ÌâÄ¿ÄÚÈÝ

(18·Ö)A¡¢B¡¢C¡¢D¡¢E¡¢FÁùÖÖÔªËØ·Ö²¼ÔÚÈý¸ö²»Í¬µÄ¶ÌÖÜÆÚ£¬ËüÃǵÄÔ­×ÓÐòÊýÒÀ´ÎÔö´ó£¬ÆäÖÐBÓëCΪͬһÖÜÆÚ£¬AÓëD¡¢CÓëF·Ö±ðΪͬһÖ÷×壬A¡¢DÁ½ÔªËصÄÔ­×ÓºËÄÚµÄÖÊ×ÓÊýÖ®ºÍÊÇC¡¢FÁ½ÔªËØÔ­×ÓºËÄÚÖÊ×ÓÊýÖ®ºÍµÄÒ»°ë¡£ÓÖÖªÁùÖÖÔªËØËùÐγɵij£¼ûµ¥ÖÊÔÚ³£Î³£Ñ¹ÏÂÓÐÈýÖÖÊÇÆøÌ壬ÈýÖÖÊǹÌÌå¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÓÉA¡¢CÁ½ÔªËØ¿ÉÒÔ×é³ÉX¡¢YÁ½ÖÖ»¯ºÏÎXÔÚÒ»¶¨Ìõ¼þÏ¿ÉÒÔ·Ö½â³ÉY£¬XµÄµç×ÓʽΪ         £¬ÆäY·Ö×ÓÊôÓÚ            £¨Ìî¡°¼«ÐÔ¡±¡¢¡°·Ç¼«ÐÔ¡±£©·Ö×Ó¡£
£¨2£©EÊǷǽðÊôÔªËØ£¬µ«ÄܱíÏÖ³öһЩ½ðÊôÔªËصÄÐÔÖÊ£¬Çëд³öµ¥ÖÊEÓëÇâÑõ»¯ÄÆÈÜÒº·´Ó¦µÄÀë×Ó·½
³Ìʽ                               
£¨3£©Ò»¶¨Ìõ¼þÏ£¬AµÄµ¥ÖÊÆøÌåÓëBµÄµ¥ÖÊÆøÌå³ä·Ö·´Ó¦Éú³É6.8g WÆøÌå[ÒÑÖªn(A):n(B)=3:1]£¬¿É·Å³ö18.44 kJÈÈÁ¿£¬Ôò¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ                  
£¨4£©AµÄµ¥ÖÊÓëCµÄµ¥ÖÊÔÚKOHµÄŨÈÜÒºÖпÉÒÔÐγÉÔ­µç³Ø£¬Èç¹ûÒÔ½ðÊôMºÍ½ðÊôNΪ¶èÐԵ缫£¬ÔÚµç³ØµÄM¼«Í¨ÈëAµÄµ¥ÖÊÆøÌ壬N¼«Í¨ÈëCµÄµ¥ÖÊÆøÌ壬ÔòN¼«µÄµç¼«·´Ó¦Ê½        ¡£
£¨5£©ÔÚ10ÉýµÄÃܱÕÈÝÆ÷ÖУ¬Í¨Èë2molµÄFC2ÆøÌåºÍ3molCµÄÆøÌåµ¥ÖÊ£¬Ò»¶¨Ìõ¼þÏ·´Ó¦ºóÉú³ÉFC3ÆøÌ壬µ±·´Ó¦´ïµ½Æ½ºâʱ£¬µ¥ÖÊCµÄŨ¶ÈΪ0.21mol/L£¬ÔòƽºâʱFC2µÄת»¯ÂÊΪ         ¡£

£¨18·Ö£¬Ã¿¿Õ3·Ö£©£¨1£©                     ¼«ÐÔ
£¨2£©Si + 2OH- + H2O =SiO32- + 2H2¡ü
£¨3£©N2(g)+3H2(g2NH3(g)£»¦¤H=£­92.2KJ/mol£¨Î´Åäƽ²»µÃ·Ö£¬Êéд²»¹æ·¶²»µÃ·Ö£©
£¨4£©2H2O + O2 + 4e-= 4OH-      £¨5£©90%

