ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿ÏÂÁÐÖ¸¶¨·´Ó¦µÄÀë×Ó·½³ÌʽÕýÈ·µÄÊÇ

A. ÓÃʯīµç¼«µç½âMgCl2ÈÜÒº:Mg2++2C1-+2H2OMg(OH)2¡ý+Cl2¡ü+H2¡ü

B. ÏòÃ÷·¯ÈÜÒºÖеμÓ̼ËáÄÆÈÜÒº:2Al3++3CO32-==Al2(CO3)3¡ý

C. ÏòCa(HCO3)2ÈÜÒºÖеμÓÉÙ×îNaOHÈÜÒº:Ca2++2HCO3-+2OH-==CaCO3¡ý+CO32-+2H2O

D. ÏòFe(NO3)3ÈÜÒºÖмÓÈë¹ýÁ¿µÄHIÈÜÒº:2NO3-+8H++6I-==3I2+2NO¡ü+4H2O

¡¾´ð°¸¡¿A

¡¾½âÎö¡¿A. ÓÃʯīµç¼«µç½âMgCl2ÈÜÒº£ºMg2++2Cl-+2H2OMg(OH)2¡ý+Cl2¡ü+H2¡ü£¬AÕýÈ·£»B. ÏòÃ÷·¯ÈÜÒºÖеμÓ̼ËáÄÆÈÜÒº·¢ÉúÇ¿ÁÒµÄË«Ë®½â£¬Éú³ÉÇâÑõ»¯ÂÁºÍ¶þÑõ»¯Ì¼ÆøÌ壬B´íÎó£»C. ÏòCa(HCO3)2ÈÜÒºÖеμÓÉÙÁ¿NaOHÈÜÒº£ºCa2++HCO3-+OH-=CaCO3¡ý+H2O£¬C´íÎó£»D. ÏòFe(NO3)3ÈÜÒºÖмÓÈë¹ýÁ¿µÄHIÈÜÒº£¬ÏõËá¡¢Èý¼ÛÌú¾ùÓëµâÀë×Ó·¢ÉúÑõ»¯»¹Ô­·´Ó¦£¬Éú³ÉÎﻹÓÐÑÇÌúÀë×ÓÉú³É£¬D´íÎó¡£´ð°¸Ñ¡A.

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿ÀûÓÃËá½â·¨ÖÆîÑ°×·Û²úÉúµÄ·ÏÒº[º¬ÓдóÁ¿FeSO4¡¢H2SO4ºÍÉÙÁ¿Fe2(SO4)3¡¢TiOSO4]£¬Éú²úÌúºìºÍ²¹Ñª¼ÁÈéËáÑÇÌú¡£ÆäÉú²ú²½ÖèÈçÏ£º

ÒÑÖª£ºTiOSO4¿ÉÈÜÓÚË®£¬ÔÚË®ÖпÉÒÔµçÀëΪTiO2+ºÍSO42-£¬TiOSO4Ë®½â³ÉTiO2xH2O³ÁµíΪ¿ÉÄæ·´Ó¦£»ÈéËá½á¹¹¼òʽΪCH3CH(OH)COOH¡£

Çë»Ø´ð£º

£¨1£©²½Öè¢ÙÖзÖÀëÁòËáÑÇÌúÈÜÒººÍÂËÔüµÄ²Ù×÷ÊÇ________________________¡£

£¨2£©¼ÓÈëÌúмµÄÄ¿µÄÒ»ÊÇ»¹Ô­ÉÙÁ¿Fe2(SO4)3£»¶þÊÇʹÉÙÁ¿TiOSO4ת»¯ÎªTiO2xH2OÂËÔü£¬ÓÃƽºâÒƶ¯µÄÔ­Àí½âÊ͵õ½ÂËÔüµÄÔ­Òò___________________________¡£

£¨3£©ÁòËáÑÇÌúÔÚ¿ÕÆøÖÐìÑÉÕÉú³ÉÌúºìºÍÈýÑõ»¯Áò£¬¸Ã·´Ó¦ÖÐÑõ»¯¼ÁºÍ»¹Ô­¼ÁµÄÎïÖʵÄÁ¿Ö®±ÈΪ_________________¡£

