ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿Ã÷´úËÎÓ¦ÐÇÖøµÄ¡¶Ì칤¿ªÎï¡·ÖÐÓйØÓÚ¡°»ð·¨¡±Ò±Á¶Ð¿µÄ¹¤ÒÕ¼ÇÔØ£º¡°Â¯¸Êʯ£¨Ö÷Òª³É·ÖΪ̼Ëáп£©Ê®½ï£¬×°ÔØÈëÒ»Äà¹ÞÄÚ£¬¡­¡­È»ºóÖð²ãÓÃú̿±ýµæÊ¢£¬Æäµ×ÆÌн£¬·¢»ðìѺ죬¡­¡­£¬Àäµí£¬»Ù¹ÞÈ¡³ö£¬¡­¡­£¬¼´ÙÁǦҲ¡£¡±ÔòÏÂÁÐ˵·¨²»ÕýÈ·µÄÊÇ

A. ¸ÃÒ±Á¶Ð¿µÄ·´Ó¦Öаüº¬ÓÐÑõ»¯»¹Ô­·´Ó¦

B. ÉÏÊö¸ßÎÂìÑÉÕʱʵÖÊÊÇCO»¹Ô­ZnO

C. ¹ÅÈË°Ñп³ÆÙÁǦÊÇÒòΪпºÍǦµÄ»¯Ñ§ÐÔÖÊÏàͬ

D. Ò±Á¶ Zn µÄ×Ü·´Ó¦·½³ÌʽΪ£º 2ZnCO3+C2Zn+3CO2¡ü

¡¾´ð°¸¡¿C

¡¾½âÎö¡¿ÓÉÌâÒ⣬̼ËáпÓë̼£¨Ì¼ÊÇ×ãÁ¿µÄ£©ÔÚ¸ßÎÂÏ·´Ó¦ÊÇ¡°»ð·¨¡±Ò±Á¶Ð¿µÄÔ­Àí£¬¿ÉÄÜ·¢ÉúµÄ·´Ó¦Îª£º¢ÙZnCO3ZnO+CO2¡ü¡¢¢ÚCO2+C2CO¡¢¢ÛZnO+COZn+CO2¡ü£¬ÒòΪ̼ÊÇ×ãÁ¿µÄ£¬ËùÒÔ×Ü·´Ó¦µÄ·½³ÌʽΪ£ºZnCO3+2CZn+3CO¡ü£»AÏ·´Ó¦¢ÚºÍ¢ÛÊÇÑõ»¯»¹Ô­·´Ó¦£¬¹ÊAÕýÈ·£»BÏÓÉ·´Ó¦¢Û¿ÉµÃ£¬ÉÏÊö¸ßÎÂìÑÉÕʱʵÖÊÊÇCO»¹Ô­ZnO£¬¹ÊBÕýÈ·£»CÏ¡¶Ì칤¿ªÎï¡·¼ÇÔØ£º¡°ÒÔÆäËÆǦ¶øÐÔÃÍ£¬¹ÊÃûÖ®Ô»ÙÁ¡±£¬Òâ˼ÊÇ£ºÓÉÓÚÖƵõÄпºÜÏñǦ¶øÓÖ±ÈǦµÄÐÔÖʸüÃÍÁÒ£¬ËùÒÔ°ÑËü½Ð×ö¡°ÙÁǦ¡±£¬Óɴ˿ɼûпºÍǦµÄ»¯Ñ§ÐÔÖʲ»Í¬£¬¹ÊC´íÎó£»DÏÓÉÉÏÃæµÄ·ÖÎö¿ÉÖªDÕýÈ·¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿³äÂúHCl£¨±ê×¼×´¿ö£©µÄÉÕÆ¿×öÍêÅçȪʵÑéºóµÃµ½µÄÏ¡ÑÎËáÈÜÒº£¬Óñê×¼ÇâÑõ»¯ÄÆÄÆÈÜÒºµÎ¶¨£¬ÒÔÈ·¶¨¸ÃÏ¡ÑÎËáµÄ׼ȷÎïÖʵÄÁ¿Å¨¶È¡£»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©¸ÃµÎ¶¨ÊµÑéÊ¢×°±ê×¼ÒºµÄÒÇÆ÷ÊÇ_______________£¬¸ÃÒÇÆ÷¶ÁÊýʱӦ¶Áµ½__________mL¡£Èô¸ÃÒÇÆ÷µÄÁ¿³ÌΪ50mL£¬µ÷ÒºÃæΪ0£¬½«¸ÃÒÇÆ÷ÖÐËùÓÐÒºÌå·Å³ö£¬Ôò·Å³öµÄÈÜÒºÌå»ý_______50mL¡££¨Ìî¡°£¾¡±£¬¡°£½¡±£¬¡°£¼¡±£©

£¨2£©Èô¸ÃµÎ¶¨ÊµÑéÓ÷Ó̪×öָʾ¼Á£¬´ïµ½µÎ¶¨ÖÕµãʱ£¬ÈÜÒºÑÕÉ«´Ó____É«±äΪ____É«ÇÒ±£³Ö30sÄÚ²»±äÉ«¡£

