K2CO3+6H2O2CH3OH+3O2+4KOH¡£ÇëÌî¿Õ£º
ͼ5
(1)³äµçʱ£¬¢ÙÔµç³ØµÄ¸º¼«ÓëµçÔ´____________¼«ÏàÁ¬¡£¢ÚÑô¼«µÄµç¼«·´Ó¦Îª________¡£
(2)·Åµçʱ£¬¸º¼«µÄµç¼«·´Ó¦Ê½Îª____________¡£
(3)Ôڴ˹ý³ÌÖÐÈôÍêÈ«·´Ó¦£¬ÒÒ³ØÖÐA¼«µÄÖÊÁ¿Éý¸ß648 g£¬Ôò¼×³ØÖÐÀíÂÛÉÏÏûºÄO2________L(±ê×¼×´¿öÏÂ)¡£
(4)ÈôÔÚ³£Î³£Ñ¹Ï£¬1 g CH3OHȼÉÕÉú³ÉCO2ºÍҺ̬H2Oʱ·ÅÈÈ22.68 kJ£¬±íʾ¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ_______________¡£
½âÎö£º(1)¢Ù¸º ¢Ú4OH--4e-====2H2O+O2
(2)CH3OH-6e-+8OH-====+6H2O
(3)¸ù¾Ýµç×ÓµÃʧÊغ㣬µÃ£º
4Ag ¡ª O2
V(O2)
V(O2)=33.6 L
(4)CH3OH(l)+O2(g)====CO2(g)+2H2O(l)£»¦¤H=725.76 kJ¡¤mol-1
´ð°¸£º(1)¸º 4OH--4e-====2H2O+O2
(2)CH3OH-6e-+8OH-====+6H2O
(3)33.6
(4)CH3OH(l)+O2(g)====CO2(g)+2H2O(l)£»¦¤H=725.76 kJ¡¤mol-1
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
·Åµç |
³äµç |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
·Åµç | ³äµç |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
ÇëÌî¿Õ£º
(1)³äµçʱ£º¢ÙÔµç³ØµÄ¸º¼«ÓëµçÔ´________¼«ÏàÁ¬¡£¢ÚÑô¼«µÄµç¼«·´Ó¦Îª_____________¡£
(2)·Åµçʱ£º¸º¼«µÄµç¼«·´Ó¦Ê½Îª__________________________________________¡£
(3)Ôڴ˹ý³ÌÖÐÈôÍêÈ«·´Ó¦£¬ÒÒ³ØÖÐA¼«µÄÖÊÁ¿Éý¸ß648 g£¬Ôò¼×³ØÖÐÀíÂÛÉÏÏûºÄO2_______L¡£(±ê×¼×´¿öÏÂ)
(4)ÈôÔÚ³£Î³£Ñ¹Ï£¬1 g CH3OHȼÁÏÉú³ÉCO2ºÍҺ̬H2Oʱ·ÅÈÈ22.68 kJ,±íʾ¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ ________________________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2010Äê±±¾©ÎåÖи߶þÏÂѧÆÚÆÚÄ©¿¼ÊÔ»¯Ñ§ÊÔÌâ ÌâÐÍ£ºÌî¿ÕÌâ
(7·Ö)¾Ý±¨µÀ£¬Ä¦ÍÐÂÞÀ¹«Ë¾¿ª·¢ÁËÒ»ÖÖÒÔ¼×´¼ÎªÔÁÏ£¬ÒÔΪµç½âÖʵÄÓÃÓÚÊÖ»úµÄ¿É³äµçµÄ¸ßЧȼÁϵç³Ø£¬³äÒ»´Îµç¿ÉÁ¬ÐøʹÓÃÒ»¸öÔ¡£ÏÂͼÊÇÒ»¸öµç»¯Ñ§¹ý³ÌµÄʾÒâͼ¡£ÒÑÖª¼×³ØµÄ×Ü·´Ó¦Ê½Îª£º2CH3OH+3O2+4KOH2K2CO3+6H2O
ÇëÌî¿Õ£º
£¨1£©·Åµçʱ£º¸º¼«µÄµç¼«·´Ó¦Ê½Îª____ _____¡£
£¨2£©³äµçʱ£º¢ÙÔµç³ØµÄ¸º¼«ÓëµçÔ´_________¼«ÏàÁ¬¡£
¢ÚÑô¼«µÄµç¼«·´Ó¦Îª__________________¡£
£¨3£©Ôڴ˹ý³ÌÖÐÈôÍêÈ«·´Ó¦£¬ÒÒ³ØÖÐB¼«µÄÖÊÁ¿Éý¸ß648g£¬Ôò¼×³ØÖÐÀíÂÛÉÏÏûºÄ_________L£¨±ê×¼×´¿öÏ£©¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2010Äê±±¾©ÎåÖи߶þÏÂѧÆÚÆÚÄ©¿¼ÊÔ»¯Ñ§ÊÔÌâ ÌâÐÍ£ºÌî¿ÕÌâ
(7·Ö)¾Ý±¨µÀ£¬Ä¦ÍÐÂÞÀ¹«Ë¾¿ª·¢ÁËÒ»ÖÖÒÔ¼×´¼ÎªÔÁÏ£¬ÒÔΪµç½âÖʵÄÓÃÓÚÊÖ»úµÄ¿É³äµçµÄ¸ßЧȼÁϵç³Ø£¬³äÒ»´Îµç¿ÉÁ¬ÐøʹÓÃÒ»¸öÔ¡£ÏÂͼÊÇÒ»¸öµç»¯Ñ§¹ý³ÌµÄʾÒâͼ¡£ÒÑÖª¼×³ØµÄ×Ü·´Ó¦Ê½Îª£º2CH3OH+3O2+4KOH2K2CO3+6H2O
ÇëÌî¿Õ£º
£¨1£©·Åµçʱ£º¸º¼«µÄµç¼«·´Ó¦Ê½Îª____ _____¡£
£¨2£©³äµçʱ£º¢ÙÔµç³ØµÄ¸º¼«ÓëµçÔ´_________¼«ÏàÁ¬¡£
¢ÚÑô¼«µÄµç¼«·´Ó¦Îª__________________¡£
£¨3£©Ôڴ˹ý³ÌÖÐÈôÍêÈ«·´Ó¦£¬ÒÒ³ØÖÐB¼«µÄÖÊÁ¿Éý¸ß648g£¬Ôò¼×³ØÖÐÀíÂÛÉÏÏûºÄ_________L£¨±ê×¼×´¿öÏ£©¡£
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