¶ÌÖÜÆÚÔªËصĵ¥ÖÊX¡¢Y¡¢ZÔÚͨ³£×´¿öϾùΪÆø̬£¬²¢ÓÐÏÂÁÐת»¯¹Øϵ£¨·´Ó¦Ìõ¼þÂÔÈ¥£©£º

ÒÑÖª£ºa.³£¼ûË«Ô­×Óµ¥ÖÊ·Ö×ÓÖУ¬X·Ö×Óº¬¹²¼Û¼ü×î¶à¡£

b.¼×·Ö×Óº¬10¸öµç×Ó£¬ÒÒ·Ö×Óº¬18¸öµç×Ó¡£

£¨1£©XµÄµç×ÓʽÊÇ______________________________¡£

£¨2£©ÊµÑéÊÒ¿ÉÓÃÏÂͼËùʾװÖã¨È±ÉÙÊÕ¼¯×°Ö㬼г̶ֹ¨×°ÖÃÂÔÈ¥£©ÖƱ¸²¢ÊÕ¼¯¼×¡£

¢ÙÔÚͼÖз½¿òÄÚ»æ³öÓÃÉÕÆ¿ÊÕ¼¯¼×µÄÒÇÆ÷×°Öüòͼ¡£

¢ÚÊÔ¹ÜÖеÄÊÔ¼ÁÊÇ£¨Ìîд»¯Ñ§Ê½£©________________________________________¡£

¢ÛÉÕ±­ÖÐÈÜÒºÓÉÎÞÉ«±äΪºìÉ«£¬ÆäÔ­ÒòÊÇ£¨ÓõçÀë·½³Ìʽ±íʾ£©

______________________________________________________________________¡£

£¨3£©Á×ÔÚZÖÐȼÉÕ¿ÉÉú³ÉÁ½ÖÖ²úÎÆäÖÐÒ»ÖÖ²úÎﶡ·Ö×ÓÖи÷Ô­×Ó×îÍâ²ã²»È«ÊÇ8µç

×ӽṹ£¬¶¡µÄ»¯Ñ§Ê½ÊÇ____________________¡£

£¨4£©n mol¶¡Óën mol±ûÔÚÒ»¶¨Ìõ¼þÏ·´Ó¦£¬Éú³É4n molÒÒºÍÁíÒ»»¯ºÏÎ¸Ã»¯ºÏÎïÕôÆøµÄÃܶÈÊÇÏàͬ״¿öÏÂÇâÆøµÄ174±¶£¬Æ仯ѧʽÊÇ____________________¡£


£¨1£©

£¨2£©¢Ù

¢ÚNH4Cl¡¢Ca£¨OH£©2¡¡£¨ºÏÀí¾ù¿É£©

¢ÛNH3¡¤H2O+OH-

£¨3£©PCl5

£¨4£©P3N3Cl6

¡¾½âÎö¡¿±¾ÌâÊÇÒ»µÀÎÞ»ú»¯Ñ§ÍƶÏÌâ¡£ÓÉÌâÒâÖª¼×ΪNH3£¬ÒÒΪHCl£¬XΪN2£¬YΪH2£¬ZΪCl2¡£ÒòΪNH3¼«Ò×ÈÜÓÚË®£¬ÃܶȱȿÕÆøС£¬ÓÃÏòÏÂÅÅÆø·¨ÊÕ¼¯¡£2NH4Cl+Ca(OH)2CaCl2+2NH3¡ü+2H2O£¬°±Ë®³Ê¼îÐÔ£¬µÎÈë·Ó̪ÈÜÒº±äºì£¬NH3¡¤H2O+OH£­¡£Á×ÔÚCl2ÖÐȼÉÕÉú³ÉPCl3ºÍPCl5£¬Óɶ¡·Ö×Ó×îÍâ²ã²»È«ÊÇ8µç×ӽṹ֪¶¡ÎªPCl5¡£¸Ã»¯ºÏÎïÕôÆøµÄÃܶÈÊÇÏàͬ״¿öÏÂÇâÆøµÄ174±¶£¬¹ÊÏà¶Ô·Ö×ÓÖÊÁ¿Îª174¡Á2=348£¬ÓÉÌâÒâ¸Ã»¯ºÏÎïÖÐN¡¢P¸öÊýÏàͬ£¬ÉèΪ(NP)xCly

