ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿Ä³Ñо¿ÐÔѧϰС×éÇëÄã²ÎÓë¡°Ñо¿ÌúÓëË®·´Ó¦ËùµÃ¹ÌÌåÎïÖʵijɷ֡¢ÐÔÖʼ°ÔÙÀûÓá±ÊµÑé̽¾¿£¬²¢¹²Í¬½â´ðÏÂÁÐÎÊÌ⣺

̽¾¿Ò»£ºÉè¼ÆÈçͼËùʾװÖýøÐС°ÌúÓëË®·´Ó¦¡±µÄʵÑé

£¨1£©Ó²Öʲ£Á§¹ÜEÖз´Ó¦µÄ»¯Ñ§·½³ÌʽΪ_________

£¨2£©·´Ó¦Ç°AÖÐͶ·ÅËé´ÉƬµÄÄ¿µÄÊÇ_________

£¨3£©×°ÖÃEÖеÄÏÖÏóÊÇ_________

̽¾¿¶þ£ºÉè¼ÆÈçÏÂʵÑé·½°¸È·¶¨·´Ó¦ºóÓ²Öʲ£Á§¹ÜÖкÚÉ«¹ÌÌåµÄ³É·Ö¡£

ʵÑé²Ù×÷£º´ýÓ²Öʲ£Á§¹ÜBÀäÈ´ºó£¬È¡ÉÙÐíÆäÖеĹÌÌåÎïÖÊÈÜÓÚÏ¡ÁòËáºó£¬½«ËùµÃÈÜÒº·Ö³ÉÁ½·Ý¡£

£¨4£©Ò»·ÝµÎ¼Ó¼¸µÎKSCNÈÜÒº¡£ÈôÈÜÒº±äѪºìÉ«£¬ÍƶÏÓ²Öʲ£Á§¹ÜBÖйÌÌåÎïÖʵijɷÖΪ________£¨Ñ¡ÌîÐòºÅ£¬ÏÂͬ£©£»Ð´³öÈÜÒº±äѪºìÉ«µÄÀë×Ó·½³Ìʽ£º____________£¬ÈôÈÜҺδ±äѪºìÉ«£¬ÍƶÏÓ²Öʲ£Á§¹ÜBÖйÌÌåÎïÖʵijɷÖΪ_________¡£

¢ÙÒ»¶¨ÓÐFe3O4 ¢ÚÒ»¶¨ÓÐFe ¢ÛÖ»ÓÐFe3O4 ¢ÜÖ»ÓÐFe

£¨5£©ÁíÒ»·ÝÓÃ_________£¨ÌîÒÇÆ÷Ãû³Æ£©£¬¼ÓÈë_________£¨ÌîÊÔ¼Á£©£¬¹Û²ì¹Û²ìµ½_________£¨ÌîʵÑéÏÖÏ󣩣¬¿ÉÒÔÖ¤Ã÷ÈÜÒºÖдæÔÚFe2+¡£

¡¾´ð°¸¡¿H2+CuOCu+H2O ·ÀÖ¹±©·Ð ºÚÉ«¹ÌÌå±äºì£¬ÓҶ˹ܱÚÓÐË®Öé ¢Ù Fe3++3SCN-=Fe£¨SCN£©3 ¢Ú ½ºÍ·µÎ¹Ü ËáÐÔKMnO4ÈÜÒº ÈÜÒº×ÏÉ«ÍÊÈ¥

¡¾½âÎö¡¿

ʵÑéAÖÐÖÆÈ¡Ë®ÕôÆø£¬BÖÐFeºÍË®ÕôÆø·´Ó¦Éú³ÉËÄÑõ»¯ÈýÌúºÍÇâÆø£¬CÊÕ¼¯ÇâÆø¡¢DÓÃÓÚ¸ÉÔïÇâÆø£¬EÖÐÇâÆø»¹Ô­CuO£¬Ö¤Ã÷ÇâÆøµÄÉú³É£»

AÖÐÖÆÈ¡Ë®ÕôÆø£¬BÖÐFeºÍË®ÕôÆø·´Ó¦Éú³ÉËÄÑõ»¯ÈýÌúºÍÇâÆø£¬CÊÕ¼¯ÇâÆø¡¢DÓÃÓÚ¸ÉÔïÇâÆø£¬EÖÐÇâÆø»¹Ô­CuO£»

£¨1£©Ó²Öʲ£Á§¹ÜEÖз´Ó¦µÄ»¯Ñ§·½³ÌʽΪH2+CuOCu+H2O£»

´ð°¸£ºH2+CuOCu+H2O

£¨2£©Ëé´ÉƬÓзÀ±©·Ð×÷Ó㬷ÀÖ¹²úÉú°²È«Ê¹ʣ»

´ð°¸£º·ÀÖ¹±©·Ð£»

