µ±Ç°Î»ÖãºÊ×Ò³ > ºþÄÏÊ¡¡¢½Î÷ʡʮËÄУ2018½ì¸ßÈýµÚÒ»´ÎÁª¿¼»¯Ñ§ÊÔ¾í£¨º¬´ð°¸£©
£¨1£©ÒÑÖª£º2Cu(s)£«
1O2(g)=Cu2O(s) ¡÷H =-169kJ¡¤mol-1 2C(s)£«
1O2(g)=CO(g) ¡÷H =-110.5kJ¡¤mol-1 2Cu(s)£«
1O2(g)=CuO(s) ¡÷H =-157kJ¡¤mol-1 2Ôò·½·¨a·¢Éú·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽÊÇ_____________________________________¡£
£¨2£©·½·¨b²ÉÓÃÀë×Ó½»»»Ä¤¿ØÖƵç½âÒºÖÐOH-µÄŨ¶È¶øÖƱ¸ÄÉÃ×Cu2O£¬×°ÖÃÈçͼËùʾ£¬¸ÃÀë×Ó½»»»Ä¤Îª______Àë×Ó½»»»Ä¤(Ìî¡°Òõ¡±»ò¡°Ñô¡±),¸Ãµç³ØµÄÑô¼«·´Ó¦Ê½Îª ______________________ ¡£
£¨3£©·½·¨cΪ¼ÓÈÈÌõ¼þÏÂÓÃҺ̬루N2H4£©»¹ÔÐÂÖÆCu(OH)2À´ÖƱ¸ÄÉÃ×¼¶Cu2O£¬Í¬Ê±·Å³öN2£¬¸ÃÖÆ·¨µÄ»¯Ñ§·½³ÌʽΪ ____________________________ ¡£
£¨4£©ÔÚÈÝ»ýΪ1LµÄºãÈÝÃܱÕÈÝÆ÷ÖУ¬ÓÃÒÔÉÏ·½·¨ÖƵõÄÈýÖÖÄÉÃ×¼¶Cu2O·Ö±ð½øÐд߻¯·Ö½âË®µÄʵÑ飺2H2O(g) 2H2(g)+O2(g)£¬¦¤H>0¡£Ë®ÕôÆøµÄŨ¶ÈcËæÊ±¼ät µÄ±ä»¯ÈçϱíËùʾ¡£
¢Ù¶Ô±ÈʵÑéµÄζȣºT2_________T1£¨Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°=¡±£© ¢Ú´ß»¯¼Á´ß»¯Ð§ÂÊ£ºÊµÑé¢Ù________ ʵÑé¢Ú£¨Ìî¡°£¾¡±»ò¡°£¼¡±£©
¢ÛÔÚʵÑé¢Û´ïµ½Æ½ºâ״̬ºó£¬Ïò¸ÃÈÝÆ÷ÖÐͨÈëË®ÕôÆøÓëÇâÆø¸÷0.1mol£¬Ôò·´Ó¦Ôٴδﵽƽºâʱ£¬ÈÝÆ÷ÖÐÑõÆøµÄŨ¶ÈΪ ____________________¡£
20¡¢£¨12 ·Ö£©ÒÔÈíÃÌ¿ó·Û£¨Ö÷Òªº¬ÓÐMnO2£¬»¹º¬ÓÐÉÙÁ¿µÄFe2O3¡¢Al2O3 µÈÔÓÖÊ£©ÎªÔÁÏÖÆ±¸¸ß´¿
MnO2µÄÁ÷³ÌÈçÏÂͼËùʾ£º
ÒÑÖª£º¢Ù³£ÎÂÏ£¬Ksp[Fe(OH)3]=8.0¡Á10-38£¬Ksp [Al(OH)3]=4.0¡Á10-34¡£
¢Ú³£ÎÂÏ£¬ÇâÑõ»¯Îï³ÁµíµÄÌõ¼þ£ºAl3+¡¢Fe3+ÍêÈ«³ÁµíµÄpH ·Ö±ðΪ4.6¡¢3.4£»Mn2+¿ªÊ¼³ÁµíµÄpH Ϊ8.1¡£
¢Û³£ÎÂÏ£¬µ±ÈÜÒºÖнðÊôÀë×ÓÎïÖʵÄÁ¿Å¨¶ÈСÓÚ»òµÈÓÚ1¡Á10-5mol¡¤L-lʱ£¬¿ÉÊÓΪ¸Ã½ðÊôÀë×ÓÒѱ»³ÁµíÍêÈ«¡£
