ÌâÄ¿ÄÚÈÝ

14£®ÓÐÏÂÁл¯Ñ§ÒÇÆ÷£º
¢ÙÍÐÅÌÌìÆ½¢Ú²£Á§°ô¢ÛÒ©³×¢ÜÉÕ±­¢ÝÁ¿Í²¢ÞÈÝÁ¿Æ¿¢ß½ºÍ·µÎ¹Ü¢àϸ¿ÚÊÔ¼ÁÆ¿¢á±êǩֽ
£¨1£©ÏÖÐèÒªÅäÖÆ480m L 1mol/L H2SO4ÈÜÒº£¬ÐèÒªÖÊÁ¿·ÖÊýΪ98%£¬ÃܶÈΪ1.84g/cm3µÄŨH2SO427.2ml£»
£¨2£©ÈôʵÑéÓöµ½ÏÂÁÐÇé¿ö£¬¶ÔÁòËáÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈÓкÎÓ°Ï죨Ìî¡°Æ«¸ß¡±¡°Æ«µÍ¡±»ò¡°²»±ä¡±£©
¢ÙÓÃÒÔÏ¡ÊÍŨÁòËáµÄÉÕ±­Î´Ï´µÓ£¬Æ«µÍ£»
¢Úδ¾­ÀäÈ´½«ÈÜҺעÈëÈÝÁ¿Æ¿ÖУ¬Æ«¸ß£»
¢ÛÒ¡ÔȺó·¢ÏÖÒºÃæÏ½µÔÙ¼ÓË®£¬Æ«µÍ£»
¢Ü¶¨ÈÝʱ¹Û²ìÒºÃæ¸©ÊÓ£¬Æ«¸ß£»
¢ÝÁ¿È¡Å¨ÁòËáʱÑöÊÓ¿´ÒºÃæÆ«¸ß£®

·ÖÎö £¨1£©ÏȼÆËãŨÁòËáµÄÎïÖʵÄÁ¿Å¨¶È£¬ÔÙ¸ù¾ÝÈÜҺϡÊÍǰºóÈÜÖʵÄÎïÖʵÄÁ¿²»±ä¼ÆËãËùÐèŨÑÎËáµÄÌå»ý£»
£¨2£©·ÖÎö²Ù×÷¶ÔÈÜÖʵÄÎïÖʵÄÁ¿ºÍÈÜÒºÌå»ýµÄÓ°Ï죬ÒÀ¾ÝC=$\frac{n}{V}$½øÐÐÎó²î·ÖÎö£®

½â´ð ½â£»£¨1£©ÖÊÁ¿·ÖÊýΪ98%£¬ÃܶÈΪ1.84g/cm3µÄŨÁòËáÎïÖʵÄÁ¿Å¨¶ÈC=$\frac{1000¡Á1.84¡Á98%}{98}$=18.4mol/L£¬ÅäÖÆ480m L 1mol/L H2SO4ÈÜÒº£¬Ó¦Ñ¡Ôñ500mLÈÝÁ¿Æ¿£¬ÉèÐèҪŨÁòËáµÄÌå»ýΪV£¬1mol/L¡Á500mL=18.4mol/L¡ÁV£¬V=27.2mL£¬
¹Ê´ð°¸Îª£º27.2£»
£¨2£©¢ÙÓÃÒÔÏ¡ÊÍŨÁòËáµÄÉÕ±­Î´Ï´µÓ£¬µ¼Ö²¿·ÖÈÜÖÊËðºÄ£¬ÈÜÖʵÄÎïÖʵÄÁ¿Æ«Ð¡£¬ÈÜҺŨ¶ÈÆ«µÍ£»
¹Ê´ð°¸Îª£ºÆ«µÍ£»
¢Úδ¾­ÀäÈ´½«ÈÜҺעÈëÈÝÁ¿Æ¿ÖУ¬ÀäÈ´ºó£¬ÒºÃæÏ½µµ¼ÖÂÈÜÒºÌå»ýƫС£¬ÈÜҺŨ¶ÈÆ«¸ß£»
¹Ê´ð°¸Îª£ºÆ«¸ß£»
¢ÛÒ¡ÔȺó·¢ÏÖÒºÃæÏ½µÔÙ¼ÓË®£¬µ¼ÖÂÈÜÒºÌå»ýÆ«´ó£¬ÈÜҺŨ¶ÈÆ«µÍ£»
¹Ê´ð°¸Îª£ºÆ«µÍ£»
¢Ü¶¨ÈÝʱ¹Û²ìÒºÃæ¸©ÊÓ£¬µ¼ÖÂÈÜÒºÌå»ýƫС£¬ÈÜҺŨ¶ÈÆ«¸ß£»
¹Ê´ð°¸Îª£ºÆ«¸ß£»
¢ÝÁ¿È¡Å¨ÁòËáʱÑöÊÓ¿´ÒºÃ棬µ¼ÖÂÁ¿È¡µÄŨÁòËáÌå»ýÆ«´ó£¬ÁòËáµÄÎïÖʵÄÁ¿Æ«´ó£¬ÈÜҺŨ¶ÈÆ«¸ß£»
¹Ê´ð°¸Îª£ºÆ«¸ß£®