½âÎö

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

(18·Ö)A¡¢B¡¢C¡¢D¡¢E¡¢FÁùÖÖÔªËØ·Ö²¼ÔÚÈý¸ö²»Í¬µÄ¶ÌÖÜÆÚ£¬ËüÃǵÄÔ­×ÓÐòÊýÒÀ´ÎÔö´ó£¬ÆäÖÐBÓëCΪͬһÖÜÆÚ£¬AÓëD¡¢CÓëF·Ö±ðΪͬһÖ÷×壬A¡¢DÁ½ÔªËصÄÔ­×ÓºËÄÚµÄÖÊ×ÓÊýÖ®ºÍÊÇC¡¢FÁ½ÔªËØÔ­×ÓºËÄÚÖÊ×ÓÊýÖ®ºÍµÄÒ»°ë¡£ÓÖÖªÁùÖÖÔªËØËùÐγɵij£¼ûµ¥ÖÊÔÚ³£Î³£Ñ¹ÏÂÓÐÈýÖÖÊÇÆøÌ壬ÈýÖÖÊǹÌÌå¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©ÓÉA¡¢CÁ½ÔªËØ¿ÉÒÔ×é³ÉX¡¢YÁ½ÖÖ»¯ºÏÎXÔÚÒ»¶¨Ìõ¼þÏ¿ÉÒÔ·Ö½â³ÉY£¬XµÄµç×ÓʽΪ         £¬ÆäY·Ö×ÓÊôÓÚ            £¨Ìî¡°¼«ÐÔ¡±¡¢¡°·Ç¼«ÐÔ¡±£©·Ö×Ó¡£

£¨2£©EÊǷǽðÊôÔªËØ£¬µ«ÄܱíÏÖ³öһЩ½ðÊôÔªËصÄÐÔÖÊ£¬Çëд³öµ¥ÖÊEÓëÇâÑõ»¯ÄÆÈÜÒº·´Ó¦µÄÀë×Ó·½

³Ìʽ                               

£¨3£©Ò»¶¨Ìõ¼þÏ£¬AµÄµ¥ÖÊÆøÌåÓëBµÄµ¥ÖÊÆøÌå³ä·Ö·´Ó¦Éú³É6.8g WÆøÌå[ÒÑÖªn(A):n(B)=3:1]£¬¿É·Å³ö18.44 kJÈÈÁ¿£¬Ôò¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ                  

£¨4£©AµÄµ¥ÖÊÓëCµÄµ¥ÖÊÔÚKOHµÄŨÈÜÒºÖпÉÒÔÐγÉÔ­µç³Ø£¬Èç¹ûÒÔ½ðÊôMºÍ½ðÊôNΪ¶èÐԵ缫£¬ÔÚµç³ØµÄM¼«Í¨ÈëAµÄµ¥ÖÊÆøÌ壬N¼«Í¨ÈëCµÄµ¥ÖÊÆøÌ壬ÔòN¼«µÄµç¼«·´Ó¦Ê½        ¡£

£¨5£©ÔÚ10ÉýµÄÃܱÕÈÝÆ÷ÖУ¬Í¨Èë2molµÄFC2ÆøÌåºÍ3molCµÄÆøÌåµ¥ÖÊ£¬Ò»¶¨Ìõ¼þÏ·´Ó¦ºóÉú³ÉFC3ÆøÌ壬µ±·´Ó¦´ïµ½Æ½ºâʱ£¬µ¥ÖÊCµÄŨ¶ÈΪ0.21mol/L£¬ÔòƽºâʱFC2µÄת»¯ÂÊΪ         ¡£

 

(18·Ö)A¡¢B¡¢C¡¢D¡¢E¡¢FÁùÖÖÔªËØ·Ö²¼ÔÚÈý¸ö²»Í¬µÄ¶ÌÖÜÆÚ£¬ËüÃǵÄÔ­×ÓÐòÊýÒÀ´ÎÔö´ó£¬ÆäÖÐBÓëCΪͬһÖÜÆÚ£¬AÓëD¡¢CÓëF·Ö±ðΪͬһÖ÷×壬A¡¢DÁ½ÔªËصÄÔ­×ÓºËÄÚµÄÖÊ×ÓÊýÖ®ºÍÊÇC¡¢FÁ½ÔªËØÔ­×ÓºËÄÚÖÊ×ÓÊýÖ®ºÍµÄÒ»°ë¡£ÓÖÖªÁùÖÖÔªËØËùÐγɵij£¼ûµ¥ÖÊÔÚ³£Î³£Ñ¹ÏÂÓÐÈýÖÖÊÇÆøÌ壬ÈýÖÖÊǹÌÌå¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©ÓÉA¡¢CÁ½ÔªËØ¿ÉÒÔ×é³ÉX¡¢YÁ½ÖÖ»¯ºÏÎXÔÚÒ»¶¨Ìõ¼þÏ¿ÉÒÔ·Ö½â³ÉY£¬XµÄµç×ÓʽΪ          £¬ÆäY·Ö×ÓÊôÓÚ             £¨Ìî¡°¼«ÐÔ¡±¡¢¡°·Ç¼«ÐÔ¡±£©·Ö×Ó¡£