£¨3£©CuO³ÊºÚÉ«¡¢Cu³ÊºìÉ«£¬ÇâÆø»¹Ô­CuOµÃµ½Cu£¬ÇÒͬʱÉú³ÉË®£¬Ôò¿´µ½µÄÏÖÏóÊǺÚÉ«¹ÌÌå±äºì£¬ÓҶ˹ܱÚÓÐË®Ö飻
´ð°¸£ººÚÉ«¹ÌÌå±äºì£¬ÓҶ˹ܱÚÓÐË®Öé

£¨4£©ÌúÀë×ÓºÍKSCN·´Ó¦Éú³ÉÂçºÏÎï¶øʹÈÜÒº³ÊѪºìÉ«£¬ÑÇÌúÀë×ÓºÍKSCN²»·´Ó¦£¬ÏòÈÜÒºÖмÓÈëKSCNÈÜÒº³ÊºìÉ«£¬ËµÃ÷º¬ÓÐËÄÑõ»¯ÈýÌú£¬¿ÉÄܺ¬ÓÐFe£»ÌúÀë×ÓºÍFe·´Ó¦Éú³ÉÑÇÌúÀë×Ó£¬²»±äºìɫ˵Ã÷¹ÌÌåÖк¬ÓÐËÄÑõ»¯ÈýÌúºÍFe£»

´ð°¸£º ¢Ù Fe3++3SCN-=Fe£¨SCN£©3 ¢Ú

£¨5£©ÑÇÌúÀë×Ó¾ßÓл¹Ô­ÐÔ£¬Äܱ»ËáÐÔ¸ßÃÌËá¼ØÈÜÒºÑõ»¯£¬ËùÒÔ¿ÉÒÔÓÃËáÐÔ¸ßÃÌËá¼ØÈÜÒº¼ìÑéÑÇÌúÀë×Ó£»

´ð°¸£º½ºÍ·µÎ¹Ü ËáÐÔKMnO4ÈÜÒº ÈÜÒº×ÏÉ«ÍÊÈ¥

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿Ãº»¯¹¤ÊÇÒÔúΪԭÁÏ£¬¾­¹ý»¯Ñ§¼Ó¹¤Ê¹Ãº×ª»¯ÎªÆøÌå¡¢ÒºÌå¡¢¹ÌÌåȼÁÏÒÔ¼°¸÷ÖÖ»¯¹¤²úÆ·µÄ¹¤Òµ¹ý³Ì¡£

(1)½«Ë®ÕôÆøͨ¹ýºìÈȵÄÌ¿¼´¿É²úÉúˮúÆø¡£·´Ó¦Îª£º

C(s)+H2O(g)CO(g)+H2(g)¡¡¦¤H= +131.3kJ/mol£¬¦¤S= +133.7J/(K¡¤mol)¡£

¢Ù¸Ã·´Ó¦ÄÜ·ñ×Ô·¢½øÐÐÓë________________Óйء£

¢ÚÒ»¶¨Î¶ÈÏ£¬ÔÚÒ»¸öÈÝ»ý¿É±äµÄÃܱÕÈÝÆ÷ÖУ¬·¢ÉúÉÏÊö·´Ó¦£¬ÏÂÁÐÄÜÅжϸ÷´Ó¦´ïµ½»¯Ñ§Æ½ºâ״̬µÄÊÇ________(Ìî×Öĸ£¬ÏÂͬ)¡£

a£®ÈÝÆ÷ÖеÄѹǿ²»±ä

b£®1mol H-H¼ü¶ÏÁѵÄͬʱ¶ÏÁÑ2mol H-O¼ü

c£®vÕý(CO)£½vÄæ(H2O)

d£®c(CO)=c(H2)

(2)½«²»Í¬Á¿µÄCO(g)ºÍH2O(g)·Ö±ðͨÈëµ½Ìå»ýΪ2 LµÄºãÈÝÃܱÕÈÝÆ÷ÖУ¬½øÐз´Ó¦CO(g)+H2O(g)CO2(g)+H2(g)£¬µÃµ½ÈçÏÂÈý×éÊý¾Ý£º

ʵÑé×é

ζÈ/¡æ

ÆðʼÁ¿/mol

ƽºâÁ¿/mol

´ïµ½Æ½ºâËùÐèʱ¼ä/min

H2O

CO

H2

CO

1

650

2

4

1.6

2.4

5

2

900

1

2

0.4

1.6

3

3

900

a

b

c

d

t

¢ÙʵÑé1ÖÐÒÔv(CO2)±íʾµÄ·´Ó¦ËÙÂÊΪ__________¡£

¢Ú¸Ã·´Ó¦µÄÄ淴ӦΪ________(Ìî¡°Îü¡±»ò¡°·Å¡±)ÈÈ·´Ó¦¡£

¢ÛÈôʵÑé3Òª´ïµ½ÓëʵÑé2ÏàͬµÄƽºâ״̬(¼´¸÷ÎïÖʵÄÖÊÁ¿·ÖÊý·Ö±ðÏàµÈ)£¬ÇÒt<3min£¬Ôòa¡¢bÓ¦Âú×ãµÄ¹ØϵÊÇ____________________(Óú¬a¡¢bµÄÊýѧʽ±íʾ)¡£

