¡¾ÌâÄ¿¡¿ÎÙÊÇÈÛµã×î¸ßµÄ½ðÊô£¬ÊÇÖØÒªµÄÕ½ÂÔÎï×Ê¡£×ÔÈ»½çÖÐÎÙÖ÷Òª´æÔÚÓÚºÚÎÙ¿óÖУ¬ÆäÖ÷Òª³É·ÖÊÇÌúºÍÃ̵ÄÎÙËáÑΣ¨FeWO4¡¢MnWO4£©£¬»¹º¬ÉÙÁ¿Si¡¢AsµÄ»¯ºÏÎï¡£ÓɺÚÎÙ¿óÒ±Á¶ÎٵŤÒÕÁ÷³ÌÈçÏ£º

ÒÑÖª£º

¢ÙÂËÔüIµÄÖ÷Òª³É·ÝÊÇFe2O3¡¢MnO2¡£

¢ÚÉÏÊöÁ÷³ÌÖУ¬ÎٵĻ¯ºÏ¼ÛÖ»ÓÐÔÚ×îºóÒ»²½·¢Éú¸Ä±ä¡£

¢Û³£ÎÂÏÂÎÙËáÄÑÈÜÓÚË®¡£

£¨1£©Çëд³öFeWO4ÔÚÈÛÈÚÌõ¼þÏ·¢Éú¼î·Ö½â·´Ó¦Éú³ÉFe2O3µÄ»¯Ñ§·½³Ìʽ£º______________________________________¡£

£¨2£©ÉÏÊöÁ÷³ÌÖÐÏò´ÖÎÙËáÄÆÈÜÒºÖмÓÁòËáÖкÍÖÁpH=10ºó£¬ÈÜÒºÖеÄÔÓÖÊÒõÀë×ÓΪSiO32¨D¡¢HAsO32¨D¡¢HAsO42¨DµÈ£¬Ôò¡°¾»»¯¡±¹ý³ÌÖУ¬¼ÓÈëH2O2ʱ·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ_________£¬ÂËÔü¢òµÄÖ÷Òª³É·ÖÊÇ______________¡£

£¨3£©ÒÑÖªÇâÑõ»¯¸ÆºÍÎÙËá¸Æ£¨CaWO4£©¶¼ÊÇ΢Èܵç½âÖÊ£¬Á½ÕßµÄÈܽâ¶È¾ùËæζÈÉý¸ß¶ø¼õС¡£ÏÂͼΪ²»Í¬Î¶ÈÏÂCa£¨OH£©2¡¢CaWO4µÄ³ÁµíÈܽâƽºâÇúÏß¡£

¢ÙT1_____T2£¨Ìî¡°>¡±»ò¡°<¡±£©T1ʱKsp£¨CaWO4£©=______________¡£

¢Ú½«ÎÙËáÄÆÈÜÒº¼ÓÈëʯ»ÒÈéµÃµ½´óÁ¿ÎÙËá¸Æ£¬·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ__________________________¡£

¡¾´ð°¸¡¿4FeWO4+O2+8NaOH2Fe2O3+4Na2WO4+4H2O H2O2+HAsO32¨D¨THAsO42¨D+H2O MgSiO3 ¡¢ MgHAsO4 < 1¡Á10-10 WO42¨D+Ca(OH)2=CaWO4+2OH¨D

¡¾½âÎö¡¿

ÓÉÁ÷³Ì¿ÉÖª£¬ÎÙËáÑÇÌúºÍÇâÑõ»¯ÄÆ¡¢ÑõÆø·´Ó¦Éú³ÉÑõ»¯ÌúºÍÎÙËáÄÆ£¬ÎÙËáÃ̺ÍÇâÑõ»¯ÄÆ·´Ó¦Éú³ÉÎÙËáÄƺÍÇâÑõ»¯ÃÌ£¬Ë®½þʱ£¬Ñõ»¯ÌúºÍÇâÑõ»¯Ã̲»ÈÜÓÚË®£¬ÎÙËáÄÆÈÜÓÚË®£¬¹Ê¹ýÂ˺óµÃµ½µÄÂËÒºº¬ÎÙËáÄÆ£¬ÂËÔüIµÄÖ÷Òª³É·ÝÊÇFe2O3¡¢MnO2£¬ÎÙËáÄƺÍŨÁòËá·´Ó¦Éú³ÉÎÙËáºÍÁòËáÄÆ£»¼ÓÈë¹ýÑõ»¯Ç⣬Ñõ»¯+5¼ÛµÄÎÙΪ+6¼Û£¬¼ÓÈëÂÈ»¯Ã¾£¬Éú³ÉÄÑÈÜÓÚË®µÄMgSiO3¡¢MgHAsO4£¬¹ýÂË£¬ÂËҺΪÎÙËáÄÆ£¬Ëữ£¬¼ÓÈÈ·Ö½â²úÉúÈýÑõ»¯ÎÙºÍË®£¬Óû¹Ô­¼Á»¹Ô­ÈýÑõ»¯ÎÙÉú³ÉÎÙ£¬¾Ý´Ë·ÖÎö½â´ð¡£

