ÁòÓжàÖÖ»¯ºÏÎÐí¶àº¬Áò»¯ºÏÎïÔÚ¹¤ÒµÉÏÓÐÖØÒªµÄÓÃ;£®

1£®¹¤ÒµÉÏ¿ÉÓûÆÌú¿ó¡¢½¹Ì¿ÔÚÓÐÏ޵ĿÕÆøÖÐȼÉÕÖƱ¸Áò»Ç£®

3FeS2£«12C£«8O2Fe3O4£«nA¡ü£«6S

AÊÇ________(д»¯Ñ§Ê½)£®ÈôµÃµ½192¿ËµÄÁò»Ç£¬Ôò²úÉú±ê×¼×´¿öϵÄAÆøÌå________L£®

2£®ÁòµÄÂÈ»¯Îï³£×öÏ𽺹¤ÒµµÄÁò»¯¼Á£®ÁòÓëÂÈÆøÔÚÒ»¶¨Ìõ¼þÏ·´Ó¦£¬µÃµ½Á½ÖÖÁòµÄÂÈ»¯ÎïBºÍD£®BÎïÖÊÖеĺ¬ÁòÁ¿Îª0.3107£¬DÎïÖÊÖеĺ¬ÂÈÁ¿Îª0.5259£¬BµÄÏà¶Ô·Ö×ÓÖÊÁ¿±ÈDС32£®¼ÆËãÈ·¶¨ÕâÁ½ÖÖÂÈ»¯ÎïµÄ·Ö×Óʽ·Ö±ðΪ________£®

3£®Áò»¯ÇâÓж¾£®ÔÚ120¡æ¡¢101 kPa£¬½«H2SºÍO2ÔÚÃܱÕÈÝÆ÷Öеãȼ£¬³ä·Ö·´Ó¦ºóÓÖ»Ö¸´µ½ÁËÔ­À´µÄζȺÍѹǿʱ£¬ÆøÌåÌå»ý¼õÉÙ30£¥£¬ÇóÔ­»ìºÏÆäÆøÌåÖÐH2SµÄÌå»ý·ÖÊý£®Ð´³öÍƵ¼¹ý³Ì£®(²»¿¼ÂÇÁò»¯ÇâµÄ·Ö½â)

4£®Áò´úÁòËáÄÆÊÇÖØÒªµÄ»¹Ô­¼Á£¬¿ÉÓÃÑÇÁòËáÄƺÍÁò·ÛÔÚË®ÈÜÒºÖмÓÈÈÖƵã®È¡15.12 g¡¡Na2SO3ÈÜÓÚ80.0 mLË®ÖУ¬¼ÓÈë5.00 gÁò·Û£¬ÓÃС»ð¼ÓÈÈÖÁ΢·Ð£¬·´Ó¦Ô¼1Сʱºó¹ýÂË£¬½«ÂËÒºÕô·¢ÖÁÌå»ýΪ30.0 mL£¬ÔÙÀäÈ´µ½10¡æ£¬ÔòÀíÂÛÉÏÎö³öNa2S2O3?5H2O¶àÉÙ¿Ë£¿Ð´³öÍƵ¼¹ý³Ì£®(ÒÑÖª£ºNa2S2O3µÄÈܽâ¶È£¬10¡æʱΪ60.0 g/100 gË®£¬100¡æʱΪ207 g/100 gË®£®100¡æʱ£¬Na2S2O3±¥ºÍÈÜÒºµÄÃܶÈΪ1.14 g/mL)£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â

¢öA×åµÄÑõ¡¢Áò¡¢Îø£¨Se£©¡¢íÚ£¨Te£©µÈÔªËØÔÚ»¯ºÏÎïÖг£±íÏÖ³ö¶àÖÖÑõ»¯Ì¬£¬º¬¢öA×åÔªËصĻ¯ºÏÎïÔÚÑо¿ºÍÉú²úÖÐÓÐÐí¶àÖØÒªÓÃ;£®Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©Sµ¥Öʵij£¼ûÐÎʽÊÇS8£¬Æä»·×´½á¹¹Èçͼ1Ëùʾ£¬SÔ­×Ó²ÉÓõĹìµÀÔÓ»¯·½Ê½ÊÇ
sp3
sp3
£»
£¨2£©Ô­×ӵĵÚÒ»µçÀëÄÜÊÇÖ¸Æø̬µçÖÐÐÔ»ù̬ԭ×Óʧȥһ¸öµç×Óת»¯ÎªÆø̬»ù̬ÕýÀë×ÓËùÐèÒªµÄ×îµÍÄÜÁ¿£¬O¡¢S¡¢SeÔ­×ӵĵÚÒ»µçÀëÄÜÓÉ´óµ½Ð¡µÄ˳ÐòΪ
O£¾S£¾Se
O£¾S£¾Se
£»
£¨3£©SeµÄÔ­×ÓÐòÊýΪ
34
34
£¬ÆäºËÍâM²ãµç×ÓµÄÅŲ¼Ê½Îª
3s23p63d10
3s23p63d10
£»
£¨4£©H2SeµÄËáÐԱȠH2S
Ç¿
Ç¿
£¨Ìî¡°Ç¿¡±»ò¡°Èõ¡±£©£®Æø̬SeO3·Ö×ÓµÄÁ¢Ìå¹¹ÐÍΪ
ƽÃæÈý½ÇÐÎ
ƽÃæÈý½ÇÐÎ
£¬
SO
2-
3
Àë×ÓµÄÁ¢Ìå¹¹ÐÍΪ
Èý½Ç׶ÐÎ
Èý½Ç׶ÐÎ
£»
£¨5£©H2SeO3 µÄK1ºÍK2·Ö±ðÊÇ2.7¡Á10-3ºÍ2.5¡Á10-8£¬H2SeO4µÄµÚÒ»²½¼¸ºõÍêÈ«µçÀ룬K2ÊÇ1.2¡Á10-2£¬Çë¸ù¾Ý½á¹¹ÓëÐÔÖʵĹØϵ½âÊÍ£º
¢ÙH2SeO3ºÍH2SeO4µÄµÚÒ»²½µçÀë³Ì¶È´óÓÚµÚ¶þ²½µçÀëµÄÔ­Òò£º
µÚÒ»²½µçÀëºóÉú³ÉµÄ¸ºÀë×Ó½ÏÄÑÔÙ½øÒ»²½µçÀë³ö´øÕýµçºÉµÄÇâÀë×Ó
µÚÒ»²½µçÀëºóÉú³ÉµÄ¸ºÀë×Ó½ÏÄÑÔÙ½øÒ»²½µçÀë³ö´øÕýµçºÉµÄÇâÀë×Ó
£»
¢ÚH2SeO4±ÈH2SeO3ËáÐÔÇ¿µÄÔ­Òò£º
H2SeO3ºÍH2SeO4¿É±íʾΪ£¨HO£©SeOºÍ£¨HO£©SeO2£®H2SeO3ÖÐSeΪ+4¼Û£¬¶øH2SeO4ÖÐSeΪ+6¼Û£¬ÕýµçÐÔ¸ü¸ß£®µ¼ÖÂSe-O-HÖеÄOÔ­×Ó¸üÏòSeÆ«ÒÆ£¬Ô½Ò×µçÀë³öH+
H2SeO3ºÍH2SeO4¿É±íʾΪ£¨HO£©SeOºÍ£¨HO£©SeO2£®H2SeO3ÖÐSeΪ+4¼Û£¬¶øH2SeO4ÖÐSeΪ+6¼Û£¬ÕýµçÐÔ¸ü¸ß£®µ¼ÖÂSe-O-HÖеÄOÔ­×Ó¸üÏòSeÆ«ÒÆ£¬Ô½Ò×µçÀë³öH+
£®
£¨6£©ZnSÔÚÓ«¹âÌå¡¢¹âµ¼Ìå²ÄÁÏ¡¢Í¿ÁÏ¡¢ÑÕÁϵÈÐÐÒµÖÐÓ¦Óù㷺£®Á¢·½ZnS¾§Ìå½á¹¹Èçͼ2Ëùʾ£¬Æ侧°û±ß³¤Îª540.0pm£¬ÃܶÈΪ
4¡Á(65+32)g?mol-1
6.02¡Á1023mol-1
(540.0¡Á10-10cm)3
=4.1
4¡Á(65+32)g?mol-1