(3)Ä¿Ç°¹¤ÒµÉÏÓÐÒ»ÖÖ·½·¨ÊÇÓÃCO2À´Éú²ú¼×´¼¡£Ò»¶¨Ìõ¼þÏ·¢Éú·´Ó¦£ºCO2(g)+3H2(g)CH3OH(g)+H2O(g)£¬Èçͼ±íʾ¸Ã·´Ó¦½øÐйý³ÌÖÐÄÜÁ¿(µ¥Î»ÎªkJ/mol)µÄ±ä»¯¡£ÔÚÌå»ýΪ1LµÄºãÈÝÃܱÕÈÝÆ÷ÖУ¬³äÈë1mol CO2ºÍ3mol H2£¬ÏÂÁдëÊ©ÖÐÄÜʹ c(CH3OH)Ôö´óµÄÊÇ___¡£

a£®Éý¸ßÎÂ¶È b£®³äÈëHe(g)£¬Ê¹ÌåϵѹǿÔö´ó

c£®½«H2O(g)´ÓÌåϵÖзÖÀë³öÀ´ ¡¡¡¡d£®ÔÙ³äÈë1mol CO2ºÍ3mol H2

¡¾ÌâÄ¿¡¿Ð¿ÔÚµç³ØÖÆÔì·½ÃæÓÐ×ÅÖØÒªµÄ×÷Óã¬Ò²ÊÇÈËÌå±ØÐèµÄ΢Á¿ÔªËØÖ®Ò»¡£ËüÓë³£¼ûµÄ·Ç½ðÊô¶¼¿ÉÒÔÐγÉÖØÒªµÄ»¯ºÏÎï¡£

£¨1£©ZnµÄºËÍâµç×ÓÅŲ¼Ê½ÊÇ[Ar]___¡£

£¨2£©ZnµÄ¸÷¼¶µçÀëÄÜÊý¾ÝÈçϱíËùʾ£º

¢ÙÇëÒÀ¾Ý±íÖÐÊý¾Ý˵Ã÷пµÄ³£¼û»¯ºÏ¼ÛΪ£«2µÄÔ­ÒòÊÇ___¡£

¢ÚÑõ¡¢Áò¡¢ÂÈÈýÖÖ³£¼û·Ç½ðÊôµÄµç¸ºÐÔ£¬ÓÉ´óµ½Ð¡µÄ˳ÐòÊÇ___¡£¶þÂÈ»¯ÁòΪÏʺìÉ«ÒºÌ壬ÈÛµã-78¡æ£¬ÔòÆ侧ÌåÖÐ΢Á£¼äµÄ×÷ÓÃÁ¦Ó¦ÊôÓÚ___£¬ËüµÄ·Ö×ÓÖÐÖÐÐÄÔ­×Ó¾ßÓеŵç×Ó¶ÔÊýÊÇ___¡£

£¨3£©ÂÈ»¯Ð¿Ò×ÈÜÓÚË®£¬ÔÚË®ÖÐÐγÉÅäºÏÎïH[ZnCl2(OH)]£¬H[ZnCl2(OH)]ÔÚË®ÖеçÀëʱµÄÀë×Ó·½³ÌʽΪ___¡£

£¨4£©¾§°ûµÄ¿Õ϶ÎÊÌâÊǾ§°ûÑо¿µÄÖØÒªÄÚÈÝ¡£

¢ÙÒÑÖªÃæÐÄÁ¢·½¾§°ûµÄËÄÃæÌå¿Õ϶ºÍ°ËÃæÌå¿Õ϶״¿öÈçͼËùʾ¡£ÃæÐÄÁ¢·½¾§°ûµÄÿ¸ö¾§°ûÖУ¬¶Ñ»ýÇòÊý£ºËÄÃæÌå¿Õ϶Êý£º°ËÃæÌå¿Õ϶Êý=___¡£

¢Ú¸ù¾ÝÁ¢·½ZnS¾§°ûʾÒâͼ£¬ÃèÊöÔÚÿ¸ö¾§°ûÖУ¬Ð¿Àë×ÓÌî³äÔÚÁòÀë×ÓµÄÁ¢·½ÃæÐľ§°û¿Õ϶Öеķ½Ê½Îª___¡£Àë×ø±ê²ÎÊýΪ(0£¬0£¬0)µÄÁòÀë×Ó×î½üµÄпÀë×Ó×ø±ê²ÎÊýΪ___ (²ÎÊýÊýÖµÏÞ¶¨ÎªÕýÖµ)¡£

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

¾«Ó¢¼Ò½ÌÍø