£¨1£©¸ù¾ÝÁ÷³Ìͼ¿ÉÖªÎÙËáÑÇÌúºÍÇâÑõ»¯ÄÆ¡¢ÑõÆø·´Ó¦Éú³ÉÑõ»¯ÌúºÍÎÙËáÄÆ£¬·´Ó¦µÄ·½³ÌʽΪ4FeWO4+O2+8NaOH2Fe2O3+4Na2WO4+4H2O£»

£¨2£©¸ù¾ÝÒÔÉÏ·ÖÎö£¬¼ÓÈëH2O2µÄÄ¿µÄÊǽ«HAsO32-Ñõ»¯³ÉHAsO42-£¬Àë×Ó·½³ÌʽΪH2O2+HAsO32-¨THAsO42-+H2O£¬ÂËÒºIÖдæÔÚSiO32-¡¢HAsO32-¡¢HAsO42-µÈÀë×Ó£¬¾­¹ýµ÷½âpHÖµºó£¬¼ÓÈëÂÈ»¯Ã¾£¬Mg2+³ÁµíSiO32-¡¢HAsO32-¡¢HAsO42-µÈÀë×Ó£¬ÂËÔü¢òµÄÖ÷Òª³É·ÖÊÇMgSiO3¡¢MgHAsO4£»

£¨3£©¢Ù¸ù¾ÝͼÏó¿ÉÖª£¬ÇâÑõ»¯¸ÆºÍÎÙËá¸ÆÔÚ¸ÆÀë×ÓŨ¶ÈÏàͬʱ£¬T1ζÈÏÂÒõÀë×ÓŨ¶È´óÓÚT2£¬ËµÃ÷T1ʱµÄÈܶȻý´óÓÚT2£¬ÈܶȻýÔ½´ó£¬ÔòÈܽâ¶ÈÔ½´ó£¬ËùÒÔT1ʱÈܽâ¶È½Ï´ó£¬ÓÉÓÚ¡°ÒÑÖªÇâÑõ»¯¸ÆºÍÎÙËá¸Æ£¨CaWO4£©¶¼ÊÇ΢Èܵç½âÖÊ£¬Á½ÕßµÄÈܽâ¶È¾ùËæζÈÉý¸ß¶ø¼õС¡±£¬ÔòT1£¼T2£»T1ʱKsp£¨CaWO4£©=c£¨Ca2+£©c£¨WO42-£©=1¡Á10-5¡Á1¡Á10-5=1¡Á10-10£»

¢Ú½«ÎÙËáÄÆÈÜÒº¼ÓÈëʯ»ÒÈ飬·¢Éú¸´·Ö½â·´Ó¦£¬ÇâÑõ»¯¸ÆºÍÎÙËá¸ùÀë×Ó·´Ó¦Éú³ÉÎÙËá¸Æ³Áµí£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪWO42¨D+Ca(OH)2=CaWO4+2OH¨D¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿½ðÊôNa¡¢Mg¡¢AlÓй㷺µÄÓ¦Óá£

£¨1£©AlÔÚÔªËØÖÜÆÚ±íÖеÄλÖÃÊÇ__________________¡£

£¨2£©½ðÊôþ¿ÉÒÔÓÃÀ´Éú²ú½ðÊôÓË£¨ÔªËØ·ûºÅÊÇU£©£ºUF4+2MgU+2MgF2£¬¸Ã·´Ó¦ÖУ¬×÷Ϊ»¹Ô­¼ÁµÄÎïÖÊÊÇ_________£¨Ìѧʽ£¬ÏÂͬ£©£¬±»»¹Ô­µÄÎïÖÊÊÇ_________¡£

