¡¾ÌâÄ¿¡¿A£¬B£¬C£¬DÊÇËÄÖÖ¶ÌÖÜÆÚÔªËØ£¬EÊǹý¶ÉÔªËØ¡£A£¬B£¬CͬÖÜÆÚ£¬C£¬DͬÖ÷×壬AµÄÔ×ӽṹʾÒâͼΪ£¬BÊÇͬÖÜÆÚµÚÒ»µçÀëÄÜ×îСµÄÔªËØ£¬CµÄ×îÍâ²ãÓÐÈý¸öδ³É¶Ôµç×Ó£¬EµÄÍâΧµç×ÓÅŲ¼Ê½Îª3d64s2¡£»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Ð´³öÏÂÁÐÔªËصķûºÅ£ºA________£¬B________£¬C________£¬D________¡£
£¨2£©Óû¯Ñ§Ê½±íʾÉÏÊöÎåÖÖÔªËØÖÐ×î¸ß¼ÛÑõ»¯Îï¶ÔӦˮ»¯ÎïËáÐÔ×îÇ¿µÄÊÇ________£¬¼îÐÔ×îÇ¿µÄÊÇ__________¡£
£¨3£©ÓÃÔªËØ·ûºÅ±íʾDËùÔÚÖÜÆÚµÚÒ»µçÀëÄÜ×î´óµÄÔªËØÊÇ________£¬µç¸ºÐÔ£¨³ýÏ¡ÓÐÆøÌåÍ⣩×î´óµÄÔªËØÊÇ_________¡£
£¨4£©EÔªËØÔ×ӵĺ˵çºÉÊýÊÇ_________£¬EÔªËØÔÚÖÜÆÚ±íµÄµÚ_______ÖÜÆÚµÚ_______×壬ÔÚ________Çø¡£
£¨5£©Ð´³öDÔªËØÔ×Ó¹¹³Éµ¥Öʵĵç×Óʽ___________£¬¸Ã·Ö×ÓÖÐÓÐ_______¸ö¦Ò¼ü£¬_______¸ö¦Ð¼ü¡£
¡¾´ð°¸¡¿Si Na P N HNO3 NaOH Ne F 26 4 VIII d 1 2
¡¾½âÎö¡¿
A£¬B£¬C£¬DÊÇËÄÖÖ¶ÌÖÜÆÚÔªËØ£¬EÊǹý¶ÉÔªËØ¡£A£¬B£¬CͬÖÜÆÚ£¬C£¬DͬÖ÷×壬AµÄÔ×ӽṹʾÒâͼΪ£¬Ôòx=2£¬AΪ14ºÅÔªËØSi£»BÊÇͬÖÜÆÚµÚÒ»µçÀëÄÜ×îСµÄÔªËØ£¬ÔòBΪNaÔªËØ£»CµÄ×îÍâ²ãÓÐÈý¸öδ³É¶Ôµç×Ó£¬ÔòCΪPÔªËØ¡¢DΪNÔªËØ£»EµÄÍâΧµç×ÓÅŲ¼Ê½Îª3d64s2£¬ÔòEΪ26ºÅÔªËØFe¡£
£¨1£©A¡¢B¡¢C¡¢DÔªËصķûºÅ·Ö±ðÊÇSi¡¢Na¡¢P¡¢N¡£
£¨2£©ÉÏÊöÎåÖÖÔªËØÖУ¬·Ç½ðÊôÐÔ×îÇ¿µÄÊÇNÔªËØ£¬½ðÊôÐÔ×îÇ¿µÄÊÇNaÔªËØ£¬Òò´Ë£¬×î¸ß¼ÛÑõ»¯Îï¶ÔӦˮ»¯ÎïËáÐÔ×îÇ¿µÄÊÇHNO3£¬¼îÐÔ×îÇ¿µÄÊÇNaOH¡£
£¨3£©NËùÔÚÖÜÆÚ¼´µÚ2ÖÜÆÚ£¬µÚÒ»µçÀëÄÜ×î´óµÄÔªËØÊÇ×îÍâ²ã´ïµ½Îȶ¨½á¹¹µÄÏ¡ÓÐÆøÌåNe£¬µç¸ºÐÔ£¨³ýÏ¡ÓÐÆøÌåÍ⣩×î´óµÄÔªËØÊÇF¡£
£¨4£©EÔªËØÊÇFe£¬Ô×ӵĺ˵çºÉÊýÊÇ26£¬EÔªËØÔÚÖÜÆÚ±íµÄµÚ4ÖÜÆÚµÚVIII×壬¸ù¾ÝÆäÍâΧµç×ÓÅŲ¼¿ÉÖªÆäÔÚÔªËØÖÜÆÚ±íÖеÄdÇø¡£
£¨5£©DÔªËØÔ×Ó¹¹³Éµ¥ÖÊΪN2£¬Æäµç×ÓʽΪ£¬¸Ã·Ö×ÓÖÐÓÐ1¸ö¦Ò¼ü£¬2¸ö¦Ð¼ü¡£
Ä꼶 | ¸ßÖÐ¿Î³Ì | Ä꼶 | ³õÖÐ¿Î³Ì |