£¨4£©ÓÃÀë×Ó·½³Ìʽ½âÊͲ½Öè¢ÝÖмÓÈéËáÄܵõ½ÈéËáÑÇÌúµÄÔ­Òò_________________¡£

£¨5£©²½Öè¢ÜµÄÀë×Ó·½³ÌʽÊÇ_________________________________________¡£

£¨6£©Îª²â¶¨²½Öè¢ÚÖÐËùµÃ¾§ÌåÖÐFeSO4¡¤7H2OµÄÖÊÁ¿·ÖÊý£¬È¡¾§ÌåÑùÆ·a g£¬ÈÜÓÚÏ¡ÁòËáÅä³É100.00 mLÈÜÒº£¬È¡³ö20.00 mLÈÜÒº£¬ÓÃKMnO4ÈÜÒºµÎ¶¨£¨ÔÓÖÊÓëKMnO4²»·´Ó¦£©¡£ÈôÏûºÄ0.1000 molL-1 KMnO4ÈÜÒº20.00 mL£¬ËùµÃ¾§ÌåÖÐFeSO4¡¤7H2OµÄÖÊÁ¿·ÖÊýΪ______________£¨ÓÃa±íʾ£©¡£

¡¾ÌâÄ¿¡¿Öйú´«Í³ÎÄ»¯ÊÇÈËÀàÎÄÃ÷µÄ¹å±¦£¬¹Å´úÎÄÏ×ÖмÇÔØÁË´óÁ¿¹Å´ú»¯Ñ§µÄÑо¿³É¹û¡£»Ø´ðÏÂÃæÎÊÌâ¡£

£¨1£©ÎÒ¹ú×îԭʼµÄÌÕ´ÉÔ¼³öÏÖÔÚ¾à½ñ12000ÄêÇ°£¬ÖÆ×÷´ÉÆ÷ËùÓõÄÔ­ÁÏÊǸßÁëÊ¿£¬Æ侧Ì廯ѧʽÊÇAl4[Si4O10](OH)8£¬ÓÃÑõ»¯Îï±íʾÆä×é³ÉΪ____¡£

£¨2£©¡¶±¾²Ý¸ÙÄ¿¡·ÖмÇÔØ£º¡°£¨»ðÒ©£©ÄËÑæÏû( KNO3)¡¢Áò»Ç¡¢É¼Ä¾Ì¿ËùºÏ£¬ÒÔΪ·éìÝ糧úÖîÒ©Õß¡£¡±·´Ó¦Ô­ÀíΪ£ºS£«2KNO3£«3CK2S£«N2¡ü£«3CO2¡ü£¬¸Ã·´Ó¦µÄÑõ»¯¼ÁÊÇ____ £¬·´Ó¦×ªÒÆ4molµç×Óʱ£¬±»SÑõ»¯µÄCÓÐ____mol¡£

£¨3£©ÎÒ¹ú¹Å´úÖÐҩѧÖø×÷¡¶ÐÂÐÞ±¾²Ý¡·¼ÇÔصÄÒ©ÎïÓÐ844ÖÖ£¬ÆäÖÐÓйء°Çà·¯¡±µÄÃèÊöΪ£º¡°±¾À´ÂÌÉ«£¬Ð³ö¿ßδ¼û·çÕߣ¬ÕýÈçè£Á§¡­ÉÕÖ®³àÉ«¡­¡£¡±ÎÒ¹úÔçÆڿƼ¼´ÔÊ顶ÎïÀíСÊÊ-½ðʯÀà¡· ¼ÇÔØÓмÓÈÈÇ෯ʱµÄ¾°Ï󣺡°Çà·¯³§ÆøѬÈË£¬Ò·þµ±Ö®Ò×Àã¬ÔØľ²»Ê¢¡£¡±Çà·¯¾ÍÊÇÂÌ·¯( FeSO4.7H2O)¡£¸ù¾ÝÒÔÉÏÐÅÏ¢£¬Ð´³ö¡°Çà·¯¡±ÊÜÈÈ·Ö½âµÄ»¯Ñ§·½³Ìʽ____

£¨4£©Öйú¹Å´úµÚÒ»²¿Ò©ÎïѧרÖø¡¶ÉñÅ©±¾¸ï¾­¡·¼ÇÔØ£º¡°Ê¯Áò»Ç(S)¡­Ö÷ÒõÊ´¾ÒÖ̶ñѪ£¬¼á½î¹Ç£¬³ýͷͺ£¬ÄÜ»¯½ðÒøÍ­ÌúÆæÎï¡£¡±¸ù¾ÝÒÔÉÏÐÅÏ¢£¬µÃ³öÁò»Ç(S)¾ßÓеÄÐÔÖÊ»òÓÃ;ÊÇ ______£¬__________¡£

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

¾«Ó¢¼Ò½ÌÍø