£¨8·Ö£©(1998ÄêÉϺ£)Âȼµç½â±¥ºÍʳÑÎË®ÖÆÈ¡NaOHµÄ¹¤ÒÕÁ÷³ÌʾÒâͼÈçÏÂ

ÒÀ¾ÝÉÏͼ£¬Íê³ÉÏÂÁÐÌî¿Õ£º
(1)ÔÚµç½â¹ý³ÌÖУ¬ÓëµçÔ´Õý¼«ÏàÁ¬µÄµç¼«ÉÏËù·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ              £¬ÓëµçÔ´¸º¼«ÏàÁ¬µÄµç¼«¸½½ü£¬ÈÜÒºpH             (Ñ¡Ìî¡°²»±ä¡±¡°Éý¸ß¡±»ò¡°Ï½µ¡±)¡£
(2)¹¤ÒµÊ³Ñκ¬Ca2+¡¢Mg2+µÈÔÓÖÊ¡£¾«Öƹý³Ì·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ
                                ¡¢                                    ¡£
(3)Èç¹û´ÖÑÎÖÐSO42-º¬Á¿½Ï¸ß£¬±ØÐëÌí¼Ó±µÊÔ¼Á³ýÈ¥SO42-£¬¸Ã±µÊÔ¼Á¿ÉÒÔ
ÊÇ          (Ñ¡Ìîa£¬b£¬c¶àÑ¡¿Û·Ö)¡£
a£®Ba(OH)2        b£®Ba(NO3)2        c£®BaCl2
(4)ΪÓÐЧ³ýÈ¥Ca2+¡¢Mg2+¡¢SO42¡ª£¬¼ÓÈëÊÔ¼ÁµÄºÏÀí˳ÐòΪ      (Ñ¡Ìîa¡¢b¡¢c¶àÑ¡¿Û·Ö)
a£®ÏȼÓNaOH,ºó¼ÓNa2CO3£¬ÔÙ¼Ó±µÊÔ¼Á
b£®ÏȼÓNaOH£¬ºó¼Ó±µÊÔ¼Á£¬ÔÙ¼ÓNa2CO3
c£®ÏȼӱµÊÔ¼Á£¬ºó¼ÓNaOH£¬ÔÙ¼ÓNa2CO3
(5)ÍÑÑι¤ÐòÖÐÀûÓÃNaOHºÍNaClÔÚÈܽâ¶ÈÉϵIJîÒ죬ͨ¹ý      ¡¢ÀäÈ´¡¢              (Ìîд²Ù×÷Ãû³Æ)³ýÈ¥NaCl
(6)ÔÚ¸ôĤ·¨µç½âʳÑÎˮʱ£¬µç½â²Û·Ö¸ôΪÑô¼«ÇøºÍÒõ¼«Çø£¬·ÀÖ¹Cl2ÓëNaOH·´Ó¦£»²ÉÓÃÎÞ¸ôĤµç½âÀäµÄʳÑÎˮʱ£¬Cl2ÓëNaOH³ä·Ö½Ó´¥£¬²úÎï½öÊÇNaClOºÍH2£¬ÏàÓ¦µÄ»¯Ñ§·½³ÌʽΪ                                                           ¡£
(1)2Cl-£­2e£ß====Cl2£»Éý¸ß(2)Ca2++CO32¡ª=====CaCO3¡ýMg2++2OH-====Mg(OH)2¡ý
(3)a¡¢c  (4)b¡¢c  (5)Õô·¢£»¹ýÂË  (6)NaCl+H2O=====NaClO+H2¡ü»ò2NaCl+2H2O=======H2¡ü+Cl2¡ü+2NaOH    Cl2+2NaOH====NaCl+NaClO+H2O
ÂÔ
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÓÃÖÊÁ¿¾ùΪ100 gµÄÍ­×÷µç¼«£¬µç½âÏõËáÒøÈÜÒº£¬µç½âÒ»¶Îʱ¼äºó£¬Á½µç¼«µÄÖÊÁ¿²îΪ28 g£¬´ËʱÁ½µç¼«µÄÖÊÁ¿·Ö±ðΪ
A£®Ñô¼«100 g£¬Òõ¼«128 gB£®Ñô¼«93.6 g£¬Òõ¼«121.6 g
C£®Ñô¼«91.0 g£¬Òõ¼«119.0 gD£®Ñô¼«86.0 g£¬Òõ¼«114.0 g

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

ÔÚÈçͼÓÃʯī×÷µç¼«µÄµç½â³ØÖУ¬·ÅÈë500mLº¬Ò»ÖÖÈÜÖʵÄijÀ¶É«ÁòËáÑÎÈÜÒº½øÐеç½â£¬¹Û²ìµ½Aµç¼«±íÃæÓкìÉ«µÄ¹Ì̬ÎïÖÊÉú³É£¬Bµç¼«ÓÐÎÞÉ«ÆøÌåÉú³É¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©Çëд³öB¼«°åµÄÃû³Æ£º            µç¼«·´Ó¦Ê½           
д³öµç½âʱ·´Ó¦µÄ×ÜÀë×Ó·½³Ìʽ                          
£¨2£©Èôµ±ÈÜÒºÖеÄÔ­ÓÐÈÜÖÊÍêÈ«µç½âºó£¬Í£Ö¹µç½â£¬È¡³öAµç¼«£¬Ï´µÓ¡¢¸ÉÔï¡¢³ÆÁ¿¡¢µç¼«ÔöÖØ1.6g¡£µç½âºóÈÜÒºµÄpHΪ          £»ÒªÊ¹µç½âºóÈÜÒº»Ö¸´µ½µç½âÇ°µÄ״̬£¬ÔòÐè¼ÓÈë       £¬ÆäÖÊÁ¿Îª         g¡££¨¼ÙÉèµç½âÇ°ºóÈÜÒºµÄÌå»ý²»±ä£©
£¨3£©ÈôÔ­ÈÜҺΪ1L  K2SO4¡¢CuSO4µÄ»ìºÏÈÜÒº£¬ÇÒc£¨SO42-£©=" 2.0mol/L" £»ÈçͼװÖõç½â£¬µ±Á½¼«¶¼ÊÕ¼¯µ½22.4LÆøÌ壨±ê×¼×´¿ö£©Ê±£¬Í£Ö¹µç½â¡£
ÔòÔ­ÈÜÒºÖеÄc(K+)£½                              

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

£¨12·Ö£©A¡¢B¡¢CÈýÖÖÇ¿µç½âÖÊ£¬ËüÃÇÔÚË®ÖеçÀë³öµÄÀë×ÓÈçϱíËùʾ£º
ÑôÀë×Ó
Na+¡¢K+¡¢Cu2+
ÒõÀë×Ó
SO42¡ª¡¢OH£­
      ÏÂͼËùʾװÖÃÖУ¬¼×¡¢ÒÒ¡¢±ûÈý¸öÉÕ±­ÒÀ´Î·Ö±ðÊ¢·Å×ãÁ¿µÄAÈÜÒº¡¢×ãÁ¿µÄBÈÜÒº¡¢×ãÁ¿µÄCÈÜÒº£¬µç¼«¾ùΪʯīµç¼«¡£

½ÓͨµçÔ´£¬¾­¹ýÒ»¶Îʱ¼äºó£¬²âµÃÒÒÖÐcµç¼«ÖÊÁ¿Ôö¼ÓÁË16g¡£³£ÎÂϸ÷ÉÕ±­ÖÐÈÜÒºµÄpHÓëµç½âʱ¼ätµÄ¹ØϵͼÈçÉÏ¡£¾Ý´Ë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©MΪµçÔ´µÄ   ¼«£¨Ìîд¡°Õý¡±»ò¡°¸º¡±£©µç¼«bÉÏ·¢ÉúµÄµç¼«·´Ó¦Îª             £»
£¨2£©¼ÆËãµç¼«eÉÏÉú³ÉµÄÆøÌåÔÚ±ê׼״̬ϵÄÌå»ý£º           £»
£¨3£©Ð´³öÒÒÉÕ±­µÄµç½â³Ø·´Ó¦                       
£¨4£©Èç¹ûµç½â¹ý³ÌÖÐBÈÜÒºÖеĽðÊôÀë×ÓÈ«²¿Îö³ö£¬´Ëʱµç½âÄÜ·ñ¼ÌÐø½øÐУ¬ÎªÊ²Ã´£¿     
£¨5£©Èô¾­¹ýÒ»¶Îʱ¼äºó£¬²âµÃÒÒÖÐcµç¼«ÖÊÁ¿Ôö¼ÓÁË16g£¬ÒªÊ¹±û»Ö¸´µ½Ô­À´µÄ״̬£¬²Ù×÷ÊÇ                 ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

£¨14·Ö£©
ÈçÏÂͼËùʾ£¬A¡¢FΪʯīµç¼«£¬B¡¢EΪÌúƬµç¼«£®°´ÒªÇó»Ø´ðÏÂÁÐÎÊÌ⣮

£¨1£©´ò¿ªK2£¬ºÏ²¢K1£®BΪ     ¼«£¬AµÄµç¼«·´Ó¦Îª                £¬×îÖտɹ۲쵽µÄÏÖÏóÊÇ                                                   £®
£¨2£©´ò¿ªK1£¬ºÏ²¢K2£®EΪ       ¼«£¬F¼«µÄµç¼«·´Ó¦Îª                       £¬
¼ìÑéF¼«²úÉúÆøÌåµÄ·½·¨ÊÇ                                       £®
£¨3£©ÈôÍùUÐ͹ÜÖеμӷÓ̪£¬½øÐУ¨1£©£¨2£©²Ù×÷ʱ£¬A¡¢B¡¢E¡¢Fµç¼«ÖÜΧÄܱäºìµÄÊÇ                       £¬Ô­ÒòÊÇ                                £®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÏÂÁйØÓÚµç½â³ØµÄÐðÊö²»ÕýÈ·µÄÊÇ£¨   £©
A£®ÔÚµç½â³ØµÄÒõ¼«·¢Éú»¹Ô­·´Ó¦
B£®ÓëµçÔ´¸º¼«ÏàÁ¬µÄÊǵç½â³ØµÄÑô¼«
C£®µç½âÖÊÈÜÒºÖеÄÒõÀë×ÓÒÆÏòµç½â³ØµÄÑô¼«
D£®µç×Ó´ÓµçÔ´µÄ¸º¼«Ñص¼ÏßÁ÷Èëµç½â³ØµÄÒõ¼«

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÓöèÐԵ缫µç½âÒ»¶¨ÖÊÁ¿µÄijŨ¶ÈµÄNaClÈÜÒº£¬Ò»¶Îʱ¼äºóÍ£Ö¹µç½â£¬´ËʱÈô¼ÓÈë100g 36.5%µÄŨÑÎËᣬËùµÃÈÜÒºÕýºÃÓëÔ­ÈÜÒºÍêÈ«Ïàͬ£¬ÔòÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ
A£®µç½â¹ý³ÌÖУ¬Á½¼«ËùµÃµ½µÄÆøÌ壬ÔÚÏàͬµÄÌõ¼þÏÂÌå»ýÏàµÈ
B£®µç½â¹ý³ÌÖУ¬ÔÚÏàͬµÄÌõ¼þÏ£¬Ñô¼«ËùµÃµ½µÄÆøÌåµÄÌå»ý±ÈÒõ¼«µÄ´ó
C£®µç½â¹ý³ÌÖÐתÒƵç×ÓµÄÎïÖʵÄÁ¿Ô¼Îª8mol
D£®Ô­Ä³Å¨¶ÈµÄNaClÈÜÒºÖÐÈÜÓÐ117g NaCl

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ
A£®µç½âÖÊÓë·Çµç½âÖʵı¾ÖÊÇø±ð£¬ÊÇÔÚË®ÈÜÒº»òÈÛ»¯×´Ì¬ÏÂÄÜ·ñµçÀë
B£®Ç¿µç½âÖÊÓëÈõµç½âÖʵı¾ÖÊÇø±ð£¬ÊÇÆäË®ÈÜÒºµ¼µçÐԵļõÈõ
C£®Ëá¡¢¼îºÍÑÎÀ඼ÊôÓÚµç½âÖÊ£¬ÆäËû»¯ºÏÎﶼÊǷǵç½âÖÊ
D£®³£¼ûµÄÇ¿Ëᡢǿ¼îºÍ´ó²¿·ÖÑζ¼ÊÇÇ¿µç½âÖÊ£¬ÆäËû»¯ºÏÎﶼÊǷǵç½âÖÊ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÈçͼËùʾµÄ×°Öã¬C¡¢D¡¢E¡¢F¡¢X¡¢Y¶¼ÊǶèÐԵ缫¡£½«µçÔ´½Óͨºó£¬ÏòÒÒÖеÎÈë·Ó̪ÈÜÒº£¬ÔÚF¼«¸½½üÏÔºìÉ«¡£ÔòÒÔÏÂ˵·¨²»ÕýÈ·µÄÊÇ

A£®µçÔ´BÊǸº¼«
B£®¼×ÒÒ×°ÖõÄC¡¢D¡¢E¡¢Fµç¼«¾ùÓе¥ÖÊÉú³É£¬ÆäÎïÖʵÄÁ¿Ö®±ÈΪ1©U2©U2©U2
C£®ÓûÓñû×°ÖøøÍ­¶ÆÒø£¬HÓ¦¸ÃÊÇAg£¬µç¶ÆҺѡÔñAgNO3ÈÜÒº
D£®×°Öö¡ÖÐX¼«¸½½üºìºÖÉ«±ädz£¬ËµÃ÷ÇâÑõ»¯Ìú½ºÁ£Îü¸½ÕýµçºÉ

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