£¨3£©Îª±È½ÏNa¡¢Mg¡¢AlµÄ½ðÊôÐÔ£¬½øÐÐÁËÈçÏÂʵÑ飨½ðÊô¹ÌÌåµÄ±íÃæ»ý¶¼Ïàͬ£©£º

ʵÑé1

ʵÑé2

ÄÆÓëË®·´Ó¦¾çÁÒ£¬Ã¾ÓëË®·´Ó¦»ºÂý

þÓëÑÎËá·´Ó¦¾çÁÒ£¬ÂÁÓëÑÎËá·´Ó¦»ºÂý

ÓÉʵÑé1ºÍʵÑé2µÃ³öµÄ½áÂÛÊÇ£º½ðÊôÐÔ_________>_________>_________£¨ÌîÔªËØ·ûºÅ£©£¬ÓÃÔ­×ӽṹÀíÂÛ½âÊÍ£ºÍ¬ÖÜÆÚÔªËØ´Ó×óµ½ÓÒ£¬_________¡£

£¨4£©°ÑþÌõ£¨È¥³ýÑõ»¯Ä¤£©Í¶È뵽ʢÓÐÑÎËáµÄ³¨¿ÚÈÝÆ÷ÖУ¬²úÉúH2µÄËÙÂÊvÓëʱ¼ätµÄ¹ØϵÈçͼËùʾ£¬AB¶ÎËÙÂÊÔö´óµÄÖ÷ÒªÔ­ÒòÊÇ__________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÏÂÁз´Ó¦ÊôÓÚÎüÈÈ·´Ó¦µÄÊÇ( )

A. C6H12O6(ÆÏÌÑÌÇaq)£«6O26CO2£«6H2O

B. ÇâÑõ»¯ÄÆÈÜÒºÓëÑÎËáµÄÖкͷ´Ó¦

C. ·´Ó¦ÎïµÄ×ÜÄÜÁ¿´óÓÚÉú³ÉÎïµÄ×ÜÄÜÁ¿

D. ÆÆ»µ·´Ó¦ÎïÈ«²¿»¯Ñ§¼üËùÐèÄÜÁ¿´óÓÚÆÆ»µÉú³ÉÎïÈ«²¿»¯Ñ§¼üËùÐèÄÜÁ¿

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÒÑÖª£º¢ÙAµÄ²úÁ¿Í¨³£ÓÃÀ´ºâÁ¿Ò»¸ö¹ú¼ÒµÄʯÓÍ»¯¹¤Ë®Æ½£»¢Ú2CH3CHO£«O22CH3COOH¡£ÏÖÒÔAΪÖ÷ÒªÔ­ÁϺϳɻ¯ºÏÎïE£¬ÆäºÏ³É·ÏßÈçͼ1Ëùʾ¡£»Ø´ðÏÂÁÐÎÊÌ⣺

ͼ1

(1)д³öÏÂÁÐÎïÖʵĹÙÄÜÍÅÃû³Æ£º

B£º____________________£»D£º____________________¡£

(2)·´Ó¦¢ÜµÄ»¯Ñ§·½³ÌʽΪ________________________________________________£¬·´Ó¦ÀàÐÍ£º________¡£

(3)ijѧϰС×éÉè¼ÆÎïÖÊB´ß»¯Ñõ»¯µÄʵÑé×°ÖÃÈçÏ£¬¸ù¾Ýͼ2×°ÖûشðÎÊÌâ¡£

¼× ÒÒ ±û ¶¡

ͼ2

¢Ù×°Öü×׶ÐÎÆ¿ÖÐÊ¢·ÅµÄ¹ÌÌåÒ©Æ·¿ÉÄÜΪ________(Ìî×Öĸ)¡£

A Na2O2 B KCl C Na2CO3 D MnO2

¢ÚʵÑé¹ý³ÌÖУ¬±û×°ÖÃÓ²Öʲ£Á§¹ÜÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ_________________________¡£

¢ÛÎïÖÊBµÄ´ß»¯Ñõ»¯²úÎïÓëÆÏÌÑÌǾßÓÐÏàͬµÄÌØÕ÷·´Ó¦£¬½«ËùµÃµÄÑõ»¯²úÎïµÎ¼Óµ½ÐÂÖÆÇâÑõ»¯Í­Ðü×ÇÒºÖмÓÈÈ£¬ÏÖÏóΪ______________________________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿Ä³»¯Ñ§ÐËȤС×éͬѧչ¿ª¶ÔƯ°×¼ÁÑÇÂÈËáÄÆ(NaClO2)µÄÑо¿¡£