£¨3£©ÅäÖÆÈýÖÖ²»Í¬Å¨¶ÈµÄ±ê×¼ÇâÑõ»¯ÄÆÈÜÒº£¬ÄãÈÏΪ×îºÏÊʵÄÊǵÚ______ÖÖ¡£

¢Ù5.000 mol/L ¢Ú0. 5000 mol/L ¢Û0.0500 mol/L

£¨4£©Èô²ÉÓÃÉÏÊöºÏÊʵıê×¼ÇâÑõ»¯ÄÆÈÜÒºµÎ¶¨Ï¡ÑÎËᣬ²Ù×÷²½ÖèºÏÀí£¬µÎ¶¨ºóµÄʵÑéÊý¾ÝÈçÏ£º

ʵÑé±àºÅ

´ý²âÑÎËáµÄÌå»ý£¨mL£©

µÎÈëÇâÑõ»¯ÄÆÈÜÒºµÄÌå»ý£¨mL£©

1

20.00

17.30

2

20.00

17.02

3

20.00

16.98

Çó²âµÃµÄÏ¡ÑÎËáµÄÎïÖʵÄÁ¿Å¨¶ÈΪ_________________________¡£

£¨5£©ÏÂÁвÙ×÷µ¼ÖµĽá¹û£¨Ìî¡°Æ«´ó¡±¡°Æ«Ð¡¡±¡°ÎÞÓ°Ï족£©

¢ÙµÎ¶¨¹ÜÏ´¾»ºó£¬Ö±½Ó×°Èë±ê×¼ÇâÑõ»¯ÄÆÈÜÒº½øÐе樣º___________

¢ÚµÎ¶¨Ç°¶ÁÊýʱÑöÊÓ£¬µÎ¶¨ºó¶ÁÊýʱ¸©ÊÓ£º___________

¢ÛÈôÓú¬ÓÐNa2OÔÓÖʵÄÇâÑõ»¯ÄƹÌÌåÅäÖƱê×¼ÈÜÒº£º___________

¢ÜµÎ¶¨Ç°£¬¼îʽµÎ¶¨¹ÜÓÐÆøÅÝ£¬µÎ¶¨ºóÏûʧ£º___________¡£

¡¾ÌâÄ¿¡¿£¨1£©ÓÃÁòËá·Ö½âÁ×β¿ó[Ö÷Òª³É·ÝΪCa5(PO4)3F]¿ÉÖƵÃÖÐÇ¿ËáÁ×ËáH3PO4¡£Çë»Ø´ð£º

¢ÙNa2HPO4ÈÜÒº³Ê¼îÐÔ£¬ÔòÔÚ¸ÃÈÜÒºÖÐc(H2PO4£­)_____c(PO43£­)£¨Ìî¡°>¡±¡¢¡°<¡±¡¢¡°£½¡±£©£¬

ÏòNa2HPO4ÈÜÒºÖмÓÈë×ãÁ¿µÄCaCl2ÈÜÒº£¬ÈÜÒºÏÔËáÐÔ£¬ÈÜÒºÏÔËáÐÔµÄÔ­ÒòÊÇ(´ÓµçÀëƽºâÒƶ¯½Ç¶È·ÖÎö)£º_____________________________¡£

¢ÚÒÑÖª£º25¡æʱ£¬H3PO4µÄµçÀëƽºâ³£Êý£ºK1£½7.52¡Á10£­3£»K2£½6.23¡Á10£­8£»K3£½6.23¡Á10£­13Ôò£ºH3PO4(aq)+OH£­(aq)H2PO4£­(aq)+H2O (l)µÄƽºâ³£ÊýK=______¡£

¢ÛCa5(PO4)3F(s) +OH£­(aq)Ca5(PO4)3(OH) (s)+F-£¬¸ÃζÈÏ£¬ÈÜÒºÖÐc(F-)ËæÈÜÒºµÄpHºÍζȣ¨T£©µÄ±ä»¯ÇúÏßÈçÓÒͼËùʾ¡£Ôò£ºpH1___pH2£¨Ìî¡°>¡±¡¢¡°<¡±¡¢¡°£½¡±£©£»A¡¢BÁ½µãµÄÈÜÒºÖÐF-µÄËÙÂÊv(A)____v(B)£¨Ìî¡°>¡±¡¢ ¡°<¡±¡¢¡°£½¡±£©¡£

£¨2£©´¦Àíº¬ÄøËáÐÔ·ÏË®¿ÉÓõç½â·¨ÊµÏÖÌúÉ϶ÆÄø£¬Æä×°ÖÃÈçÓÒͼËùʾ¡£

¢ÙµçÁ÷·½ÏòÊÇ_____£¨Ìa.̼°ô¡úÌú°ô£»b.Ìú°ô¡ú̼°ô)£»

¢ÚÖмä¸ôÊҵĿÉÒԵõ½µÄÎïÖÊXÊÇ_______£¨Ìѧʽ£©£»

¢ÛÑô¼«µÄµç¼«·´Ó¦Ê½Îª_______________£»µç½â×Ü·´Ó¦µÄÀë×Ó·½³ÌʽΪ_______________¡£

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

¾«Ó¢¼Ò½ÌÍø