15£®Ä³Ñ§Ï°Ð¡×éÓÃDISϵͳ²â¶¨Ê³Óð״×Öд×ËáµÄÎïÖʵÄÁ¿Å¨¶È£¬ÒÔÈÜÒºµÄµ¼µçÄÜÁ¦À´Åжϵζ¨Öյ㣮ʵÑé²½ÖèÈçÏ£º
£¨1£©ÓÃËáʽµÎ¶¨¹Ü£¨ÌîÒÇÆ÷Ãû³Æ£©Á¿È¡10.00mLʳÓð״ף¬ÔÚÉÕ±­£¨ÌîÒÇÆ÷Ãû³Æ£©ÖÐÓÃˮϡÊͺóתÒƵ½100mLÈÝÁ¿Æ¿£¨ÌîÒÇÆ÷Ãû³Æ£©Öж¨ÈÝ£¬È»ºó½«Ï¡ÊͺóµÄÈÜÒºµ¹ÈëÊÔ¼ÁÆ¿ÖУ®Á¿È¡20.00mLÉÏÊöÈÜÒºµ¹ÈëÉÕ±­ÖУ¬ÏòÉÕ±­ÖеμÓŨ¶ÈΪ0£¬.1000mol•L-1µÄ°±Ë®£¬¼ÆËã»úÆÁÄ»ÉÏÏÔʾ³öÈÜÒºµ¼µçÄÜÁ¦Ë氱ˮÌå»ý±ä»¯µÄÇúÏߣ¨¼ûͼ£©£®
£¨2£©Óõζ¨¹ÜÊ¢°±Ë®Ç°£¬µÎ¶¨¹ÜÒªÓð±Ë®ÈóÏ´2¡«3±é£¬ÈóÏ´µÄÄ¿µÄÊÇ·ÀÖ¹°±Ë®±»Ï¡ÊÍ£®
£¨3£©°±Ë®Óë´×Ëá·´Ó¦µÄÀë×Ó·½³ÌʽÊÇCH3COOH+NH3•H2O=CH3COO-+NH4++H2O£®
£¨4£©Ê³Óð״×Öд×ËáµÄÎïÖʵÄÁ¿Å¨¶ÈÊÇ1.000mol•L-1£®ÈôÁ¿È¡Ê³Óð״׵ÄÒÇÆ÷ûÓÐÓÃʳÓð״×ÈóÏ´£¬Ôò²â¶¨½á¹ûÆ«µÍ£¨Ñ¡Ìî¡°Æ«¸ß¡±¡¢¡°Æ«µÍ¡±¡¢¡°²»ÊÜÓ°Ï족£©£®

·ÖÎö £¨1£©¸ù¾ÝÁ¿È¡ÒºÌåÌå»ýµÄ¾«È·¶ÈÒÔ¼°ÒºÌåµÄÐÔÖÊÑ¡ÔñÒÇÆ÷£»ÔÚÉÕ±­ÖÐÏ¡ÊÍÈÜÒº£»ÓÃÈÝÁ¿Æ¿ÅäÖÆÒ»¶¨Å¨¶ÈµÄÈÜÒº£»
£¨2£©µÎ¶¨¹ÜÇåÏ´ÍêÖ®ºó¹Ü±ÚÉϲÐÁôË®£¬»áʹ´ý²âÈÜҺŨ¶ÈϽµ¾Ý´Ë·ÖÎö½â´ð£»
£¨3£©´×ËáÓëһˮºÏ°±·´Ó¦Éú³ÉÑκÍË®£»
£¨4£©¸ù¾Ýͼ±í¿ÉÖªµ±»ìºÏÒºµ¼µçÄÜÁ¦×îǿʱ£¬´×ËáÇ¡ºÃÓ백ˮ·´Ó¦ÍêÈ«£¬·Ö±ðÇó³ö·´Ó¦ÏûºÄµÄ´×ËáµÄÎïÖʵÄÁ¿ºÍ°±Ë®µÄÎïÖʵÄÁ¿£¬ÒÀ¾ÝËá¼îÖкͷ´Ó¦n£¨CH3COOH£©=n£¨NH3•H2O£©¼ÆËãÇó½â£»Á¿È¡Ê³Óð״׵ÄÒÇÆ÷ûÓÐÓÃʳÓð״×ÈóÏ´£¬°×´×±»Ï¡ÊÍ£®

