½üÄêÀ´£¬ÓÉCO2´ß»¯¼ÓÇâºÏ³É¼×´¼µÄÏà¹ØÑо¿Êܵ½Ô½À´Ô½¶àµÄ¹Ø×¢¡£¸Ã·½·¨¼È¿É½â¾öCO2·ÏÆøµÄÀûÓÃÎÊÌ⣬ÓÖ¿É¿ª·¢Éú²ú¼×´¼µÄÐÂ;¾¶£¬¾ßÓÐÁ¼ºÃµÄÓ¦ÓÃÇ°¾°¡£ÒÑÖª4.4 g CO2ÆøÌåÓëH2¾´ß»¯¼ÓÇâÉú³ÉCH3OHÆøÌåºÍË®ÕôÆøʱ·Å³ö4.95 kJµÄÄÜÁ¿¡£
(1)¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ£º__________________________________________.
(2)ÔÚ270¡æ¡¢8 MPaºÍÊʵ±´ß»¯¼ÁµÄÌõ¼þÏ£¬CO2µÄת»¯ÂÊ´ïµ½22%£¬Ôò4.48 m3(ÒÑÕÛºÏΪ±ê×¼×´¿ö)µÄCO2ÔںϳÉCH3OHÆøÌå¹ý³ÌÖÐÄܷųöÈÈÁ¿________ kJ.
(3)ÓÖÒÑÖªH2O(g)===H2O(l)¡¡¦¤H£½£44 kJ/mol£¬ÔòCO2ÆøÌåÓëH2ÆøÌå·´Ó¦Éú³ÉCH3OHÆøÌå16gºÍҺ̬ˮʱ·Å³öÈÈÁ¿Îª kJ£®
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
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
½üÄêÀ´£¬ÓÉCO2´ß»¯¼ÓÇâºÏ³É¼×´¼µÄÏà¹ØÑо¿Êܵ½Ô½À´Ô½¶àµÄ¹Ø×¢¡£¸Ã·½·¨¼È¿É½â¾öCO2·ÏÆøµÄÀûÓÃÎÊÌ⣬ÓÖ¿É¿ª·¢Éú²ú¼×´¼µÄÐÂ;¾¶£¬¾ßÓÐÁ¼ºÃµÄÓ¦ÓÃÇ°¾°¡£ÒÑÖª4.4 g CO2ÆøÌåÓëH2¾´ß»¯¼ÓÇâÉú³ÉCH3OHÆøÌåºÍË®ÕôÆøʱ·Å³ö4.95 kJµÄÄÜÁ¿¡£
(1)¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ£º__________________________________________.
(2)ÔÚ270¡æ¡¢8 MPaºÍÊʵ±´ß»¯¼ÁµÄÌõ¼þÏ£¬CO2µÄת»¯ÂÊ´ïµ½22%£¬Ôò4.48 m3(ÒÑÕÛºÏΪ±ê×¼×´¿ö)µÄCO2ÔںϳÉCH3OHÆøÌå¹ý³ÌÖÐÄܷųöÈÈÁ¿________ kJ.
(3)ÓÖÒÑÖªH2O(g)===H2O(l)¡¡¦¤H£½£44 kJ/mol£¬ÔòCO2ÆøÌåÓëH2ÆøÌå·´Ó¦Éú³ÉCH3OHÆøÌå16gºÍҺ̬ˮʱ·Å³öÈÈÁ¿Îª kJ£®
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2010Äêºþ±±Ê¡¾£ÖÝÖÐѧ¸ß¶þÉÏѧÆÚÆÚÖп¼ÊÔ»¯Ñ§ÊÔ¾í ÌâÐÍ£ºÌî¿ÕÌâ
½üÄêÀ´£¬ÓÉCO2´ß»¯¼ÓÇâºÏ³É¼×´¼µÄÏà¹ØÑо¿Êܵ½Ô½À´Ô½¶àµÄ¹Ø×¢¡£¸Ã·½·¨¼È¿É½â¾öCO2·ÏÆøµÄÀûÓÃÎÊÌ⣬ÓÖ¿É¿ª·¢Éú²ú¼×´¼µÄÐÂ;¾¶£¬¾ßÓÐÁ¼ºÃµÄÓ¦ÓÃÇ°¾°¡£ÒÑÖª4.4 g CO2ÆøÌåÓëH2¾´ß»¯¼ÓÇâÉú³ÉCH3OHÆøÌåºÍË®ÕôÆøʱ·Å³ö4.95 kJµÄÄÜÁ¿¡£
(1)¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ£º__________________________________________.
(2)ÔÚ270¡æ¡¢8 MPaºÍÊʵ±´ß»¯¼ÁµÄÌõ¼þÏ£¬CO2µÄת»¯ÂÊ´ïµ½22%£¬Ôò4.48 m3(ÒÑÕÛºÏΪ±ê×¼×´¿ö)µÄCO2ÔںϳÉCH3OHÆøÌå¹ý³ÌÖÐÄܷųöÈÈÁ¿________ kJ.
(3)ÓÖÒÑÖªH2O(g)===H2O(l)¡¡¦¤H£½£44 kJ/mol£¬ÔòCO2ÆøÌåÓëH2ÆøÌå·´Ó¦Éú³ÉCH3OHÆøÌå16gºÍҺ̬ˮʱ·Å³öÈÈÁ¿Îª kJ£®
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2010Äêºþ±±Ê¡¸ß¶þÉÏѧÆÚÆÚÖп¼ÊÔ»¯Ñ§ÊÔ¾í ÌâÐÍ£ºÌî¿ÕÌâ
½üÄêÀ´£¬ÓÉCO2´ß»¯¼ÓÇâºÏ³É¼×´¼µÄÏà¹ØÑо¿Êܵ½Ô½À´Ô½¶àµÄ¹Ø×¢¡£¸Ã·½·¨¼È¿É½â¾öCO2·ÏÆøµÄÀûÓÃÎÊÌ⣬ÓÖ¿É¿ª·¢Éú²ú¼×´¼µÄÐÂ;¾¶£¬¾ßÓÐÁ¼ºÃµÄÓ¦ÓÃÇ°¾°¡£ÒÑÖª4.4 g CO2ÆøÌåÓëH2¾´ß»¯¼ÓÇâÉú³ÉCH3OHÆøÌåºÍË®ÕôÆøʱ·Å³ö4.95 kJµÄÄÜÁ¿¡£
(1)¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ£º__________________________________________.
(2)ÔÚ270¡æ¡¢8 MPaºÍÊʵ±´ß»¯¼ÁµÄÌõ¼þÏ£¬CO2µÄת»¯ÂÊ´ïµ½22%£¬Ôò4.48 m3(ÒÑÕÛºÏΪ±ê×¼×´¿ö)µÄCO2ÔںϳÉCH3OHÆøÌå¹ý³ÌÖÐÄܷųöÈÈÁ¿________ kJ.
(3)ÓÖÒÑÖªH2O(g)===H2O(l)¡¡¦¤H£½£44 kJ/mol£¬ÔòCO2ÆøÌåÓëH2ÆøÌå·´Ó¦Éú³ÉCH3OHÆøÌå16gºÍҺ̬ˮʱ·Å³öÈÈÁ¿Îª kJ£®
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