ʵÑé¢ñ£ºÖÆÈ¡NaClO2¾§Ìå¡£

ÒÑÖª£ºNaClO2±¥ºÍÈÜÒºÔÚζȵÍÓÚ38 ¡æʱÎö³öµÄ¾§ÌåÊÇNaClO2¡¤3H2O£¬¸ßÓÚ38 ¡æʱÎö³öµÄ¾§ÌåÊÇNaClO2£¬¸ßÓÚ60 ¡æʱNaClO2·Ö½â³ÉNaClO3ºÍNaCl¡£ÀûÓÃÏÂͼËùʾװÖýøÐÐʵÑé¡£

£¨1£©×°ÖâٵÄ×÷ÓÃÊÇ_______________

£¨2£©×°ÖâÚÖвúÉúClO2ÆøÌåµÄ»¯Ñ§·½³ÌʽΪ____________¡£

£¨3£©´Ó×°Öâܷ´Ó¦ºóµÄÈÜÒº»ñµÃ¾§ÌåNaClO2µÄ²Ù×÷²½ÖèΪ£º

¢Ù¼õѹ£¬55 ¡æÕô·¢½á¾§£»

¢Ú³ÃÈȹýÂË£»

¢Û____________________________________£»

¢ÜµÍÓÚ60 ¡æ¸ÉÔµÃµ½³ÉÆ·¡£

ʵÑé¢ò£º²â¶¨Ä³ÑÇÂÈËáÄÆÑùÆ·µÄ´¿¶È¡£Éè¼ÆÈçÏÂʵÑé·½°¸£¬²¢½øÐÐʵÑé¡£

¢Ù׼ȷ³ÆÈ¡ËùµÃÑÇÂÈËáÄÆÑùÆ·m gÓÚÉÕ±­ÖУ¬¼ÓÈëÊÊÁ¿ÕôÁóË®ºÍ¹ýÁ¿µÄµâ»¯¼Ø¾§Ì壬ÔÙµÎÈëÊÊÁ¿µÄÏ¡ÁòËᣬ³ä·Ö·´Ó¦(ÒÑÖª£ºClO2£­£«4I£­£«4H£«===2H2O£«2I2£«Cl£­)¡£½«ËùµÃ»ìºÏÒºÅä³É100 mL´ý²âÈÜÒº¡£

¢ÚÒÆÈ¡25.00 mL´ý²âÈÜÒºÓÚ׶ÐÎÆ¿ÖУ¬ÓÃc mol¡¤L£­1 Na2S2O3±ê×¼ÒºµÎ¶¨£¬ÖÁµÎ¶¨Öյ㡣Öظ´2´Î£¬²âµÃÏûºÄ±ê×¼ÈÜÒºµÄÌå»ýµÄƽ¾ùֵΪV mL(ÒÑÖª£ºI2 £«2S2O32£­===2I£­£«S4O62£­)¡£

£¨4£©ÑùÆ·ÖÐNaClO2µÄÖÊÁ¿·ÖÊýΪ__________(Óú¬m¡¢c¡¢VµÄ´úÊýʽ±íʾ)¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿Ä³Ê³Æ·¸ÉÔï¼ÁµÄÖ÷Òª³É·ÖÊÇÑõ»¯¸Æ£®Ñõ»¯¸Æ£¨CaO£©Ó¦ÊôÓÚ£¨¡¡¡¡£©

A. Ëá B. ¼î C. ÑÎ D. Ñõ»¯Îï

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÎªÂäʵ¡°ÎåË®¹²ÖΡ±£¬Ä³¹¤³§Äâ×ۺϴ¦Àíº¬NH4+·ÏË®ºÍ¹¤Òµ·ÏÆø(Ö÷Òªº¬N2¡¢CO2¡¢SO2¡¢NO¡¢CO£¬²»¿¼ÂÇÆäËû³É·Ö)£¬Éè¼ÆÁËÈçÏÂÁ÷³Ì£º

»Ø´ðÏÂÁÐÎÊÌ⣺£®

(1)¹ÌÌå1º¬ÓеÄÎïÖÊÊÇ_________¡¢CaCO3¡¢Ca(OH)2£»²¶»ñ¼ÁËù²¶»ñµÄÆøÌåÖ÷ÒªÊÇ_______________¡£

(2)ÈôXÊÇ¿ÕÆø£¬ÔòÉÏÊö·´Ó¦ÖÐNOºÍO2µÄÎïÖʵÄÁ¿Ö®±È×îºÃΪ___________£¬Èô¿ÕÆø¹ýÁ¿£¬µ¼ÖµĽá¹û»áÊÇ____________________________¡£

