ʹÓÃÊý×Ö»¯ÊµÑéÊÒ£¨ÓɵçÄÔ¡¢Êý¾Ý²É¼¯Æ÷¡¢×¨Óô«¸ÐÆ÷¡¢Êý×Ö»¯Ì½Í·ÒÔ¼°Æ½³£Ê¹ÓõĻ¯Ñ§ÒÇÆ÷ºÍÊÔ¼ÁµÈ×é³É£©²âµÃ¹ØÓÚ0.1mol/LµÄCH3COONaÈÜÒºµÄ¼¸×éÊý¾Ý£ºÊÒÎÂʱ0.1mol/LµÄCH3COONaÈÜÒºµÄpHΪ8.75£»ÔÚÊÒν«Æä¼ÓˮϡÊ͵½0.01mol/L, ÆäpHΪ8.27£»Èô½«0.1mol/LµÄCH3COONaÈÜÒº¼ÓÈÈ£¬¼ÓÈȹý³ÌÖÐpHµÄ±ä»¯ÈçÏÂͼ£¬

 ÔòÏÂÁÐ˵·¨´íÎóµÄÊÇ£º

   A£®ÒÔÉÏʵÑéÊý¾Ý¶¼¿ÉÒÔ˵Ã÷CH3COONaÈÜÒºÖдæÔÚË®½âƽºâ

B£®CH3COONaË®½âƽºâ³£Êý

   C£®ÔÚÊÒν«0.1mol/L CH3COONaÈÜÒº¼ÓˮϡÊ͵½0.01mol/L£¬pHÓÉ8.75±ä»¯µ½8.27£¬¿ÉÒÔ˵Ã÷0.1mol/LCH3COONaÈÜҺˮ½â³Ì¶È¸ü´ó

   D£®¼ÓÈÈ0.1mol/LµÄCH3COONaÈÜÒº£¬pHµÄ±ä»¯¿É˵Ã÷ζÈÉý¸ß£¬KÔö´ó

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


¼È¿ÉÓÃÀ´¼ø±ð¼×ÍéÓëÒÒÏ©£¬ÓֿɳýÈ¥¼×ÍéÖлìÓÐÒÒÏ©µÄ×î¼Ñ·½·¨ÊÇ£¨    £©                                             

A¡¢Í¨ÈëËáÐÔ¸ßÃÌËá¼ØÈÜÒºÖР      B¡¢Í¨Èë×ãÁ¿äåË®ÖÐ

C¡¢Ò»¶¨Ìõ¼þÏÂͨÈëH2            D¡¢µãȼ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÏÂÁи÷»¯ºÏÎïÖУ¬ÄÜ·¢Éúõ¥»¯¡¢»¹Ô­¡¢¼Ó³É¡¢ÏûÈ¥ËÄÖÖ·´Ó¦µÄÊÇ          £¨     £©

A£®CH3£­CH£½CH£­CHO             B£®HOCH2£­CH2£­CH£½CH£­CHO

C£®CH3£­£­£­CH3               D£®HOCH2£­£­CH2£­CHO

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÈçÓÒͼËùʾװÖýøÐÐʵÑ飬½«ÒºÌåAÖðµÎ¼ÓÈëµ½¹ÌÌåBÖУ¬ ÏÂÁÐÐðÊöÖдíÎóµÄÊÇ£¨   £©                                               

A£®ÊµÑéÖÐÒÇÆ÷C¿ÉÆðµ½·ÀÖ¹µ¹ÎüµÄ×÷ÓÃ

B. ÈôAΪ´×ËᣬBΪ±´¿Ç£¨·Û×´£©£¬DÖÐÊ¢C6H5ONa

ÈÜÒº£¬ÔòDÖÐÈÜÒº±ä»ë×Ç

C. ÈôAΪŨ°±Ë®£¬BΪÉúʯ»Ò£¬DÖÐÊ¢AgNO3ÈÜÒº£¬

ÔòDÖÐÎÞÏÖÏó

D. ÈôAΪʳÑÎË®£¬BΪµçʯ£¬DÖÐÊ¢KMnO4ËáÐÔÈÜÒº£¬ÔòD  

ÖÐÈÜÒº×ϺìÉ«ÍÊÉ«

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÏòÒ»ÖÖÈÜÒºÖеμÓÁíÒ»ÖÖÈÜÒººó£¬ÈÜÒºµÄÑÕÉ«²»·¢ÉúÏÔÖø±ä»¯µÄÊÇ

A£®ÂÈ»¯ÑÇÌúÈÜÒºÖмÓÈëË«ÑõË®        B£®ÁòËáÌúÈÜÒºÖеμÓÁòÇèËá¼ØÈÜÒº

   C£®ÁòËáÍ­ÈÜÒºÖеμÓÏõËá±µÈÜÒº      D£®¸ßÃÌËá¼ØËáÐÔÈÜÒºÖеμÓÑÇÁòËáÄÆÈÜÒº 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


¢ñ.ÏÂÁÐÓйØ˵·¨²»ÕýÈ·µÄÊÇ________¡£

A£®ÏàͬÀàÐ͵ÄÀë×Ó¾§Ì壬¾§¸ñÄÜÔ½´ó£¬Ðγɵľ§ÌåÔ½Îȶ¨

B£®NH3ºÍH3O+Êǵȵç×ÓÌ壬Òò´Ë½á¹¹¶¼ÊÇÈý½Ç׶ÐÎ

C£®ÁÚôÇ»ù±½¼×È©·ÐµãµÍÓÚ¶ÔôÇ»ù±½¼×È©£¬Ô­ÒòÊÇÇ°Õß´æÔÚ·Ö×ÓÄÚÇâ¼üºóÕß´æÔÚ

·Ö×Ó¼äÇâ¼ü

D£®H3O£«¡¢HF2£­ºÍ[Ag(NH3)2]£«Öоù´æÔÚÅäλ¼ü

¢ò.̼¼°Æ仯ºÏÎïÔÚ×ÔÈ»½çÖй㷺´æÔÚ¡£

(1)»ù̬̼ԭ×ӵļ۵ç×ÓÅŲ¼Í¼¿É±íʾΪ              ¡£µÚËÄÖÜÆÚÓëÆäÓÐÏàͬ

δ³É¶Ôµç×ÓÊýµÄ¹ý¶É½ðÊôÓР                 £¨ÌîÔªËØ·ûºÅ£©

(2)µÚÒ»µçÀëÄÜ£ºC¡¢N¡¢O¡¢FËÄÖÖÔªËØÓÉ´óµ½Ð¡Ë³Ðò___    _                  £¬

Ô­ÒòÊÇ                                                               £¬

HCN¡¢NF3·Ö×Ó¹¹ÐÍ·Ö±ðÊÇ                                             ¡£

(3)±ù¾§°ûÖÐË®·Ö×ӵĿռäÅÅÁз½Ê½Óë½ð¸Õʯ¾§°ûÀàËÆ¡£Ã¿¸ö±ù¾§°ûƽ¾ùÕ¼ÓÐ________¸öË®·Ö×Ó£¬±ù¾§°ûÓë½ð¸Õʯ¾§°ûÅÅÁз½Ê½ÏàͬµÄÔ­ÒòÊÇ__________________________¡£

(4)C60µÄ¾§ÌåÖУ¬·Ö×ÓΪÃæÐÄÁ¢·½¶Ñ»ý£¬ÒÑÖª¾§°ûÖÐC60·Ö×ÓÖÐÐļäµÄ×î¶Ì¾àÀëΪ

d cm£¬¿É¼ÆËãC60¾§ÌåµÄÃܶÈΪ________g/cm3¡£

(5)Çëд³öÒ»¸ö·´Ó¦·½³ÌʽÒÔ±í´ï³ö·´Ó¦Ç°Ì¼Ô­×ÓµÄÔÓ»¯·½Ê½Îªsp2£¬·´Ó¦ºó±äΪsp3£º________________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÏÂÁÐÀë×Ó·½³ÌʽÊéдÕýÈ·µÄÊÇ