¸ßÒ» | ¸ßÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ | ³õÒ» | ³õÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ |
¸ß¶þ | ¸ß¶þÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õ¶þ | ³õ¶þÃâ·Ñ¿Î³ÌÍƼö£¡ |
¸ßÈý | ¸ßÈýÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õÈý | ³õÈýÃâ·Ñ¿Î³ÌÍƼö£¡ |
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ï®(Li)¡ª¿ÕÆøµç³ØµÄ¹¤×÷ÔÀíÈçͼËùʾÏÂÁÐ˵·¨²»ÕýÈ·µÄÊÇ
A. ½ðÊôï®×÷¸º¼«£¬·¢ÉúÑõ»¯·´Ó¦
B. Li+ͨ¹ýÓлúµç½âÖÊÏòË®ÈÜÒº´¦Òƶ¯
C. Õý¼«µÄµç¼«·´Ó¦£ºO2+4e¡ª==2O2¡ª
D. µç³Ø×Ü·´Ó¦£º4Li+O2+2H2O==4LiOH
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÒÑÖª£º¢ÙC2H6£¨g£© C2H4£¨g£©+H2£¨g£© H1 £¾0¡£
¢ÚC2H6£¨g£©+=2CO2£¨g£©+3H2O£¨l£© H 2 £½-1559.8 kJ¡¤mol-1
¢ÛC2H4£¨g£©+3O2£¨g£©=2CO2£¨g£©+2H2O£¨l£© H 3£½-1411.0 kJ¡¤mol-1
ÏÂÁÐÐðÊöÕýÈ·µÄÊÇ
A.Éýλò¼Óѹ¾ùÄÜÌá¸ß¢ÙÖÐÒÒÍéµÄת»¯ÂÊ
B.¢ÙÖжϼüÎüÊÕµÄÄÜÁ¿ÉÙÓڳɼü·Å³öµÄÄÜÁ¿
C.ÓÃH 2ºÍH 3¿É¼ÆËã³ö¢ÙÖеÄH
D.ÍƲâ1 mol C2H2£¨g£©ÍêȫȼÉշųöµÄÈÈÁ¿Ð¡ÓÚ1411.0 kJ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÈçͼËùʾװÖã¬Á½¸öÏàͬµÄ²£Á§¹ÜÖÐÊ¢ÂúNaClÏ¡ÈÜÒº£¨µÎÓзÓ̪£©£¬a¡¢bΪ¶à¿×ʯīµç¼«¡£±ÕºÏS1Ò»¶Îʱ¼äºó£¬a¸½½üÈÜÒºÖð½¥±äºì£»¶Ï¿ªS1£¬±ÕºÏS2£¬µçÁ÷±íÖ¸Õë·¢Éúƫת¡£
ÏÂÁзÖÎö²»ÕýÈ·µÄÊÇ
A.±ÕºÏS1ʱ£¬a¸½½üµÄºìÉ«Öð½¥ÏòÏÂÀ©É¢
B.±ÕºÏS1ʱ£¬ a¸½½üÒºÃæ±Èb¸½½üµÄµÍ
C.¶Ï¿ªS1¡¢±ÕºÏS2ʱ£¬b¸½½ü»ÆÂÌÉ«±ädz
D.¶Ï¿ªS1¡¢±ÕºÏS2ʱ£¬aÉÏ·¢Éú·´Ó¦£ºH2 2e- = 2H+
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿Ñо¿À´Ô´ÓÚÕæ¾úµÄÌìÈ»²úÎïLµÄºÏ³É¶Ô¿¹Ö×ÁöÒ©ÎïÑз¢ÓÐ×ÅÖØÒªÒâÒ壬ÆäºÏ³É·ÏßÖ÷Òª·ÖΪÁ½¸ö½×¶Î£º
I£®ºÏ³ÉÖмäÌåF
ÒÑÖª£º¢¡£®TBSClΪ