½â´ð ½â£º£¨1£©µÎ¶¨¹Ü¾«È·¶ÈΪ0.01ml£¬´×Ëá¾ßÓÐËáÐÔÄܸ¯Ê´Ï𽺹ܣ¬ËùÒÔӦѡÓÃËáʽµÎ¶¨¹ÜÁ¿È¡°×´×£»È»ºó°Ñ°×´×£¬×ªÒƵ½ÉÕ±­ÖмÓˮϡÊÍÈÜÒº£»ÔÙ°ÑÈÜҺתÈë100mLÈÝÁ¿Æ¿ÖÐÖж¨ÈÝ£»
¹Ê´ð°¸Îª£ºËáʽµÎ¶¨¹Ü£»ÉÕ±­£»ÈÝÁ¿Æ¿£»
£¨2£©Îª·ÀÖ¹µÎ¶¨¹ÜÇåÏ´ÍêÖ®ºó¹Ü±ÚÉϲÐÁôË®½«´ý²âҺϡÊÍ£¬Ó¦ÓÃËùÊ¢ÒºÌåÈóÏ´2-3´Î£¬
¹Ê´ð°¸Îª£º°±Ë®£»·ÀÖ¹°±Ë®±»Ï¡ÊÍ£»
£¨3£©´×ËáÓëһˮºÏ°±·´Ó¦Éú³É´×Ëá狀ÍË®£ºCH3COOH+NH3•H2O=CH3COONH4+H2O£¬
¹Ê´ð°¸Îª£ºCH3COOH+NH3•H2O=CH3COO-+NH4++H2O£»
£¨4£©Éè°×´×µÄŨ¶ÈΪC£¬Ôò·´Ó¦ÏûºÄµÄ´×ËáµÄÎïÖʵÄÁ¿Îªn£¨CH3COOH£©=C¡Á10.00mL¡Á$\frac{20}{100}$£»
·´Ó¦ÏûºÄµÄ°±Ë®µÄÎïÖʵÄÁ¿Îª£ºn£¨NH3•H2O£©=0.1000mol•L-1¡Á20ml£¬
¸ù¾Ýͼ±í¿ÉÖªµ±»ìºÏÒºµ¼µçÄÜÁ¦×îǿʱ£¬´×ËáÇ¡ºÃÓ백ˮ·´Ó¦ÍêÈ«£¬
ËùÒÔn£¨CH3COOH£©=n£¨NH3•H2O£©£¬C¡Á10.00mL¡Á$\frac{20}{100}$=0.1000mol•L-1¡Á20ml£¬C=1.000mol•L-1£»
Á¿È¡Ê³Óð״׵ÄÒÇÆ÷ûÓÐÓÃʳÓð״×ÈóÏ´£¬°×´×±»Ï¡ÊÍ£¬Ó백ˮ·´Ó¦Ê±ÏûºÄµÄ°±Ë®µÄÌå»ý»áƫС£¬ËùÒÔÇó³öµÄ°×´×µÄŨ¶È»áÆ«µÍ£»
¹Ê´ð°¸Îª£º1.000mol•L-1£»Æ«µÍ£®

µãÆÀ ±¾Ì⿼²éÈÜÒºµÄÅäÖÆ¡¢Öк͵ζ¨ÊµÑ飬ÌâÄ¿ÄѶÈÖеȣ¬×¢ÒâµÎ¶¨¹ÜµÄÑ¡ÔñºÍʹÓõÄ×¢ÒâÊÂÏ²àÖØÓÚ¿¼²éѧÉúµÄʵÑé²Ù×÷ÄÜÁ¦£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

