ÌâÄ¿ÄÚÈÝ

13£®Õý¶¡È©ÊÇÒ»ÖÖ»¯¹¤Ô­ÁÏ£®Ä³ÊµÑéС×éÀûÓÃÈçÓÒ×°ÖúϳÉÕý¶¡È©£®
·¢ÉúµÄ·´Ó¦ÈçÏ£ºCH3CH2CH2CH2OH$¡ú_{H_{2}SO_{4}¡÷}^{Na_{2}Cr_{2}O_{7}}$CH3CH2CH2CHO
·´Ó¦ÎïºÍ²úÎïµÄÏà¹ØÊý¾ÝÁбíÈçÏ£º
·Ðµã/¡æÃܶÈ/£¨g•cm-3£©Ë®ÖÐÈܽâÐÔ
Õý¶¡´¼117.20.810 9΢ÈÜ
Õý¶¡È©75.70.801 7΢ÈÜ
ʵÑé²½ÖèÈçÏ£º
½«6.0g Na2Cr2O7·ÅÈë100mLÉÕ±­ÖУ¬¼Ó30mLË®Èܽ⣬ÔÙ»ºÂý¼ÓÈë5mLŨÁòËᣬ½«ËùµÃÈÜҺСÐÄתÒÆÖÁBÖУ¬ÔÚAÖмÓÈë4.0gÕý¶¡´¼ºÍ¼¸Á£·Ðʯ£¬¼ÓÈÈ£®µ±ÓÐÕôÆø³öÏÖʱ£¬¿ªÊ¼µÎ¼ÓBÖÐÈÜÒº£®µÎ¼Ó¹ý³ÌÖб£³Ö·´Ó¦Î¶ÈΪ90¡«95¡æ£¬ÔÚEÖÐÊÕ¼¯90¡æÒÔϵÄÁó·Ö£®½«Áó³öÎïµ¹Èë·ÖҺ©¶·ÖУ¬·Öȥˮ²ã£¬Óлú²ã¸ÉÔïºóÕôÁó£¬ÊÕ¼¯75¡«77¡æÁó·Ö£¬²úÁ¿2.0g£®
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Èô¼ÓÈȺó·¢ÏÖδ¼Ó·Ðʯ£¬Ó¦²ÉÈ¡µÄÕýÈ··½·¨ÊÇÀäÈ´ºó²¹¼Ó£®
£¨2£©ÉÏÊö×°ÖÃͼÖУ¬DÒÇÆ÷µÄÃû³ÆÊÇÖ±ÐÎÀäÄý¹Ü£®
£¨3£©½«Õý¶¡È©´Ö²úÆ·ÖÃÓÚ·ÖҺ©¶·ÖзÖˮʱ£¬Ë®ÔÚÏ£¨Ìî¡°ÉÏ¡±»ò¡°Ï¡±£©²ã£®
£¨4£©·ÖҺ©¶·Ê¹ÓÃÇ°±ØÐë½øÐеIJÙ×÷ÊÇc£¨ÌîÕýÈ·´ð°¸±êºÅ£©£®
A£®Èóʪ¡¡¡¡  B£®¸ÉÔï¡¡¡¡¡¡  C£®¼ì©¡¡¡¡D£®±ê¶¨£®

·ÖÎö ±¾ÌâÊÇÀûÓô¼µÄ´ß»¯Ñõ»¯ÖƱ¸Õý¶¡È©µÄʵÑé²Ù×÷Ì⣬Ö÷ÒªÉæ¼°·´Ó¦»ìºÏ»ìºÏ¼ÓÈÈ·À±©·Ð´ëÊ©¼°Ó¦¼±´¦Àí·½·¨¡¢²Ù×÷¹ý³ÌÖÐÉú³ÉÎïµÄÀäÄýÒÔ¼°´Ö²úÆ·µÄ·ÖÀëÌá´¿£»
£¨1£©»ìºÏ¼ÓÈÈΪ·ÀÖ¹·ÀÖ¹±©·ÐÒª¼ÓÈë·Ðʯ£¬Èô¼ÓÈȺó·¢ÏÖδ¼Ó·Ðʯ£¬Ó¦¸ÃÀäÈ´ºó²¹¼Ó£»
£¨2£©ÀûÓÃÒÇÆ÷D¶ÔÕôÆû½øÐÐÀäÄý£¬´ËDÒÇÆ÷µÄÃû³ÆÖ±ÐÎÀäÄý¹Ü£»
£¨3£©ÓɱíÖÐÊý¾Ý¿ÉÖª£¬Õý¶¡È©ÃܶÈСÓÚË®µÄÃܶȣ¬¾Ý´ËÅжϣ»
£¨4£©·ÖҺ©¶·Ê¹ÓÃÇ°±ØÐë½øÐеĵÚÒ»²½²Ù×÷ÊǼì©£®

½â´ð ½â£º£¨1£©·ÀÖ¹±©·ÐÒª¼ÓÈë·Ðʯ£¬Èô¼ÓÈȺó·¢ÏÖδ¼Ó·Ðʯ£¬Ó¦¸ÃÀäÈ´ºó²¹¼Ó£¬¹Ê´ð°¸Îª£º·ÀÖ¹±©·Ð£»ÀäÈ´ºó²¹¼Ó£»
£¨2£©DÒÇÆ÷ÊÇÀäÄý×°Öã¬Ãû³ÆΪֱÐÎÀäÄý¹Ü£¬¹Ê´ð°¸Îª£ºÖ±ÐÎÀäÄý¹Ü£»
£¨3£©Õý¶¡È©ÃܶÈΪ0.8017 g•cm-3£¬Ð¡ÓÚË®µÄÃܶȣ¬Õñµ´¡¢¾²ÖᢷֲãºóË®²ãÔÚÏ·½£¬¹Ê´ð°¸Îª£ºÏ£»
£¨4£©·ÖҺ©¶·ÒòÓлîÈû£¬Ê¹ÓÃÇ°±ØÐë½øÐеĵÚÒ»Ïî²Ù×÷ÊǼì©£¬¹Ê´ð°¸Îª£ºc£®

µãÆÀ ±¾Ì⿼²éÓлú»¯Ñ§ÊµÑé»ù±¾²Ù×÷µÈ£¬ÄѶȲ»´ó£¬×¢Òâ¶Ô»ù´¡ÖªÊ¶µÄÀí½âÕÆÎÕͬʱ³ä·ÖÀûÓÃÌâÖÐÐÅÏ¢·ÖÎöÎÊÌ⣮

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
15£®ÓÃ50mL0.50mol/LÑÎËáÓë50mL0.55mol/LNaOHÈÜÒºÔÚÈçͼËùʾµÄ×°ÖÃÖнøÐÐÖкͷ´Ó¦£®Í¨¹ý²â¶¨·´Ó¦¹ý³ÌÖÐËù·Å³öµÄÈÈÁ¿¿É¼ÆËãÖкÍÈÈ£®»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©´ÓʵÑé×°ÖÃÉÏ¿´£¬Í¼ÖÐÉÐȱÉÙµÄÒ»ÖÖ²£Á§ÓÃÆ·ÊÇ»·Ðβ£Á§½Á°è°ô£®
£¨2£©ÉÕ±­¼äÌîÂúËéÖ½ÌõµÄ×÷ÓÃÊǼõÉÙʵÑé¹ý³ÌÖеÄÈÈÁ¿Ëðʧ£®
£¨3£©´óÉÕ±­ÉÏÈç²»¸ÇÓ²Ö½°å£¬ÇóµÃµÄÖкÍÈÈÊýֵƫС£¨Ìî¡°Æ«´ó¡¢Æ«Ð¡¡¢ÎÞÓ°Ï족£©
£¨4£©ÈôÉÏÊöHCl¡¢NaOHÈÜÒºµÄÃܶȶ¼½üËÆΪ1g/cm3£¬ÖкͺóÉú³ÉµÄÈÜÒºµÄ±ÈÈÈÈÝC=4.18J/£¨g•¡æ£©£¬·´Ó¦ºóζÈÉý¸ßÁË¡÷t£¬Éú³É1molˮʱµÄ·´Ó¦ÈÈ¡÷H=-$\frac{0.418¡÷t}{0.025}$ kJ/mol£¨Ìî±í´ïʽ£©£®
£¨5£©Èç¹ûÓÃ60mL0.50mol/LÑÎËáÓë50mL0.55mol/LNaOHÈÜÒº½øÐз´Ó¦£¬ÓëÉÏÊöʵÑéÏà±È£¬Ëù·Å³öµÄÈÈÁ¿²»ÏàµÈ£¨Ìî¡°ÏàµÈ¡¢²»ÏàµÈ¡±£©£¬ËùÇóÖкÍÈÈÏàµÈ£¨Ìî¡°ÏàµÈ¡¢²»ÏàµÈ¡±£©£¬¼òÊöÀíÓÉÒòΪÖкÍÈÈÊÇÖ¸Ëá¸ú¼î·¢ÉúÖкͷ´Ó¦Éú³É1molH2OËù·Å³öµÄÈÈÁ¿£¬ÓëËá¼îµÄÓÃÁ¿ÎÞ¹Ø
£¨6£©ÓÃÏàͬŨ¶ÈºÍÌå»ýµÄ°±Ë®£¨NH3•H2O£©´úÌæNaOHÈÜÒº½øÐÐÉÏÊöʵÑ飬²âµÃµÄÖкÍÈȵÄÊýÖµ»áƫС£»£¨Ìî¡°Æ«´ó¡±¡¢¡°Æ«Ð¡¡±¡¢¡°ÎÞÓ°Ï족£©£®
2£®Ä¿Ç°³ÇÊпÕÆøÖÊÁ¿¶ñ»¯Ô­ÒòÖ®Ò»ÊÇ»ú¶¯³µÎ²ÆøºÍȼú²úÉúµÄÑÌÆø£¬NOºÍCO¾ùΪÆû³µÎ²ÆøµÄ³É·Ö£¬ÕâÁ½ÖÖÆøÌåÔÚ´ß»¯×ª»»Æ÷Öз¢ÉúÈçÏ·´Ó¦£º
2NO£¨g£©+2CO£¨g£©?2CO2£¨g£©+N2£¨g£©¡÷H=-akJ•mol-1£¨a£¾0£©
ÔÚÒ»¶¨Î¶ÈÏ£¬½«2.0mol NO¡¢2.4mol COÆøÌåͨÈëµ½¹Ì¶¨ÈÝ»ýΪ2LµÄÃܱÕÈÝÆ÷ÖУ¬·´Ó¦¹ý³ÌÖв¿·ÖÎïÖʵÄŨ¶È±ä»¯ÈçͼËùʾ£¨¼ÆËã½á¹û±£ÁôÈýλÓÐЧÊý×Ö£©£º
£¨1£©0¡«15minNOµÄƽ¾ùËÙÂÊv£¨NO£©=0.0267mol/£¨L•min£©£»COµÄת»¯ÂÊΪ40.0%£®
£¨2£©15minºóÈô½«´ËʱµÃµ½µÄCO2ÆøÌåÓú¬ÓÐ0.8mol KOHÈÜÒºÎüÊÕ£¬Ôòµç½âÖÊÈÜÒºÖи÷ÖÖÀë×ÓµÄÎïÖʵÄÁ¿Å¨¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòÊÇc£¨K+£©£¾c£¨CO32-£©£¾c£¨OH-£©£¾c£¨HCO3-£©£¾c£¨H+£©£®
£¨3£©20minʱ£¬Èô¸Ä±äijһ·´Ó¦Ìõ¼þ£¬µ¼ÖÂCOŨ¶ÈÔö´ó£¬Ôò¸Ä±äµÄÌõ¼þ¿ÉÄÜÊÇae£¨Ñ¡ÌîÐòºÅ£©£®
a£®ËõСÈÝÆ÷Ìå»ý     b£®À©´óÈÝÆ÷Ìå»ý     c£®Ôö¼ÓNOµÄÁ¿
d£®½µµÍζȠ       e£®Ôö¼ÓCOµÄÁ¿       f£®Ê¹Óô߻¯¼Á
£¨4£©Èô±£³Ö·´Ó¦ÌåϵµÄζȲ»±ä£¬20minʱÔÙÏòÈÝÆ÷ÖгäÈëCO¡¢CO2¸÷0.4mol£¬»¯Ñ§Æ½ºâ½«Ïò×óÒƶ¯£¨Ñ¡Ìî¡°Ïò×󡱡¢¡°ÏòÓÒ¡±»ò¡°²»¡±£©£¬ÖØдﵽƽºâºó£¬¸Ã·´Ó¦µÄ»¯Ñ§Æ½ºâ³£ÊýΪ0.139L/mol£®

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

¾«Ó¢¼Ò½ÌÍø