6.02¡Á1023mol-1
(540.0¡Á10-10cm)3
=4.1
g?cm-3£¨ÁÐʽ²¢¼ÆË㣩£¬aλÖÃS2-Àë×ÓÓëbλÖÃZn2+Àë×ÓÖ®¼äµÄ¾àÀëΪ
270.0
1-cos109¡ã28¡ä
»ò
135.0¡Á
2
sin
109¡ã28¡ä
2
»ò135
3
270.0
1-cos109¡ã28¡ä
»ò
135.0¡Á
2
sin
109¡ã28¡ä
2
»ò135
3
pm£¨ÁÐʽ±íʾ£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÁòÓжàÖÖ»¯ºÏÎÐí¶àº¬Áò»¯ºÏÎïÔÚ¹¤ÒµÉÏÓÐÖØÒªµÄÓÃ;£®
£¨1£©¹¤ÒµÉÏ¿ÉÓûÆÌú¿ó¡¢½¹Ì¿ÔÚÓÐÏ޵ĿÕÆøÖÐȼÉÕÖƱ¸Áò»Ç£®
3FeS2+12C+8O2
¡÷
 Fe3O4+nA¡ü+6S
AÊÇ
CO
CO
£¨Ð´»¯Ñ§Ê½£©£®ÈôµÃµ½192¿ËµÄÁò»Ç£¬Ôò²úÉú±ê×¼×´¿öϵÄAÆøÌå
268.8
268.8
L£®
£¨2£©ÁòµÄÂÈ»¯Îï³£×öÏ𽺹¤ÒµµÄÁò»¯¼Á£®ÁòÓëÂÈÆøÔÚÒ»¶¨Ìõ¼þÏ·´Ó¦£¬µÃµ½Á½ÖÖÁòµÄÂÈ»¯ÎïBºÍD£®BÎïÖÊÖеĺ¬ÁòÁ¿Îª0.3107£¬DÎïÖÊÖеĺ¬ÂÈÁ¿Îª0.5259£¬BµÄÏà¶Ô·Ö×ÓÖÊÁ¿±ÈDС32£®¼ÆËãÈ·¶¨ÕâÁ½ÖÖÂÈ»¯ÎïµÄ·Ö×Óʽ·Ö±ðΪ
SCl2¡¢S2Cl2
SCl2¡¢S2Cl2
£®
£¨3£©Áò»¯ÇâÓж¾£®ÔÚ120¡æ¡¢101kPa£¬½«H2SºÍO2ÔÚÃܱÕÈÝÆ÷Öеãȼ£¬³ä·Ö·´Ó¦ºóÓÖ»Ö¸´µ½ÁËÔ­À´µÄζȺÍѹǿʱ£¬ÆøÌåÌå»ý¼õÉÙ30%£¬ÇóÔ­»ìºÏÆäÆøÌåÖÐH2SµÄÌå»ý·ÖÊý£®Ð´³öÍƵ¼¹ý³Ì£®£¨²»¿¼ÂÇÁò»¯ÇâµÄ·Ö½â£©
£¨4£©Áò´úÁòËáÄÆÊÇÖØÒªµÄ»¹Ô­¼Á£¬¿ÉÓÃÑÇÁòËáÄƺÍÁò·ÛÔÚË®ÈÜÒºÖмÓÈÈÖƵã®È¡15.12g Na2SO3ÈÜÓÚ80.0mLË®ÖУ¬¼ÓÈë5.00gÁò·Û£¬ÓÃС»ð¼ÓÈÈÖÁ£¬Î¢·Ð£¬·´Ó¦Ô¼1Сʱºó¹ýÂË£¬½«ÂËÒºÕô·¢ÖÁÌå»ýΪ30.0mL£¬ÔÙÀäÈ´µ½10¡æ£¬ÔòÀíÂÛÉÏÎö³öNa2S2O3?5H2O¶àÉÙ¿Ë£¿Ð´³öÍƵ¼¹ý³Ì£®
£¨ÒÑÖª£ºNa2S2O3µÄÈܽâ¶È£¬10¡æʱΪ60.0g/100gË®£¬100¡æʱΪ207g/100gË®£®100¡æʱ£¬Na2S2O3±¥ºÍÈÜÒºµÄÃܶÈΪ1.14g/mL£©£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2012-2013ѧÄêÉϺ£ÊÐÑîÆÖÇø¸ßÈýÉÏѧÆÚѧҵÖÊÁ¿µ÷Ñл¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£º¼ÆËãÌâ

ÁòÓжàÖÖ»¯ºÏÎÐí¶àº¬Áò»¯ºÏÎïÔÚ¹¤ÒµÉÏÓÐÖØÒªµÄÓÃ;¡£

£¨1£©¹¤ÒµÉÏ¿ÉÓûÆÌú¿ó¡¢½¹Ì¿ÔÚÓÐÏ޵ĿÕÆøÖÐȼÉÕÖƱ¸Áò»Ç¡£

3FeS2 + 12C+ 8O2 Fe3O4 + nA¡ü+ 6S

AÊÇ     £¨Ð´»¯Ñ§Ê½£©¡£ÈôµÃµ½192¿ËµÄÁò»Ç£¬Ôò²úÉú±ê×¼×´¿öϵÄAÆøÌå     L¡£    

£¨2£©ÁòµÄÂÈ»¯Îï³£×öÏ𽺹¤ÒµµÄÁò»¯¼Á¡£ÁòÓëÂÈÆøÔÚÒ»¶¨Ìõ¼þÏ·´Ó¦£¬µÃµ½Á½ÖÖÁòµÄÂÈ»¯ÎïBºÍD¡£BÎïÖÊÖеĺ¬ÁòÁ¿Îª0.3107£¬DÎïÖÊÖеĺ¬ÂÈÁ¿Îª0.5259£¬BµÄÏà¶Ô·Ö×ÓÖÊÁ¿±ÈDС32¡£¼ÆËãÈ·¶¨ÕâÁ½ÖÖÂÈ»¯ÎïµÄ·Ö×Óʽ·Ö±ðΪ                   ¡£

£¨3£©Áò»¯ÇâÓж¾¡£ÔÚ120¡æ¡¢101kPa£¬½«H2SºÍO2ÔÚÃܱÕÈÝÆ÷Öеãȼ£¬³ä·Ö·´Ó¦ºóÓÖ»Ö¸´µ½ÁËÔ­À´µÄζȺÍѹǿʱ£¬ÆøÌåÌå»ý¼õÉÙ30%£¬ÇóÔ­»ìºÏÆäÆøÌåÖÐH2SµÄÌå»ý·ÖÊý¡£Ð´³öÍƵ¼¹ý³Ì¡££¨²»¿¼ÂÇÁò»¯ÇâµÄ·Ö½â£©

£¨4£©Áò´úÁòËáÄÆÊÇÖØÒªµÄ»¹Ô­¼Á£¬¿ÉÓÃÑÇÁòËáÄƺÍÁò·ÛÔÚË®ÈÜÒºÖмÓÈÈÖƵá£È¡15.12g Na2SO3ÈÜÓÚ80.0mLË®ÖУ¬¼ÓÈë5.00gÁò·Û£¬ÓÃС»ð¼ÓÈÈÖÁ΢·Ð£¬·´Ó¦Ô¼1Сʱºó¹ýÂË£¬½«ÂËÒºÕô·¢ÖÁÌå»ýΪ30.0mL£¬ÔÙÀäÈ´µ½10¡æ£¬ÔòÀíÂÛÉÏÎö³öNa2S2O3∙5H2O¶àÉÙ¿Ë£¿Ð´³öÍƵ¼¹ý³Ì¡££¨ÒÑÖª£ºNa2S2O3µÄÈܽâ¶È£¬10¡æʱΪ60.0g/100gË®£¬100¡æʱΪ207g/100gË®¡£100¡æʱ£¬Na2S2O3±¥ºÍÈÜÒºµÄÃܶÈΪ1.14g/mL£©¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