£¨2£©EÊǷǽðÊôÔªËØ£¬µ«ÄܱíÏÖ³öһЩ½ðÊôÔªËصÄÐÔÖÊ£¬Çëд³öµ¥ÖÊEÓëÇâÑõ»¯ÄÆÈÜÒº·´Ó¦µÄÀë×Ó·½

³Ìʽ                               

£¨3£©Ò»¶¨Ìõ¼þÏ£¬AµÄµ¥ÖÊÆøÌåÓëBµÄµ¥ÖÊÆøÌå³ä·Ö·´Ó¦Éú³É6.8g WÆøÌå[ÒÑÖªn(A):n(B)=3:1]£¬¿É·Å³ö18.44 kJÈÈÁ¿£¬Ôò¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ                   

£¨4£©AµÄµ¥ÖÊÓëCµÄµ¥ÖÊÔÚKOHµÄŨÈÜÒºÖпÉÒÔÐγÉÔ­µç³Ø£¬Èç¹ûÒÔ½ðÊôMºÍ½ðÊôNΪ¶èÐԵ缫£¬ÔÚµç³ØµÄM¼«Í¨ÈëAµÄµ¥ÖÊÆøÌ壬N¼«Í¨ÈëCµÄµ¥ÖÊÆøÌ壬ÔòN¼«µÄµç¼«·´Ó¦Ê½         ¡£

£¨5£©ÔÚ10ÉýµÄÃܱÕÈÝÆ÷ÖУ¬Í¨Èë2molµÄFC2ÆøÌåºÍ3molCµÄÆøÌåµ¥ÖÊ£¬Ò»¶¨Ìõ¼þÏ·´Ó¦ºóÉú³ÉFC3ÆøÌ壬µ±·´Ó¦´ïµ½Æ½ºâʱ£¬µ¥ÖÊCµÄŨ¶ÈΪ0.21mol/L£¬ÔòƽºâʱFC2µÄת»¯ÂÊΪ          ¡£

 

(18·Ö) A¡¢B¡¢C¡¢DÊÇËÄÖÖ¶ÌÖÜÆÚÔªËØ£¬ËüÃǵÄÔ­×ÓÐòÊýÒÀ´ÎÔö´ó£¬ÆäÖУ¬A¡¢C¼°B¡¢D·Ö±ðÊÇͬÖ÷×åÔªËØ£»AÔªËصÄÔ­×Ӱ뾶ÊÇËùÓÐÖ÷×åÔªËØÖÐÔ­×Ӱ뾶×îСµÄ£»B¡¢DÁ½ÔªËصÄÔ­×ÓºËÖÐÖÊ×ÓÊýÖ®ºÍÊÇA¡¢CÁ½ÔªËØÔ­×ÓºËÖÐÖÊ×ÓÊýÖ®ºÍµÄ2±¶£»ËÄÖÖÔªËØËùÐγɵĵ¥ÖÊÖÐA¡¢Bµ¥ÖÊÊÇÆøÌ壬C¡¢Dµ¥ÖÊÊǹÌÌå¡£

£¨1£©Ð´³öÒÔÏÂÔªËصÄÃû³Æ£º B____________£»C___________¡£

£¨2£©Ð´³öBµÄÔ­×ӽṹʾÒâͼ               ÈôÒ»¸öBÔ­×ÓÖÐÓÐ8¸öÖÐ×Ó£¬Ð´³öBµÄÔ­×Ó·ûºÅ               

£¨3£©ÓÉB¡¢CÁ½ÔªËØËùÐγɵÄÔ­×Ó¸öÊý±ÈΪ1£º1µÄ»¯ºÏÎïÊôÓÚ__________£¨Ìî¡°Àë×Ó¡± »ò¡°¹²¼Û¡±£©»¯ºÏÎ´æÔڵĻ¯Ñ§¼üµÄÖÖÀàÓÐ________£¨ÈôÓй²¼Û¼üÐèҪд³ö¾ßÌåµÄ¹²¼Û¼üÀàÐÍ£©£¬Ð´³öËüÓëË®·´Ó¦µÄÀë×Ó·½³Ìʽ____                       ¡£

£¨4£©Óõç×Óʽ±íʾC2DµÄÐγɹý³Ì£º____                        _______¡£

£¨5£©ÊµÑéÊÒ³£ÓÃA¡¢BÁ½ÔªËØËùÐγɵÄÔ­×Ó¸öÊýΪ1£º1µÄ»¯ºÏÎïÀ´ÖƱ¸Ò»ÖÖ³£¼ûÆøÌ壬д³öʵÑéÊÒÖÐÓùÌÌåÒ©Æ·ÖƱ¸¸ÃÆøÌåµÄµÄ»¯Ñ§·½³Ìʽ_____              _________¡£

