ÌâÄ¿ÄÚÈÝ

13£®ÒÑÖª£ºS£¨g£©+O2£¨g£©¨TSO2£¨g£©£»¡÷H1
S£¨s£©+O2£¨g£©¨TSO2£¨g£©£»¡÷H2
2H2S£¨g£©+O2£¨g£©¨T2S£¨s£©+2H2O£¨l£©£»¡÷H3
2H2S£¨g£©+3O2£¨g£©¨T2SO2£¨g£©+2H2O£¨l£©£»¡÷H4
SO2£¨g£©+2H2S£¨g£©¨T3S£¨s£©+2H2O£¨l£©£»¡÷H5
ÏÂÁйØÓÚÉÏÊö·´Ó¦ìʱäµÄÅжϲ»ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®¡÷H1£¼¡÷H2B£®¡÷H3£¼¡÷H4C£®¡÷H5=¡÷H3-¡÷H2D£®2¡÷H5=3¡÷H3-¡÷H4

·ÖÎö A¡¢Áò´Ó¹ÌÌå±äΪÆøÌåµÄ¹ý³ÌÊÇÎüÈȵĹý³Ì£»
B¡¢Áò»¯ÇâÍêÈ«Éú³É¶þÑõ»¯Áò·Å³öµÄÈÈÁ¿Òª±È²»ÍêȫȼÉÕÉú³ÉÁòµ¥ÖʷųöµÄÈÈÁ¿¶à£»
C¡¢·½³ÌʽSO2£¨g£©+2H2S£¨g£©¨T3S£¨s£©+2H2O£¨l£©¿ÉÒÔ¸ù¾Ý2H2S£¨g£©+O2£¨g£©¨T2S£¨s£©+2H2O£¨l£©¼õÈ¥S£¨s£©+O2£¨g£©¨TSO2£¨g£©µÃµ½£¬¾Ý¸Ç˹¶¨ÂÉÀ´»Ø´ð£»
D¡¢·½³ÌʽSO2£¨g£©+2H2S£¨g£©¨T3S£¨s£©+2H2O£¨l£©¿ÉÒÔ¸ù¾Ý2H2S£¨g£©+O2£¨g£©¨T2S£¨s£©+2H2O£¨l£©¡¢2H2S£¨g£©+3O2£¨g£©¨T2SO2£¨g£©+2H2O£¨l£©µÃµ½£¬¾Ý¸Ç˹¶¨ÂÉÀ´»Ø´ð£®

½â´ð ½â£ºA¡¢Áò´Ó¹ÌÌå±äΪÆøÌåµÄ¹ý³ÌÊÇÎüÈȵĹý³Ì£¬Áòµ¥ÖʵÄȼÉÕ·ÅÈÈ£¬ìʱä¾ùСÓÚÁ㣬ËùÒÔ¡÷H1£¼¡÷H2£¬¹ÊAÕýÈ·£»
B¡¢Áò»¯ÇâÍêÈ«Éú³É¶þÑõ»¯Áò·Å³öµÄÈÈÁ¿Òª±È²»ÍêȫȼÉÕÉú³ÉÁòµ¥ÖʷųöµÄÈÈÁ¿¶à£¬Áò»¯ÇâµÄȼÉÕ¶¼ÊÇ·ÅÈȵģ¬ìʱäÊǸºÖµ£¬ËùÒÔ¡÷H3£¾¡÷H4£¬¹ÊB´íÎó£»
C¡¢·½³ÌʽSO2£¨g£©+2H2S£¨g£©¨T3S£¨s£©+2H2O£¨l£©¿ÉÒÔ¸ù¾Ý2H2S£¨g£©+O2£¨g£©¨T2S£¨s£©+2H2O£¨l£©¼õÈ¥S£¨s£©+O2£¨g£©¨TSO2£¨g£©µÃµ½£¬¾Ý¸Ç˹¶¨ÂÉ¡÷H5=¡÷H3-¡÷H2£¬¹ÊCÕýÈ·£»