6£®µ¥ÖÊÅðÊÇÒ±½ð¡¢½¨²Ä¡¢»¯¹¤¡¢ºË¹¤ÒµµÈÁìÓòµÄÖØÒªÔ­ÁÏ£¬¶ø½ðÊôþÊǵ±½ñÊÀ½ç¸ß¿Æ¼¼ÁìÓòÓÃ;×î¹ã·ºµÄÓÐÉ«²ÄÁÏ£¬¹ã·ºÓ¦ÓÃÓÚº½¿Õº½ÌìµÈ¹ú·À¾üʹ¤Òµ£®ÒÔÌúÅð¿óΪԭÁÏ£¬ÀûÓÃ̼¼î·¨¹¤ÒÕÉú²úÅðºÍþ£¬Æ乤ÒÕÁ÷ͼÈçÏ£º
ÒÑÖª£ºÌúÅð¿óµÄÖ÷Òª³É·ÖΪMg2B2O5•H2OºÍFe3O4£¬º¬ÓÐÉÙÁ¿Al2O3ºÍSiO2µÈ£»ÅðÉ°µÄ»¯Ñ§Ê½ÎªNa2B4O7•10H2O£®
£¨1£©ÌúÅðÉ°¡°ÑÐÄ¥·ÛË顱µÄÄ¿µÄÊÇÔö´ó½Ó´¥Ãæ»ý£¬Ìá¸ß·´Ó¦ËÙÂÊ£»²Ù×÷¢ñµÄÃû³ÆΪ¹ýÂË£»¡°ÅðÄࡱÖгýÔÓÖÊSiO2ºÍAl2O3Í⣬ÆäËû³É·ÖΪMgCO3¡¢Fe3O4£¨Ìѧʽ£©£®
£¨2£©Ð´³öÁ÷³ÌͼÖС°Ì¼½â¡±¹ý³ÌµÄ»¯Ñ§·½³Ìʽ£º2Mg2B2O5•H2O+Na2CO3+3CO2=Na2B4O7+2H2O+4MgCO3£»½ðÊôMgÓëA·´Ó¦ÖƱ¸ÅðµÄ»¯Ñ§·½³ÌʽΪ3Mg+B2O3$\frac{\underline{\;Ò»¶¨Ìõ¼þ\;}}{\;}$2B+3MgO£®
£¨3£©Ì¼¼î·¨¹¤ÒÕÐèÁ½´Îµ÷½ÚpH£º¢Ù³£ÓÃH2SO4ÈÜÒºµ÷½ÚpH=2¡«3ÖÆÈ¡H3BO3£¬ÆäÀë×Ó·½³ÌʽΪB4O72-+2H++5H2O=4H3BO3¡ý£»¢ÚÖÆÈ¡MgCl2•6H2OµÄ¡°¾»»¯³ýÔÓ¡±¹ý³ÌÖУ¬ÐèÏȼÓH2O2ÈÜÒº£¬Æä×÷ÓÃÊǽ«Fe2+Ñõ»¯ÎªFe3+£¬È»ºóÓÃMgOµ÷½ÚpHԼΪ5£¬ÆäÄ¿µÄÊÇʹFe3+Ë®½âת»¯ÎªFe£¨OH£©3£®
£¨4£©MgCl2•6H2OÖÆÈ¡MgCl2¹ý³ÌÖÐΪ·ÀÖ¹MgCl2Ë®½âÐèÔÚ¡°Ò»¶¨Ìõ¼þÏ¡±£¬Ôò¸ÃÌõ¼þÊÇÖ¸HClÆøÌå·ÕΧÏÂ
£¨5£©ÊµÑéʱ²»Ð¡ÐÄ´¥Åöµ½NaOH£¬¿ÉÓôóÁ¿ÇåË®³åÏ´ÔÙͿĨÅðËᣬÓйػ¯Ñ§·½³ÌʽΪNaOH+H3BO3¨TNa[B£¨OH£©4]£¬ÏÂÁйØÓÚNa[B£¨OH£©4]ÈÜÒºÖÐ΢Á£Å¨¶È¹ØϵÕýÈ·µÄÊÇAC£¨Ìî×Öĸ£©£®
A£®c£¨Na+£©+c£¨H+£©=c{[B£¨OH£©4]-}+c£¨OH-£©
B£®c£¨Na+£©£¾c{[B£¨OH£©4]-}£¾c£¨H+£©£¾c£¨OH-£©
C£®c£¨Na+£©=c{[B£¨OH£©4]-}+c£¨H3BO3£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