(3)Á÷³ÌÖд¦Àíº¬NH4+·Ïˮʱ·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ____________________________¡£ÆøÌå1Ò²¿Éͨ¹ý´ß»¯¼ÁÖ±½Óת»¯³ÉÎÞÎÛȾÆøÌ壬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ_____________________________________¡£

(4)Èô¹¤Òµ·ÏÆøÖÐÖ÷Òªº¬N2¡¢SO2£¬²»¿¼ÂÇÆäËû³É·Ö£¬Ôò¿ÉÓÃÏÂÁй¤ÒÕÉú²úNaHSO3¹ý±¥ºÍÈÜÒº[NaHSO3¹ý±¥ºÍÈÜÒºÊÇÉú²ú½¹ÑÇÁòËáÄÆ(Na2S2O5)µÄÔ­ÁÏ]¡£

pH=4.1ʱ£¬IÖз´Ó¦µÄ»¯Ñ§·½³ÌʽΪ_________________________£»¹¤ÒÕÖмÓÈëNa2CO3

¹ÌÌå¡¢²¢ÔÙ´ÎͨÈë·ÏÆøµÄÄ¿µÄÊÇ_______________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÃܱÕÈÝÆ÷ÖнøÐеķ´Ó¦Îª X(g)£«3Y(g) 2Z(g)£¬X¡¢Y¡¢Z µÄÆðʼŨ¶ÈÒÀ´ÎΪ0.1 mol / L£¬0.3 mol / L£¬0.2 mol / L£¬µ±·´Ó¦´ïƽºâʱ£¬¸÷ÎïÖʵÄŨ¶È¿ÉÄÜÊÇ ( )

A. X£½0.2 mol / L£¬Y£½0.6 mol / L B. Y£½0.5 mol / L»òY£½0.1 mol / L

C. Y£½0.6 mol / L D. Z£½0.4 mol / L

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿25 ¡æʱ£¬½«Ìå»ýΪVa¡¢pH£½aµÄijһԪËáHAÈÜÒºÓëÌå»ýΪVb¡¢pH£½bµÄijһԪ¼îBOHÈÜÒº»ìºÏ£¬Çë»Ø´ð¡£

£¨1£©Èôa + b£½14£¬2Va£½Vb£¬·´Ó¦ºóËùµÃÈÜÒºpH£½7¡£ÔòÉú³ÉµÄÑÎÈÜÒºÖУ¬Ò»¶¨·¢ÉúË®½âµÄÀë×Ó·½ ³ÌʽΪ_______

£¨2£©Èôa + b£½12£¬ËáÊÇÑÎËᣬ¼îÊÇKOH£¬·´Ó¦ºóËùµÃÈÜÒºpH£½7£¬ÔòVaÓëVbµÄ¹ØϵÊÇ_______

£¨3£©ÈôËáÊÇÑÎËᣬ¼îÊÇ°±Ë®£¬·´Ó¦ºóËùµÃÈÜÒºÖÐÀë×ÓŨ¶È´óС¹Øϵ²»¿ÉÄÜÊÇ_________£¨ÌîÐòºÅ£©

A£®c(Cl£­)£¾c(NH)£¾c(H+)£¾c(OH£­) B£®c(H+)£¾c(OH£­)£¾c(Cl£­)£¾c(NH)

C£®c(NH)£¾c(Cl£­)£¾c(OH£­)£¾c(H+) D£®c(Cl£­)£¾c(H+)£¾c(NH)£¾c(OH£­)

E£®c(Cl£­)£½c(NH)£¾c(H+)£½c(OH£­)

£¨4£©ÈôËáÊÇ´×Ëᣬ¼îÊÇNaOH£¬ÇÒ·´Ó¦ºó»ìºÏÈÜÒºÖÐc(CH3COO£­)£¾c(H+)£¬Ôò»ìºÏÈÜÒº¿ÉÄܳÊ_____£¨ÌîÐòºÅ£©

A£®ËáÐÔ B£®¼îÐÔ C£®ÖÐÐÔ D£®¶¼ÓпÉÄÜ

£¨5£©25 ¡æʱ£¬½«Ìå»ýVa£½200 mL£¬pH£½2µÄH2SO4ÈÜÒºÓëÌå»ýVb£½10 mL¡¢pH£½11µÄ°±Ë®ÈÜÒº»ìºÏ£¬Ç¡ºÃÍêÈ«·´Ó¦¡£Ôò´ËÌõ¼þÏ£¬°±Ë®µÄµçÀëƽºâ³£ÊýÊÇ____________

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