£¨15·Ö£©£¨1£©µç½âÔ­ÀíÔÚ»¯Ñ§¹¤ÒµÖÐÓй㷺µÄÓ¦Óá£ÏÖ½«ÄãÉè¼ÆµÄÔ­µç³ØÍ¨¹ýµ¼ÏßÓëͼÖеç½â³ØÏàÁ¬£¬ÆäÖÐaΪµç½âÒº£¬XºÍYÊÇÁ½¿éµç¼«°â£¬Ôò

  ¢ÙÈôXºÍY¾ùΪ¶èÐԵ缫£¬aΪ±¥ºÍʳÑÎË®£¬Ôòµç½âʱ¼ìÑéYµç¼«·´Ó¦²úÎïµÄ·½·¨ÊÇ                                       ¡£

  ¢ÚÈôX¡¢Y·Ö±ðΪʯīºÍÌú£¬aÈÔΪ±¥ºÍµÄNaClÈÜÒº£¬Ôòµç½â¹ý³ÌÖлáÉú³É°×É«¹ÌÌ壬·¢ÉúµÄ×Ü·´Ó¦»¯Ñ§·½³ÌʽΪ

                                                        ¡£

¸Ã°×É«¹ÌÌå¶ÖÃÔÚ¿ÕÆøÖУ¬¿É¹Û²ìµ½µÄÏÖÏóÊÇ                

                         ¡£

  ¢ÛÈôXºÍY¾ùΪ¶èÐԵ缫£¬aΪһ¶¨Å¨¶ÈµÄÁòËáÍ­ÈÜÒº£¬Í¨µçºó£¬·¢ÉúµÄ×Ü·´Ó¦»¯Ñ§·½³ÌʽΪ                                                                      ¡£Í¨µçÒ»¶Îʱ¼äºó£¬ÏòËùµÃÈÜÒºÖмÓÈë0£®05 mol Cu(OH)2£¬Ç¡ºÃ»Ö¸´µç½âǰµÄŨ¶ÈºÍpH£¬Ôòµç½â¹ý³ÌÖеç×Ó×ªÒÆµÄÎïÖʵÄÁ¿Îª              mol¡£

£¨2£©ÀûÓù¤ÒµÉÏÀë×Ó½»»»Ä¤·¨ÖÆÉÕ¼îµÄÔ­Àí£¬ÓÃÈçͼËùʾװÖõç½âK2SO­4ÈÜÒº¡£

¢Ù¸Ãµç½â²ÛµÄÑô¼«·´Ó¦Ê½Îª                               £¬Í¨¹ýÒõÀë×Ó½»»»Ä¤µÄÀë×ÓÊý       £¨Ìî¡°>¡±¡¢¡°<¡±»ò¡°=¡±£©Í¨¹ýÑôÀë×Ó½»»»Ä¤µÄÀë×ÓÊý£»

¢ÚͼÖÐa¡¢b¡¢c¡¢d·Ö±ð±íʾÓйØÈÜÒºµÄpH£¬Ôòa¡¢b¡¢c¡¢dÓÉСµ½´óµÄ˳ÐòΪ            £»

¢Ûµç½âÒ»¶Îʱ¼äºó£¬B¿ÚÓëC¿Ú²úÉúÆøÌåµÄÖÊÁ¿±ÈΪ                  ¡£

 

