ÌâÄ¿ÄÚÈÝ

²Ýľ»ÒÊÇÒ»ÖÖÅ©¼Ò·Ê£¬ËüµÄÖ÷Òª³É·ÖÊÇ̼Ëá¼Ø£¬»¹ÓÐÁòËá¼Ø£¬ÂÈ»¯¼ØµÈ¡£»¯Ñ§ÐËȤС×éΪ²â¶¨Ä³²Ýľ»ÒÑùÆ·ÖеÄÓÐЧ³É·Ö£¬È¡50g²Ýľ»ÒÓÚÉÕ±­ÖУ¬²»¶ÏµÎÈëÁòËáÈÜÒº£¬µ±µÎÈë30gÁòËáÈÜҺʱ£¬²»ÔÙÓÐÆøÅݲúÉú£¬´ËʱÉÕ±­ÖеIJÐÓà»ìºÏÎïµÄÖÊÁ¿Îª77.8g¡££¨¼ÙÉè²Ýľ»ÒµÄÆäËû³É·Ö²»º¬¼ØÔªËØÇÒ²»ÓëËá·´Ó¦£©
Çë¼ÆËã»Ø´ð£º
£¨1£©²Ýľ»ÒµÄÖ÷Òª³É·ÖËùÊôµÄÎïÖÊÀà±ðΪ____________£¨Ìî¡°Ëá¡¢¼î»òÑΡ±£©¡£
£¨2£©Èçͼ±íʾ·´Ó¦¹ý³ÌÖзųöµÄÆøÌåÖÊÁ¿Óë¼ÓÈëÁòËáÈÜÒºµÄ¹ØϵÇúÏߣ¬ÇëÇó³öͼÖÐ×Ý×ø±êÉÏaµãÊýÖµ£ºa=________________g£»

£¨3£©¼ÆËã²Ýľ»ÒÑùÆ·ÖÐ̼Ëá¼ØµÄÖÊÁ¿·ÖÊý¡££¨ÒªÇóд³ö¼ÆËã¹ý³Ì£©
£¨4£©Í¨¹ýÆäËûʵÑ飬²âµÃ¸Ã50g²Ýľ»ÒÖл¹º¬µÄÁòËá¼ØÖÊÁ¿Îª8.7g£¬ÂÈ»¯¼ØµÄÖÊÁ¿Îª1.49g£¬Ôò¸Ã50g²Ýľ»ÒÑùÆ·ÖмØÔªËصÄÖÊÁ¿Îª__________g¡£
£¨1£©ÑΣ»£¨2£©2.2£¨3£©13.8%£¨4£©8.58