45¡¤x+35.5¡¤y=348  x=3  y=6

»¯Ñ§Ê½ÎªP3N3Cl6

·´Ó¦Ê½Îª3PCl5+3NH4ClP3N3Cl6+12HCl


Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


°×Á×£¨P4£©ÊÇÁ׵ĵ¥ÖÊÖ®Ò»£¬Ò×Ñõ»¯£¬Óë±Ëص¥ÖÊ·´Ó¦Éú³É±»¯Áס£Â±»¯Á×ͨ³£ÓÐÈý±»¯Á×»òÎå±»¯Á×£¬Îå±»¯Á×·Ö×ӽṹ£¨ÒÔPCl5ΪÀý£©ÈçÓÒͼËùʾ¡£¸Ã½á¹¹ÖÐÂÈÔ­×ÓÓÐÁ½ÖÖ²»Í¬Î»Öá£

1£©6.20g°×Á×ÔÚ×ãÁ¿ÑõÆøÖÐÍêȫȼÉÕÉú³ÉÑõ»¯Î·´Ó¦ËùÏûºÄµÄÑõÆøÔÚ±ê×¼×´¿öϵÄÌå»ýΪ             L¡£

ÉÏÊöȼÉÕ²úÎïÈÜÓÚË®Åä³É50.0mLÁ×ËᣨH3PO4£©ÈÜÒº£¬¸ÃÁ×ËáÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈΪ                 mol¡¤L-1¡£

2£©º¬0.300mol H3PO4µÄË®ÈÜÒºµÎ¼Óµ½º¬0.500mol Ca(OH)2µÄÐü¸¡ÒºÖУ¬·´Ó¦Ç¡ºÃÍêÈ«£¬Éú³ÉlÖÖÄÑÈÜÑκÍ16.2g H2O¡£¸ÃÄÑÈÜÑεĻ¯Ñ§Ê½¿É±íʾΪ                    ¡£

3£©°×Á׺ÍÂÈ¡¢äå·´Ó¦£¬Éú³É»ìºÏ±»¯Á×£¨£¬ÇÒxΪÕûÊý£©¡£

Èç¹ûij»ìºÏ±»¯Á×¹²ÓÐ3ÖÖ²»Í¬½á¹¹£¨·Ö×ÓÖÐäåÔ­×ÓλÖò»ÍêÈ«ÏàͬµÄ½á¹¹£©£¬¸Ã»ìºÏ±»¯Á×µÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª                                    ¡£

4£©Á×ë滯ºÏÎﺬÓÐ3ÖÖÔªËØ£¬ÇÒ·Ö×ÓÖÐÔ­×Ó×ÜÊýСÓÚ20¡£0.10mol PCl5ºÍ0.10mol NH4ClÇ¡ºÃÍêÈ«·´Ó¦£¬Éú³ÉÂÈ»¯ÇâºÍ0.030molÁ×ë滯ºÏÎï¡£ÍÆËãÁ×ë滯ºÏÎïµÄÏà¶Ô·Ö×ÓÖÊÁ¿£¨Ìáʾ£ºM>300£©¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÏÂͼ±íʾij¹Ì̬µ¥ÖÊA¼°Æ仯ºÏÎïÖ®¼äµÄת»¯¹Øϵ£¨Ä³Ð©²úÎïºÍ·´Ó¦Ìõ¼þÒÑÂÔÈ¥£©¡£»¯ºÏÎïBÔÚ³£Î³£Ñ¹ÏÂΪÆøÌ壬BºÍCµÄÏà¶Ô·Ö×ÓÖÊÁ¿Ö®±ÈΪ4£º5£¬»¯ºÏÎïDÊÇÖØÒªµÄ¹¤ÒµÔ­ÁÏ¡£

£¨1£©Ð´³öAÔÚ¼ÓÈÈÌõ¼þÏÂÓëH2·´Ó¦µÄ»¯Ñ§·½³Ìʽ

                                                                      

£¨2£©Ð´³öEÓëAµÄÇ⻯Îï·´Ó¦Éú³ÉAµÄ»¯Ñ§·½³Ìʽ                              

£¨3£©Ð´³öÒ»¸öÓÉDÉú³ÉBµÄ»¯Ñ§·½³Ìʽ                                       ;

