²Ù×÷¼òÊö | |
ÏÖÏó»òÊý¾Ý | |
½áÂÛ | H2SO3µÚÒ»²½²»ÍêÈ«µçÀë |
·ÖÎö £¨1£©Na2S2ÊÇÀë×Ó»¯ºÏÎÁòÔ×Ó¼äÐγÉÒ»¶Ô¹²Óõç×Ó¶Ô£»
£¨2£©¢Ù¸ù¾Ý»¯ºÏ¼Û´úÊýºÍΪ0·ÖÎö£»
¢Ú¸ù¾ÝÌâÒ⣬¼×ËáºÍNaOHÈÜÒº»ìºÏ£¬ÔÙͨÈëSO2ÆøÌ壬¼×Ëá±»Ñõ»¯ÎªCO2£¬SO2ÆøÌå±»»¹Ô³ÉNa2S2O4£¬¾Ý´ËÊéд·½³Ìʽ£»
¢ÛNa2S2O4±©Â¶ÓÚ¿ÕÆøÖÐÒ×±»ÑõÆøÑõ»¯£¬ÔòÑõ»¯¼ÁΪÑõÆø£¬»¹Ô¼ÁΪNa2S2O4£¬¸ù¾ÝµÃʧµç×ÓÊغã¼ÆËã·´Ó¦ºó²úÎ
£¨3£©¢Ù½¹ÑÇÁòËáÄÆ£¨Na2S2O5£©½«Cr2O72-ת»¯Îª¶¾ÐԽϵ͵ÄCr3+£¬Í¬Ê±½¹ÑÇÁòËáÄƱ»Ñõ»¯ÎªÁòËáÄÆ£¬¸ù¾Ýµç×ÓÊغãºÍÔ×ÓÊغãÊéд·½³Ìʽ£»
¢Ú¸ù¾Ý·ÖÀë³öÎÛÄàºó²âµÃ·ÏË®ÖÐCr3+Ũ¶ÈΪ0.52mg/L¼ÆËãCr3+µÄÎïÖʵÄÁ¿Å¨¶È£»
£¨4£©¢ÙÒÀ¾ÝÁò»¯ÇâµÄµçÀëƽºâ³£Êý¼ÆË㣻
¢ÚÖ¤Ã÷ÑÇÁòËáµÚÒ»²½µçÀë´æÔÚ»¯Ñ§Æ½ºâ£¬¿ÉÒԲⶨ0.1mol/LÑÇÁòËáÈÜÒºPH·ÖÎöÅжϣ®
½â´ð ½â£º£¨1£©Na2S2ÊÇÀë×Ó»¯ºÏÎÁòÔ×Ó¼äÐγÉÒ»¶Ô¹²Óõç×Ó¶Ô£¬Ðγɵĵç×ÓʽÀàËƹýÑõ»¯ÄƵĵç×Óʽ£¬µç×ÓʽΪ£º£¬
¹Ê´ð°¸Îª£º£»
£¨2£©¢ÙÒòΪNaΪ+1¼Û£¬OΪ-2¼Û£¬¸ù¾Ý»¯ºÏ¼Û´úÊýºÍΪ0£¬ÔòSΪ+3¼Û£¬
¹Ê´ð°¸Îª£º+3£»
¢Ú¸ù¾ÝÌâÒ⣬¼×ËáºÍNaOHÈÜÒº»ìºÏ£¬ÔÙͨÈëSO2ÆøÌ壬¼×Ëá±»Ñõ»¯ÎªCO2£¬SO2ÆøÌå±»»¹Ô³ÉNa2S2O4£¬Ôò·½³ÌʽΪHCOOH+2SO2+2NaOH=Na2S2O4+CO2+H2O£¬
¹Ê´ð°¸Îª£ºHCOOH+2SO2+2NaOH=Na2S2O4+CO2+H2O£»
¢ÛNa2S2O4±©Â¶ÓÚ¿ÕÆøÖÐÒ×±»ÑõÆøÑõ»¯£¬ÔòÑõ»¯¼ÁΪÑõÆø£¬»¹Ô¼ÁΪNa2S2O4£¬ÓÖÑõ»¯¼ÁºÍ»¹Ô¼ÁµÄÎïÖʵÄÁ¿Ö®±ÈΪl£º2£¬ÉèNa2S2O4±»Ñõ»¯ºóÁòµÄ»¯ºÏ¼ÛΪ+x¼Û£¬Ôò¸ù¾ÝµÃʧµç×ÓÊغ㣬1¡Á4=2¡Á2¡Á£¨x-3£©£¬½âµÃx=4£¬ÓÖÓÐË®ÕôÆø²ÎÓë·´Ó¦£¬ËùÒÔ²úÎïΪNaHSO3£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º2Na2S2O4+O2+2H2O=4NaHSO3£¬
¹Ê´ð°¸Îª£º2Na2S2O4+O2+2H2O=4NaHSO3£»
£¨3£©¢Ù½¹ÑÇÁòËáÄÆ£¨Na2S2O5£©½«Cr2O72-ת»¯Îª¶¾ÐԽϵ͵ÄCr3+£¬Í¬Ê±½¹ÑÇÁòËáÄƱ»Ñõ»¯ÎªÁòËáÄÆ£¬Æä·´Ó¦µÄÀë×Ó·½³ÌʽΪ£º3S2O52-+2Cr2O72-+10H+=6SO42-+4Cr3++5H2O£»
¹Ê´ð°¸Îª£º3S2O52-+2Cr2O72-+10H+=6SO42-+4Cr3++5H2O£»
¢Ú·ÖÀë³öÎÛÄàºó²âµÃ·ÏË®ÖÐCr3+Ũ¶ÈΪ0.52mg/L£¬c£¨Cr3+£©=$\frac{0.52¡Á1{0}^{-3}g/L}{52g/mol}$=1¡Á10-5mol/L£»
¹Ê´ð°¸Îª£º1¡Á10-5£»
£¨4£©¢ÙH2S?H++HS-£¬K1=$\frac{c£¨{H}^{+}£©c£¨H{S}^{-}£©}{c£¨{H}_{2}S£©}$
HS-?H++S2-£¬K2=$\frac{c£¨{H}^{+}£©c£¨{S}^{2-}£©}{c£¨H{S}^{-}£©}$£¬
K1¡ÁK2=$\frac{{c}^{2}£¨{H}^{+}£©c£¨{S}^{2-}£©}{c£¨{H}_{2}S£©}$£¬ºöÂÔH2SºÍH2OµçÀë²úÉúµÄH+£¬c£¨H+£©=0.1mol/L£¬c£¨H2S£©=0.1mol/L£¬
c£¨S2-£©=1.0¡Á10-22£¬
¹Ê´ð°¸Îª£º1.0¡Á10-22£»
¢ÚʵÑéÖ¤Ã÷ÑÇÁòËáµÚÒ»²½²»ÍêÈ«µçÀ룬¿ÉÒÔÑ¡ÔñÓÃpH¼Æ²â¶¨0.1mol/LÑÇÁòËáÈÜÒºµÄpH£¬ÈôPH£¾1£¬ËµÃ÷ÑÇÁòËáµÚÒ»²½²»ÍêÈ«µçÀ룬
´ð£ºÓÃpH¼Æ²â¶¨0.1mol/LÑÇÁòËáÈÜÒºµÄpH£¬PH£¾1£®
µãÆÀ ±¾Ì⿼²éÁËÎïÖÊÐÔÖÊ¡¢Ñõ»¯»¹Ô·´Ó¦¡¢Èõµç½âÖʵçÀëƽºâ¡¢Æ½ºâ³£Êý¼ÆËãµÈ£¬ÕÆÎÕ»ù´¡ÊǽâÌâ¹Ø¼ü£¬ÌâÄ¿ÄѶÈÖеȣ®
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£ºÑ¡ÔñÌâ
A£® | H2ºÍD2»¥ÎªÍ¬Î»ËØ | B£® | ºÍ»¥ÎªÍ¬·ÖÒì¹¹Ìå | ||
C£® | Õý¶¡ÍéºÍÒ춡ÍéÊÇͬϵÎï | D£® | ºÍ ÊÇͬһÖÖÎïÖÊ |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ
A£® | ËáÓê¡¢ÎÂÊÒЧӦ¡¢¹â»¯Ñ§ÑÌÎí | B£® | ËáÓê¡¢³ôÑõ²ã¿Õ¶´¡¢¹â»¯Ñ§ÑÌÎí | ||
C£® | Ë®»ª¡¢³ôÑõ²ã¿Õ¶´¡¢¹â»¯Ñ§ÑÌÎí | D£® | ËáÓê¡¢³ôÑõ²ã¿Õ¶´¡¢°×É«ÎÛȾ |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ
A£® | ±´úÌþµÄÏûÈ¥·´Ó¦±ÈË®½â·´Ó¦¸üÈÝÒ×½øÐÐ | |
B£® | ÔÚäåÒÒÍéÖмÓÈë¹ýÁ¿NaOHÈÜÒº£¬³ä·ÖÕñµ´£¬È»ºó¼ÓÈ뼸µÎAgNO3ÈÜÒº£¬³öÏÖ»ÆÉ«³Áµí | |
C£® | ÏûÈ¥·´Ó¦ÊÇÒýÈë̼̼˫¼üµÄΨһ;¾¶ | |
D£® | ÔÚäåÒÒÍéÓëNaCN¡¢CH3C¡ÔCNaµÄ·´Ó¦ÖУ¬¶¼ÊÇÆäÖеÄÓлú»ùÍÅÈ¡´úäåÔ×ÓµÄλÖà |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ
ÆðʼÎïÖʵÄÁ¿ | ¼× | ÒÒ | ±û |
n£¨H2£©/mol | 1 | 2 | 2 |
n£¨I2£©/mol | 1 | 1 | 2 |
A£® | ƽºâʱ£¬±ûÖлìºÏÆøÌåµÄƽ¾ùĦ¶ûÖÊÁ¿ÊǼ×ÖеÄ2±¶ | |
B£® | ƽºâʱ£¬ÈÝÆ÷ÖÐHIµÄÎïÖʵÄÁ¿n£¨±û£©£¼2n£¨ÒÒ£© | |
C£® | ƽºâʱ£¬ÒÒÖÐI2µÄת»¯ÂÊ´óÓÚ80% | |
D£® | ·´Ó¦¿ªÊ¼Ê±£¬¼×Öеķ´Ó¦ËÙÂÊ×îÂý |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ
A£® | CO2µÄµç×Óʽ£º | B£® | NH3µÄ½á¹¹Ê½Îª | ||
C£® | CH4µÄ±ÈÀýÄ£ÐÍ£º | D£® | Cl¡¥Àë×ӵĽṹʾÒâͼ£º |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ
A£® | ¢Ù¢Ú¢Ü¢Ý¢ß¢á | B£® | ¢Ù¢Ü¢Ý¢Þ¢â | C£® | ¢Ù¢Û¢Ü¢Ý¢ß¢à¢á¢â | D£® | ¢Ù¢Ü¢Ý¢á¢â |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ
A£® | ½«0.1 mol CO2ͨÈë×ãÁ¿NaOHÈÜÒº£¬ÔòÈÜÒºÖÐÒõÀë×ÓÊýĿΪ0.1 NA | |
B£® | ±ê×¼×´¿öÏ£¬½«11.2 LµÄCl2ͨÈë×ãÁ¿µÄNaOHÈÜÒºÖУ¬×ªÒƵĵç×ÓÊýΪNA | |
C£® | 0.1 mol N2Óë×ãÁ¿H2·´Ó¦£¬×ªÒƵĵç×ÓÊýΪ0.6 NA | |
D£® | 4.6gÓÉNO2ºÍN2O4×é³ÉµÄ»ìºÏÎïÖк¬ÓÐÑõÔ×ÓµÄÊýĿΪ0.2 NA |
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