ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿Ã¾¡¢Åð¼°Æ仯ºÏÎïÔÚ¹¤Å©ÒµÉú²úÖÐÓ¦Óù㷺¡£ÒÑÖª£ºÅð¡¢ÂÁͬÖ÷×壻ÅðÉ°³£Ö¸ËÄÅðËáÄƵÄʮˮºÏÎ»¯Ñ§Ê½ÎªNa2B4O7¡¤10H2O¡£ÀûÓÃÅðþ¿ó£¨Ö÷Òª³É·ÖΪMg2B2O5¡¤H2O£©ÖÆÈ¡½ðÊôþ¼°´ÖÅðµÄ¹¤ÒÕÁ÷³ÌͼÈçͼËùʾ£º

£¨1£©Mg2B2O5¡¤H2O ÖÐBµÄ»¯ºÏ¼ÛΪ______¼Û¡£

£¨2£©²Ù×÷¢ÚµÄÖ÷Òª²½ÖèÊÇ_______________ºÍ¹ýÂË£¬´Ó¶ø»ñµÃMgCl26H2O¡£

£¨3£©MgCl26H2OÔÚHCl·ÕΧÖмÓÈȲÅÄÜÖƵÃÎÞË®ÂÈ»¯Ã¾¡£ÆäÔ­ÒòÊÇ____________________£»ÈôÓÃÇéÐԵ缫µç½âMgCl2ÈÜÒº£¬ÆäÒõ¼«·´Ó¦Ê½Îª_____________________________¡£

£¨4£©ÅðÉ°ÈÜÓÚ90¡æÈÈË®ºó£¬³£ÓÃÏ¡ÁòËáµ÷pHÖÁ2~3ÖÆÈ¡H3BO3¾§Ì壬¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽΪ______£»XΪÅðËᾧÌå¼ÓÈÈÍêÈ«ÍÑË®ºóµÄ²úÎÆäÓëMg·´Ó¦ÖÆÈ¡´ÖÅðµÄ»¯Ñ§·½ñÎʽΪ_____________¡£

£¨5£©Ã¾-H2O2ËáÐÔȼÁϵç³ØµÄ·´Ó¦»úÀíΪMg+H2O2+2H+=Mg2++2H2O¡£³£ÎÂÏ£¬ÈôÆðʼµç½âÖÊÈÜÒºpH=1£¬Ôòµ±pH =2ʱ£¬ÈÜÒºÖÐMg2+µÄŨ¶ÈΪ___________¡£

£¨6£©ÖƵõĴÖÅðÔÚÒ»¶¨Ìõ¼þÏ¿ɷ´Î»Éú³ÉBI3£¬BI3¼ÓÈȷֽ⼴¿ÉµÃµ½´¿¾»µÄµ¥ÖÊÅð¡£ÏÖ½«mg´ÖÅðÖƳɵÄBI3³äÈ«·Ö½â£¬Éú³ÉµÄI2ÓÃc mol L-1Na2S2O3ÈÜÒºµÎ¶¨ÖÁÖյ㣬Èý´ÎÏûºÄNa2S2O3ÈÜÒºµÄÌå»ýƽ¾ùֵΪ VmL¡£Ôò¸Ã´ÖÅðÑùÆ·µÄ´¿¶ÈΪ___________¡££¨Ìáʾ£ºI2+2S2O32-=2I-+S4O62-£©

¡¾´ð°¸¡¿ +3 ¼ÓÈÈŨËõ¡¢ÀäÈ´½á¾§ ÒÖÖÆMgCl2Ë®½â 2H2O+2e-+Mg2+=H2¡ü+Mg(OH)2¡ý B4O72-+2H++5H2O=4H3BO3 3Mg+B2O32B+3MgO 0.045mol¡¤L-1 %£¨ÆäËûºÏÀí±í´ïʽҲ¸ø·Ö£©

¡¾½âÎö¡¿£¨1£©ÄÆÔªËØ»¯ºÏ¼ÛΪ+1¼Û£¬ÑõÔªËØ»¯ºÏ¼Û-2¼Û£¬ÒÀ¾Ý»¯ºÏ¼Û´úÊýºÍΪ0¼ÆËãµÃ£ºÅðÔªËØ»¯ºÏ¼ÛΪ+3¼Û¡£

£¨2£©´ÓÈÜÒºÖеõ½½á¾§Ë®ºÏÎïΪ·Àֹʧȥ½á¾§Ë®£¬²»ÄÜÖ±½ÓÕô·¢½á¾§£¬Ó¦²ÉÓÃÕô·¢Å¨Ëõ¡¢ÀäÈ´½á¾§µÄ·½·¨¡£

£¨3£©MgCl2ÊÇÇ¿ËáÈõ¼îÑΣ¬Ë®½âÏÔËáÐÔ£¬¼ÓÈÈ»á´Ù½øMg2+Ë®½â£¬ÔÚHCl·ÕΧÖмÓÈÈ£¬¿ÉÒÖÖÆMgCl2Ë®½â£»ÇéÐԵ缫µç½âMgCl2ÈÜÒº£¬Òõ¼«ÉÏÓÉË®µçÀëµÄH+µÃµç×Ó·¢Éú»¹Ô­·´Ó¦Éú³ÉH2£¬Í¬Ê±Mg2+»á½áºÏOH-Éú³É³Áµí£¬·´Ó¦Ê½Îª£º2H2O+2e-+Mg2+=H2¡ü+Mg(OH)2¡ý¡£

£¨4£©ÓÉ»¯Ñ§Ê½¿ÉÖª£¬ÅðÉ°ÈÜÓÚ90¡æÈÈË®ºó£¬ÓÃÏ¡ÁòËáµ÷pHÖÁ2~3ÖÆÈ¡H3BO3£¬BµÄ»¯ºÏ¼Û²¢Ã»Óз¢Éú±ä»¯£¬¸ù¾ÝÔ­×ÓÊغãºÍµçºÉÊغ㣬Àë×Ó·½³ÌʽΪ£ºB4O72-+2H++5H2O=4H3BO3£»XΪH3BO3¾§Ìå¼ÓÈÈÍÑË®µÄ²úÎÔòΪB2O3£¬ÓëMg·´Ó¦Éú³É´ÖÅðºÍÑõ»¯Ã¾£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ3Mg+B2O32B+3MgO¡£

£¨5£©Ã¾-H2O2ËáÐÔȼÁϵç³Ø£ºMg+H2O2+2H+=Mg2++2H2O£¬Õý¼«ÉÏÊǹýÑõ»¯ÇâÖÐ-1¼ÛOµÃµ½µç×ÓÉú³ÉË®£¬Õý¼«·´Ó¦Ê½Îª£ºH2O2+2H++2e-=2H2O£»ÈôÆðʼpH=1£¬ÔòpH=2ʱÈÜÒºÖУ¬ÇâÀë×ÓŨ¶È¼õС0.1molL-1-0.01 molL-1=0.09 molL-1£¬ÒÀ¾Ý×Ü·´Ó¦Ê½µÃc(Mg2+)=0.045 molL-1¡£

£¨6£©ÓÉÌâÒâ¸ù¾Ý¹Øϵʽ£ºB¡«BI3¡«I2¡«3S2O32-£¬µÃ£ºn£¨B£©=n£¨S2O32-£©=c¡ÁV¡Á10-3mol£¬ËùÒԸôÖÅðÑùÆ·µÄ´¿¶ÈΪ£º c¡ÁV¡Á10-3mol¡Á11gmol-1¡Âm g¡Á100%=¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø