ëÂ(N2H4)ÓÖ³ÆÁª°±£¬ÊÇÒ»ÖÖ¿ÉȼÐÔÒºÌ壬ÓëÑõÆø»òµªÑõ»¯Îï·´Ó¦¾ù¿ÉÉú³ÉµªÆøºÍË®£®ÇâÆøÊÇÒ»ÖÖÇå½àÄÜÔ´£¬ÒºÇâºÍë¾ù¿ÉÓÃ×÷»ð¼ýȼÁÏ£®
¢ñÇâÆøµÄÖÆÈ¡Óë´¢´æÊÇÇâÄÜÔ´ÀûÓÃÁìÓòµÄÑо¿Èȵ㣮
ÒÑÖª£ºCH4(g)£«H2O(g)£½CO(g)£«3H2(g)¡¡¦¤H£½£«206.2 kJ¡¤mol£1
CH4(g)£«CO2(g)£½2CO(g)£«2H2(g)¦¤H£½£«247.4 kJ¡¤mol£1
(1)ÇâÆø×÷ΪÐÂÄÜÔ´µÄÓŵã________£®(´ð2µã)
(2)ÒÔ¼×ÍéΪÔÁÏÖÆÈ¡ÇâÆøÊǹ¤ÒµÉϳ£ÓõÄÖÆÇâ·½·¨£®CH4(g)ÓëH2O(g)·´Ó¦Éú³ÉCO2(g)ºÍH2(g)µÄÈÈ»¯Ñ§·½³ÌʽΪ________£®
(3)H2OµÄÈÈ·Ö½âÒ²¿ÉµÃµ½H2£¬¸ßÎÂÏÂË®·Ö½âÌåϵÖÐÖ÷ÒªÆøÌåµÄÌå»ý·ÖÊýÓëζȵĹØϵÈçͼËùʾ£®Í¼ÖÐA¡¢B±íʾµÄÎïÖÊÒÀ´ÎÊÇ________¡¢________£®
¢ò(4)ëÂÒ»¿ÕÆøȼÁϵç³ØÊÇÒ»ÖÖ¼îÐÔȼÁϵç³Ø£¬µç½âÖÊÈÜÒºÊÇ20£¥£30£¥µÄKOHÈÜÒº£®¸Ãµç³Ø·Åµçʱ£¬¸º¼«µÄµç¼«·´Ó¦Ê½ÊÇ________£®
(5)ÏÂͼÊÇÒ»¸öµç»¯Ñ§×°ÖÃʾÒâͼ£®ÓÃëÂÒ»¿ÕÆøȼÁϵç³Ø×ö´Ë×°ÖõĵçÔ´£®
¢ÙÈç¹ûAÊDz¬µç¼«£¬BÊÇʯīµç¼«£¬CÊÇÁòËá£ÁòËá泥¬Òõ¼«µÄµç¼«·´Ó¦Ê½ÊÇ________£®
¢ÚÀûÓøÃ×°ÖÿÉÖƵÃÉÙÁ¿¹ýÑõ»¯Ç⣺ÔÚÑô¼«ÉÏSO42£±»Ñõ»¯³ÉS2O82£(¹ý¶þÁòËá¸ùÀë×Ó)£¬S2O82£ÓëH2O·´Ó¦Éú³ÉH2O2£¬S2O82££«2H2O£½2SO42££«H2O2£«2H+£®ÈôÒªÖÆÈ¡2 molH2O2£¬¸ÃȼÁϵç³ØÀíÂÛÉÏÐèÏûºÄ________molN2H4£®
(6)ÓÉA¡¢B¡¢C¡¢DËÄÖÖ½ðÊô°´Ï±íÖÐ×°ÖýøÐÐʵÑ飮
ʵÑé×°ÖÃÓëÏÖÏó
¸ù¾ÝʵÑéÏÖÏó»Ø´ðÏÂÁÐÎÊÌ⣺
¢Ù×°ÖñûÖÐÈÜÒºµÄPH________£®(Ìî¡°±ä´ó¡±¡°±äС¡±»ò¡°²»±ä¡±)
¢ÚËÄÖÖ½ðÊô»îÆÃÐÔÓÉÈõµ½Ç¿µÄ˳ÐòÊÇ________£®
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
(1)ëÂ(N2H4)ÓÖ³ÆÁª°±£¬ÊÇÒ»ÖÖ¿ÉȼÐÔµÄÒºÌ壬¿ÉÓÃ×÷»ð¼ýȼÁÏ¡£ÒÑÖªÔÚ101 kPaʱ£¬32.0 g N2H4ÔÚÑõÆøÖÐÍêȫȼÉÕÉú³ÉµªÆø£¬·Å³öÈÈÁ¿624 kJ(25¡æʱ)£¬N2H4ÍêȫȼÉÕ·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽÊÇ____________________________________________________________________¡£
(2)ëÂ-¿ÕÆøȼÁϵç³ØÊÇÒ»ÖÖ¼îÐÔȼÁϵç³Ø£¬µç½âÖÊÊÇ20%¡ª30%µÄKOHÈÜÒº¡£Ð´³öëÂ-¿ÕÆøȼÁϵç³Ø·ÅµçʱÕý¡¢¸º¼«µÄµç¼«·´Ó¦Ê½¡£
Õý¼«£º________________________________£¬
¸º¼«£º________________________________
(3)ͼ2-2-5ÊÇÒ»¸öµç»¯Ñ§¹ý³ÌʾÒâͼ¡£
ͼ2-2-5
¢ÙпƬÉÏ·¢ÉúµÄµç¼«·´Ó¦ÊÇ________________________________________________¡£
¢Ú¼ÙÉèʹÓÃëÂ?¿ÕÆøȼÁϵç³Ø×÷Ϊ±¾¹ý³ÌÖеĵçÔ´¡¢ÍƬµÄÖÊÁ¿±ä»¯128 g£¬ÔòëÂ-¿ÕÆøȼÁϵç³ØÀíÂÛÉÏÏûºÄ±ê±ê×¼×´¿öϵĿÕÆø___________L(¼ÙÉè¿ÕÆøÖÐÑõÆøÌå»ýº¬Á¿Îª20%)
(4)´«Í³ÖƱ¸ëµķ½·¨£¬ÊÇÒÔNaClOÑõ»¯NH3£¬ÖƵÃëµÄÏ¡ÈÜÒº¡£¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ________________________________________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2013-2014ѧÄêºþÄÏÊ¡ÖêÖÞÊиßÈýµÚËÄ´ÎÔ¿¼»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÑ¡ÔñÌâ
ëÂ(N2H4)ÓÖ³ÆÁª°±£¬ÊÇÒ»ÖÖ¿ÉȼÐÔµÄÒºÌ壬¿ÉÓÃ×÷»ð¼ýȼÁÏ¡£ÒÑÖª£º
¢ÙN2(g) + 2O2(g) =2 NO2(g) ¦¤H = +67£®7kJ/mol
¢Ú2N2H4(g) + 2NO2(g) =3N2(g) + 4H2O (g) ¦¤H = £1135£®7kJ/mol
ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ
A£®N2H4(g) + O2(g) = N2(g) + 2H2O(g) ¦¤H = £1068 kJ/mol
B£®ëÂÊÇÓë°±ÀàËƵÄÈõ¼î£¬ËüÒ×ÈÜÓÚË®£¬ÆäµçÀë·½³Ìʽ£ºN2H4 + H2O = N2H5+ + OH-
C£®²¬×öµç¼«£¬ÒÔKOHÈÜҺΪµç½âÖÊÈÜÒºµÄ롪¡ª¿ÕÆøȼÁϵç³Ø£¬·ÅµçʱµÄ¸º¼«·´Ó¦Ê½£ºN2H4 £4e£ + 4OH£ = N2 + 4H2O
D£®²¬×öµç¼«£¬ÒÔKOHÈÜҺΪµç½âÖÊÈÜÒºµÄ롪¡ª¿ÕÆøȼÁϵç³Ø£¬¹¤×÷Ò»¶Îʱ¼äºó£¬KOHÈÜÒºµÄpH½«Ôö´ó
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2013-2014ѧÄêºþÄÏÊ¡ÖêÖÞÊиßÈýµÚËÄ´ÎÔ¿¼»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÑ¡ÔñÌâ
ëÂ(N2H4)ÓÖ³ÆÁª°±£¬ÊÇÒ»ÖÖ¿ÉȼÐÔµÄÒºÌ壬¿ÉÓÃ×÷»ð¼ýȼÁÏ¡£ÒÑÖª£º
¢ÙN2(g) + 2O2(g) =2 NO2(g) ¦¤H = +67£®7kJ/mol
¢Ú2N2H4(g) + 2NO2(g) =3N2(g) + 4H2O (g) ¦¤H = £1135£®7kJ/mol
ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ
A£®N2H4(g) + O2(g) = N2(g) + 2H2O(g) ¦¤H = £1068 kJ/mol
B£®ëÂÊÇÓë°±ÀàËƵÄÈõ¼î£¬ËüÒ×ÈÜÓÚË®£¬ÆäµçÀë·½³Ìʽ£ºN2H4 + H2O = N2H5+ + OH-
