ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿ÇâÆøÊÇÒ»ÖÖÇå½àÄÜÔ´£¬ÇâÆøµÄÖÆÈ¡Óë´¢´æÊÇÇâÄÜÔ´ÀûÓÃÁìÓòµÄÑо¿Èȵ㡣

(1)ijͬѧÔÚÓÃÏ¡ÁòËáÓëпÖÆÈ¡ÇâÆøµÄʵÑéÖУ¬·¢ÏÖ¼ÓÈëÉÙÁ¿CuSO4ÈÜÒº¿É¼Ó¿ìÇâÆø µÄÉú³ÉËÙÂÊ¡£µ«µ±¼ÓÈëµÄCuSO4ÈÜÒº³¬¹ýÒ»¶¨Á¿Ê±£¬Éú³ÉÇâÆøµÄËÙÂÊ·´¶ø»áϽµ¡£ Çë·ÖÎöÇâÆøÉú³ÉËÙÂÊϽµµÄÖ÷ÒªÔ­Òò___________________________¡£

(2)Óü×ÍéÖÆÈ¡ÇâÆøµÄÁ½²½·´Ó¦µÄÄÜÁ¿±ä»¯ÈçÏÂͼËùʾ£¬Ôò¼×ÍéºÍË®ÕôÆø·´ Ó¦Éú³É¶þ Ñõ»¯Ì¼ºÍÇâÆøµÄÈÈ»¯Ñ§·½³ÌʽÊÇ___________________________¡£

(3)¿Æѧ¼Ò×î½üÑо¿³öÒ»ÖÖ»·±££¬°²È«µÄ´¢Çâ·½·¨£¬ÆäÔ­Àí¿É±íʾΪ£ºNaHCO3 H2HCOONa H2O¡£ÏÂÁÐÓйØ˵·¨ÕýÈ·µÄÊÇ_____________¡£

A£®´¢Çâ¡¢ÊÍÇâ¹ý³Ì¾ùÎÞÄÜÁ¿±ä»¯

B£®NaHCO3¡¢HCOONa¾ù¾ßÓÐÀë×Ó¼üºÍ¹²¼Û¼ü

C£®´¢Çâ¹ý³ÌÖУ¬NaHCO3±»Ñõ»¯

D£®ÊÍÇâ¹ý³ÌÖУ¬Ã¿ÏûºÄ0.1 mol H2O·Å³ö2.24 LµÄH2

(4)Mg2CuÊÇÒ»ÖÖ´¢ÇâºÏ½ð¡£350 ¡æʱ£¬Mg2CuÓëH2·´Ó¦£¬Éú³ÉMgCu2 ºÍ½öº¬Ò»ÖÖ ½ðÊôÔªËصÄÇ⻯Îï(¸ÃÇ⻯ÎïÖÐÇâµÄÖÊÁ¿·ÖÊýΪ0.077)¡£Mg2CuÓëH2·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ______________¡£

¡¾´ð°¸¡¿µ±¼ÓÈëÒ»¶¨Á¿µÄCuSO4ºó£¬Éú³ÉµÄµ¥ÖÊCu»á³Á»ýÔÚZnµÄ±íÃ棬½µµÍÁËZnÓëÈÜ ÒºµÄ½Ó´¥Ãæ»ý CH4(g) + 2H2O(g) £½ 4H2(g) + CO2(g) ¦¤H£½£­136.5 kJ/mol B 2Mg2Cu£«3H2MgCu2£«3MgH2

¡¾½âÎö¡¿

£¨1£©Éú³ÉµÄµ¥ÖÊÍ­»á³Á»ýÔÚпµÄ±íÃ棬½µµÍÁËпÓëÈÜÒºµÄ½Ó´¥Ãæ»ý£»

£¨2£©¸ù¾ÝµÚÒ»²½·´Ó¦¹ý³Ì¿ÉÒԵóö£º¢ÙCH4£¨g£©+H2O£¨g£©=3H2£¨g£©+CO£¨g£©£¬¡÷H=-103.3KJ/mol£¬¸ù¾ÝµÚ¶þ²½·´Ó¦¹ý³Ì¿ÉÒԵóö£º¢ÚCO£¨g£©+H2O£¨g£©=H2£¨g£©+CO2£¨g£©£¬¡÷H=-33.2KJ/mol£¬¸ù¾Ý¸Ç˹¶¨Âɽ«¢Ù+¢Ú¿ÉµÃ¼×ÍéºÍË®ÕôÆø·´Ó¦Éú³É¶þÑõ»¯Ì¼ºÍÇâÆøµÄ·´Ó¦£¬¾Ý´ËÊéдÈÈ»¯Ñ§·½³Ìʽ£»