µãÆÀ ±¾Ì⿼²éÁËÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈÈÜÒºµÄÅäÖÆ¼°ÓйØÎïÖʵÄÁ¿Å¨¶È¼ÆË㣬Ã÷È·ÅäÖÆÔ­Àí¼°²Ù×÷²½ÖèÊǽâÌâ¹Ø¼ü£¬×¢ÒâÎó²î·ÖÎöµÄ·½·¨£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
4£®¹¤ÒµºÏ³É°±·´Ó¦Îª£ºN2£¨g£©+3H2£¨g£© $?_{¸ßθßѹ}^{´ß»¯¼Á}$ 2NH3£¨g£©£¬¶ÔÆäÑо¿ÈçÏ£º
£¨1£©ÒÑÖªH-H¼üÄÜΪ436kJ•mol-1£¬N-H¼üÄÜΪ391kJ•mol-1£¬N¡ÔN¼üµÄ¼üÄÜÊÇ945.6kJ•mol-1£¬ÔòÉÏÊö·´Ó¦µÄ¡÷H=-92.46kJ•mol-1£®
£¨2£©ÉÏÊö·´Ó¦µÄƽºâ³£ÊýKµÄ±í´ïʽΪ£º$\frac{{c}^{2}£¨N{H}_{3}£©}{c£¨{N}_{2}£©{c}^{3}£¨{H}_{2}£©}$£®Èô·´Ó¦·½³Ìʽ¸ÄдΪ£º$\frac{1}{2}$N2£¨g£©+$\frac{3}{2}$H2£¨g£©?NH3£¨g£©£¬ÔÚ¸ÃζÈÏÂµÄÆ½ºâ³£Êý£ºK1=$\sqrt{K}$£¨ÓÃK±íʾ£©£®
£¨3£©ÔÚ773Kʱ£¬·Ö±ð½«2molN2ºÍ6molH2³äÈëÒ»¸ö¹Ì¶¨ÈÝ»ýΪ1LµÄÃܱÕÈÝÆ÷ÖУ¬Ëæ×Å·´Ó¦µÄ½øÐУ¬ÆøÌå»ìºÏÎïÖÐn£¨H2£©¡¢n£¨NH3£©Ó뷴Ӧʱ¼ätµÄ¹ØÏµÈçÏÂ±í£º
t/min051015202530
n£¨H2£©/mol6.004.503.603.303.033.003.00
n£¨NH3£©/mol01.001.601.801.982.002.00
¢Ù¸ÃζÈÏ£¬ÈôÏòͬÈÝ»ýµÄÁíÒ»ÈÝÆ÷ÖÐͶÈëµÄN2¡¢H2¡¢NH3Ũ¶È·Ö±ðΪ3mol/L£¬3mol/L¡¢3mol/L£¬´ËʱvÕý´óÓÚvÄæ£¨Ìî¡°´óÓÚ¡±¡¢¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡±£©
¢ÚÓÉÉÏÊö±íÖеÄʵÑéÊý¾Ý¼ÆËãµÃµ½¡°Å¨¶È¡«Ê±¼ä¡±µÄ¹ØÏµ¿ÉÓÃÈçͼµÄÇúÏß±íʾ£¬±íʾ    c£¨N2£©¡«tµÄÇúÏßÊÇÒÒ£®ÔÚ´ËζÈÏ£¬ÈôÆðʼ³äÈë4molN2ºÍ12molH2£¬·´Ó¦¸Õ´ïµ½Æ½ºâʱ£¬±íʾc£¨H2£©µÄÇúÏßÉÏÏàÓ¦µÄµãΪB£®

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

¾«Ó¢¼Ò½ÌÍø