C£®²¬×öµç¼«£¬ÒÔKOHÈÜҺΪµç½âÖÊÈÜÒºµÄ롪¡ª¿ÕÆøȼÁϵç³Ø£¬·ÅµçʱµÄ¸º¼«·´Ó¦Ê½£ºN2H4 £4e£ + 4OH£ = N2 + 4H2O
D£®²¬×öµç¼«£¬ÒÔKOHÈÜҺΪµç½âÖÊÈÜÒºµÄ롪¡ª¿ÕÆøȼÁϵç³Ø£¬¹¤×÷Ò»¶Îʱ¼äºó£¬KOHÈÜÒºµÄpH½«Ôö´ó
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2013-2014ѧÄêºþÄÏÊ¡ÖêÖÞÊиßÈýµÚËÄ´ÎÔ¿¼»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÑ¡ÔñÌâ
ëÂ(N2H4)ÓÖ³ÆÁª°±£¬ÊÇÒ»ÖÖ¿ÉȼÐÔµÄÒºÌ壬¿ÉÓÃ×÷»ð¼ýȼÁÏ¡£ÒÑÖª£º
¢ÙN2(g) + 2O2(g) =2 NO2(g) ¦¤H = +67£®7kJ/mol
¢Ú2N2H4(g) + 2NO2(g) =3N2(g) + 4H2O (g) ¦¤H = £1135£®7kJ/mol
ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ
A£®N2H4(g) + O2(g) = N2(g) + 2H2O(g) ¦¤H = £1068 kJ/mol
B£®ëÂÊÇÓë°±ÀàËƵÄÈõ¼î£¬ËüÒ×ÈÜÓÚË®£¬ÆäµçÀë·½³Ìʽ£ºN2H4 + H2O = N2H5+ + OH-
C£®²¬×öµç¼«£¬ÒÔKOHÈÜҺΪµç½âÖÊÈÜÒºµÄ롪¡ª¿ÕÆøȼÁϵç³Ø£¬·ÅµçʱµÄ¸º¼«·´Ó¦Ê½£ºN2H4 £4e£ + 4OH£ = N2 + 4H2O
D£®²¬×öµç¼«£¬ÒÔKOHÈÜҺΪµç½âÖÊÈÜÒºµÄ롪¡ª¿ÕÆøȼÁϵç³Ø£¬¹¤×÷Ò»¶Îʱ¼äºó£¬KOHÈÜÒºµÄpH½«Ôö´ó
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2013-2014ѧÄêºþÄÏÊ¡ÖêÖÞÊиßÈýµÚËÄ´ÎÔ¿¼»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÑ¡ÔñÌâ
ëÂ(N2H4)ÓÖ³ÆÁª°±£¬ÊÇÒ»ÖÖ¿ÉȼÐÔµÄÒºÌ壬¿ÉÓÃ×÷»ð¼ýȼÁÏ¡£ÒÑÖª£º
¢ÙN2(g) + 2O2(g) =2 NO2(g) ¦¤H = +67£®7kJ/mol
¢Ú2N2H4(g) + 2NO2(g) =3N2(g) + 4H2O (g) ¦¤H = £1135£®7kJ/mol
ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ
A£®N2H4(g) + O2(g) = N2(g) + 2H2O(g) ¦¤H = £1068 kJ/mol
B£®ëÂÊÇÓë°±ÀàËƵÄÈõ¼î£¬ËüÒ×ÈÜÓÚË®£¬ÆäµçÀë·½³Ìʽ£ºN2H4 + H2O = N2H5+ + OH-
C£®²¬×öµç¼«£¬ÒÔKOHÈÜҺΪµç½âÖÊÈÜÒºµÄ롪¡ª¿ÕÆøȼÁϵç³Ø£¬·ÅµçʱµÄ¸º¼«·´Ó¦Ê½£ºN2H4 £4e£ + 4OH£ = N2 + 4H2O
D£®²¬×öµç¼«£¬ÒÔKOHÈÜҺΪµç½âÖÊÈÜÒºµÄ롪¡ª¿ÕÆøȼÁϵç³Ø£¬¹¤×÷Ò»¶Îʱ¼äºó£¬KOHÈÜÒºµÄpH½«Ôö´ó
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