ÏÂÁи÷×éÀë×ÓÄÜÔÚÖ¸¶¨»·¾³ÖдóÁ¿¹²´æµÄÊÇ£¨   £©

A£®ÔÚc(HCO3£­)=0.1 mol·L-1µÄÈÜÒºÖУºNH4+¡¢AlO2£­¡¢Cl£­¡¢NO3£­

B£®ÔÚÓÉË®µçÀë³öµÄc(H+)=1¡Á10-12 mol·L-1µÄÈÜÒºÖУºFe2+¡¢ClO£­¡¢Na+¡¢SO42£­

C£®ÔÚ¼ÓÈëÂÁ·Û²úÉúH2µÄÈÜÒºÖУºSO42£­¡¢NO3£­¡¢Na+¡¢NH4+

D£®ÔÚʹºìɫʯÈïÊÔÖ½±äÀ¶µÄÈÜÒºÖУºSO32£­¡¢CO32£­¡¢Na+¡¢K+

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


»¯Ñ§ÒÑÉøÍ¸µ½ÈËÀàÉú»îµÄ¸÷¸ö·½Ãæ¡£ÏÂÁÐ˵·¨²»ÕýÈ·µÄÊÇ

A£®°¢Ë¾Æ¥ÁÖ¾ßÓнâÈÈÕòÍ´×÷ÓÃ

B£®¿ÉÒÔÓÃSi3N4¡¢Al2O3ÖÆ×÷¸ßνṹÌÕ´ÉÖÆÆ·

C£®ÔÚÈ뺣¿ÚµÄ¸ÖÌúÕ¢ÃÅÉÏ×°Ò»¶¨ÊýÁ¿µÄÍ­¿é¿É·ÀÖ¹Õ¢Ãű»¸¯Ê´

D£®½ûֹʹÓÃËÄÒÒ»ùǦ×÷ÆûÓÍ¿¹±¬Õð¼Á£¬¿É¼õÉÙÆû³µÎ²ÆøÎÛȾ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÏÂÁÐÌþÖо­´ß»¯¼ÓÇ⣬²»Äܵõ½2-¼×»ù¶¡ÍéµÄÊÇ

A£®2-¼×»ù-1-¶¡Ï©       B£®2-¼×»ù-2-¶¡Ï©   

C£®3-¼×»ù-1-¶¡È²       D£®3£¬3-¶þ¼×»ù-1-¶¡È²

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÏÂÁйØÓÚµç½â¾«Á¶Í­µÄÐðÊöÖв»ÕýÈ·µÄÊÇ£¨    £©

A.´ÖÍ­°å£ºÑô¼«

B.µç½âʱ£¬Ñô¼«·¢ÉúÑõ»¯·´Ó¦£¬¶øÒõ¼«·¢ÉúµÄ·´Ó¦ÎªCu2++2e-====Cu

C.´ÖÍ­ÖÐËùº¬Na¡¢Fe¡¢ZnµÈÔÓÖÊ£¬µç½âºóÒÔµ¥ÖÊÐÎʽ³Á»ý²Ûµ×£¬ÐγÉÑô¼«Äà

D.µç½âÍ­µÄ´¿¶È¿É´ï99.95 %¡ª99.98 %

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


º¬µÄ¹¤ÒµËáÐÔ·ÏË®»áÔì³É¸õÎÛȾ£¬ÅÅ·ÅǰҪ½øÐÐÈçÏ´¦Àí£º£¨¢ñ£©Íù¹¤Òµ·ÏË®ÖмÓÈëÊÊÁ¿µÄNaCl½Á°è¾ùÔÈ£»£¨¢ò£©ÒÔFe×öÁ½µç¼«½øÐеç½â£¬´Ó¶øÊ¹ÈÜÒºµÄpH²»¶ÏÉý¸ß£¬·ÏË®ÓÉËáÐÔת»¯Îª¼îÐÔ£¬¾­¹ýÒ»¶Îʱ¼äÓÐCr£¨OH£©3ºÍFe£¨OH£©3³Áµí²úÉú£º(¢ó)¹ýÂË»ØÊÕ³Áµí£¬·ÏË®´ïµ½Åŷűê×¼¡£