ÁòÓжàÖÖ»¯ºÏÎÐí¶àº¬Áò»¯ºÏÎïÔÚ¹¤ÒµÉÏÓÐÖØÒªµÄÓÃ;£®
£¨1£©¹¤ÒµÉÏ¿ÉÓûÆÌú¿ó¡¢½¹Ì¿ÔÚÓÐÏ޵ĿÕÆøÖÐȼÉÕÖƱ¸Áò»Ç£®
3FeS2+12C+8O2Êýѧ¹«Ê½ Fe3O4+nA¡ü+6S
AÊÇ______£¨Ð´»¯Ñ§Ê½£©£®ÈôµÃµ½192¿ËµÄÁò»Ç£¬Ôò²úÉú±ê×¼×´¿öϵÄAÆøÌå______L£®
£¨2£©ÁòµÄÂÈ»¯Îï³£×öÏ𽺹¤ÒµµÄÁò»¯¼Á£®ÁòÓëÂÈÆøÔÚÒ»¶¨Ìõ¼þÏ·´Ó¦£¬µÃµ½Á½ÖÖÁòµÄÂÈ»¯ÎïBºÍD£®BÎïÖÊÖеĺ¬ÁòÁ¿Îª0.3107£¬DÎïÖÊÖеĺ¬ÂÈÁ¿Îª0.5259£¬BµÄÏà¶Ô·Ö×ÓÖÊÁ¿±ÈDС32£®¼ÆËãÈ·¶¨ÕâÁ½ÖÖÂÈ»¯ÎïµÄ·Ö×Óʽ·Ö±ðΪ______£®
£¨3£©Áò»¯ÇâÓж¾£®ÔÚ120¡æ¡¢101kPa£¬½«H2SºÍO2ÔÚÃܱÕÈÝÆ÷Öеãȼ£¬³ä·Ö·´Ó¦ºóÓÖ»Ö¸´µ½ÁËÔ­À´µÄζȺÍѹǿʱ£¬ÆøÌåÌå»ý¼õÉÙ30%£¬ÇóÔ­»ìºÏÆäÆøÌåÖÐH2SµÄÌå»ý·ÖÊý£®Ð´³öÍƵ¼¹ý³Ì£®£¨²»¿¼ÂÇÁò»¯ÇâµÄ·Ö½â£©
£¨4£©Áò´úÁòËáÄÆÊÇÖØÒªµÄ»¹Ô­¼Á£¬¿ÉÓÃÑÇÁòËáÄƺÍÁò·ÛÔÚË®ÈÜÒºÖмÓÈÈÖƵã®È¡15.12g Na2SO3ÈÜÓÚ80.0mLË®ÖУ¬¼ÓÈë5.00gÁò·Û£¬ÓÃС»ð¼ÓÈÈÖÁ£¬Î¢·Ð£¬·´Ó¦Ô¼1Сʱºó¹ýÂË£¬½«ÂËÒºÕô·¢ÖÁÌå»ýΪ30.0mL£¬ÔÙÀäÈ´µ½10¡æ£¬ÔòÀíÂÛÉÏÎö³öNa2S2O3?5H2O¶àÉÙ¿Ë£¿Ð´³öÍƵ¼¹ý³Ì£®
£¨ÒÑÖª£ºNa2S2O3µÄÈܽâ¶È£¬10¡æʱΪ60.0g/100gË®£¬100¡æʱΪ207g/100gË®£®100¡æʱ£¬Na2S2O3±¥ºÍÈÜÒºµÄÃܶÈΪ1.14g/mL£©£®

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