ÌâÄ¿ÄÚÈÝ

ÊÔÍê³ÉÏÂÁÐÁ½Ð¡Ì⣺

£¨1£©»¯Ñ§ÊµÑé±ØÐë×¢Òⰲȫ£¬ÏÂÁÐ×ö·¨´æÔÚ°²È«Òþ»¼µÄÊÇ___________£¨Ñ¡Ìî×Öĸ£©¡£

A.ÇâÆø»¹Ô­Ñõ»¯Í­ÊµÑéÖУ¬ÏȼÓÈÈÑõ»¯Í­ºóͨÇâÆø

B.ÕôÁóʯÓÍʱ£¬¼ÓÈÈÒ»¶Îʱ¼äºó·¢ÏÖδ¼ÓËé´ÉƬ£¬Á¢¿Ì°Î¿ªÏðƤÈû²¢Í¶ÈëËé´ÉƬ

C.ʵÑéÊÒ×öÄƵÄʵÑéʱ£¬ÓàϵÄÄÆмͶÈëµ½·ÏÒº¸×ÖÐ

D.ÅäÖÆŨÁòËáÓë¾Æ¾«»ìºÏҺʱ£¬½«1Ìå»ýµÄ¾Æ¾«µ¹Èë3Ìå»ýµÄŨÁòËáÖÐ

E.Ƥ·ôÉÏÕ´ÓÐÉÙÁ¿Å¨ÏõËáʱ£¬Á¢¿ÌÓôóÁ¿Ë®³åÏ´£¬ÔÙÍ¿ÉÏϡ̼ËáÇâÄÆÈÜÒº

F.ÖÆÒÒϩʱ£¬ÓÃÁ¿³ÌΪ300 ¡æµÄζȼƴúÌæÁ¿³ÌΪ200 ¡æµÄζȼƣ¬²â·´Ó¦ÒºµÄζÈ

£¨2£©¼ä½ÓµâÁ¿·¨²â¶¨µ¨·¯ÖÐÍ­º¬Á¿µÄÔ­ÀíºÍ·½·¨ÈçÏ£º

ÒÑÖª£ºÔÚÈõËáÐÔÌõ¼þÏ£¬µ¨·¯ÖÐCu2+ÓëI-×÷Óö¨Á¿Îö³öI2£¬I2ÈÜÓÚ¹ýÁ¿µÄKIÈÜÒºÖУº

I2+I-====£¬ÓÖÖªÑõ»¯ÐÔ£ºFe3+£¾Cu2+£¾I2£¾¡£

Îö³öµÄI2¿ÉÓÃc mol¡¤L-1 Na2S2O3±ê×¼ÈÜÒºµÎ¶¨£º2+====+3I-¡£

׼ȷ³ÆÈ¡a gµ¨·¯ÊÔÑù£¬ÖÃÓÚ250 mLµâÁ¿Æ¿£¨´øÄ¥¿ÚÈûµÄ׶ÐÎÆ¿£©ÖУ¬¼Ó50 mLÕôÁóË®¡¢5 mL 3 mol¡¤L-1 H2SO4ÈÜÒº£¬¼ÓÉÙÁ¿NaF£¬ÔÙ¼ÓÈë×ãÁ¿µÄ10% KIÈÜÒº£¬Ò¡ÔÈ¡£¸ÇÉϵâÁ¿Æ¿Æ¿¸Ç£¬ÖÃÓÚ°µ´¦5 min£¬³ä·Ö·´Ó¦ºó£¬¼ÓÈë1¡ª2 mL 05%µÄµí·ÛÈÜÒº£¬ÓÃNa2S2O3±ê×¼ÈÜÒºµÎ¶¨µ½À¶É«ÍÊȥʱ£¬¹²ÓÃÈ¥V mL±ê×¼Òº¡£

¢ÙʵÑéÖУ¬ÔÚ¼ÓKIÇ°Ðè¼ÓÈëÉÙÁ¿NaF£¬ÍƲâÆä×÷ÓÿÉÄÜÊÇ£º______________________¡£

¢ÚʵÑéÖмÓÈë5 mL 3 mol¡¤L-1 H2SO4ÈÜÒº£¬ÄãÈÏΪÁòËáµÄ×÷ÓÃÊÇ£º_______________________¡£

¢Û±¾ÊµÑéÖÐÓõâÁ¿Æ¿¶ø²»ÓÃÆÕͨ׶ÐÎÆ¿ÊÇÒòΪ£º_____________________________________¡£

¢ÜÁòËáÍ­ÈÜÒºÓëµâ»¯¼ØÈÜÒº·´Ó¦Éú³É°×É«³Áµí£¨µâ»¯ÑÇÍ­£©²¢Îö³öµâ£¬¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽΪ£º____________________________________________________________________¡£