£¨1£©ÔÚµç½â¹ý³ÌÖУ¬ÈÜÒºpH²»¶ÏÉý¸ßµÄÔ­Òò¿ÉÄÜÊÇ£¨    £©

¢Ùµç½âʱ·ÏË®µÄÌå»ý²»¶Ï¼õС  ¢Úµç½âʱH+ÔÚÒõ¼«±»»¹Ô­  ¢Û×÷ΪÑô¼«µÄFe²»¶ÏÈÜ½â  ¢Üת»¯ÎªCr3+ʱÏûºÄÁËH+  ¢ÝNaClÔÚµç½âʱת»¯³ÉÁËNaOH

A.¢Ý          B.¢Ú¢Ü               C.¢Ú¢Û¢Ü               D.¢Ù¢Ú¢Û¢Ü

£¨2£©Á½¼«·¢Éú·´Ó¦µÄµç¼«·´Ó¦Ê½Îª

£¨3£©Ð´³ö±ä³ÉCr3+µÄÀë×Ó·½³Ìʽ__________________________________¡£

£¨4£©______________£¨Ìî¡°ÄÜ¡±»ò¡°²»ÄÜ¡±£©¸ÄÓÃʯīµç¼«½øÐеç½â¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


°´ÒÔÏÂʵÑé·½°¸¿É´Óº£Ñó¶¯Îï±úº£ÇÊÖÐÌáÈ¡¾ßÓп¹Ö×Áö»îÐÔµÄÌìÈ»²úÎï¡£

ÏÂÁÐ˵·¨´íÎóµÄÊÇ£¨   £©

A£®²½Öè(1)ÐèÒª¹ýÂË×°Öà           B£®²½Öè(2)ÐèÒªÓõ½·ÖҺ©¶·

C£®²½Öè(3)ÐèÒªÓõ½ÛáÛö            D£®²½Öè(4)ÐèÒªÕôÁó×°ÖÃ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


½«51.2gCuÍêÈ«ÈÜÓÚÊÊÁ¿Å¨ÏõËáÖУ¬ÊÕ¼¯µ½µªµÄÑõ»¯Îº¬NO¡¢N2O4¡¢NO2£©µÄ»ìºÏÎï¹²0.8mol£¬ÕâÐ©ÆøÌåÇ¡ºÃÄܱ»500ml 2mol/LNaOHÈÜÒºÍêÈ«ÎüÊÕ£¬Éú³Éº¬NaNO3ºÍNaNO2µÄÑÎÈÜÒº£¬ÆäÖÐNaNO2µÄÎïÖʵÄÁ¿Îª£¨   £©

A£®0.2mol              B£®0.6mol             C£®0.8mol             D£®1.0mol

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÏÂÁйØÓÚ¹¤ÒµÉú²úµÄ˵·¨ÖУ¬²»ÕýÈ·µÄÊÇ(¡¡¡¡)

A£®¹¤ÒµÉÏ£¬Óý¹Ì¿ÔÚµç¯Öл¹Ô­¶þÑõ»¯¹èµÃµ½º¬ÔÓÖʵĴֹè

B£®Éú²úÆÕͨˮÄàµÄÖ÷ÒªÔ­ÁÏÓÐʯ»Òʯ¡¢Ê¯Ó¢ºÍ´¿¼î

C£®¹¤ÒµÉϽ«´ÖÍ­½øÐо«Á¶£¬Ó¦½«´ÖÍ­Á¬½ÓÔÚµçÔ´µÄÕý¼«

D£®Ôڸ߯Á¶ÌúµÄ·´Ó¦ÖУ¬Ò»Ñõ»¯Ì¼×÷»¹Ô­¼Á

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


Na2S2O3ÊÇÖØÒªµÄ»¯¹¤Ô­ÁÏ£¬Ò×ÈÜÓÚË®£¬ÔÚÖÐÐÔ»ò¼îÐÔ»·¾³ÖÐÎȶ¨¡£

¢ñ.ÖÆ±¸Na2S2O3¡¤5H2O

·´Ó¦Ô­Àí£ºNa2SO3(aq)£«S(s)Na2S2O3(aq)

ʵÑé²½Ö裺

¢Ù³ÆÈ¡15 g Na2SO3¼ÓÈëÔ²µ×ÉÕÆ¿ÖУ¬ÔÙ¼ÓÈë80 mLÕôÁóË®¡£ÁíÈ¡5 gÑÐϸµÄÁò·Û£¬ÓÃ3 mLÒÒ´¼Èóʪ£¬¼ÓÈëÉÏÊöÈÜÒºÖС£

¢Ú°²×°ÊµÑé×°ÖÃ(ÈçͼËùʾ£¬²¿·Ö¼Ð³Ö×°ÖÃÂÔÈ¥)£¬Ë®Ô¡¼ÓÈÈ£¬Î¢·Ð60 min¡£

¢Û³ÃÈȹýÂË£¬½«ÂËҺˮԡ¼ÓÈÈŨËõ£¬ÀäÈ´Îö³öNa2S2O3¡¤5H2O£¬¾­¹ýÂË¡¢Ï´µÓ¡¢¸ÉÔµÃµ½²úÆ·¡£

»Ø´ðÎÊÌ⣺

(1)Áò·ÛÔÚ·´Ó¦Ç°ÓÃÒÒ´¼ÈóʪµÄÄ¿µÄÊÇ__________________________¡£

(2)ÒÇÆ÷aµÄÃû³ÆÊÇ________£¬Æä×÷ÓÃÊÇ____________________¡£

(3)²úÆ·ÖгýÁËÓÐδ·´Ó¦µÄNa2SO3Í⣬×î¿ÉÄÜ´æÔÚµÄÎÞ»úÔÓÖÊÊÇ______________¡£¼ìÑéÊÇ·ñ´æÔÚ¸ÃÔÓÖʵķ½·¨ÊÇ____________________________¡£

(4)¸ÃʵÑéÒ»°ã¿ØÖÆÔÚ¼îÐÔ»·¾³Ï½øÐУ¬·ñÔò²úÆ··¢»Æ£¬ÓÃÀë×Ó·´Ó¦·½³Ìʽ±íʾÆäÔ­Òò£º________________________________________________________________________

________________________________________________________________________¡£

¢ò.²â¶¨²úÆ·´¿¶È

׼ȷ³ÆÈ¡W g²úÆ·£¬ÓÃÊÊÁ¿ÕôÁóË®Èܽ⣬ÒÔµí·Û×÷ָʾ¼Á£¬ÓÃ0.100 0 mol¡¤L£­1µâµÄ±ê×¼ÈÜÒºµÎ¶¨¡£

·´Ó¦Ô­ÀíΪ2S2O£«I2===S4O£«2I£­

(5)µÎ¶¨ÖÁÖÕµãʱ£¬ÈÜÒºÑÕÉ«µÄ±ä»¯£º____________________________________________¡£

(6)µÎ¶¨ÆðʼºÍÖÕµãµÄÒºÃæÎ»ÖÃÈçͼ£¬ÔòÏûºÄµâµÄ±ê×¼ÈÜÒºÌå»ýΪ__________mL¡£²úÆ·µÄ´¿¶ÈΪ(ÉèNa2S2O3¡¤5H2OÏà¶Ô·Ö×ÓÖÊÁ¿ÎªM)______________¡£

¢ó.Na2S2O3µÄÓ¦ÓÃ

(7)Na2S2O3»¹Ô­ÐÔ½ÏÇ¿£¬ÔÚÈÜÒºÖÐÒ×±»Cl2Ñõ»¯³ÉSO£¬³£ÓÃ×÷ÍÑÂȼÁ£¬¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽΪ____________________________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