D¡¢·½³Ìʽ2SO2£¨g£©+4H2S£¨g£©¨T6S£¨s£©+4H2O£¨l£©¿ÉÒÔ¸ù¾Ý·´Ó¦3H2S£¨g£©+3O2£¨g£©¨T6S£¨s£©+6H2O£¨l£©¼õÈ¥2H2S£¨g£©+3O2£¨g£©¨T2SO2£¨g£©+2H2O£¨l£©µÃµ½£¬¾Ý¸Ç˹¶¨ÂÉ£¬2¡÷H5=3¡÷H3-¡÷H4£¬¹ÊDÕýÈ·£®
¹ÊÑ¡B£®

µãÆÀ ±¾Ì⿼²éÁËìʱ䡢ìرäµÄÅжϣ¬Ã÷È··´Ó¦ÈÈЧӦÓëìʱäÖµ´óСµÄ¹ØϵÊǽâÌâ¹Ø¼ü£¬×¢Òâ¸Ç˹¶¨ÂɵÄÓ¦Óã¬ÌâÄ¿ÄѶÈÖеȣ®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
1£®2014ŵ±´¶ûÎïÀíѧ½±ÊÚÓ迪·¢À¶É«·¢¹â¶þ¼«¹Ü£¨LED£©µÄÈýλ¿Æѧ¼Ò£®³¤ÆÚÒÔÀ´£¬ÈËÃÇ·¢Ã÷Á˺ìÉ«¼°ÂÌÉ«LED£¬µ«ÈýÔ­É«Ö®Ò»µÄÀ¶É«LEDÈ´ÒòÔÚ²ÄÁÏ»·½ÚÓö×è¶ø±»¶ÏÑÔ¡°ÄÑÒÔÔÚ20ÊÀ¼ÍʵÏÖ¡±£®Í¨¹ý¿Æѧ¼ÒÂþ³¤µÄʵÑ飬·¢ÏÖµª»¯ïؾ§Ìå¿ÉÒÔ²úÉúÀ¶É«LED£®×ÊÁÏÏÔʾ£º
¢ÙÔÚÊÒÎÂÏ£¬µª»¯ïز»ÈÜÓÚË®¡¢ËáºÍ¼î£®µ«ÔÚ¼ÓÈÈʱµª»¯ïØ¿ÉÈÜÓÚNaOH¡¢H2SO4£®
¢Úµª»¯ïؾ§ÌåµÄÖƱ¸Í¨³£ÓÐÁ½ÖÖ·½·¨£º
a¡¢Í¨³£ÒÔ¼×»ùïØGa£¨CH3£©3×÷ΪïØÔ´£¬NH3×÷ΪµªÔ´£¬ÔÚÒ»¶¨Ìõ¼þÏ·´Ó¦ÖÆÈ¡µª»¯ïØ
b¡¢Í¨¹ý¼×»ùïØGa£¨CH3£©3·Ö½â²úÉúµ¥ÖÊïØ£¬ÀûÓ÷ֽâ³öµÄGaÓëNH3µÄ»¯Ñ§·´Ó¦ÊµÏֵģ¬ÆäµÄ·´Ó¦·½³ÌʽΪ£º
2Ga£¨s£©+2NH3£¨g£©?2GaN£¨s£©+3H2£¨g£©
¢ÛÇâÑõ»¯ïØGa£¨OH£©3ÓëÇâÑõ»¯ÂÁÀàËÆ£¬¾ùÓÐÁ½ÐÔ£®
£¨1£©ÏÂÁÐÓйØïؼ°Æ仯ºÏÎï˵·¨ÕýÈ·µÄÊÇB£®
A¡¢µÚ¢óA×åÔªËØî÷µÄÐÔÖÊÓëïØÏàËÆ£¬µª»¯î÷µÄ»¯Ñ§Ê½ÎªIn2N3
B¡¢¼×»ùïØÓë°±Æø·´Ó¦Ê±³ýÉú³Éµª»¯ïØÍ⣬»¹Éú³ÉCH4
C¡¢ÔªËØÖÜÆÚ±íÖÐïصĽðÊôÐÔ±ÈÂÁÈõ
D¡¢ïØλÓÚÔªËØÖÜÆÚ±íµÚËÄÖÜÆÚµÚ¢òA×å