ÊÔÌâ·ÖÎö£º£¨1£©¸ù¾ÝÌâÒâ¿ÉÖª£¬¡°²Ýľ»ÒÊÇÒ»ÖÖÅ©¼Ò·Ê£¬ËüµÄÖ÷Òª³É·ÖÊÇ̼Ëá¼Ø£¬»¹ÓÐÁòËá¼Ø£¬ÂÈ»¯¼ØµÈ¡±¡£ÆäÖеÄ̼Ëá¼Ø£¬»¹ÓÐÁòËá¼Ø£¬ÂÈ»¯¼Ø¾ùÊÇÓɽðÊôÀë×ÓºÍËá¸ùÀë×Ó¹¹³ÉµÄ»¯ºÏÎÊôÓÚÑÎÀà¡£
£¨2£©aµã±íʾÉú³ÉÆøÌåµÄÖÊÁ¿µÄ×î´óÖµ¡£¸ù¾ÝÌâÒâ¿ÉÖª£º¸ÃÆøÌåΪ̼Ëá¼ØÓëÁòËá·´Ó¦Éú³ÉµÄ¶þÑõ»¯Ì¼ÆøÌå¡£ÆäÖÊÁ¿Îª£º50g£¨²Ýľ»Ò£©+30g£¨ÁòËáÈÜÒº£©¡ª77.8g£¨²ÐÓà»ìºÏÎ="2.2" g¡£ËùÒÔa="2.2" g£»
£¨3£©¸ù¾ÝÌâÒâ¿ÉÖª£ºÒÑÖªÁ¿ÎªÉú³É¶þÑõ»¯Ì¼µÄÖÊÁ¿£»Î´ÖªÁ¿ÎªÑùÆ·ÖÐ̼Ëá¼ØµÄÖÊÁ¿·ÖÊý¡£
½âÌâ˼·£º¿É¸ù¾Ý¶þÑõ»¯Ì¼Óë̼Ëá¼ØÔÚ·´Ó¦ÖеÄÖÊÁ¿£¬Çó³ö̼Ëá¼ØµÄÖÊÁ¿£¬½øÒ»²½Çó³öÆäÔÚÑùÆ·ÖеÄÖÊÁ¿·ÖÊý¡£¾ßÌå¹ý³ÌÈçÏ£º
½â£ºÉè²Ýľ»ÒÑùÆ·ÖÐ̼Ëá¼ØµÄÖÊÁ¿Îªx
K2CO3+H2SO4=K2SO4+CO2¡ü+H2O
138                44
x                  2.2g
138£º44=x£º2.2g
½âµÃ£ºx=6.9g
̼Ëá¼ØµÄÖÊÁ¿·ÖÊý=¡Á100%=13.8%
´ð£º²Ýľ»ÒÑùÆ·ÖÐ̼Ëá¼ØµÄÖÊÁ¿·ÖÊýΪ13.8%
£¨4£©²Ýľ»ÒÖеļØÔªËØÊÇ̼Ëá¼Ø¡¢ÁòËá¼ØºÍÂÈ»¯¼ØÈýÖÖÎïÖÊÖеļØÔªËصÄÖÊÁ¿ºÍ¡£¿É·Ö±ð¸ù¾ÝÈýÖÖÎïÖʵĻ¯Ñ§Ê½Çó³öÆäÖÐËùº¬¼ØÔªËصÄÖÊÁ¿¡£¾ßÌå½âÌâ¹ý³ÌÈçÏ£º
6.9g̼Ëá¼Ø£¨K2 CO3£©Öк¬¼ØÔªËصÄÖÊÁ¿Îª£º6.9g¡Á£¨ ¡Á100%£©=3.9g£»
8.7gÁòËá¼Ø£¨K2 SO4£©Öк¬¼ØÔªËصÄÖÊÁ¿Îª£º8.7g¡Á£¨ ¡Á100%£©=3.9g£»
1.49gÂÈ»¯¼Ø£¨KCl£©Öк¬¼ØÔªËصÄÖÊÁ¿Îª£º1.49g¡Á£¨ ¡Á100%£©=0.78g£»
²Ýľ»ÒÑùÆ·ÖмØÔªËصÄÖÊÁ¿Îª3.9g+3.9g+0.78g=8.58g¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
£¨14·Ö£©»¯Ñ§ÊÇÑо¿ÎïÖʵÄ×é³É¡¢½á¹¹¡¢ÐÔÖʼ°±ä»¯¹æÂɵĿÆѧ¡£
£¨1£©¢ÙÍ­¡¢¢Ú¸É±ù¡¢¢ÛÂÈ»¯ÄÆÈýÖÖÎïÖÊÖУ¬ÓÉÀë×Ó¹¹³ÉµÄÊÇ        £¨ÌîдÐòºÅ£¬ÏÂͬ£©£¬¿ÉÓÃÓÚÈ˹¤½µÓêµÄÊÇ        ¡£
£¨2£©Ê³´×Öк¬Óд×ËᣨCH3COOH£©£¬´×ËáÓÉ        ÖÖÔªËØ×é³É£¬Æä·Ö×ÓÖÐÇâ¡¢ÑõÔ­×Ó¸öÊý±ÈΪ        ¡£
£¨3£©ÆøÌåAºÍÆøÌåB½Ó´¥¿É·¢Éú·´Ó¦£¬Éú³É¹ÌÌåCºÍÒºÌåD£¬Æä΢¹Û¹ý³ÌÈçÏÂËùʾ¡£
    
ͼ1                                         Í¼2
¢Ù ¸Ã·´Ó¦¹ý³ÌÖУ¬ÊôÓÚÑõ»¯ÎïµÄÊÇ        £¨ÌîÃû³Æ£©¡£
¢Ú ½«µÈÌå»ýµÄÁ½Ö»¼¯ÆøÆ¿ÖзֱðÊ¢ÂúA¡¢B£¬Èçͼ2Ëùʾ½øÐÐʵÑé¡£ÒÑÖªÏàͬÌõ¼þÏ£¬ÆøÌåµÄÌå»ý±ÈµÈÓÚ·Ö×Ó¸öÊý±È£¬ Ôò³ä·Ö·´Ó¦ºó£¬Ê£ÓàµÄÆøÌåÊÇ        £¨Ìѧʽ£©¡£
£¨4£©ÔìÖ½»á²úÉú´óÁ¿º¬NaOHµÄ¼îÐÔ·ÏË®£¬Ðè¾­´¦Àí³ÊÖÐÐÔºóÅÅ·Å¡£
¢Ù ÓÃpHÊÔÖ½¼ì²â·ÏË®³Ê¼îÐԵķ½·¨ÊÇ        ¡£
¢Ú ÈôijÔìÖ½³§·ÏË®Öк¬NaOHµÄÖÊÁ¿·ÖÊýΪl.6%£¬ÏÖÓзÏÁòËá9.8t£¨H2SO4µÄÖÊÁ¿·ÖÊýΪ20%£©£¬¿ÉÒÔ´¦ÀíµÄ·ÏË®ÖÊÁ¿ÊǶàÉÙ£¿£¨Ð´³ö¼ÆËã¹ý³Ì£©          

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

¾«Ó¢¼Ò½ÌÍø