9£®Ä³³ÎÇåÈÜÒºÖпÉÄܺ¬ÓÐÏÂÁÐÀë×Ó£ºK+¡¢NH4+¡¢Fe2+¡¢Fe3+¡¢Ba2+¡¢SO42¡¥¡¢HCO3¡¥¡¢Cl¡¥£¬
Ϊ¼ø±ðÆäÖк¬ÓеÄÀë×Ó£¬½øÐÐÈçÏÂʵÑ飺
¢ÙÓò£Á§°ôպȡԭÈÜÒºµÎÔÚpHÊÔÖ½ÉÏ£¬ÊÔÖ½ÏÔºìÉ«£»
¢ÚÁíÈ¡ÉÙÁ¿Ô­ÈÜÒº¼ÓÈëBaCl2ÈÜÒº£¬Éú³É²»ÈÜÓÚÏ¡ÏõËáµÄ°×É«³Áµí£»
¢ÛÈ¡¢ÚÖÐÉϲãÇåÒº¼ÓÈëËữµÄÏõËáÒøÈÜÒº£¬Éú³É°×É«³Áµí£®
ÏÂÁйØÓÚÔ­ÈÜÒºµÄ˵·¨ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®Ô­ÈÜÒºÖÐÒ»¶¨²»´æÔÚBa2+ºÍHCO3-
B£®È¡¢ÛÖÐÂËÒº¼ÓÈëKSCN£¬ÈÜÒºÏÔºìÉ«£¬ÔòÔ­ÈÜÒºÒ»¶¨ÓÐFe3+
C£®Ô­ÈÜÒºÖÐÒ»¶¨´æÔÚSO42-ºÍCl-
D£®ÎªÈ·¶¨Ô­ÈÜÒºÖÐÊÇ·ñº¬ÓÐK+£¬¿Éͨ¹ýÑæÉ«·´Ó¦Ö±½Ó¹Û²ìÑæÉ«ÊÇ·ñΪ×ÏÉ«À´È·¶¨

·ÖÎö ¢ÙÓò£Á§°ôպȡԭÈÜÒºµÎÔÚpHÊÔÖ½ÉÏ£¬ÊÔÖ½ÏÔºìÉ«£¬ÔòÈÜÒº³ÊËáÐÔ£¬ÔòÒ»¶¨²»´æÔÚHCO3-£¬ÈÜÒºÖпÉÄÜ´æÔÚNH4+¡¢Fe2+¡¢Fe3+£»
¢ÚÁíÈ¡ÉÙÁ¿Ô­ÈÜÒº¼ÓÈëBaCl2ÈÜÒº£¬Éú³É²»ÈÜÓÚÏ¡ÏõËáµÄ°×É«³Áµí£¬¸Ã°×É«³ÁµíΪÁòËá±µ£¬ÔòÈÜÒºÖÐÒ»¶¨´æÔÚSO42-£¬¸ù¾ÝÀë×Ó¹²´æÔòÒ»¶¨²»´æÔÚBa2+£»
¢ÛÈ¡¢ÚÖÐÉϲãÇåÒº¼ÓÈëËữµÄÏõËáÒøÈÜÒº£¬Éú³É°×É«³Áµí£¬ÓÉÓÚ¢ÚÖмÓÈëÂÈ»¯±µÈÜÒºÒý½øÁËÂÈÀë×Ó£¬ÔòÎÞ·¨ÅжÏÔ­ÈÜÒºÖÐÊÇ·ñº¬ÓÐCl-£»
¸ù¾Ý·ÖÎö¿ÉÖª£¬ÈÜÒºÖÐÒ»¶¨²»´æÔÚBa2+ºÍHCO3-£¬Ò»¶¨´æÔÚSO42-£¬ÎÞ·¨È·¶¨ÊÇ·ñº¬ÓÐCl-£¬
¾Ý´Ë½øÐнâ´ð£®

½â´ð ½â£º¸ù¾Ý¢Ù¿ÉÖªÈÜÒº³ÊËáÐÔ£¬ÔòÒ»¶¨²»´æÔÚHCO3-£¬ÈÜÒºÖпÉÄÜ´æÔÚNH4+¡¢Fe2+¡¢Fe3+£»¸ù¾Ý¢Ú¿ÉÖªÈÜÒºÖÐÒ»¶¨´æÔÚSO42-£¬¸ù¾ÝÀë×Ó¹²´æÔòÒ»¶¨²»´æÔÚBa2+£»
ÓÉÓÚ¢ÚÖмÓÈëÂÈ»¯±µÈÜÒºÒý½øÁËÂÈÀë×Ó£¬Ôò¸ù¾Ý¢ÛÎÞ·¨ÅжÏÔ­ÈÜÒºÖÐÊÇ·ñº¬ÓÐCl-£¬
A£®¸ù¾Ý·ÖÎö¿ÉÖª£¬ÈÜÒºÖÐÒ»¶¨²»´æÔÚBa2+ºÍHCO3-£¬¹ÊAÕýÈ·£»
B£®È¡¢ÛÖÐÂËÒº¼ÓÈëKSCN£¬ÈÜÒºÏÔºìÉ«£¬ËµÃ÷¢ÛÖÐÂËÒºÖÐÒ»¶¨ÓÐFe3+£¬µ«ÊÇÓÉÓÚ¼ÓÈëÏõËᣬÑÇÌúÀë×ӻᱻÑõ»¯³ÉÌúÀë×Ó£¬ËùÒÔÎÞ·¨ÅжÏÔ­ÈÜÒºÖÐÊÇ·ñº¬ÓÐFe3+£¬¹ÊB´íÎó£»
C£®Ô­ÈÜÒºÖÐÒ»¶¨´æÔÚSO42-£¬ÎÞ·¨È·¶¨ÊÇ·ñº¬ÓÐCl-£¬¹ÊC´íÎó£»
D£®È·¶¨Ô­ÈÜÒºÖÐÊÇ·ñº¬ÓÐK+£¬¿Éͨ¹ýÑæÉ«·´Ó¦Í¸¹ýÀ¶É«îܲ£Á§¹Û²ìÑæÉ«ÊÇ·ñΪ×ÏÉ«À´È·¶¨£¬¹ÊD´íÎó£»
¹ÊÑ¡A£®

µãÆÀ ±¾Ì⿼²éÁ˳£¼ûÀë×ÓµÄÐÔÖʼ°¼ìÑé·½·¨£¬ÌâÄ¿ÄѶÈÖеȣ¬Ã÷È·³£¼ûÀë×ÓµÄÐÔÖÊΪ½â´ð¹Ø¼ü£¬È·¶¨ÊÇ·ñº¬ÓÐÂÈÀë×ÓΪÒ×´íµã£¬×¢ÒâÃ÷È·¼ìÑéÀë×Ó´æÔÚʱ£¬±ØÐëÅųý¸ÉÈÅÀë×Ó£¬È·±£¼ìÑé·½°¸µÄÑÏÃÜÐÔ£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

19£®ÓöèÐԵ缫µç½âÏÂÁÐ×ãÁ¿µÄÈÜÒº£¬Ò»¶Îʱ¼äºó£¬ÔÙ¼ÓÈëÒ»¶¨Á¿µÄÁíÒ»ÎïÖÊ£¨À¨ºÅÄÚ£©ºó£¬ÈÜÒºÄÜÓëÔ­À´ÈÜҺŨ¶ÈÒ»ÑùµÄÊÇ£¨¡¡¡¡£©
A£®CuCl2£¨CuCl2ÈÜÒº£©B£®AgNO3£¨Ag2O£©C£®NaCl£¨HClÈÜÒº£©D£®CuSO4£¨Cu£¨OH£©2£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

20£®ÏÂÁÐ״̬Ï£¬²»Äܵ¼µçµÄÇ¿µç½âÖÊÊÇ£¨¡¡¡¡£©
A£®±ù´×ËáB£®´¿ÁòËáC£®ÕáÌÇÈÜÒºD£®ÇâÑõ»¯ÄÆÈÜÒº

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

17£®º£Ñó×ÊÔ´µÄ¿ª·¢ÓëÀûÓþßÓйãÀ«µÄÇ°¾°£®º£Ë®µÄpHÒ»°ãÔÚ7.5¡«8.6Ö®¼ä£¬Ä³µØº£Ë®ÖÐÖ÷ÒªÀë×ӵĺ¬Á¿ÈçÏÂ±í£º
³É·ÖNa+K+Ca2+Mg2+Cl-SO42-HCO3-
º¬Á¿/mg•L-19360832001100160001200118
£¨1£©º£Ë®ÏÔÈõ¼îÐÔµÄÔ­ÒòÊÇ£¨ÓÃÀë×Ó·½³Ìʽ±íʾ£©HCO3-+H2O?H2CO3+OH-£®¸Ãº£Ë®ÖÐCa2+µÄÎïÖʵÄÁ¿Å¨¶ÈΪ5¡Á10-3 mol•L-1£®
£¨2£©µçÉøÎö·¨ÊǽüÄê·¢Õ¹ÆðÀ´µÄÒ»ÖֽϺõĺ£Ë®µ­»¯¼¼Êõ£¬ÆäÔ­ÀíÈçͼ1Ëùʾ£®ÆäÖÐÒõ£¨Ñô£©Àë×Ó½»»»Ä¤Ö»ÔÊÐíÒõ£¨Ñô£©Àë×Óͨ¹ý£®

¢ÙÒõ¼«µÄµç¼«·´Ó¦Ê½Îª2H++2e-=H2¡ü£®
¢Úµç½âÒ»¶Îʱ¼ä£¬Òõ¼«Çø»á²úÉúË®¹¸£¬Æä³É·ÖΪCaCO3ºÍMg£¨OH£©2£¬Ð´³öÉú³ÉCaCO3µÄÀë×Ó·½³Ìʽ
Ca2++OH-+HCO3-=CaCO3¡ý+H2O£®
¢Ûµ­Ë®µÄ³ö¿ÚΪa¡¢b¡¢cÖеÄb³ö¿Ú£®
£¨3£©º£Ë®ÖÐï®ÔªËØ´¢Á¿·Ç³£·á¸»£¬´Óº£Ë®ÖÐÌáȡ﮵ÄÑо¿¼«¾ßDZÁ¦£®ï®ÊÇÖÆÔ컯ѧµçÔ´µÄÖØÒªÔ­ÁÏ£¬ÈçLiFePO4µç³Øijµç¼«µÄ¹¤×÷Ô­ÀíÈçͼ2Ëùʾ£º
¸Ãµç³Øµç½âÖÊΪÄÜ´«µ¼Li+µÄ¹ÌÌå²ÄÁÏ£®·Åµçʱ¸Ãµç¼«Êǵç³ØµÄÕý¼«£¨Ìî¡°Õý¡±»ò¡°¸º¡±£©£¬µç¼«·´Ó¦Ê½ÎªFePO4+e-+Li+=LiFePO4£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

4£®°±ÊÇ×îÖØÒªµÄ»¯¹¤²úÆ·Ö®Ò»£®
£¨1£©ºÏ³É°±ÓõÄÇâÆø¿ÉÒÔÒÔ¼×ÍéΪԭÁÏÖƵã®Óйػ¯Ñ§·´Ó¦µÄÄÜÁ¿±ä»¯Èçͼ¼×¡¢ÒÒËùʾ£®·´Ó¦¼×ÊÇ·ÅÈÈ·´Ó¦£¨Ìî¡°·ÅÈÈ¡±»ò¡°ÎüÈÈ¡±£©£®ÅжÏÒÀ¾ÝÊÇ·´Ó¦ÎïµÄ×ÜÄÜÁ¿´óÓÚÉú³ÉÎïµÄ×ÜÄÜÁ¿£®CH4£¨g£©ÓëH2O£¨g£©·´Ó¦Éú³ÉCO£¨g£©ºÍH2 £¨g£©µÄÈÈ»¯Ñ§·½³ÌʽÊÇCH4£¨g£©+H2O£¨g£©=CO£¨g£©+3H2 £¨g£©¡÷H=+161.1kJ•mol-1£®