¢¢£®
£¨1£©AÖк¬Ñõ¹ÙÄÜÍÅÃû³Æ__________¡£
£¨2£©BµÄ½á¹¹¼òʽÊÇ__________¡£
£¨3£©ÊÔ¼ÁaÊÇ__________¡£
£¨4£©TBSClµÄ×÷ÓÃÊÇ__________¡£
II. ºÏ³ÉÓлúÎïL
ÒÑÖª£º
£¨5£©HÖк¬ÓÐÁ½¸öõ¥»ù£¬HµÄ½á¹¹¼òʽÊÇ__________¡£
£¨6£©I¡úJµÄ·´Ó¦·½³ÌʽÊÇ__________¡£
£¨7£©K¡úLµÄת»¯ÖУ¬Á½²½·´Ó¦µÄ·´Ó¦ÀàÐÍÒÀ´ÎÊÇ__________¡¢__________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿µÈÌå»ý¡¢µÈÎïÖʵÄÁ¿Å¨¶ÈµÄÁòËá¡¢ÇâÑõ»¯ÄÆÈÜÒº·Ö±ð·ÅÔڼס¢ÒÒÁ½ÉÕ±ÖУ¬¸÷¼ÓÈëµÈÖÊÁ¿µÄÂÁ£¬Éú³ÉÇâÆøµÄÌå»ý±ÈΪ5:6£¬Ôò¼×¡¢ÒÒÁ½ÉÕ±Öеķ´Ó¦Çé¿ö¿ÉÄÜ·Ö±ðÊÇ
A. ¼×¡¢ÒÒÖж¼ÊÇÂÁ¹ýÁ¿ B. ¼×ÖÐÂÁ¹ýÁ¿£¬ÒÒÖмî¹ýÁ¿
C. ¼×ÖÐËá¹ýÁ¿£¬ÒÒÖÐÂÁ¹ýÁ¿ D. ¼×ÖÐËá¹ýÁ¿£¬ÒÒÖмî¹ýÁ¿
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿(1)ÄÆÓëÑõÆøµÄ·´Ó¦»áÒòÌõ¼þ²»Í¬¶øµ¼ÖÂÏÖÏó²»Í¬£¬²úÎﲻͬ£¬·´Ó¦µÄʵÖÊÒ²²»Í¬¡£¾Ý´Ë»Ø´ðÏÂÁÐÎÊÌ⣺
¢Ù³£ÎÂÏ£¬ÔÚ¿ÕÆøÖÐÇпª½ðÊôÄÆ£¬ÄƵĶÏÃæÓÉÒø°×É«Ö𽥱䰵¶øʧȥ½ðÊô¹âÔó£¬ÇëÓû¯Ñ§·½³Ìʽ½âÊÍÕâÖÖÏÖÏó²úÉúµÄÔÒò£º__________________¡£
¢ÚÄÆÔÚ¿ÕÆøÖÐÊÜÈÈËù·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ__________________¡£
¢Û½«4.6¿ËÄÆͶÈë×ãÁ¿Ë®ÖУ¬±ê¿öÏÂÉú³ÉÆøÌåµÄÌå»ýÊÇ__________¡£
(2)ÈËÌåθҺÖÐÓÐθËá(0.2%¡«0.4%µÄÑÎËá)£¬Æðɱ¾ú¡¢°ïÖúÏû»¯µÈ×÷Ó㬵«Î¸ËáµÄÁ¿²»Äܹý¶à»ò¹ýÉÙ£¬Ëü±ØÐë¿ØÖÆÔÚÒ»¶¨·¶Î§ÄÚ£¬µ±Î¸Ëá¹ý¶àʱ£¬Ò½Éúͨ³£Óá°Ð¡ËÕ´òƬ¡±»ò¡°Î¸Êæƽ¡±¸ø²¡ÈËÖÎÁÆ¡£
¢ÙÓÃСËÕ´òƬ(NaHCO3)ÖÎÁÆθËá¹ý¶àµÄÀë×Ó·½³ÌʽΪ____________¡£
¢ÚÈç¹û²¡ÈËͬʱ»¼ÓÐθÀ£Ññ£¬´Ëʱ×îºÃ·þÓÃθÊæƽ[Ö÷Òª³É·ÖÊÇAl(OH)3]£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪ___________________________________¡£
¢ÛʵÑéÊÒÖƱ¸Al(OH)3µÄ³£Ó÷½·¨ÊÇÏòAl2(SO4)3ÈÜÒºÖÐÖðµÎµÎ¼Ó°±Ë®ÖÁ¹ýÁ¿£¬Çëд³ö¶ÔÓ¦µÄ»¯Ñ§·½³Ìʽ£º___________________________________¡£