£¨2£©Èý¼×»ùïØÔÚ³£Î³£Ñ¹ÏÂΪÎÞɫ͸Ã÷Óж¾ÒºÌ壮ÔÚ¿ÕÆøÖÐÒ×Ñõ»¯£¬ÔÚÊÒÎÂ×Ôȼ£¬È¼ÉÕʱ²úÉú½ðÊôÑõ»¯Îï°×ÑÌ£¬ÆäËûÔªËؾùת»¯ÎªÎȶ¨µÄÑõ»¯ÎÇëд³öÈý¼×»ùïØÔÚ¿ÕÆøÖÐ×ÔȼµÄ»¯Ñ§·½³Ìʽ£º2Ga£¨CH3£©3+12O2=Ga2O3+6CO2+9H2O£®
£¨3£©¹¤ÒµÉÏÀûÓÃGaÓëNH3ºÏ³Éµª»¯ïصķ´Ó¦·½³ÌʽΪ£º
2Ga£¨s£©+2NH3£¨g£©?2GaN£¨s£©+3H2£¨g£©¡÷H£¼0
¢ÙÇëд³öÉÏÊö·´Ó¦µÄ»¯Ñ§Æ½ºâ³£Êý±í´ïʽ£º$\frac{{c}^{3}£¨{H}_{2}£©}{{c}^{2}£¨N{H}_{3}£©}$£®
¢ÚÔÚºãκãÈݵÄÃܱÕÌåϵÄÚ½øÐÐÉÏÊö¿ÉÄæ·´Ó¦£¬ÏÂÁÐÓйرí´ïÕýÈ·µÄÊÇA

A¡¢¢ÙͼÏóÖÐÈç¹û×Ý×ø±êΪÕý·´Ó¦ËÙÂÊ£¬Ôòtʱ¿Ì¸Ä±äµÄÌõ¼þ¿ÉÒÔΪÉýλò¼Óѹ
B¡¢¢ÚͼÏóÖÐ×Ý×ø±ê¿ÉÒÔΪïصÄת»¯ÂÊ
C¡¢¢ÛͼÏóÖÐ×Ý×ø±ê¿ÉÒÔΪ»¯Ñ§·´Ó¦ËÙÂÊ
D¡¢¢ÜͼÏóÖÐ×Ý×ø±ê¿ÉÒÔΪÌåϵÄÚ»ìºÏÆøÌåƽ¾ùÏà¶Ô·Ö×ÓÖÊÁ¿
£¨4£©¹¤ÒµÉÏÌá´¿ïصķ½·¨ºÜ¶à£¬ÆäÖÐÒÔµç½â¾«Á¶·¨Îª¶à£®¾ßÌåÔ­ÀíÈçÏ£ºÒÔ´ýÌá´¿µÄ´ÖïØ£¨ÄÚº¬Zn¡¢Fe¡¢CuÔÓÖÊ£©ÎªÑô¼«£¬ÒԸߴ¿ïØΪÒõ¼«£¬ÒÔNaOHË®ÈÜҺΪµç½âÖÊÈÜÒº£®ÔÚµçÁ÷×÷ÓÃÏÂʹ´ÖïØÔÚÑô¼«Èܽâ½øÈëµç½âÖÊÈÜÒº£¬Í¨¹ýijÖÖÌØÊâµÄÀë×ÓǨÒƼ¼Êõµ½´ïÒõ¼«²¢ÔÚÒõ¼«·ÅµçÎö³ö¸ß´¿ïØ£®µç½âʱÑô¼«·´Ó¦·½³ÌʽΪ£ºGa-3e-+4OH-=GaO2-+2H2O
¢ÙÇëд³öÔÚÑô¼«Éú³ÉµÄGaO2-ͨ¹ýÀë×ÓǨÒƵ½Òõ¼«Ê±£¬Òõ¼«µç¼«·´Ó¦·½³Ìʽ
GaO2-+3e-+2H2O=Ga+4OH-£®
¢ÚÒÑÖªÀë×ÓÑõ»¯ÐÔ˳ÐòΪ£ºZn2+£¼Ga3+£¼Fe2+£¼Cu2+£¬Çëд³öµç½â¾«Á¶ïØʱÑô¼«ÄàµÄ³É·ÖFe¡¢Cu£®£¨Ìѧʽ£©
17£®Ë®ºÏ루N2H4•H2O£©ÊÇÒ»ÖÖÎÞÉ«Ò×ÈÜÓÚË®µÄÓÍ×´ÒºÌ壬¾ßÓмîÐԺͼ«Ç¿µÄ»¹Ô­ÐÔ£¬ÔÚ¹¤ÒµÉú²úÖÐÓ¦Ó÷dz£¹ã·º£®
£¨1£©Ä¿Ç°ÕýÔÚÑз¢µÄ¸ßÄÜÁ¿ÃܶÈȼÁϵç³Ø³µÊÇÒÔË®ºÏëÂȼÁϵç³Ø×÷Ϊ¶¯Á¦À´Ô´£¬µç³Ø½á¹¹Èçͼ1Ëùʾ£®
