ÌâÄ¿ÄÚÈÝ
15£®Ä³ÉÕ¼îÑùÆ·Öк¬ÓÐÉÙÁ¿²»ÓëËá×÷ÓõĿÉÈÜÐÔÔÓÖÊ£¬ÎªÁ˲ⶨÆä´¿¶È£¬½øÐÐÒÔÏµζ¨²Ù×÷£ºA£®ÔÚ250mLµÄÈÝÁ¿Æ¿Öж¨ÈݳÉ250mLÉÕ¼îÈÜÒº£»
B£®ÓÃÒÆÒº¹ÜÒÆÈ¡25mLÉÕ¼îÈÜÒºÓÚ׶ÐÎÆ¿Öв¢µÎ¼Ó¼¸µÎ¼×»ù³Èָʾ¼Á£»
C£®ÔÚÌìƽÉÏ׼ȷ³ÆÈ¡ÉÕ¼îÑùÆ·Wg£¬ÔÚÉÕ±ÖмÓÕôÁóË®Èܽ⣻
D£®½«ÎïÖʵÄÁ¿Å¨¶ÈΪMmol/LµÄ±ê×¼H2SO4ÈÜҺװÈëËáʽµÎ¶¨¹Ü£¬µ÷ÕûÒºÃ棬¼ÇÏ¿ªÊ¼Ê±µÄ¿Ì¶ÈÊýΪV1mL£»
E£®ÔÚ׶ÐÎƿϵæÒ»ÕÅ°×Ö½£¬µÎ¶¨µ½Öյ㣬¼Ç¼ÖÕµãʱÏûºÄËáµÄÌå»ýΪV2mL£®
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÕýÈ·µÄ²Ù×÷²½ÖèµÄ˳ÐòÊÇ£¨ÓÃ×ÖĸÌîд£©C¡úA¡úB¡úD¡úE£»
£¨2£©µÎ¶¨¹Ü¶ÁÊýӦעÒâµÎ¶¨¹Ü´¹Ö±£¬ÒºÃæ²»Ôٱ仯ʱ²Å¿É¶ÁÊý£¬ÊÓÏßÓëÈÜÒº°¼ÒºÃæ×îµÍµãÏàƽ£¬¶ÁÊý¾«È·µ½0.01ml£®
£¨3£©²Ù×÷EÖÐÔÚ׶ÐÎƿϵæÒ»ÕÅ°×Ö½µÄ×÷ÓÃÊDZãÓÚ׼ȷÅжÏÖÕµãʱÑÕÉ«µÄ±ä»¯Çé¿ö£®
£¨4£©²Ù×÷DÖÐÒºÃæÓ¦µ÷Õûµ½Áã¿Ì¶È»òÁãÉÔϵÄijһ¿Ì¶È£¬¼â×첿·ÖÓ¦³äÂú±ê×¼H2SO4ÈÜÒº£®
£¨5£©µÎ¶¨ÖÕµãʱ׶ÐÎÆ¿ÄÚÈÜÒºµÄpHԼΪ4£¬ÖÕµãʱÈÜÒºÑÕÉ«µÄ±ä»¯ÊÇÈÜÒºÓÉ»ÆÉ«±äΪ³ÈÉ«£¬ÇÒ°ë·ÖÖÓÄÚ²»»Ö¸´ÎªÔÀ´µÄÑÕÉ«£®
£¨6£©ÈôËáʽµÎ¶¨¹ÜûÓÐÓñê×¼H2SOÈÜÒºÈóÏ´£¬»á¶Ô²â¶¨½á¹ûÓкÎÓ°Ï죿ƫ¸ß£¨Ìî¡°Æ«¸ß¡±¡¢¡°Æ«µÍ¡±»ò¡°ÎÞÓ°Ï족£©£¨ÆäËû²Ù×÷¾ùÕýÈ·£©£®
£¨7£©¸ÃÉÕ¼îÑùÆ·µÄ´¿¶È¼ÆËãʽÊÇ$\frac{80m£¨{V}_{2}-{V}_{1}£©¡Á1{0}^{-2}g}{Wg}$¡Á100%£®
·ÖÎö £¨1£©ÊµÑéʱӦÏȳÆÁ¿Ò»¶¨ÖÊÁ¿µÄ¹ÌÌ壬ÈܽâºóÅäÖƳÉÈÜÒº£¬Á¿È¡´ý²âÒºÓë׶ÐÎÆ¿ÖУ¬È»ºóÓñê×¼Òº½øÐе樣»
£¨2£©µÎ¶¨¹Ü´¹Ö±£¬ÒºÃæ²»Ôٱ仯ʱ²Å¿É¶ÁÊý£¬ÊÓÏßÓëÈÜÒº°¼ÒºÃæ×îµÍµãÏàƽ£¬×¢ÒâµÎ¶¨¹ÜµÄ¾«È·Öµ£»
£¨3£©Òò׶ÐÎƿϵæÒ»ÕÅ°×ֽʹµÎ¶¨ÖÕµãÑÕÉ«±ä»¯¸üÃ÷ÏÔ£¬±ãÓڷֱ棻
£¨4£©µÎ¶¨¹Ü0¿Ì¶ÈÔÚÉÏ£¬µÎ¶¨Ç°Ó¦µ÷½Úµ½Áã¿Ì¶È»òÁãÉÔϵÄijһ¿Ì¶È£¬Îª¼õСÎó²î£¬¼â×첿·ÖÓ¦³äÂúÒºÌ壬ÎÞÆøÅÝ£»
£¨5£©Ö¸Ê¾¼ÁΪ¼×»ù³È£¬±äÉ«·¶Î§Îª3.1-4.4£»
£¨6£©Òò±ê׼ҺŨ¶È±äС£¬ÓÃÁ¿±ä´ó£»
£¨7£©¸ù¾Ý·´Ó¦ÏûºÄµÄÁòËᣬÇó³öÇâÑõ»¯ÄÆ£¬½øÒ»²½Çó³öÑùÆ·µÄ´¿¶È£®
½â´ð ½â£º£¨1£©ÊµÑéʱӦÏȳÆÁ¿Ò»¶¨ÖÊÁ¿µÄ¹ÌÌ壬ÈܽâºóÅäÖƳÉÈÜÒº£¬Á¿È¡´ý²âÒºÓë׶ÐÎÆ¿ÖУ¬È»ºóÓñê×¼Òº½øÐе樣¬
¹Ê´ð°¸Îª£ºC£»A£»B£»D£»E£»
£¨2£©µÎ¶¨¹Ü´¹Ö±£¬ÒºÃæ²»Ôٱ仯ʱ²Å¿É¶ÁÊý£¬ÊÓÏßÓëÈÜÒº°¼ÒºÃæ×îµÍµãÏàƽ£¬µÎ¶¨¹Ü¶ÁÊý׼ȷµ½0.01ml£¬
¹Ê´ð°¸Îª£ºµÎ¶¨¹Ü´¹Ö±£¬ÒºÃæ²»Ôٱ仯ʱ²Å¿É¶ÁÊý£¬ÊÓÏßÓëÈÜÒº°¼ÒºÃæ×îµÍµãÏàƽ£¬¶ÁÊý¾«È·µ½0.01ml£»
£¨3£©Òò׶ÐÎƿϵæÒ»ÕÅ°×ֽʹµÎ¶¨ÖÕµãÑÕÉ«±ä»¯¸üÃ÷ÏÔ£¬±ãÓڷֱ棬¹Ê´ð°¸Îª£º±ãÓÚ׼ȷÅжÏÖÕµãʱÑÕÉ«µÄ±ä»¯Çé¿ö£»
£¨4£©µÎ¶¨¹Ü0¿Ì¶ÈÔÚÉÏ£¬µÎ¶¨Ç°Ó¦µ÷½Úµ½Áã¿Ì¶È»òÁãÉÔϵÄijһ¿Ì¶È£¬Îª¼õСÎó²î£¬¼â×첿·ÖÓ¦³äÂúÒºÌ壬ÎÞÆøÅÝ£¬
¹Ê´ð°¸Îª£ºÁã¿Ì¶È»òÁãÉÔϵÄijһ¿Ì¶È£»³äÂú±ê×¼H2SO4ÈÜÒº£»
£¨5£©Ö¸Ê¾¼ÁΪ¼×»ù³È£¬±äÉ«·¶Î§Îª3.1-4.4£¬ÖÕµãʱpHԼΪ4£¬ÈÜÒºÓÉ»ÆÉ«±äΪ³ÈÉ«£¬ÇÒ°ë·ÖÖÓÄÚ²»»Ö¸´ÎªÔÀ´µÄÑÕÉ«£»
¹Ê´ð°¸Îª£º4£»ÈÜÒºÓÉ»ÆÉ«±äΪ³ÈÉ«£¬ÇÒ°ë·ÖÖÓÄÚ²»»Ö¸´ÎªÔÀ´µÄÑÕÉ«£»
£¨6£©Òò±ê׼ҺŨ¶È±äС£¬ÓÃÁ¿±ä´ó£¬¹Ê´ð°¸Îª£ºÆ«¸ß£»
£¨7£©µÎµ½ÏûºÄµÄÁòËáΪ£ºn£¨ÁòËᣩ=cV=£¨V2-V1£©¡Á10-3L¡Ám mol/L£¬¸ù¾Ý·´Ó¦·½³Ì¿ÉÖª£¬n£¨NaOH£©=2n£¨ÁòËᣩ=2m£¨V2-V1£©¡Á10-3mol£¬
ËùÒÔÔÀ´ÑùÆ·ÖÐÇâÑõ»¯ÄƵÄÎïÖʵÄÁ¿Îª£º2m£¨V2-V1£©¡Á10-3mol¡Á$\frac{250}{25}$=2m£¨V2-V1£©¡Á10-2mol£¬ÔòÑùÆ·ÖÐÇâÑõ»¯ÄƵÄÖÊÁ¿Îªm£¨NaOH£©=nM=80m£¨V2-V1£©¡Á10-2g£¬Ôò¸ÃÉÕ¼îÑùÆ·µÄ´¿¶ÈΪ£º$\frac{80m£¨{V}_{2}-{V}_{1}£©¡Á1{0}^{-2}g}{Wg}$¡Á100%£¬
¹Ê´ð°¸Îª£º$\frac{80m£¨{V}_{2}-{V}_{1}£©¡Á1{0}^{-2}g}{Wg}$¡Á100%£®
µãÆÀ ±¾Ìâ×ۺϿ¼²éËá¼îÖк͵樣¬²àÖØÓÚ»¯Ñ§ÊµÑé»ù±¾²Ù×÷ÒÔ¼°ÎïÖʵĺ¬Á¿µÄ²â¶¨µÈÎÊÌ⣬ÌâÄ¿ÄѶÈÖеȣ¬½¨ÒéÔÚѧϰÖаÑÎÕÏà¹Ø»ù±¾ÊµÑé·½·¨£¬Ñ§Ï°ÖÐ×¢Òâ»ýÀÛ£®
¢ÙNH4ClÈÜÒº¿É×÷º¸½Ó½ðÊôÖеijýÐâ¼Á
¢ÚÓÃNaHCO3ÓëAl2£¨SO4£©3Á½ÖÖÈÜÒº¿É×÷ÅÝÄÃð»ð¼Á
¢Û²Ýľ»ÒÓëï§Ì¬µª·Ê²»ÄÜ»ìºÏÊ©ÓÃ
¢Ü½«ÂÈ»¯Ìú±¥ºÍÈÜÒºµÎÈë·ÐË®ÖÐÖÆÈ¡½ºÌå
¢Ý¼ÓÈÈÂÈ»¯ÌúÈÜÒºÑÕÉ«±äÉî
¢ÞÌúÔÚ³±ÊªµÄ»·¾³ÏÂÉúÐ⣮
A£® | ¢Ù¢Ú¢Û | B£® | ¢Ú¢Û¢Ü | C£® | ¢Ù¢Ü¢Ý | D£® | ¢Þ |
A£® | ̼ËáÇâï§ÈÜÒºÓë×ãÁ¿µÄNaOHÈÜÒº»ìºÏºó¼ÓÈÈ£ºNH${\;}_{4}^{+}$+OH-$\frac{\underline{\;\;¡÷\;\;}}{\;}$NH3¡ü+H2O | |
B£® | Ca£¨HCO3£©2ÈÜÒºÖеμÓÉÙÁ¿NaOHÈÜÒº Ca2++HCO3-+OH-=CaCO3¡ý+H2O | |
C£® | NaHSO4ÈÜÒººÍBa£¨OH£©2ÈÜÒº³ä·Ö·´Ó¦ºóÈÜÒº³ÊÖÐÐÔ£ºBa2++OH-+H++SO42-=BaSO4¡ý+H2O | |
D£® | ÏòFe£¨OH£©2ÖмÓÈëÏ¡ÏõË᣺3Fe2++4H++NO${\;}_{3}^{-}$¨T3Fe3++NO¡ü+2H2O |
A£® | »¯ºÏÎHT¡¢CaCl2¡¢NaOH¡¢ÑÎËá | |
B£® | ´¿¾»Îˮ²£Á§¡¢µ¨·¯¾§Ìå¡¢ÒºÂÈ¡¢ÁòËá | |
C£® | ÌìÈ»¸ß·Ö×Ó»¯ºÏÎµí·Û¡¢ÏËάËØ¡¢µ°°×ÖÊ¡¢¾ÛÂÈÒÒÏ© | |
D£® | ͬ·ÖÒì¹¹Ì壺CH3CH2CH2CH2CH3¡¢CH3CH2CH£¨CH3£©2¡¢C£¨CH3£©4 |
A£® | +2£¨b-a£© kJ•mol-1 | B£® | +£¨b+c-a£© kJ•mol-1 | ||
C£® | +£¨a+b£© kJ•mol-1 | D£® | +2£¨a-b-c£© kJ•mol-1 |
A£® | Cu2++2OH-¨TCu£¨OH£©2CuCO3+2NaOH¨TCu£¨OH£©2¡ý+Na2CO3 | |
B£® | Ba2++SO${\;}_{4}^{2-}$¨TBaSO4¡ý Ba£¨OH£©2+H2SO4¨TBaSO4¡ý+2H2O | |
C£® | Ag++Cl-¨TAgCl¡ý AgNO3+NaCl¨TAgCl¡ý+NaNO3 | |
D£® | Cu+2Ag+¨TCu2++2Ag¡ý Cu+2AgCl¨T2Ag+CuCl2 |