£¨4£©½«5mL0.10mol¡¤L-1µÄEÈÜÒºÓë10mL0.10mol¡¤L-1µÄNaOHÈÜÒº»ìºÏ¡£

¢Ùд³ö·´Ó¦µÄÀë×Ó·½³Ìʽ                                               ;

¢Ú·´Ó¦ºóÈÜÒºµÄpH          7£¨Ìî¡°´óÓÚ¡±¡¢¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡±£©£¬ÀíÓÉÊÇ                                  ;

¢Û¼ÓÈÈ·´Ó¦ºóµÄÈÜÒº£¬ÆäpH           £¨Ìî ¡°Ôö´ó¡±¡¢¡°²»±ä¡±»ò¡°¼õС¡±£©£¬ÀíÓÉÊÇ

                                                        ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÏÂÁÐÓйع¤ÒµÉú²úµÄÐðÊöÕýÈ·µÄÊÇ

A.ºÏ³É°±Éú²ú¹ý³ÌÖн«NH3Òº»¯·ÖÀ룬¿É¼Ó¿ìÕý·´Ó¦ËÙÂÊ£¬Ìá¸ßN2¡¢H2µÄת»¯ÂÊ

B.ÁòËṤҵÖУ¬ÔÚ½Ó´¥ÊÒ°²×°ÈȽ»»»Æ÷ÊÇΪÁËÀûÓÃSO3ת»¯ÎªH2SO4ʱ·Å³öµÄÈÈÁ¿

C.µç½â±¥ºÍʳÑÎË®ÖÆÉÕ¼î²ÉÓÃÀë×Ó½»»»Ä¤·¨£¬¿É·ÀÖ¹Òõ¼«ÊÒÉú²úµÄCl2½øÈëÑô¼«ÊÒ

D.µç½â¾«Á¶Í­Ê±£¬Í¬Ò»Ê±¼äÄÚÑô¼«ÈܽâÍ­µÄÖÊÁ¿±ÈÒõ¼«Îö³öÍ­µÄÖÊÁ¿Ð¡

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


´¿¼îÊÇÒ»ÖÖÖØÒªµÄ»¯¹¤Ô­ÁÏ¡£Ä¿Ç°ÖƼҵÖ÷ÒªÓС°°±¼î·¨¡±ºÍ¡°ÁªºÏÖƼ¡±Á½ÖÖ¹¤ÒÕ¡£Çë°´ÒªÇó»Ø´ðÎÊÌ⣺

£¨1£©¡°°±¼î·¨¡±²úÉú´óÁ¿CaCl2·ÏÆúÎÇëд³ö¸Ã¹¤ÒÕÖвúÉúCaCl2µÄ»¯Ñ§·½³Ìʽ£º

                                                                       £»

£¨2£©Ð´³ö¡°ÁªºÏÖƼ¡±Óйط´Ó¦µÄ»¯Ñ§·½³Ìʽ£º

                                                                       £»

                                                                       £»

£¨3£©CO2ÊÇÖƼҵµÄÖØÒªÔ­ÁÏ£¬¡°ÁªºÏÖƼ¡±Óë¡°°±¼î·¨¡±ÖÐCO2µÄÀ´Ô´Óкβ»Í¬£¿

                                                                       £»

£¨4£©ÂÌÉ«»¯Ñ§µÄÖØÒªÔ­ÔòÖ®Ò»ÊÇÌá¸ß·´Ó¦µÄÔ­×ÓÀûÓÃÂÊ¡£¸ù¾Ý¡°ÁªºÏÖƼ¡±×Ü·´Ó¦£¬Áгö¼ÆËãÔ­×ÓÀûÓÃÂʵıí´ïʽ£º

Ô­×ÓÀûÓÃÂÊ£¨%£©=                                                       ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÔÚʵÑéÊÒÀï¿ÉÓÃÏÂͼËùʾװÖÃÖÆÈ¡ÂÈËá¼Ø¡¢´ÎÂÈËáÄƺÍ̽¾¿ÂÈË®µÄÐÔÖÊ¡£

ͼÖУº¢ÙΪÂÈÆø·¢Éú×°Ö㻢ڵÄÊÔ¹ÜÀïÊ¢ÓÐ15mL30%KOH ÈÜÒº.²¢ÖÃÓÚˮԡÖУ» ¢Û µÄÊÔ¹ÜÀïÊ¢ÓÐ15mL 8 % NaOH ÈÜÒº.²¢ÖÃÓÚ±ùˮԡÖУ» ¢Ü µÄÊÔ¹ÜÀï¼ÓÓÐ×ÏɫʯÈïÊÔÒº£» ¢Ý ΪβÆøÎüÊÕ×°Öá£

ÇëÌîдÏÂÁпհףº

£¨1£©ÖÆÈ¡ÂÈÆøʱ£¬ÔÚÉÕÆ¿Àï¼ÓÈËÒ»¶¨Á¿µÄ¶þÑõ»¯ÃÌ.ͨ¹ý________________£¨ÌîдÒÇÆ÷Ãû³Æ£©ÏòÉÕÆ¿ÖмÓÈËÊÊÁ¿µÄŨÑÎËᡣʵÑéʱΪÁ˳ýÈ¥ÂÈÆøÖеÄÂÈ»¯ÇâÆøÌ壬¿ÉÔÚ¢Ù Óë¢Ú Ö®¼ä°²×°Ê¢ÓÐ___________£¨ÌîдÏÂÁбàºÅ×Öĸ£©µÄ¾»»¯×°Öá£

A.¼îʯ»Ò  B.±¥ºÍʳÑÎË®  C.ŨÁòËá  D.±¥ºÍ̼ËáÇâÄÆÈÜÒº

£¨2£©±È½ÏÖÆÈ¡ÂÈËá¼ØºÍ´ÎÂÈËáÄƵÄÌõ¼þ£¬¶þÕߵIJîÒìÊÇ£º_______________________________

·´Ó¦Íê±Ï¾­ÀäÈ´ºó£¬¢ÚµÄÊÔ¹ÜÖÐÓдóÁ¿¾§ÌåÎö³ö¡£ÏÂͼÖзûºÏ¸Ã¾§ÌåÈܽâ¶ÈÇúÏßµÄÊÇ________£¨Ìîд±àºÅ×Öĸ£©£»´Ó¢ÚµÄÊÔ¹ÜÖзÖÀë³ö¸Ã¾§ÌåµÄ·½·¨ÊÇ________________(ÌîдʵÑé²Ù×÷Ãû³Æ£©

(3)±¾ÊµÑéÖÐÖÆÈ¡´ÎÂÈËáÄƵÄÀë×Ó·½³ÌʽÊÇ£º_____________________________

(4)ʵÑéÖпɹ۲쵽¢ÜµÄÊÔ¹ÜÀïÈÜÒºµÄÑÕÉ«·¢ÉúÁËÈçϱ仯£¬ÇëÌîдϱíÖеĿհףº

        ʵÑéÏÖÏó

                     Ô­Òò

ÈÜÒº×î³õ´Ó×ÏÉ«Öð½¥±äΪ____É«

ÂÈÆøÓëË®·´Ó¦Éú³ÉµÄH+ʹʯÈï±äÉ«

ËæºóÈÜÒºÖð½¥±äΪÎÞÉ«

_______________________________________________

È»ºóÈÜÒº´ÓÎÞÉ«Öð½¥±äΪ____É«

_______________________________________________

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


U¡¢V¡¢W¡¢X¡¢Y¡¢ZÊÇÔ­×ÓÐòÊýÒÀ´ÎÔö´óµÄÁùÖÖ³£¼ûÔªËØ¡£YµÄµ¥ÖÊÔÚW2ÖÐȼÉյIJúÎï¿ÉʹƷºìÈÜÒºÍÊÉ«¡£ZºÍWÔªËØÐγɵĻ¯ºÏÎïZ3W4¾ßÓдÅÐÔ¡£UµÄµ¥ÖÊÔÚW2ÖÐȼÉÕ¿ÉÉú³ÉUWºÍUW2Á½ÖÖÆøÌå¡£XµÄµ¥ÖÊÊÇÒ»ÖÖ½ðÊô£¬¸Ã½ðÊôÔÚUW2ÖоçÁÒȼÉÕÉú³ÉºÚ¡¢°×Á½ÖÖ¹ÌÌå¡£

Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©VµÄµ¥ÖÊ·Ö×ӵĽṹʽΪ                              £»XWµÄµç×ÓʽΪ                             £»ZÔªËØÔÚÖÜÆÚ±íÖеÄλÖÃÊÇ                                ¡£

£¨2£©UÔªËØÐγɵÄͬËØÒìÐÎÌåµÄ¾§ÌåÀàÐÍ¿ÉÄÜÊÇ£¨ÌîÐòºÅ£©                                  ¡£

¢ÙÔ­×Ó¾§Ìå    ¢ÚÀë×Ó¾§Ìå    ¢Û·Ö×Ó¾§Ìå    ¢Ü½ðÊô¾§Ìå

£¨3£©U¡¢V¡¢WÐγɵÄ10µç×ÓÇ⻯ÎïÖУ¬U¡¢VµÄÇ⻯Îï·Ðµã½ÏµÍµÄÊÇ£¨Ð´»¯Ñ§Ê½£©
                                   £»V¡¢WµÄÇ⻯Îï·Ö×Ó½áºÏH+ÄÜÁ¦½ÏÇ¿µÄÊÇ£¨Ð´»¯Ñ§Ê½£©                                £»ÓÃÒ»¸öÀë×Ó·½³Ìʽ¼ÓÒÔÖ¤Ã÷                                                      ¡£

£¨4£©YW2ÆøÌåͨÈëBaCl2ºÍHNO3µÄ»ìºÏÈÜÒº£¬Éú³É°×É«³ÁµíºÍÎÞÉ«ÆøÌåVW£¬Óйط´Ó¦µÄÀë×Ó·½³ÌʽΪ                                                  £¬ÓÉ´Ë¿ÉÖªVWºÍYW2»¹Ô­ÐÔ½ÏÇ¿µÄÊÇ£¨Ð´»¯Ñ§Ê½£©                                 ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÏÂͼÊDz¿·Ö¶ÌÖÜÆÚÔªËصĵ¥Öʼ°Æ仯ºÏÎïµÄת»¯¹Øϵͼ£¨Óйط´Ó¦µÄÌõ¼þ¼°Éú³ÉµÄH2OÒÑÂÔÈ¥£©£¬ÒÑÖª£º(a)A¡¢B¡¢C¡¢DÊǷǽðÊôµ¥ÖÊ£¬ÆäÖÐB¡¢C¡¢DÔÚ³£Î³£Ñ¹ÏÂÊÇÆøÌå¡£(b)·´Ó¦¢Ù¡¢¢ÚÊÇ»¯¹¤Éú²úÖеÄÖØÒª·´Ó¦¡£(c)»¯ºÏÎïEÊÇÐγÉËáÓêµÄÎÛȾÎïÖ®Ò»£¬»¯ºÏÎïKÊdz£Óõĵª·Ê¡£(d)»¯ºÏÎïL¾ßÓÐƯ°×ÐÔ£¬¿ÉÓÉCl2ÓëNaOHÈÜÒº·´Ó¦¶øÖƵá£(e)»¯ºÏÎïJÓÉÁ½ÖÖÔªËØ×é³É£¬ÆäÏà¶Ô·Ö×ÓÖÊÁ¿Îª32¡£

Çë°´ÒªÇóÌî¿Õ£º

¢Å·´Ó¦¢ÛµÄ»¯Ñ§·½³Ìʽ__________________________¡£

¢ÆCµÄ½á¹¹Ê½______________£»HµÄ»¯Ñ§Ê½______________¡£

¢ÇLµÄÈÜÒºÓ뻯ºÏÎïE·´Ó¦µÄÀë×Ó·½³Ìʽ__________________________¡£

¢È»¯ºÏÎïJµÄ»¯Ñ§Ê½______________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÏÂÁз´Ó¦ÖÐÌúÔªËر»»¹Ô­µÄÊÇ£¨¡¡¡¡£©

¡¡

A£®

2Fe£¨OH£©3Fe2O3+3H2O

B£®

Fe+CuSO4=Cu+FeSO4

¡¡

C£®

Fe£¨OH£©3+3HCl=FeCl3+3H2O

D£®

2Fe2O3+3C4Fe+3CO2¡ü

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