ÌâÄ¿ÄÚÈÝ

ÇâÑõ»¯±µÊÇÒ»Öֹ㷺ʹÓõĻ¯Ñ§ÊÔ¼Á£¬ÏÖͨ¹ýÏÂÁÐʵÑé²â¶¨Ä³ÊÔÑùÖÐBa(OH)2¡¤nH2OµÄº¬Á¿£®

(1)³ÆÈ¡3.50 gÊÔÑùÈÜÓÚÕôÁóË®Åä³É100 mlÈÜÒº£¬´ÓÖÐÈ¡³ö10.0 mlÈÜÒºÓÚ×¶ÐÎÆ¿ÖУ¬¼Ó2µÎָʾ¼Á£¬ÓÃ0.1000 mol/LµÄ±ê×¼ÑÎËáÈÜÒºµÎ¶¨ÖÁÖյ㣬¹²ÏûºÄ±ê×¼Òº20.0 ml(ÔÓÖʲ»ÓëËá·´Ó¦)£¬¸ÃÊÔÑùÖÐÇâÑõ»¯±µµÄÎïÖʵÄÁ¿Îª________£»

(2)È¡5.25 gÊÔÑù¼ÓÈÈÖÁºãÖØ(ÔÓÖÊÊÜÈȲ»·Ö½â)£¬³ÆµÃÖÊÁ¿Îª3.09 g£¬ÇóBa(OH)2¡¤nH2OÖеÄnµÄÖµ(д³ö¼ÆËã¹ý³Ì)£®

´ð°¸£º
½âÎö£º

(1)0.01 mol¡¡(2)8


Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
ÄÆ¡¢Ã¾¡¢ÂÁ¡¢ÌúµÈ½ðÊôÔÚÈÕ³£Éú»î¡¢¹¤ÒµÉú²ú¡¢º½ÌìÊÂÒµÖÐÓÐ׏㷺µÄÓÃ;£®
£¨1£©4.8gAlÔÚ¿ÕÆøÖзÅÖÃÒ»¶Îʱ¼äÖ®ºó£¬ÖÊÁ¿±äΪ5.28g£¬Ôòδ·¢Éú±ä»¯µÄÂÁµÄÖÊÁ¿Îª
 
g£®
£¨2£©Ïò10mL 0.2mol/LµÄÂÈ»¯ÂÁÈÜÒºÖÐÖðµÎ¼ÓÈëδ֪Ũ¶ÈµÄÇâÑõ»¯±µÈÜÒº£¬²âµÃµÎ¼Ó10mLÓëµÎ¼Ó30mLʱËùµÃ³ÁµíÒ»Ñù¶à£®ÇóÇâÑõ»¯±µÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶È£®
£¨3£©½«Ã¾¡¢ÂÁºÏ½ð6.3g»ìºÏÎï¼ÓÈ뵽ijŨ¶ÈµÄÏõËáÖУ¬²úÉú±ê¿öÏÂµÄÆøÌå8.96L£¨¼ÙÉè·´Ó¦¹ý³ÌÖл¹Ô­²úÎïÊÇNOºÍNO2µÄ»ìºÏÆøÌ壩£¬ÏòËùµÃÈÜÒºÖмÓÈ백ˮÖÁ³Áµí´ïµ½×î´óÁ¿£¬³Áµí¸ÉÔïºó²âµÃµÄÖÊÁ¿±ÈÔ­ºÏ½ðµÄÖÊÁ¿Ôö¼Ó10.2g£¬¼ÆË㻹ԭ²úÎïÖÐNOºÍNO2µÄÎïÖʵÄÁ¿Ö®±È£®
£¨4£©ÏÖÓÐÒ»°üÂÁÈȼÁÊÇÂÁ·ÛºÍÑõ»¯Ìú·ÛÄ©µÄ»ìºÏÎÔÚ¸ßÎÂÏÂʹ֮³ä·Ö·´Ó¦£¬½«·´Ó¦ºóµÄ¹ÌÌå·ÖΪÁ½µÈ·Ý£¬½øÐÐÈçÏÂʵÑ飨¼ÆËãpHʱ¼Ù¶¨ÈÜÒºÌå»ýûÓб仯£©£º¢ÙÏòÆäÖÐÒ»·Ý¹ÌÌåÖмÓÈë50mL 3.0mol/LµÄNaOHÈÜÒº£¬¼ÓÈÈʹÆä³ä·Ö·´Ó¦ºó¹ýÂË£¬²âµÃÂËÒºµÄc£¨OH-£©Îª1.0mol/L£»¢ÚÏòÁíÒ»·Ý¹ÌÌåÖмÓÈë100mL 4.0mol/LµÄHClÈÜÒº£¬Ê¹¹ÌÌåÈ«²¿Èܽ⣬²âµÃ·´Ó¦ºóËùµÃÈÜÒºÖÐÖ»ÓÐH+¡¢Fe2+ºÍAl3+ÈýÖÖÑôÀë×ÓÇÒc£¨H+£©Îª0.2mol/L£¬¼ÆËãÕâ°üÂÁÈȼÁÖÐÂÁµÄÖÊÁ¿ºÍÑõ»¯ÌúµÄÖÊÁ¿£®

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

¾«Ó¢¼Ò½ÌÍø