6£®ÒÑ֪ˮÔÚ25¡æºÍ95¡æʱ£¬ÆäµçÀëƽºâÇúÏßÈçͼËùʾ£º
£¨1£©Ôò25¡æʱˮµÄµçÀëƽºâÇúÏßӦΪA£¨Ìî¡°A¡±»ò¡°B¡±£©£¬Çë˵Ã÷ÀíÓÉ
Ë®µÄµçÀëÊÇÎüÈȹý³Ì£¬Î¶ȵÍʱ£¬Ë®µÄµçÀë³Ì¶ÈС£¬c£¨H+£©¡¢c£¨OH-£©£®
£¨2£©25¡æʱ£¬½«pH=9µÄNaOHÈÜÒºÓëpH=4µÄH2SO4ÈÜÒº»ìºÏ£¬ÈôËùµÃ»ìºÏÈÜÒºµÄpH=7£¬ÔòNaOHÈÜÒºÓëH2SO4ÈÜÒºµÄÌå»ý±ÈΪ10£º1£®
£¨3£©95¡æʱ£¬Èô100Ìå»ýpH1=a µÄijǿËáÈÜÒºÓë1Ìå»ýpH2=bµÄijǿ¼îÈÜÒº»ìºÏºóÈÜÒº³ÊÖÐÐÔ£¬Ôò»ìºÏÇ°£¬¸ÃÇ¿ËáµÄpH1ÓëÇ¿¼îµÄpH2Ö®¼äÓ¦Âú×ãµÄ¹ØϵÊÇpHa+pHb=14£®
£¨4£©95¡æÏ£¬½«pH=11µÄNaOHÈÜÒºV1LÓëpH=1µÄÏ¡ÁòËáV2L»ìºÏ£¨Éè»ìºÏºóÈÜÒºµÄÌå»ýΪԭÁ½ÈÜÒºÌå»ýÖ®ºÍ£©£¬ËùµÃ»ìºÏÈÜÒºµÄpH=2£¬ÔòV1£ºV2=9£º11£®
£¨5£©ÇúÏßB¶ÔӦζÈÏ£¬pH=2µÄijHAÈÜÒººÍpH=10µÄNaOHÈÜÒºµÈÌå»ý»ìºÏºó£¬»ìºÏÈÜÒºµÄpH=5£®Çë·ÖÎöÆäÔ­Òò£ºÇúÏßB¶ÔÓ¦95¡æ£¬´ËʱˮµÄÀë×Ó»ýΪ10-12£¬HAΪÈõËᣬHAÖкÍNaOHºó£¬»ìºÏÈÜÒºÖл¹Ê£Óà½Ï¶àµÄHA·Ö×Ó£¬¿É¼ÌÐøµçÀë³öH+£¬Ê¹ÈÜÒºpH=5£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

3£®ÏÂÁйØÓÚÆøÌåĦ¶ûÌå»ýµÄÐðÊöÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®1 molÈκÎÆøÌåµÄÌå»ý¶¼Îª22.4 L
B£®1 molÈκÎÎïÖÊÔÚ±ê×¼×´¿öÏÂËùÕ¼µÄÌå»ý¶¼Îª22.4 L
C£®±ê×¼×´¿öÏ£¬1 molËÄÂÈ»¯Ì¼ËùÕ¼µÄÌå»ýÊÇ22.4 L
D£®±ê×¼×´¿öÏ£¬22.4 LµÄÈκÎÆøÌåµÄÎïÖʵÄÁ¿¶¼ÊÇ1 mol