¢ÙÆðʼʱÕý¼«ÇøÓ븺¼«ÇøNaOHÈÜҺŨ¶ÈÏàͬ£¬¹¤×÷Ò»¶Îʱ¼äºó£¬NaOHŨ¶È½Ï´óµÄÊÇÕý£¨Ìî¡°Õý¡±»ò¡°¸º¡±£©¼«Çø£®
¢Ú¸Ãµç³Ø¸º¼«µÄµç¼«·´Ó¦Ê½ÎªN2H4+4OH--4e-=4H2O+N2¡ü£®
£¨2£©ÒÑ֪ˮºÏëÂÊǶþÔªÈõ¼î£¨25¡æ£¬K1=5¡Á10-7£¬K2=5¡Á10-15£©£¬0.1mol•L-1Ë®ºÏëÂÈÜÒºÖÐËÄÖÖÀë×Ó£º¢ÙH+¡¢¢ÚOH-¡¢¢ÛN2H5+¡¢¢ÜN2H52+µÄŨ¶È´Ó´óµ½Ð¡µÄ˳ÐòΪ¢Ú¢Û¢Ü¢Ù£¨ÌîÐòºÅ£©£®
£¨3£©ÔÚÈõËáÐÔÌõ¼þÏÂË®ºÏë¿ɴ¦Àíµç¶Æ·ÏË®£¬½«Cr2O72-»¹Ô­ÎªCr£¨OH£©3³Áµí¶ø³ýÈ¥£¬Ë®ºÏë±»Ñõ»¯ÎªN2£¬¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽΪ2Cr2O72-+4H++3N2H4•H2O=4Cr£¨OH£©3¡ý+3N2+5H2O£®
£¨4£©ëÂÊÇÒ»ÖÖÓÅÁ¼µÄÖüÇâ²ÄÁÏ£¬ÆäÔÚ²»Í¬Ìõ¼þÏ·ֽⷽʽ²»Í¬£®
¢ÙÔÚ¸ßÎÂÏ£¬N2H4¿ÉÍêÈ«·Ö½âΪNH3¡¢N2¼°H2£¬ÊµÑé²âµÃ·Ö½â²úÎïÖÐN2ÓëH2ÎïÖʵÄÁ¿±ä»¯Èçͼ2Ëùʾ£¬¸Ã·Ö½â·´Ó¦·½³ÌʽΪ7N2H4$\frac{\underline{\;¸ßÎÂ\;}}{\;}$8NH3+3N2+2H2£®
¢ÚÔÚ303K£¬NiPt´ß»¯Ï£¬Ôò·¢ÉúN2H4£¨l£©?N2£¨g£©+2H2£¨g£©£®ÔÚ1LÃܱÕÈÝÆ÷ÖмÓÈë0.1mol N2H4£¬²âµÃÈÝÆ÷ÖÐ $\frac{n£¨H{\;}_{2}£©+n£¨N{\;}_{2}£©}{n£¨{N}_{2}{H}_{4}£©}$Óëʱ¼ä¹ØϵÈçͼ3Ëùʾ£®Ôò0¡«4minµªÆøµÄƽ¾ùËÙ
ÂÊv£¨N2£©=0.0125mol•L-1•min-1£®
£¨5£©Ì¼õ£ë£¨·Ö×Óʽ£ºCH6N4O£©ÊÇÓÉDEC£¨õ¥£©ÓëN2H4·¢ÉúÈ¡´ú·´Ó¦µÃµ½£¬ÒÑÖªDECµÄ·Ö×ÓʽΪC5H10O3£¬DECµÄºË´Å¹²ÕñÇâÆ×Èçͼ4Ëùʾ£¬ÔòDECµÄ½á¹¹¼òʽΪ£®

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø