£¨10·Ö£©½¹ÑÇÁòËáÄÆ£¨Na2S2O5£©Îª°×É«»ò»ÆÉ«½á¾§·ÛÄ©»òС½á¾§¡£ÆäÐÔÖÊ»îÆÃ£¬¾ßÓÐÇ¿»¹Ô­ÐÔ£¬ÔÚʳƷ¼Ó¹¤ÖÐ×÷·À¸¯¼Á¡¢Æ¯°×¼Á¡¢ÊèËɼÁ¡£Ä³ÊµÑéС×éÄâ²ÉÓÃÈçͼ1×°Öã¨ÊµÑéǰÒѳý¾¡×°ÖÃÄÚµÄ¿ÕÆø£©À´ÖÆÈ¡½¹ÑÇÁòËáÄÆ£¨Na2S2O5£©£®

£¨1£©×°ÖÃIÊÇÓÃÑÇÁòËáÄÆ¹ÌÌåºÍŨÁòËáÖÆ±¸¶þÑõ»¯ÁòÆøÌ壬¸Ã×°ÖÃÖз´Ó¦µÄ»¯Ñ§·½³ÌʽΪ £®Èç¹ûÏë¿ØÖÆ·´Ó¦ËÙ¶È£¬Èçͼ2ÖпÉÑ¡Óõķ¢Éú×°ÖÃÊÇ £¨Ìîд×Öĸ£©£®

£¨2£©×°ÖâòÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º2 NaHSO3=Na2S2O5 + H2O£¬µ±ÓÐNa2S2O5¾§ÌåÎö³ö£¬Òª»ñµÃÒÑÎö³öµÄ¾§Ìå¿É²ÉÈ¡µÄ·ÖÀë·½·¨ÊÇ £»Ä³Í¬Ñ§ÐèÒª420mL0.1mol/L½¹ÑÇÁòËáÄÆÈÜÒºÀ´Ñо¿ÆäÐÔÖÊ£¬ÅäÖÆÊ±Ðè³ÆÁ¿½¹ÑÇÁòËáÄÆµÄÖÊÁ¿Îª £»ÅäÖÆÊ±³ýÓõ½ÍÐÅÌÌìÆ½¡¢Ò©³×¡¢ÉÕ±­¡¢²£Á§°ôµÈÒÇÆ÷Í⣬»¹±ØÐëÓõ½µÄ²£Á§ÒÇÆ÷ÊÇ £®

£¨3£©×°ÖâóÓÃÓÚ´¦ÀíÎ²Æø£¬¿ÉÑ¡ÓÃÈçͼ3µÄ×îºÏÀí×°Ö㨼гÖÒÇÆ÷ÒÑÂÔÈ¥£©Îª £¨ÌîÐòºÅ£©£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2015-2016ѧÄêºþ±±Ê¡»ÆÊ¯ÊиßÒ»ÉÏѧÆÚ10ÔÂÔ¿¼»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÌî¿ÕÌâ

ÏÖÓÐm gÄ³ÆøÌ壬ËüÓÉ˫ԭ×Ó·Ö×Ó¹¹³É£¬ËüµÄĦ¶ûÖÊÁ¿ÎªM g / mol¡£Èô°¢·ü¼ÓµÂÂÞ³£ÊýµÄÖµÓÃNA±íʾ£¬Ôò£º£¨×¢ÒâÌîдµ¥Î»£©

£¨1£©¸ÃÆøÌåµÄÎïÖʵÄÁ¿Îª______________¡£

£¨2£©¸ÃÆøÌåËùº¬Ô­×Ó×ÜÊýΪ________________¡£

£¨3£©¸ÃÆøÌåÔÚ±ê×¼×´¿öϵÄÌå»ýΪ__________________¡£

£¨4£©¸ÃÆøÌåÈÜÓÚ1 LË®ÖÐ(²»¿¼ÂÇ·´Ó¦)£¬ÆäÈÜÒºÖÐÈÜÖʵÄÖÊÁ¿·ÖÊýΪ_ ______ __¡£

£¨5£©¸ÃÆøÌåÈÜÓÚË®ºóÐγÉV LÈÜÒº£¬ÆäÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈΪ ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2016½ìºþ±±Ê¡ÔæÑôÊиßÈýÉÏѧÆÚµÚÒ»´ÎÔ¿¼»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÍƶÏÌâ

£¨15·Ö£©³£¼ûµÄÎåÖÖÑÎX¡¢Y¡¢Z¡¢M¡¢N£¬ËüÃǵÄÒõÀë×Ó¿ÉÄÜÊÇSO42-¡¢Cl£­¡¢NO3-¡¢CO32-£¬ÑôÀë×Ó¿ÉÄÜÊÇAg£«¡¢NH4+¡¢Na£«¡¢Al3£«¡¢Cu2£«¡¢Ba2£«¡¢Fe3£«£¬ÒÑÖª£º