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

10£®ÊÒÎÂÏ£¬½«0.10mol•L-1ÑÎËáµÎÈë20.00mL 0.10mol•L-1°±Ë®ÖУ¬ÈÜÒºÖÐpHºÍpOHËæ¼ÓÈëÑÎËáÌå»ý±ä»¯ÇúÏßÈçͼËùʾ£®ÒÑÖª£ºpOH=-lg c£¨OH-£©£¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®MµãËùʾÈÜÒºÖÐc£¨NH4+£©+c£¨NH3•H2O£©=c£¨Cl-£©
B£®NµãËùʾÈÜÒºÖÐc£¨NH4+£©£¾c£¨Cl-£©
C£®QµãÏûºÄÑÎËáµÄÌå»ýµÈÓÚ°±Ë®µÄÌå»ý
D£®MµãºÍNµãËùʾÈÜÒºÖÐË®µÄµçÀë³Ì¶ÈÏàͬ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

20£®µ±¹âÊøͨ¹ýÏÂÁзÖɢϵʱ£¬²»¿ÉÄܲúÉú¶¡´ï¶ûЧӦµÄÊÇ£¨¡¡¡¡£©
A£®Ä«Ë®B£®Fe£¨OH£©3½ºÌåC£®CuSO4ÈÜÒºD£®Ï¡¶¹½¬

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

7£®ÏÂÁи÷×é»ìºÏÎïÖУ¬ÒÔÈÎÒâ±ÈÀý»ìºÏ£¬Ö»Òª×ÜÖÊÁ¿¹Ì¶¨£¬¾­³ä·ÖȼÉÕºó²úÉúCO2µÄÁ¿ÎªÒ»ºã¶¨ÖµµÄÊÇ£¨¡¡¡¡£©
A£®¼×ÍéºÍÒÒÍéB£®ÒÒ´¼ºÍÒÒËáC£®±ûÏ©ºÍ±ûÍéD£®ÒÒȲºÍ±½ÕôÆø

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

4£®ÂÌÉ«»¯Ñ§Êǽ«·´Ó¦ÎïÈ«²¿×ª»¯ÎªÆÚÍûµÄ²úÎʹԭ×ÓµÄÀûÓÃÂÊ´ïµ½100%£¬ÒÔ¼õÉÙÓк¦ÎïÖÊÉú³ÉµÄ»¯Ñ§¹¤ÒÕÀíÄÏÂÁзûºÏ¡°ÂÌÉ«»¯Ñ§¡±ÒªÇóµÄÊÇ£¨¡¡¡¡£©
A£®Êª·¨Ò±Í­£ºFe+CuSO4¨TFeSO4+Cu
B£®¸ß¯Á¶Ìú£ºFe2O3+3CO¨T2Fe+3CO2
C£®ÖÆÑõÆø£º2KMnO4$\frac{\underline{\;\;¡÷\;\;}}{\;}$K2MnO4+MnO2+O2¡ü
D£®ÖÆ»·ÑõÒÒÍ飺2C2H2£¨ ÒÒȲ£©+O2$\frac{\underline{\;´ß»¯¼Á\;}}{¼ÓÈÈ}$2C2H4O £¨ »·ÑõÒÒÍ飩

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º¶àÑ¡Ìâ

5£®ÏÂÁи÷×éÌþµÄ»ìºÏÎֻҪ×ÜÖÊÁ¿Ò»¶¨£¬ÎÞÂÛËüÃÇ°´Ê²Ã´±ÈÀý»ìºÏ£¬ÍêȫȼÉÕºóÉú³ÉCO2ºÍH2O¶¼ÊǺãÁ¿µÄÊÇ£¨¡¡¡¡£©
A£®C3H6¡¡C2H4B£®C2H4¡¡C2H6C£®C3H8¡¡C3H6D£®C6H6¡¡C2H2

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