A£®½«ÉÙÁ¿SO2ÆøÌåͨÈëNaClOÈÜÒºÖР SO2 +2ClO£­+H2O = SO32£­+2HClO

B£®ÏòKHSO4ÈÜÒºÖмÓÈëBa(OH)2ÈÜÒºÖÁËùµÃÈÜÒºµÄpH=7   Ba2+ +OH£­+ H++ SO42£­ =  BaSO4¡ý+ 2H2O

C£®ÏòCa(H2PO4)2ÈÜÒºÖеÎÈë¹ýÁ¿µÄNaOHÈÜÒº    3Ca2+ + 6H2PO4£­+12OH£­= Ca3(PO4)2¡ý+ 4PO43£­+12H2O

D£®112mL£¨±ê¿ö£©Cl2ͨÈë10mL1mol/LµÄFeBr2ÈÜÒºÖР 2Fe2+ + 4Br£­+3Cl2 = 2Fe3+ + 6Cl£­+2Br2

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


µÈÎïÖʵÄÁ¿µÄÇâÆøºÍº¤ÆøÔÚͬÎÂͬѹϾßÓÐÏàµÈµÄ

    A£®Ô­×ÓÊý        B£®ÃܶȠ        C£®ÖÊ×ÓÊý        D£®ÖÊÁ¿

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


[»¯Ñ§¡ªÎïÖʽṹÓëÐÔÖÊ]ʯīϩ[Èçͼ(a)Ëùʾ]ÊÇÒ»ÖÖÓɵ¥²ã̼ԭ×Ó¹¹³ÉµÄƽÃæ½á¹¹ÐÂÐÍ̼²ÄÁÏ£¬Ê¯Ä«Ï©Öв¿·Ö̼ԭ×Ó±»Ñõ»¯ºó£¬ÆäƽÃæ½á¹¹»á·¢Éú¸Ä±ä£¬×ª»¯ÎªÑõ»¯Ê¯Ä«Ï©[Èçͼ(b)Ëùʾ]¡£

¡¡

(a)ʯīϩ½á¹¹¡¡¡¡¡¡¡¡¡¡(b)Ñõ»¯Ê¯Ä«Ï©½á¹¹

(1)ͼ(a)ÖУ¬1ºÅCÓëÏàÁÚCÐγɦҼüµÄ¸öÊýΪ________¡£

(2)ͼ(b)ÖУ¬1ºÅCµÄÔÓ»¯·½Ê½ÊÇ________£¬¸ÃCÓëÏàÁÚCÐγɵļü½Ç________(Ìî¡°>¡±¡°<¡±»ò¡°£½¡±)ͼ(a)ÖÐ1ºÅCÓëÏàÁÚCÐγɵļü½Ç¡£

(3)Èô½«Í¼(b)ËùʾµÄÑõ»¯Ê¯Ä«Ï©·ÖÉ¢ÔÚH2OÖУ¬ÔòÑõ»¯Ê¯Ä«Ï©ÖпÉÓëH2OÐγÉÇâ¼üµÄÔ­×ÓÓÐ________(ÌîÔªËØ·ûºÅ)¡£

(4)ʯīϩ¿Éת»¯Îª¸»ÀÕÏ©(C60)£¬Ä³½ðÊôMÓëC60¿ÉÖƱ¸Ò»ÖÖµÍ㬵¼²ÄÁÏ£¬¾§°ûÈçͼËùʾ£¬MÔ­×ÓλÓÚ¾§°ûµÄÀâÉÏÓëÄÚ²¿¡£¸Ã¾§°ûÖÐMÔ­×ӵĸöÊýΪ________£¬¸Ã²ÄÁϵĻ¯Ñ§Ê½Îª________¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