¢ÙMµÄÑæÉ«·´Ó¦³Ê»ÆÉ«¡£

¢ÚÎåÖÖÑξùÈÜÓÚË®£¬Ë®ÈÜÒº¾ùΪÎÞÉ«¡£

¢ÛXµÄÈÜÒº³ÊÖÐÐÔ£¬Y¡¢Z¡¢NµÄÈÜÒº³ÊËáÐÔ£¬MµÄÈÜÒº³Ê¼îÐÔ¡£

¢ÜÈôÔÚÕâÎåÖÖÑεÄÈÜÒºÖзֱð¼ÓÈëBa(NO3)2ÈÜÒº£¬Ö»ÓÐX¡¢ZµÄÈÜÒº²»²úÉú³Áµí¡£

¢ÝÈôÔÚÕâÎåÖÖÑεÄÈÜÒºÖУ¬·Ö±ð¼ÓÈ백ˮ£¬NºÍZµÄÈÜÒºÖÐÉú³É³Áµí£¬¼ÌÐø¼Ó°±Ë®£¬ZÖгÁµíÏûʧ¡£

¢Þ°ÑXµÄÈÜÒº·Ö±ð¼ÓÈëµ½Y¡¢Z¡¢NµÄÈÜÒºÖУ¬¾ùÄÜÉú³É²»ÈÜÓÚÏ¡ÏõËáµÄ³Áµí¡£

Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©ÎåÖÖÑÎÖУ¬Ò»¶¨²»º¬ÓеÄÑôÀë×ÓÊÇ_________£»Ëùº¬ÒõÀë×ÓÏàͬµÄÁ½ÖÖÑεĻ¯Ñ§Ê½ÊÇ_____________¡£

£¨2£©MµÄ»¯Ñ§Ê½Îª_______________£¬MÈÜÒºÏÔ¼îÐÔµÄÔ­ÒòÊÇ________________(ÓÃÀë×Ó·½³Ìʽ±íʾ)¡£

£¨3£©XºÍZµÄÈÜÒº·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ______________£»NºÍ°±Ë®·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ__________________¡£

£¨4£©ÈôÒª¼ìÑéYÖÐËùº¬µÄÑôÀë×Ó£¬ÕýÈ·µÄʵÑé·½·¨ÊÇ______________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2015-2016ѧÄêºÓ±±Ê¡¸ß¶þÉÏѧÆÚµÚ¶þ´ÎÔ¿¼»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÑ¡ÔñÌâ

¶ÔÓÚ¿ÉÄæ·´Ó¦A(g)£«3B(s)2C(g)£«2D(g)£¬ÔÚ²»Í¬Ìõ¼þÏµĻ¯Ñ§·´Ó¦ËÙÂÊÈçÏ£¬ÆäÖбíʾµÄ·´Ó¦ËÙÂÊ×î¿ìµÄÊÇ

A£®v(A)£½0.5 mol¡¤L£­1¡¤min£­1 B£®v(B)£½1.2 mol¡¤L£­1¡¤s£­1

C£®v(D)£½0.4 mol¡¤L£­1¡¤min£­1 D£®v(C)£½0.1 mol¡¤L£­1¡¤s£­1

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2016½ì¸£½¨Ê¡¸ßÈýÉÏѧÆÚ10ÔµÚÒ»´ÎÔ¿¼»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÑ¡ÔñÌâ

ÓÐA¡¢BÁ½¸öÍêÈ«ÏàͬµÄ×°Öã¬Ä³Ñ§Éú·Ö±ðÔÚËüÃǵIJà¹ÜÖÐ×°Èë1.06 g Na2CO3ºÍ0.84gNaHCO3£¬A¡¢BÖзֱðÓÐ10 mLÏàͬŨ¶ÈµÄÑÎËᣬ½«Á½¸ö²à¹ÜÖеÄÎïÖÊͬʱµ¹Èë¸÷×ÔµÄÊÔ¹ÜÖÐ

ÏÂÁÐÐðÊöÕýÈ·µÄÊÇ

A£®A×°ÖÃµÄÆøÇòÅòÕÍËÙÂÊ´ó

B£®Èô×îÖÕÁ½ÆøÇòÌå»ýÏàͬ£¬ÔòÑÎËáµÄŨ¶ÈÒ»¶¨´óÓÚ»òµÈÓÚ2 mol/L

C£®Èô×îÖÕÁ½ÆøÇòÌå»ý²»Í¬£¬ÔòÑÎËáµÄŨ¶ÈÒ»¶¨Ð¡ÓÚ»òµÈÓÚ1 mol/L

D£®×îÖÕÁ½ÊÔ¹ÜÖÐNa+¡¢Cl-µÄÎïÖʵÄÁ¿Ò»¶¨Ïàͬ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2015-2016ѧÄêºÓÄÏÊ¡¸ßÒ»ÉϵÚÒ»´ÎÁª¿¼»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÌî¿ÕÌâ

