ÅÈÈ¡±)
(2)ÉèÒº»¯Ê¯ÓÍÆøµÄÖ÷Òª³É·ÖΪ±ûÍ飬Æä³ä·ÖȼÉÕºó²úÎïΪºÍ£®ÊԱȽÏÍêȫȼÉÕµÈÖÊÁ¿µÄ¼°COËùÐèµÄÑõÆøµÄÖÊÁ¿±ÈΪ________£®
(3)ÇâÆøÊÇδÀ´µÄÄÜÔ´£¬È¼ÉÕʱ³ý²úÉúÈÈÁ¿¶àÍ⣬»¹¾ßÓеÄÓŵãÊÇ_________£®
(4)ʹÓùܵÀúÆø±ØÐë·ÀÖ¹ÒòúÆø驶øÒýÆðµÄÖж¾ºÍ±¬Õ¨£®Ä³×¡·¿ÒòúÆø驶ø·¢Éú±¬Õ¨Ê¹ʣ®µ÷²é·¢ÏÖúÆøÊÇÔÚ´ïµ½ÊÒÄÚ¿ÕÆøÖÊÁ¿µÄ6£¥Ê±·¢Éú±¬Õ¨µÄ£¬¸Ãס»§ÃºÆøй©ËÙ¶ÈΪ0.6g¡¤£¬Ôò±¬Õ¨Ê±ÃºÆøÒÑй©Լ________h(¼ÙÉè³ø·¿Ìå»ýΪ25£¬¿ÕÆøÃܶÈԼΪ1.29g¡¤)
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
£¨8·Ö£©³ÇÊÐʹÓõÄȼÁÏ£¬ÏÖ´ó¶àÓÃúÆø¡¢Òº»¯Ê¯ÓÍÆø¡£ÃºÆøµÄÖ÷Òª³É·ÖÊÇCOºÍH2µÄ»ìºÏÆøÌ壬ËüÓÉú̿ºÍË®ÕôÆø·´Ó¦ÖƵ㬹ÊÓÖ³ÆˮúÆø¡£
£¨1£©ÊÔд³öÖÆȡˮúÆøµÄÖ÷Òª»¯Ñ§·½³Ìʽ____________________________.
£¨2£©Òº»¯Ê¯ÓÍÆøµÄÖ÷Òª³É·ÖÊDZûÍ飬±ûÍéȼÉÕµÄÈÈ»¯Ñ§·½³ÌʽΪ
C3H8(g) +5O2(g)== 3CO2(g) + 4H2O(l)£»¡÷H=¨C2220.0 kJ¡¤,
ÓÖÖªCOÆøÌåȼÉÕµÄÈÈ»¯Ñ§·½³ÌʽΪ£º
CO(g) +1/2O2(g) == CO2(g) £»¡÷H=¨C282.57kJ¡¤,
ÊԱȽÏͬÖÊÁ¿µÄC3H8ºÍCOȼÉÕ£¬²úÉúµÄÈÈÁ¿±ÈֵԼΪ_________£º1¡£
£¨3£©ÒÑÖªÇâÆøȼÉÕµÄÈÈ»¯Ñ§·½³ÌʽΪ2H2(g)+ O2(g) == 2H2O(l)£»¡÷H=¨C571.6 kJ¡¤,ÊԱȽÏͬÖÊÁ¿µÄH2ºÍC3H8ȼÉÕ£¬²úÉúµÄÈÈÁ¿±ÈֵԼΪ____________£º1¡£
£¨4£©ÇâÆøÊÇδÀ´µÄÄÜÔ´£¬³ýÀ´Ô´·á¸»Ö®Í⣬»¹¾ßÓеÄÓŵãÊÇ____________________
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2010ÄêÁÉÄþÊ¡±¾ÏªÏظßÖи߶þÊîÆÚ²¹¿Î½×¶Î¿¼ÊÔ»¯Ñ§ÊÔÌâ ÌâÐÍ£ºÌî¿ÕÌâ
£¨8·Ö£©³ÇÊÐʹÓõÄȼÁÏ£¬ÏÖ´ó¶àÓÃúÆø¡¢Òº»¯Ê¯ÓÍÆø¡£ÃºÆøµÄÖ÷Òª³É·ÖÊÇCOºÍH2µÄ»ìºÏÆøÌ壬ËüÓÉú̿ºÍË®ÕôÆø·´Ó¦ÖƵ㬹ÊÓÖ³ÆˮúÆø¡£
£¨1£©ÊÔд³öÖÆȡˮúÆøµÄÖ÷Òª»¯Ñ§·½³Ìʽ____________________________.
£¨2£©Òº»¯Ê¯ÓÍÆøµÄÖ÷Òª³É·ÖÊDZûÍ飬±ûÍéȼÉÕµÄÈÈ»¯Ñ§·½³ÌʽΪ
C3H8(g) +5O2(g) == 3CO2(g) + 4H2O(l)£»¡÷H=¨C2220.0 kJ¡¤,
ÓÖÖªCOÆøÌåȼÉÕµÄÈÈ»¯Ñ§·½³ÌʽΪ£º
CO(g) +1/2O2(g) == CO2(g) £»¡÷H=¨C282.57kJ¡¤,
ÊԱȽÏͬÖÊÁ¿µÄC3H8ºÍCOȼÉÕ£¬²úÉúµÄÈÈÁ¿±ÈֵԼΪ_________£º1¡£
£¨3£©ÒÑÖªÇâÆøȼÉÕµÄÈÈ»¯Ñ§·½³ÌʽΪ2H2(g) + O2(g) == 2H2O(l)£»¡÷H=¨C571.6 kJ¡¤,ÊԱȽÏͬÖÊÁ¿µÄH2ºÍC3H8ȼÉÕ£¬²úÉúµÄÈÈÁ¿±ÈֵԼΪ____________£º1¡£
£¨4£©ÇâÆøÊÇδÀ´µÄÄÜÔ´£¬³ýÀ´Ô´·á¸»Ö®Í⣬»¹¾ßÓеÄÓŵãÊÇ____________________
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2010ÄêÁÉÄþÊ¡¸ß¶þÊîÆÚ²¹¿Î½×¶Î¿¼ÊÔ»¯Ñ§ÊÔÌâ ÌâÐÍ£ºÌî¿ÕÌâ
£¨8·Ö£©³ÇÊÐʹÓõÄȼÁÏ£¬ÏÖ´ó¶àÓÃúÆø¡¢Òº»¯Ê¯ÓÍÆø¡£ÃºÆøµÄÖ÷Òª³É·ÖÊÇCOºÍH2µÄ»ìºÏÆøÌ壬ËüÓÉú̿ºÍË®ÕôÆø·´Ó¦ÖƵ㬹ÊÓÖ³ÆˮúÆø¡£
£¨1£©ÊÔд³öÖÆȡˮúÆøµÄÖ÷Òª»¯Ñ§·½³Ìʽ____________________________.
£¨2£©Òº»¯Ê¯ÓÍÆøµÄÖ÷Òª³É·ÖÊDZûÍ飬±ûÍéȼÉÕµÄÈÈ»¯Ñ§·½³ÌʽΪ
C3H8(g) +5O2(g) == 3CO2(g) + 4H2O(l)£»¡÷H=¨C2220.0 kJ¡¤,
ÓÖÖªCOÆøÌåȼÉÕµÄÈÈ»¯Ñ§·½³ÌʽΪ£º
CO(g) +1/2O2(g) == CO2(g) £»¡÷H=¨C282.57kJ¡¤,
ÊԱȽÏͬÖÊÁ¿µÄC3H8ºÍCOȼÉÕ£¬²úÉúµÄÈÈÁ¿±ÈֵԼΪ_________£º1¡£
£¨3£©ÒÑÖªÇâÆøȼÉÕµÄÈÈ»¯Ñ§·½³ÌʽΪ2H2(g) + O2(g) == 2H2O(l)£»¡÷H=¨C571.6 kJ¡¤,ÊԱȽÏͬÖÊÁ¿µÄH2ºÍC3H8ȼÉÕ£¬²úÉúµÄÈÈÁ¿±ÈֵԼΪ____________£º1¡£
£¨4£©ÇâÆøÊÇδÀ´µÄÄÜÔ´£¬³ýÀ´Ô´·á¸»Ö®Í⣬»¹¾ßÓеÄÓŵãÊÇ____________________
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£ºÍ¬²½Ìâ ÌâÐÍ£ºÅжÏÌâ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
³ÇÊÐʹÓõÄȼÁÏ£¬ÏÖ´ó¶àÓÃúÆø¡¢Òº»¯Ê¯ÓÍÆø¡£ÃºÆøµÄÖ÷Òª³É·ÖÊÇCOºÍH2µÄ»ìºÏÆøÌ壬ËüÓÉú̿ºÍË®ÕôÆø·´Ó¦ÖƵ㬹ÊÓÖ³ÆˮúÆø¡£
£¨1£©ÊÔд³öÖÆȡˮúÆøµÄÖ÷Òª»¯Ñ§·½³Ìʽ____________________________.
£¨2£©Òº»¯Ê¯ÓÍÆøµÄÖ÷Òª³É·ÖÊDZûÍ飬±ûÍéȼÉÕµÄÈÈ»¯Ñ§·½³ÌʽΪ
C3H8(g) +5O2(g) == 3CO2(g) + 4H2O(l)£»¡÷H=¨C2220.0 kJ¡¤,
ÓÖÖªCOÆøÌåȼÉÕµÄÈÈ»¯Ñ§·½³ÌʽΪ£º
CO(g) +1/2O2(g) == CO2(g) £»¡÷H=¨C282.57kJ¡¤,
ÊԱȽÏͬÖÊÁ¿µÄC3H8ºÍCOȼÉÕ£¬²úÉúµÄÈÈÁ¿±ÈֵԼΪ_________£º1¡£
£¨3£©ÒÑÖªÇâÆøȼÉÕµÄÈÈ»¯Ñ§·½³ÌʽΪ2H2(g) + O2(g) == 2H2O(l)£»¡÷H=¨C571.6 kJ¡¤,ÊԱȽÏͬÖÊÁ¿µÄH2ºÍC3H8ȼÉÕ£¬²úÉúµÄÈÈÁ¿±ÈֵԼΪ____________£º1¡£
£¨4£©ÇâÆøÊÇδÀ´µÄÄÜÔ´£¬³ýÀ´Ô´·á¸»Ö®Í⣬»¹¾ßÓеÄÓŵãÊÇ____________________
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