¡¾´ð°¸¡¿

£¨1£©¢Ù½«ÊªÈóµÄµí·Ûµâ»¯¼ØÊÔÖ½¿¿½üY¼«Ö§¹Ü¿Ú£¬ÊÔÖ½±äÀ¶¡££¨1·Ö£©

        ¢ÚFe+2H2O = Fe(OH)2¡ý+H2¡ü £¨2·Ö£©

         °×É«¹ÌÌåѸËÙ±ä»ÒÂÌÉ«£¬×îÖÕ±ä³ÉºìºÖÉ«¡££¨1·Ö£©

        ¢Û2CuSO4+2H2O = 2Cu+O2+2H2SO4£¨2·Ö£© 0.2£¨2·Ö£©

   £¨2£©¢Ù4OH--4e-=O2 ¡ü+2H2O£¨2·Ö£©£¬  <£¨1·Ö£©

           ¢Ú b<a<c<d£¨2·Ö£©           ¢Û 8£º1  £¨2·Ö£©

¡¾½âÎö¡¿

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

µç½âÔ­ÀíÔÚ»¯Ñ§ÖÐÓй㷺ӦÓá£ÏÂͼ±íʾһ¸öµç½â³Ø£¬aΪµç½âÒº£»X¡¢YÊÇÁ½¿éµç¼«°å£¬Í¨¹ýµ¼ÏßÓëÖ±Á÷µçÔ´ÏàÁ¬¡£ÇëÍê³ÉÒÔÏÂÎÊÌ⣺

                                            

£¨1£©ÈôX¡¢Y¶¼ÊǶèÐԵ缫£¬aÊDZ¥ºÍNaClÈÜÒº£¬ÊµÑ鿪ʼʱ£¬Í¬Ê±ÔÚÁ½±ß¸÷µÎÈ뼸µÎ·Ó̪ÊÔÒº£¬Ôò

¢Ùµç½â³ØÖÐX¼«Éϵĵ缫·´Ó¦Ê½Îª________________£¬ÔÚX¼«¸½½ü¹Û²ìµ½µÄÏÖÏóÊÇ______

_______________________£»

¢ÚYµç¼«Éϵĵ缫·´Ó¦Ê½Îª________________________£¬¼ìÑé¸Ãµç¼«·´Ó¦²úÎïµÄ·½·¨ÊÇ___

______________________¡£

£¨2£©Èç¹ûÓõç½â·½·¨¾«Á¶´ÖÍ­£¬µç½âÒºaÑ¡ÓÃCuSO4ÈÜÒº£¬Ôò£º

¢ÙXµç¼«µÄ²ÄÁÏÊÇ________________________£¬µç¼«·´Ó¦Ê½Îª______________________£»

¢ÚYµç¼«µÄ²ÄÁÏÊÇ________________________£¬µç¼«·´Ó¦Ê½Îª_______________________¡£

£¨ËµÃ÷£ºÔÓÖÊ·¢ÉúµÄµç¼«·´Ó¦²»±ØÐ´³ö£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨15·Ö£©£¨1£©µç½âÔ­ÀíÔÚ»¯Ñ§¹¤ÒµÖÐÓй㷺µÄÓ¦Óá£ÏÖ½«ÄãÉè¼ÆµÄÔ­µç³ØÍ¨¹ýµ¼ÏßÓëͼÖеç½â³ØÏàÁ¬£¬ÆäÖÐaΪµç½âÒº£¬XºÍYÊÇÁ½¿éµç¼«°â£¬Ôò

  ¢ÙÈôXºÍY¾ùΪ¶èÐԵ缫£¬aΪ±¥ºÍʳÑÎË®£¬Ôòµç½âʱ¼ìÑéYµç¼«·´Ó¦²úÎïµÄ·½·¨ÊÇ                                      ¡£

  ¢ÚÈôX¡¢Y·Ö±ðΪʯīºÍÌú£¬aÈÔΪ±¥ºÍµÄNaClÈÜÒº£¬Ôòµç½â¹ý³ÌÖлáÉú³É°×É«¹ÌÌ壬·¢ÉúµÄ×Ü·´Ó¦»¯Ñ§·½³ÌʽΪ

                                                       ¡£

¸Ã°×É«¹ÌÌå¶ÖÃÔÚ¿ÕÆøÖУ¬¿É¹Û²ìµ½µÄÏÖÏóÊÇ                

                        ¡£

  ¢ÛÈôXºÍY¾ùΪ¶èÐԵ缫£¬aΪһ¶¨Å¨¶ÈµÄÁòËáÍ­ÈÜÒº£¬Í¨µçºó£¬·¢ÉúµÄ×Ü·´Ó¦»¯Ñ§·½³ÌʽΪ                                                                     ¡£Í¨µçÒ»¶Îʱ¼äºó£¬ÏòËùµÃÈÜÒºÖмÓÈë0£®05 molCu(OH)2£¬Ç¡ºÃ»Ö¸´µç½âǰµÄŨ¶ÈºÍpH£¬Ôòµç½â¹ý³ÌÖеç×Ó×ªÒÆµÄÎïÖʵÄÁ¿Îª             mol¡£

£¨2£©ÀûÓù¤ÒµÉÏÀë×Ó½»»»Ä¤·¨ÖÆÉÕ¼îµÄÔ­Àí£¬ÓÃÈçͼËùʾװÖõç½âK2SO­4ÈÜÒº¡£

¢Ù¸Ãµç½â²ÛµÄÑô¼«·´Ó¦Ê½Îª                              £¬Í¨¹ýÒõÀë×Ó½»»»Ä¤µÄÀë×ÓÊý       £¨Ìî¡°>¡±¡¢¡°<¡±»ò¡°=¡±£©Í¨¹ýÑôÀë×Ó½»»»Ä¤µÄÀë×ÓÊý£»

¢ÚͼÖÐa¡¢b¡¢c¡¢d·Ö±ð±íʾÓйØÈÜÒºµÄpH£¬Ôòa¡¢b¡¢c¡¢dÓÉСµ½´óµÄ˳ÐòΪ            £»

¢Ûµç½âÒ»¶Îʱ¼äºó£¬B¿ÚÓëC¿Ú²úÉúÆøÌåµÄÖÊÁ¿±ÈΪ                  ¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

µç½âÔ­ÀíÔÚ»¯Ñ§£º¹¤ÒµÖÐÓй㷺ӦÓá£ÏÂͼËùʾΪһÖÖµç½â×°Öã¬GΪ×ÔÁ÷µçÔ´£¬a¡¢bΪµç¼«£¬UÐιÜÖÐ×°Óеç½âÖÊÈÜÒºW(500 mL£¬Éèµç½âǰºóÈÜÒºµÄÌå»ý²»±ä)£¬X¡¢YÊÇÁ½¿éµç¼«°å£¬Í¨¹ýµ¼ÏßÓëÖ±Á÷µçÔ´ÏàÁ¬¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺

(1)ÀûÓõç½âÔ­ÀíÔÚÈçͼװÖÃÖÐÍê³É´ÖÍ­Ìá´¿£¬Ôòµç½âÖÊÈÜÒºWΪ_________£¬Ñô¼«²ÄÁÏXΪ_________£»

(2)ÒÑÖªÖ±Á÷µçÔ´GΪ¸ßÌúµç³Ø£¬¸ßÌúµç³ØÊÇÒ»ÖÖÐÂÐͿɳäµçµç³Ø£¬ÓëÆÕͨ¸ßÄÜµç³ØÏà±È£¬¸Ãµç³ØÄܳ¤Ê±¼ä±£³ÖÎȶ¨µÄ·Åµçµçѹ£¬¸ßÌúµç³ØµÄ×Ü·´Ó¦Ê½Îª£º

3Zn+2K2FeO4+8H2O3Zn(OH)2+2Fe(OH)3+4KOHÔòbµç¼«·´Ó¦Ê½Îª_________£»·ÅµçÊ±Ã¿×ªÒÆ3 molµç×Ó£¬Õý¼«ÓÐ_________mol_________±»»¹Ô­£»