£¨6·Ö£©Ò»¶¨Á¿ÇâÆøÔÚÂÈÆøÖÐȼÉÕ£¬ËùµÃ»ìºÏÆøÌåÓÃ100 mL 2.00 moL¡¤L-1NaOHÈÜҺǡºÃÍêÈ«ÎüÊÕ£¬²âµÃÈÜÒºÖк¬ÓÐNaClOµÄÎïÖʵÄÁ¿Îª0.05mol¡£ÊÔ¼ÆËãÏÂÁÐÎÊÌ⣺

£¨1£©ËùµÃÈÜÒºÖÐCl-µÄÎïÖʵÄÁ¿Îª £»

£¨2£©ËùÓÃÂÈÆøºÍ²Î¼Ó·´Ó¦µÄÇâÆøµÄÎïÖʵÄÁ¿Ö®±Èn(C12)£ºn(H2)Ϊ £»

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2015-2016ѧÄêɽ¶«Ê¡¸ßÒ»ÉÏѧÆÚµÚÒ»´ÎÔ¿¼»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÑ¡ÔñÌâ

¶ÔÓÚÎïÖʵÄÁ¿ÏàͬµÄÁòËáºÍÁ×ËᣨH3PO4£©£¬ÏÂÁÐ˵·¨²»ÕýÈ·µÄÊÇ

A£®Ô­×Ó×ÜÊý²»Ïàͬ B£®ÑõÔ­×ÓÊýÏàͬ

C£®·Ö×ÓÊýÏàͬ D£®ÇâÔ­×ÓÊýÏàͬ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2016½ìºþÄÏÊ¡³£µÂÊиßÈýÉÏѧÆÚ10ÔÂÔ¿¼»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÑ¡ÔñÌâ

½ñÓÐÒ»»ìºÏÎïµÄË®ÈÜÒº£¬¿ÉÄÜ´óÁ¿º¬ÓÐÒÔÏÂÀë×ÓÖеÄÈô¸ÉÖÖ£ºK+¡¢NH4+¡¢Cl¡ª¡¢Mg2+¡¢Ba2+¡¢CO32-¡¢SO42-£¬ÏÖÈ¡Èý·Ý100mLÈÜÒº½øÐÐÈçÏÂʵÑ飺

£¨1£©µÚÒ»·Ý¼ÓÈëAgNO3ÈÜÒºÓгÁµí²úÉú

£¨2£©µÚ¶þ·Ý¼Ó×ãÁ¿NaOHÈÜÒº¼ÓÈȺó£¬ÊÕ¼¯µ½ÆøÌå0.04mol

£¨3£©µÚÈý·Ý¼Ó×ãÁ¿BaCl2ÈÜÒººó£¬µÃ¸ÉÔï³Áµí6.27g£¬¾­×ãÁ¿ÑÎËáÏ´µÓ¡¢¸ÉÔïºó£¬³ÁµíÖÊÁ¿Îª2.33g¡£

¸ù¾ÝÉÏÊöʵÑ飬ÒÔÏÂÕë¶ÔÔ­ÈÜÒºµÄÍÆ²âÕýÈ·µÄÊÇ

A£®Cl¡ªÒ»¶¨²»´æÔÚ B£®K+Ò»¶¨´æÔÚ

C£®Mg2+Ò»¶¨´æÔÚ D£®Ba2+¿ÉÄÜ´æÔÚ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2016½ì°²»ÕÊ¡°²ÇìÊиßÈýÉÏѧÆÚµÚÒ»´ÎÖʼ컯ѧÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÑ¡ÔñÌâ

½üÄêÀ´£¬¼Ó¡°µâ¡±Ê³Ñν϶àµÄʹÓÃÁ˵âËá¼Ø£¨KIO3£©£¬µâËá¼ØÔÚ¹¤ÒµÉÏ¿ÉÓõç½â·¨ÖÆÈ¡¡£ÒÔʯīºÍ²»Ðâ¸ÖΪµç¼«£¬ÒÔKIÈÜҺΪµç½âÒº£¬ÔÚÒ»¶¨Ìõ¼þϵç½â£¬·´Ó¦·½³ÌʽΪ£ºKI+3H2O==KIO3+3H2¡ü¡£ÏÂÁÐÓйØËµ·¨ÕýÈ·µÄÊÇ

A£®µç½âʱ£¬Ê¯Ä«×÷Òõ¼«£¬²»Ðâ¸Ö×÷Ñô¼«

B£®µç½âʱ£¬Ñô¼«·´Ó¦ÊÇ£ºI--6e-+3H2O=IO-3+6H+

C£®ÈÜÒºµ÷½ÚÖÁÇ¿ËáÐÔ£¬¶ÔÉú²úÓÐÀû

D£®µç½â¹ý³ÌÖÐÈÜÒºµÄpHÖð½¥¼õС

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