(3)ÌúÊÇÈËÀà½ÏÔçʹÓõĽðÊôÖ®Ò»¡£ÔËÓÃÌú¼°Æ仯ºÏÎïµÄ֪ʶ£¬Íê³ÉÏÂÁÐÎÊÌâ¡£
¢ÙÖйú¹Å´úËÄ´ó·¢Ã÷Ö®Ò»µÄÖ¸ÄÏÕëÊÇÓÉÌìÈ»´ÅʯÖƳɵģ¬ÆäÖ÷Òª³É·ÖÊÇ_______¡£
¢Úд³ö´ÅʯµÄÖ÷Òª³É·ÖºÍÑÎËá·´Ó¦µÄÀë×Ó·½³Ìʽ______________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÒÑ֪ij´¿¼îÊÔÑùÖк¬ÓÐ NaCl ÔÓÖÊ£¬Îª²â¶¨ÊÔÑùÖд¿¼îµÄÖÊÁ¿·ÖÊý£¬¿ÉÓÃÏÂͼÖеÄ×°ÖýøÐÐʵÑé¡£
Ö÷ҪʵÑé²½ÖèÈçÏ£º
¢Ù°´Í¼×é×°ÒÇÆ÷£¬²¢¼ì²é×°ÖõÄÆøÃÜÐÔ£»
¢Ú½« a g ÊÔÑù·ÅÈë׶ÐÎÆ¿ÖУ¬¼ÓÊÊÁ¿ÕôÁóË®Èܽ⣬µÃµ½ÊÔÑùÈÜÒº£»
¢Û³ÆÁ¿Ê¢Óмîʯ»ÒµÄ U ÐιܵÄÖÊÁ¿£¬µÃµ½ b g£»
¢Ü´Ó·ÖҺ©¶·µÎÈë 6 molL-1µÄÁòËᣬֱµ½²»ÔÙ²úÉúÆøÌåʱΪֹ£»
¢Ý´Óµ¼¹Ü A ´¦»º»º¹ÄÈëÒ»¶¨Á¿µÄ¿ÕÆø£»
¢ÞÔٴγÆÁ¿Ê¢Óмîʯ»ÒµÄ U Ð͹ܵÄÖÊÁ¿£¬µÃµ½ c g£»
¢ßÖظ´²½Öè¢ÝºÍ¢ÞµÄ²Ù×÷£¬Ö±µ½ U Ð͹ܵÄÖÊÁ¿»ù±¾²»±ä£¬Îª d g£»
ÇëÌî¿ÕºÍ»Ø´ðÎÊÌ⣺
£¨1£©µÚÒ»¸öÏ´ÆøÆ¿ÖÐÇâÑõ»¯ÄÆÈÜÒºµÄ×÷ÓÃÊÇ____________________
£¨2£©×°ÖÃÖиÉÔï¹Ü B µÄ×÷ÓÃÊÇ_______________________________
£¨3£©Èç¹û½«·ÖҺ©¶·ÖеÄÁòËá»»³ÉŨ¶ÈÏàͬµÄÑÎËᣬ²âÊԵĽá¹û_________£¨ÌîÆ«¸ß¡¢Æ«µÍ»ò²»±ä£©¡£
£¨4£©²½Öè¢ÝµÄÄ¿µÄÊÇ_______________________________________
£¨5£©²½Öè¢ßµÄÄ¿µÄÊÇ_________________________________________
£¨6£©¸ÃÊÔÑùÖд¿¼îµÄÖÊÁ¿·ÖÊýµÄ¼ÆËãʽΪ________________________¡£
²é¿´´ð°¸ºÍ½âÎö>>
°Ù¶ÈÖÂÐÅ - Á·Ï°²áÁбí - ÊÔÌâÁбí
ºþ±±Ê¡»¥ÁªÍøÎ¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨Æ½Ì¨ | ÍøÉÏÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | µçÐÅթƾٱ¨×¨Çø | ÉæÀúÊ·ÐéÎÞÖ÷ÒåÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | ÉæÆóÇÖȨ¾Ù±¨×¨Çø
Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com