¢Ý¸ù¾Ý±¾´ÎʵÑé½á¹û£¬¸Ãµ¨·¯ÊÔÑùÖÐÍ­ÔªËصÄÖÊÁ¿·ÖÊýΪw(Cu)=______________¡£

£¨1£©ABCD

£¨2£©¢ÙÑÚ±ÎFe3+£¬·ÀÖ¹Ôì³ÉÆ«´óµÄÎó²î  ¢ÚÌṩÈõËáÐÔ»·¾³Ìõ¼þ²¢·ÀֹͭÀë×ÓË®½â  ¢Û·ÀÖ¹¿ÕÆøÖеÄÑõÆøÓëµâ»¯¼Ø·´Ó¦£»·ÀÖ¹Éú³ÉµÄµâÉý»ª  ¢Ü2Cu2++4I-====2CuI¡ý+I2  ¢Ý¡Á100%

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
ÊÔÍê³ÉÏÂÁÐÁ½Ð¡Ì⣺

£¨1£©C1»¯Ñ§ÊÇÖ¸´ÓÒ»¸ö̼ԭ×ӵĻ¯ºÏÎÈçCH4£¬CO£¬CO2£¬CH3OH£¬HCHOµÈ£©³ö·¢ºÏ³É¸÷ÖÖ»¯Ñ§Æ·µÄ¼¼Êõ¡£´Óú¡¢ÌìÈ»ÆøÖƺϳÉÆøÔÙ½øÒ»²½ÖƱ¸¸÷ÖÖ»¯¹¤²úÆ·ºÍ½à¾»È¼ÁÏ£¬ÒѳÉΪµ±½ñ»¯Ñ§¹¤Òµ·¢Õ¹µÄ±ØÈ»Ç÷ÊÆ¡£ÆäÖм״¼ÊÇC1»¯Ñ§µÄ»ù´¡¡£

¢ÙCOÓëH2°´Ò»¶¨±ÈÀý¿ÉÉú³ÉÒÒ¶þ´¼£¬Ôòn(CO)/n(H2)=_____________£¨ÌîÊý×Ö£©¡£

¢ÚÈôÆûÓÍƽ¾ù×é³ÉÓÃCmHn±íʾ£¬ÔòºÏ³ÉÆûÓÍ¿ØÖÆn(CO)/n(H2)=(ÓÃm¡¢n±íʾ)¡£

¢Û¼×´¼ÔÚÒ»¶¨Ìõ¼þÏÂÓëCO¡¢H2×÷ÓÃÉú³ÉÓлúÎïA£¬A·¢Éú¼Ó¾Û¿ÉÉú³É¸ß·Ö×Óд³öÉú³ÉAµÄ»¯Ñ§·½³Ìʽ£º_________________________________¡£

£¨2£©ÒÑÖª´¼È©ÔÚÒ»¶¨Ìõ¼þÏÂÄÜ·¢ÉúËõºÏ·´Ó¦£¬Ê¾ÀýÈçÏ£º

ÒÑÖª£º

¢Ù1827ÄêÈËÃǾͷ¢ÏÖÓлúÎïA£¬ËüµÄ·Ö×ÓʽΪC13H18O7£¬ÓëÒ»·Ö×ÓË®×÷Óã¬Ë®½âÉú³ÉBºÍC¡£

¢ÚBÄÜ·¢ÉúÒø¾µ·´Ó¦£¬BÒ²¿ÉÓɵí·ÛË®½âµÃµ½£¬BµÄ·Ö×ÓʽΪC6H12O6¡£

¢ÛCÓöÂÈ»¯ÌúÈÜÒºÄÜ·¢ÉúÏÔÉ«·´Ó¦£¬1 mol CÓë×ãÁ¿ÄÆ·´Ó¦¿É²úÉú1 mol H2¡£

¢ÜCÔÚÊʵ±µÄÌõ¼þÏÂÓÃÊʵ±Ñõ»¯¼ÁÑõ»¯£¬¿ÉµÃD£¬DµÄ·Ö×ÓʽΪC7H6O3£¬Ïà¶Ô·Ö×ÓÖÊÁ¿D±ÈC´ó14¡£

¢ÝDÓÐÁ½¸öÈ¡´ú»ù£¬µ«²»ÊǼä룬ËüÓëBr2ÔÚ´ß»¯¼Á×÷ÓÃÏ·¢ÉúÒ»äåÈ¡´ú£¬²úÎïÓÐËÄÖÖ£¬DÄÜÓë̼ËáÇâÄÆÈÜÒº·´Ó¦¡£

¢ÞDÓëÒÒËáôû¡²(CH3CO)2O¡³·´Ó¦£¬¿ÉµÃ³£¼ûÒ©ÎïEºÍÒÒËᣬEÄÜÓë̼ËáÇâÄÆ·´Ó¦·Å³ö¶þÑõ»¯Ì¼¡£

ÊԻشðÏÂÁÐÎÊÌ⣺

¢Ùд³ö½á¹¹¼òʽ£ºC_________________£¬E_________________¡£

¢Úд³öÓëD»¥ÎªÍ¬·ÖÒì¹¹Ì壬º¬Óб½»·ÇÒº¬ÓÐõ¥½á¹¹µÄ½á¹¹¼òʽ£º_________________£¨Ö»ÐèдһÖÖ£©¡£

¢ÛBͨ³£ÒÔÁùÔª»·×´½á¹¹´æÔÚ£¬Ð´³öBµÄ»·×´½á¹¹¼òʽ£º_________________________¡£

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

¾«Ó¢¼Ò½ÌÍø