£¨3£©ÓÉ·½³ÌʽNaHCO3+H2HCOONa+H2O¿ÉÖª£¬·´Ó¦ÖÐ̼ԪËØ»¯ºÏ¼Û½µµÍ±»»¹Ô­£¬ÇâÔªËØ»¯ºÏ¼ÛÉý¸ß±»Ñõ»¯£»

£¨4£©ÏÈÒÀ¾ÝÇ⻯ÎïµÄÖÊÁ¿·ÖÊý¼ÆËãÇ⻯Îﻯѧʽ£¬ÔÚÒÀ¾Ý·´Ó¦ÎïºÍÉúÎïд³ö»¯Ñ§·½³Ìʽ¡£

£¨1£©ÒòΪп»áÏÈÓëÁòËáÍ­·´Ó¦£¬Ö±ÖÁÁòËáÍ­·´Ó¦Íê²ÅÓëÁòËá·´Ó¦Éú³ÉÇâÆø£¬ÁòËáÍ­Á¿½Ï¶àʱ£¬·´Ó¦Ê±¼ä½Ï³¤£¬¶øÇÒÉú³ÉµÄÍ­»á¸½×ÅÔÚпƬÉÏ£¬»á×谭пƬÓëÁòËá¼ÌÐø·´Ó¦£¬ÇâÆøÉú³ÉËÙÂÊϽµ£¬¹Ê´ð°¸Îª£ºµ±¼ÓÈëÒ»¶¨Á¿µÄÁòËáÍ­ºó£¬Éú³ÉµÄµ¥ÖÊÍ­»á³Á»ýÔÚпµÄ±íÃ棬½µµÍÁËпÓëÈÜÒºµÄ½Ó´¥Ãæ»ý£»

£¨2£©¸ù¾ÝµÚÒ»²½·´Ó¦¹ý³Ì¿ÉÒԵóö£ºCH4£¨g£©+H2O£¨g£©=3H2£¨g£©+CO£¨g£©£¬¡÷H=-103.3KJ/mol£»¸ù¾ÝµÚ¶þ²½·´Ó¦¹ý³Ì¿ÉÒԵóö£ºCO£¨g£©+H2O£¨g£©=H2£¨g£©+CO2£¨g£©£¬¡÷H=-33.2KJ/mol£»¸ù¾Ý¸Ç˹¶¨ÂÉ£¬ÉÏÏÂÁ½Ê½Ïà¼Ó¿ÉµÃ£ºCH4£¨g£©+2H2O£¨g£©=4H2£¨g£©+CO2£¨g£©¡÷H=-136.5kJ/mol£¬¹Ê´ð°¸Îª£ºCH4£¨g£©+2H2O£¨g£©=4H2£¨g£©+CO2£¨g£©¡÷H=-136.5kJ/mol£»

£¨3£©A¡¢´¢Çâ¡¢ÊÍÇâ¹ý³Ì¶¼ÊÇ»¯Ñ§±ä»¯£¬»¯Ñ§±ä»¯ÖÐÒ»¶¨°éËæÄÜÁ¿±ä»¯£¬¹Ê´íÎó£»

B¡¢NaHCO3¡¢HCOONa¾ùÊÇÀë×ÓÀë×Ó»¯ºÏÎ»¯ºÏÎïÖк¬ÓÐÀë×Ó¼üºÍ¹²¼Û¼ü£¬¹ÊÕýÈ·£»

C¡¢´¢Çâ¹ý³ÌÖÐCÔªËصĻ¯ºÏ¼ÛÓÉ+4½µµÍΪ+2£¬NaHCO3±»»¹Ô­£¬¹Ê´íÎó£»

D¡¢ÊÍÇâ¹ý³ÌÖУ¬Ã¿ÏûºÄ0.1molH2O·Å³ö0.1molH2£¬µ«Ã»ËµÃ÷״̬£¬ÎÞ·¨ÖªµÀÆäÌå»ý£¬¹Ê´íÎó£»

¹ÊÑ¡B£¬¹Ê´ð°¸Îª£ºB£»

£¨4£©Áî½ðÊôÇ⻯ÎïΪRHx£¬½ðÊôRµÄÏà¶Ô·Ö×ÓÖÊÁ¿Îªa£¬Ôòx/(a+x)=0.077£¬¼´923x=77a£¬XΪ½ðÊôµÄ»¯ºÏ¼Û£¬ÌÖÂۿɵÃx=2£¬a=24£¬¹Ê¸Ã½ðÊôÇ⻯ÎïΪMgH2,Ôò·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ2Mg2Cu£«3H2MgCu2£«3MgH2£¬¹Ê´ð°¸Îª£º2Mg2Cu£«3H2MgCu2£«3MgH2¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿ËÄÂÈ»¯Îý¿ÉÓÃ×÷ýȾ¼Á¡£ÀûÓÃÈçͼËùʾװÖÿÉÒÔÖƱ¸ËÄÂÈ»¯Îý£¨²¿·Ö¼Ð³Ö×°ÖÃÒÑÂÔÈ¥£©£»

ÓйØÐÅÏ¢ÈçÏÂ±í£º

»¯Ñ§Ê½

SnCl2

SnCl4

ÈÛµã/¡æ

246

33

·Ðµã/¡æ

652

144

ÆäËûÐÔÖÊ

ÎÞÉ«¾§Ì壬Ò×Ñõ»¯

ÎÞÉ«ÒºÌ壬Ò×Ë®½â

»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©¼××°ÖÃÖÐÒÇÆ÷AµÄÃû³ÆΪ_______¡£

£¨2£©Óü××°ÖÃÖÆÂÈÆø£¬MnO4 ±»»¹Ô­ÎªMn2+£¬¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽΪ________¡£

£¨3£©½«×°ÖÃÈçͼÁ¬½ÓºÃ£¬¼ì²éÆøÃÜÐÔ£¬ÂýÂýµÎÈëŨÑÎËᣬ´ý¹Û²ìµ½_______£¨ÌîÏÖÏ󣩺󣬿ªÊ¼¼ÓÈȶ¡×°Öã¬ÎýÈÛ»¯ºóÊʵ±Ôö´óÂÈÆøÁ÷Á¿£¬¼ÌÐø¼ÓÈȶ¡×°Ö㬴Ëʱ¼ÌÐø¼ÓÈȶ¡×°ÖõÄÄ¿µÄÊÇ£º

¢Ù´Ù½øÂÈÆøÓëÎý·´Ó¦£»¢Ú__________¡£

£¨4£©Èç¹ûȱÉÙÒÒ×°Ö㬶¡×°ÖÃÖпÉÄܲúÉúSnCl2ÔÓÖÊ£¬²úÉúSnCl2ÔÓÖʵĻ¯Ñ§·½³ÌʽΪ______£¬ÒÔÏÂÊÔ¼ÁÖпÉÓÃÓÚ¼ì²âÊÇ·ñ²úÉúSnCl2µÄÓÐ_____

A£®H2O2ÈÜÒº B£®FeCl3ÈÜÒº£¨µÎÓÐKSCN£© C£®AgNO3ÈÜÒº D£®äåË®

£¨5£©¼º×°ÖõÄ×÷ÓÃÊÇ_____¡£

A£®³ýȥδ·´Ó¦µÄÂÈÆø£¬·ÀÖ¹ÎÛȾ¿ÕÆø

B£®·ÀÖ¹¿ÕÆøÖÐCO2ÆøÌå½øÈëÎì×°ÖÃ

C£®·ÀֹˮÕôÆø½øÈëÎì×°ÖõÄÊÔ¹ÜÖÐʹ²úÎïË®½â

D£®·ÀÖ¹¿ÕÆøÖÐO2½øÈëÎì×°ÖõÄÊÔ¹ÜÖÐʹ²úÎïÑõ»¯

£¨6£©·´Ó¦ÖÐÓÃÈ¥ÎýÁ£1.19g£¬·´Ó¦ºóÔÚÎì×°ÖõÄÊÔ¹ÜÖÐÊÕ¼¯µ½2.35gSnCl4£¬ÔòSnCl4µÄ²úÂÊΪ____£¨±£Áô3λÓÐЧÊý×Ö£©¡£

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

¾«Ó¢¼Ò½ÌÍø