£¨2£©Í¼±ûÊÇ°±ÑõȼÁϵç³ØµÄʾÒâͼ£¬»Ø´ðÏÂÁÐÎÊÌ⣺
¢Ùaµç¼«×÷¸º¼«£¨Ìî¡°Õý¡±¡°¸º¡±¡°Òõ¡±»ò¡°Ñô¡±£©£¬Æäµç¼«·´Ó¦Ê½Îª2NH3-6e-+6OH-=N2+6H2O£®
¢Ú·´Ó¦Ò»¶Îʱ¼äºó£¬µç½âÖÊÈÜÒºµÄpH½«¼õС£¨Ìî¡°Ôö´ó¡±¡°¼õС¡±»ò¡°²»±ä¡±£©£®
¢ÛÈôÓøÃȼÁϵç³Ø²úÉúµÄµçÄÜÔÚÌúƤÉ϶Æп£¨ÖÆ°×ÌúƤ£©£¬Ä³ÌúƤÉÏÏÖÐèÒª¶ÆÉÏ 97.5gп£¬ÀíÂÛÉÏÖÁÉÙÐèÒªÏûºÄ°±Æø22.4 L£¨±ê×¼×´¿ö£©£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

14£®ÏÂÁл¯Ñ§±íÊöÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®ÂÈ»¯ÄƵľ§ÌåÄ£ÐÍ£º
B£®îëÔ­×Ó×îÍâ²ãµÄµç×ÓÔÆͼ£º
C£®´ÎÂÈËáµÄ½á¹¹Ê½£ºH-O-Cl
D£®CH3CHOHCH£¨CH3£©2Ãû³Æ£º2-3-¶þ¼×»ù±û´¼

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

1£®¿É¼òдΪ£®´ï·ÆÊÇÊÀ½çÎÀÉú×éÖ¯ÍƼöµÄ¿¹ÇÝÁ÷¸ÐÒ©ÎÆä½á¹¹¼òʽÈçͼ£¨£©£¬ÓйØËüµÄ˵·¨´íÎóµÄÊÇ£¨¡¡¡¡£©
A£®´ï·ÆÊÇÒ»ÖÖÁ×ËáÑÎ
B£®´ï·Æ·Ö×ÓÖк¬ÓÐëļü
C£®´ï·ÆµÄ·Ö×ÓʽΪC16H31N2O8P
D£®1mol´ï·Æ¿ÉÓë2mol H2·¢Éú¼Ó³É·´Ó¦

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

18£®ÉèNA´ú±í°¢·ü¼ÓµÂÂÞ³£ÊýµÄÖµ£¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®28g N2¡¢N2O¡¢N2O4µÄ»ìºÏÆøÌ庬ÓеªÔ­×ÓÊýΪ2NA
B£®ÖÊÁ¿¾ùΪ7.8 g Na2S¡¢Na2O2µÄ¹ÌÌåÖк¬ÓеÄÒõÀë×ÓÊý¾ùΪ0.1NA
C£®0.5 mol•L-1µÄÒÒËáÈÜÒºÖУ¬ÒÒËá·Ö×ÓµÄÊýĿСÓÚ0.5NA
D£®78g±½·Ö×ÓÖк¬ÓÐ̼̼˫¼üÊýΪ3 NA

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

19£®¶ÔÓÚ»¯Ñ§·´Ó¦A+B¨TC+DµÄÏÂÁÐ˵·¨ÖÐÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®ÈôÉú³ÉÎïCºÍD·Ö±ðΪÑκÍË®£¬Ôò¸Ã·´Ó¦Ò»¶¨ÊÇÖкͷ´Ó¦
B£®ÈôAºÍCÊǵ¥ÖÊ£¬BºÍDÊÇ»¯ºÏÎÔò¸Ã·´Ó¦Ò»¶¨ÊÇÖû»·´Ó¦
C£®ÈôAÊÇ¿ÉÈÜÐԼBÊÇ¿ÉÈÜÐÔÑΣ¬ÔòCºÍD²»¿ÉÄÜÊÇÁ½ÖÖ³Áµí
D£®ÈôAÊÇ¿ÉÈÜÐԼBÊÇ¿ÉÈÜÐÔÑΣ¬ÔòCºÍDÒ»¶¨ÊÇÁíÒ»ÖÖ¼îºÍÁíÒ»ÖÖÑÎ

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