9£®ÏÂÁÐÅжÏÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®Í¬Ìå»ý¡¢Í¬Å¨¶ÈµÄNaFºÍNaClÈÜÒºÖУ¬ËùÓÐÀë×Ó×ÜÊý£¬N£¨NaCl£©£¾N£¨NaF£©£»Í¬Ìå»ýͬŨ¶ÈµÄNa2CO3ºÍNa2SO4ÈÜÒºÖУ¬ËùÓÐÀë×Ó×ÜÊý£¬N£¨Na2CO3£©£¾N£¨Na2SO4£©
B£®ÒÔÏÂÊÇijζÈϸßÂÈËá¡¢ÁòËá¡¢ÏõËáºÍÑÎËáÔÚ±ù´×ËáÖеĵçÀë³£Êý£¬ÔòÔÚ±ù´×ËáÖÐÁòËáµÄµçÀë·½³Ìʽ¿É±íʾΪH2SO4?2H++SO42-
ËáHClO4H2SO4HClOHNO3
Ka1.6¡Á10-56.3¡Á10-91.6¡Á10-94.2¡Á10-10
C£®ÒÑÖªËáÐÔHF£¾CH3COOH£¬pHÏàµÈµÄNaFÓëCH3COOKÈÜÒºÖУ¬[c£¨Na+£©-c£¨F-£©]£¼[c£¨K+£©-c£¨CH3COO-£©]
D£®Æ½ºâÌåϵCaCO3£¨s£©?CaO£¨s£©+CO2Öнöº¬ÓÐ̼Ëá¸Æ¡¢Ñõ»¯¸Æ¼°¶þÑõ»¯Ì¼ÆøÌ壮ijʱ¿Ì£¬±£³ÖζȲ»±ä£¬½«ÈÝÆ÷Ìå»ýËõСΪԭÀ´µÄÒ»°ë²¢±£³Ö²»±ä£¬Ôò¶þÑõ»¯Ì¼Å¨¶ÈÔö´ó

·ÖÎö A¡¢¸ù¾ÝF-ºÍCO32-ÄÜË®½â²¢½áºÏµçºÉÊØºãÀ´·ÖÎö£»
B¡¢´æÔÚµçÀëÆ½ºâµÄËáÔÚµçÀëʱÊDz»ÍêÈ«µçÀëµÄ£¬ÇÒ¶àÔªËáÊÇ·Ö²½µçÀëµÄ£»
C¡¢NaFÈÜÒºÖÐc£¨Na+£©-c£¨F-£©ºÍCH3COOKÈÜÒºÖÐc£¨K+£©-c£¨CH3COO-£©¾ùµÈÓÚc£¨OH-£©-c£¨H+£©£»
D¡¢Î¶Ȳ»±ä£¬K=c£¨CO2£©£¬Æ½ºâ³£Êý½öÓëζÈÓйØÀ´Åжϣ®

½â´ð ½â£ºA¡¢¸ù¾ÝµçºÉÊØºã¿ÉÖªNaF£¨aq£©ÖÐc£¨Na+£©+c£¨H+£©=c£¨OH-£©+c£¨F-£©£¬ËùÒÔNaFÈÜÒºÖÐÀë×Ó×ÜŨ¶ÈΪ2[c£¨Na+£©+c£¨H+£©]£¬¸ù¾ÝµçºÉÊØºã¿ÉÖªÂÈ»¯ÄÆÈÜÒºÖÐÀë×Ó×ÜŨ¶ÈҲΪ2[c£¨Na+£©+c£¨H+£©]£¬NaF·¢ÉúË®½â£¬ÈÜÒºÖÐF-Ë®½â£¬ÈÜÒº³Ê¼îÐÔ£¬ËùÒÔc£¨OH-£©£¾c£¨H+£©£¬µÈÎïÖʵÄÁ¿Å¨¶ÈµÄNaNaF£¨aq£©ÓëNaCl£¨aq£©ÖУ¬NaFÈÜÒºÖÐc£¨H+£©Ð¡ÓÚNaCl£¨aq£©ÖÐc£¨H+£©£¬Á½ÈÜÒºÖÐc£¨Na+£©Ïàͬ£¬ËùÒÔ£®µÈÌå»ýµÈÎïÖʵÄÁ¿Å¨¶ÈµÄNaF£¨aq£©ÓëNaCl£¨aq£©ÖÐÀë×Ó×ÜÊý¶àÉÙ£ºÀë×Ó×ÜÊýNaCl£¾NaF£»
CO32-µÄµÚÒ»²½Ë®½âCO32-+H2O?HCO3-+OH-»áµ¼ÖÂÀë×ÓÊýÄ¿µÄÔö¶à£¬¹ÊͬÌå»ýͬŨ¶ÈµÄNa2CO3ºÍNa2SO4ÈÜÒºÖУ¬ËùÓÐÀë×Ó×ÜÊý£¬N£¨Na2CO3£©£¾N£¨Na2SO4£©£¬¹ÊAÕýÈ·£»
B¡¢ÔÚ±ù´×ËáÖÐÁòËá´æÔÚµçÀëÆ½ºâ£¬ËùÒÔÆäµçÀë·½³ÌʽΪH2SO4?H++HSO4-£¬¹ÊB´íÎó£»
C¡¢NaFÈÜÒºÖÐc£¨Na+£©-c£¨F-£©ºÍCH3COOKÈÜÒºÖÐc£¨K+£©-c£¨CH3COO-£©¾ùµÈÓÚc£¨OH-£©-c£¨H+£©£¬¶øÁ½ÈÜÒºÖеÄpHÏàµÈ£¬¹Êc£¨H+£©ÏàµÈ£¬c£¨OH-£©Ò²ÏàµÈ£¬¹ÊÁ½ÈÜÒºÖеÄ[c£¨OH-£©-c£¨H+£©]ÏàµÈ£¬¼´[c£¨Na+£©-c£¨F-£©]=[c£¨K+£©-c£¨CH3COO-£©]£¬¹ÊC´íÎó£»
D¡¢CaCO3£¨s£©?CaO£¨s£©+CO2£¨g£©£¬K=c£¨CO2£©£¬ËõСÌå»ýºó£¬Æ½ºâÄæÏòÒÆ¶¯£¬µ«ÒòΪζÈû±ä£¬ÔòK=c£¨CO2£©²»±ä£¬ËùÒÔCO2Ũ¶È²»±ä£¬¹ÊD´íÎó£®
¹ÊÑ¡A£®