£¨6£©Ð´³ö¾ùÓÉA¡¢B¡¢C¡¢DËÄÖÖÔªËØ×é³ÉµÄÁ½ÖÖÎïÖÊÖ®¼ä·¢Éú·´Ó¦µÄÀë×Ó·½³Ìʽ£º

                                                             

£¨7£©ÓÉA¡¢BÁ½ÔªËØÐγɵĵ¥ÖÊÔÚÒÔ²¬×÷µç¼«Ï¡ÁòËá×÷µç½âÒºÐγɵÄȼÁϵç³ØÖУ¬Õý¼«µÄµç¼«·½³ÌʽΪ                          £¬ÈôÒÔÇâÑõ»¯¼ØΪµç½âÒºÔò¸º¼«µÄµç¼«·½³ÌʽΪ                       ¡£

 

(18·Ö) A¡¢B¡¢C¡¢DÊÇËÄÖÖ¶ÌÖÜÆÚÔªËØ£¬ËüÃǵÄÔ­×ÓÐòÊýÒÀ´ÎÔö´ó£¬ÆäÖУ¬A¡¢C¼°B¡¢D·Ö±ðÊÇͬÖ÷×åÔªËØ£»AÔªËصÄÔ­×Ӱ뾶ÊÇËùÓÐÖ÷×åÔªËØÖÐÔ­×Ӱ뾶×îСµÄ£»B¡¢DÁ½ÔªËصÄÔ­×ÓºËÖÐÖÊ×ÓÊýÖ®ºÍÊÇA¡¢CÁ½ÔªËØÔ­×ÓºËÖÐÖÊ×ÓÊýÖ®ºÍµÄ2±¶£»ËÄÖÖÔªËØËùÐγɵĵ¥ÖÊÖÐA¡¢Bµ¥ÖÊÊÇÆøÌ壬C¡¢Dµ¥ÖÊÊǹÌÌå¡£

£¨1£©Ð´³öÒÔÏÂÔªËصÄÃû³Æ£º B____________£»C___________¡£

£¨2£©Ð´³öBµÄÔ­×ӽṹʾÒâͼ              ÈôÒ»¸öBÔ­×ÓÖÐÓÐ8¸öÖÐ×Ó£¬Ð´³öBµÄÔ­×Ó·ûºÅ               

£¨3£©ÓÉB¡¢CÁ½ÔªËØËùÐγɵÄÔ­×Ó¸öÊý±ÈΪ1£º1µÄ»¯ºÏÎïÊôÓÚ__________£¨Ìî¡°Àë×Ó¡± »ò¡°¹²¼Û¡±£©»¯ºÏÎ´æÔڵĻ¯Ñ§¼üµÄÖÖÀàÓÐ________£¨ÈôÓй²¼Û¼üÐèҪд³ö¾ßÌåµÄ¹²¼Û¼üÀàÐÍ£©£¬Ð´³öËüÓëË®·´Ó¦µÄÀë×Ó·½³Ìʽ____                      ¡£

£¨4£©Óõç×Óʽ±íʾC2DµÄÐγɹý³Ì£º____                       _______¡£

£¨5£©ÊµÑéÊÒ³£ÓÃA¡¢BÁ½ÔªËØËùÐγɵÄÔ­×Ó¸öÊýΪ1£º1µÄ»¯ºÏÎïÀ´ÖƱ¸Ò»ÖÖ³£¼ûÆøÌ壬д³öʵÑéÊÒÖÐÓùÌÌåÒ©Æ·ÖƱ¸¸ÃÆøÌåµÄµÄ»¯Ñ§·½³Ìʽ_____             _________¡£

£¨6£©Ð´³ö¾ùÓÉA¡¢B¡¢C¡¢DËÄÖÖÔªËØ×é³ÉµÄÁ½ÖÖÎïÖÊÖ®¼ä·¢Éú·´Ó¦µÄÀë×Ó·½³Ìʽ£º

                                                            

£¨7£©ÓÉA¡¢BÁ½ÔªËØÐγɵĵ¥ÖÊÔÚÒÔ²¬×÷µç¼«Ï¡ÁòËá×÷µç½âÒºÐγɵÄȼÁϵç³ØÖУ¬Õý¼«µÄµç¼«·½³ÌʽΪ                         £¬ÈôÒÔÇâÑõ»¯¼ØΪµç½âÒºÔò¸º¼«µÄµç¼«·½³ÌʽΪ                      ¡£

 

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

¾«Ó¢¼Ò½ÌÍø