£¨1£©¡°Ëá½þ¡±¹ý³ÌÖв»ÄÜÓÃŨÑÎËáÌæ´úÁòËᣬÔÒòÊÇ_____________________¡£¡°Ëá½þ¡±Ê±¼ÓÈëÒ»¶¨Á¿µÄÁòËᣬÁòËá²»Äܹý¶à»ò¹ýÉÙ¡£¡°ÁòËᡱ¹ý¶àÔì³É°±µÄËðʧ£»ÁòËá¹ýÉÙʱ£¬¡°Ëá½þ¡±Ê±»áÓкìºÖÉ«Ôü³öÏÖ£¬ÔÒòÊÇ___________________________________¡£
£¨2£©¼ÓÈ백ˮӦµ÷½ÚpHµÄ·¶Î§Îª______ £¬µ±Fe3+Ç¡ºÃ³ÁµíÍêȫʱ£¬c(Al3+)=________mol¡¤L-l¡£ £¨3£©¡°¹ýÂË¡±ËùµÃÂËÔüΪMnCO3£¬ÂËÒºÖÐÈÜÖʵÄÖ÷Òª³É·ÖÊÇ____________£¨Ìѧʽ£©£¬Ð´³öÆäÑôÀë×ӵĵç×Óʽ£º_______________________________¡£ £¨4£©¼ÓÈë̼ËáÇâï§²úÉú³ÁµíµÄ¹ý³Ì³ÆÎª¡°³ÁÃÌ¡±¡£
¢Ù¡°³ÁÃÌ¡±¹ý³ÌÖзųöCO2£¬·´¸çÓ¦µÄÀë×Ó·½³ÌʽΪ_______________________________¡£
¢Ú¡°³ÁÃÌ¡±¹ý³ÌÖгÁÃÌËÙÂÊÓëζȵĹØÏµÈçͼËùʾ¡£µ±Î¶ȸßÓÚ60¡æÊ±£¬³ÁÃÌËÙÂÊËæ×ÅζÈÉý¸ß¶ø¼õÂýµÄÔÒò¿ÉÄÜÊÇ_______________________________¡£
21¡¢£¨12 ·Ö£©Ì¼Ì¼Ë«¼üÓÐÈçÏÂËùʾµÄ¶ÏÁÑ·½Ê½£º
£»
¡£
¸ß·Ö×Óµ¥ÌåA(C6H10O3)¿É½øÐÐÈçÏ·´Ó¦£¨·´Ó¦¿òͼ£©
ÒÑÖª£º
¢ñ.¶Ô¿òͼÖÐijЩ»¯ºÏÎïÐÔÖʵÄ˵Ã÷£ºA ÔÚÊÒÎÂϲ»ÓëNaHCO3ÈÜÒº·´Ó¦£¬µ«¿ÉÓëNa·´Ó¦·Å³öH2£»B¿ÉÓëNaHCO3ÈÜÒº·´Ó¦·Å³öCO2£»C¿ÉÓëNa·´Ó¦·Å³öH2 ¶øD²»ÄÜ£»GÔÚÊÒÎÂϼȲ»ÓëNaHCO3 ÈÜÒº·´Ó¦£¬Ò²²»ÓëNa ·´Ó¦·Å³öH2¡£ ¢ò.Á½¸öÒ»OHÁ¬ÔÚͬһ¸öCÔ×ÓÉϵĽṹ²»Îȶ¨¡£
£¨1£©Ð´³ö·´Ó¦¢ÙÖÐ(1)µÄ·´Ó¦ÀàÐÍ£º___________£»Ð´³öDµÄ¼üÏßʽ£º____________________¡£ £¨2£©Ð´³öÎïÖÊEËùº¬¹ÙÄÜÍŵÄÃû³Æ£º ____________________¡£
£¨3£©BÔÚŨH2SO4 ´æÔÚÏÂÓë¼×´¼¹²ÈÈ·´Ó¦Éú³ÉµÄÓлúÎïµÄϵͳÃû³ÆÎª____________________¡£ £¨4£©Ð´³öF ¡úGµÄ»¯Ñ§·´Ó¦·½³Ìʽ£º____________________________________¡£
£¨5£©ÓëB»¥ÎªÍ¬·ÖÒì¹¹Ì壬·Ö×ÓΪÁ´×´µÄõ¥ÀàÎïÖʹ²ÓÐ______________ÖÖ£¨²»¿¼ÂÇÁ¢ÌåÒì¹¹£©¡£ £¨6£©ÇëÉè¼ÆºÏÀí·½°¸£¬ÓÃÎïÖÊEÖÆ±¸ÒÒ¶þËá(ÆäËûÔÁÏ×ÔÑ¡£¬Ó÷´Ó¦Á÷³Ìͼ±íʾ£¬²¢×¢Ã÷±ØÒªµÄ·´Ó¦Ìõ¼þ)¡£ÀýÈ磺
¹²·ÖÏí92ƪÏà¹ØÎĵµ