(3)ÈôX¡¢XΪʯī°å£¬WΪCuSO4ÈÜÒº£¬µç½âÒ»¶Îʱ¼äºó£¬Ïòµç½âºóµÄ²ÐÁôÒºÖмÓÈë×ãÁ¿Ìú·Û³ä·Ö·´Ó¦£¬¹ýÂË¡¢Õô¸É¡¢³ÆÖØ£¬·¢ÏÖÌú·ÛÔöÖØ3.2g£»Ï´¾»¡¢ºæ¸É¡¢³ÆÖØ£¬·¢ÏÖY°åÔöÖØ1.6 g¡£Ôò´ÓÀíÂÛÉϽ²£¬µç½â¹ý³ÌÖиßÌúµç³ØµÄпµç¼«ÖÊÁ¿¼õÉÙ_________g£»µç½âºóÈÜÒºpHΪ_________£»Ô­CuSO4ÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈΪ_________mol¡¤L-1¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨1£©µç½âÔ­ÀíÔÚ»¯Ñ§¹¤ÒµÖÐÓй㷺µÄÓ¦Óá£ÏÖ½«ÄãÉè¼ÆµÄÔ­µç³ØÍ¨¹ýµ¼ÏßÓëͼÖеç½â³ØÏàÁ¬£¬ÆäÖÐaΪµç½âÒº£¬XºÍYÊÇÁ½¿éµç¼«°â£¬Ôò

  ¢ÙÈôXºÍY¾ùΪ¶èÐԵ缫£¬aΪ±¥ºÍʳÑÎË®£¬Ôòµç½âʱ¼ìÑéYµç¼«·´Ó¦²úÎïµÄ·½·¨ÊÇ                                       ¡£

  ¢ÚÈôX¡¢Y·Ö±ðΪʯīºÍÌú£¬aÈÔΪ±¥ºÍµÄNaClÈÜÒº£¬Ôòµç½â¹ý³ÌÖлáÉú³É°×É«¹ÌÌ壬·¢ÉúµÄ×Ü·´Ó¦»¯Ñ§·½³ÌʽΪ

                                                        ¡£

¸Ã°×É«¹ÌÌå¶ÖÃÔÚ¿ÕÆøÖУ¬¿É¹Û²ìµ½µÄÏÖÏóÊÇ                

                         ¡£

  ¢ÛÈôXºÍY¾ùΪ¶èÐԵ缫£¬aΪһ¶¨Å¨¶ÈµÄÁòËáÍ­ÈÜÒº£¬Í¨µçºó£¬·¢ÉúµÄ×Ü·´Ó¦»¯Ñ§·½³ÌʽΪ                                                                      ¡£Í¨µçÒ»¶Îʱ¼äºó£¬ÏòËùµÃÈÜÒºÖмÓÈë0£®05 mol Cu(OH)2£¬Ç¡ºÃ»Ö¸´µç½âǰµÄŨ¶ÈºÍpH£¬Ôòµç½â¹ý³ÌÖеç×Ó×ªÒÆµÄÎïÖʵÄÁ¿Îª              mol¡£¸ß¡î¿¼¡á×Ê¡âÔ´*Íø

£¨2£©ÀûÓù¤ÒµÉÏÀë×Ó½»»»Ä¤·¨ÖÆÉÕ¼îµÄÔ­Àí£¬ÓÃÈçͼËùʾװÖõç½âK2SO??4ÈÜÒº¡£

¢Ù¸Ãµç½â²ÛµÄÑô¼«·´Ó¦Ê½Îª                               £¬Í¨¹ýÒõÀë×Ó½»»»Ä¤µÄÀë×ÓÊý       £¨Ìî¡°>¡±¡¢¡°<¡±»ò¡°=¡±£©Í¨¹ýÑôÀë×Ó½»»»Ä¤µÄÀë×ÓÊý£»

¢ÚͼÖÐa¡¢b¡¢c¡¢d·Ö±ð±íʾÓйØÈÜÒºµÄpH£¬Ôòa¡¢b¡¢c¡¢dÓÉСµ½´óµÄ˳ÐòΪ            £»

¢Ûµç½âÒ»¶Îʱ¼äºó£¬B¿ÚÓëC¿Ú²úÉúÆøÌåµÄÖÊÁ¿±ÈΪ                  ¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