µãÆÀ ±¾ÌâÖ÷Òª¿¼²éÁËÆ½ºâÒÆ¶¯¼°Æäƽºâ³£ÊýµÄ±í´ïʽ¼°ÆäÓ°ÏìÒòËØ£¬ÄѶȲ»´ó£¬×¥×¡Æ½ºâ³£Êý½öÓëζÈÓйؼ´¿É½âÌ⣮

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

4£®Ä³»¯Ñ§ÐËȤС×éΪ̽¾¿Í­ÓëŨÁòËáµÄ·´Ó¦£¬ÓÃÏÂͼËùʾװÖýøÐÐÓйØÊµÑ飮 ¼×ͬѧȡag Cu Æ¬ºÍ12ml 18mol/LŨH2SO4·ÅÈëÔ²µ×ÉÕÆ¿ÖмÓÈÈ£¬Ö±µ½·´Ó¦Íê±Ï£¬×îºó·¢ÏÖÉÕÆ¿Öл¹ÓÐÒ»¶¨Á¿µÄH2SO4ºÍCuÊ£Ó࣮
£¨1£©CuÓëŨH2SO4·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇCu+2H2SO4£¨Å¨£©$\frac{\underline{\;\;¡÷\;\;}}{\;}$CuSO4+2H2O+SO2¡ü£®
£¨2£©×°ÖÃDÊ¢ÓÐÆ·ºìÈÜÒº£¬µ±CÖÐÆøÌ弯Âúºó£¬DÖÐÓпÉÄܹ۲쵽µÄÏÖÏóÊÇÆ·ºìÈÜÒºÍÊÉ«£¬
£¨3£©×°ÖÃDÖÐÊԹܿڷÅÖõÄÃÞ»¨ÖÐÓ¦½þÒ»ÖÖÒºÌ壬ÕâÖÖÒºÌåÊÇÇâÑõ»¯ÄÆÈÜÒº£¬Æä×÷ÓÃÊÇÔÚÊԹܿÚÎüÊÕ¶þÑõ»¯Áò£®
£¨4£©×°ÖÃBµÄ×÷ÓÃÊÇÖü´æ¶àÓàµÄÆøÌ壮µ±D´¦ÓÐÃ÷ÏÔÏÖÏóºó£¬¹Ø±ÕÐýÈûK£¬ÒÆÈ¥¾Æ¾«µÆ£¬µ«ÓÉÓÚÓàÈȵÄ×÷Óã¬A´¦ÈÔÓÐÆøÌå²úÉú£¬´ËʱBÖÐÊÔ¼ÁÆ¿ÖÐÒºÃæÏ½µ£¬³¤¾±Â©¶·ÖÐÒºÃæÉÏÉý£®BÖÐÓ¦·ÅÖõÄÒºÌåÊÇd£¨ÌîÐòºÅ£©£®
a£®±¥ºÍNa2SO3ÈÜÒº   b£®ËáÐÔ KMnO4ÈÜÒº   c£®Å¨äåË®  d£®±¥ºÍNaHSO3ÈÜÒº
£¨5£©·´Ó¦Íê±Ïºó£¬ÉÕÆ¿Öл¹ÓÐÒ»¶¨Á¿µÄÓàËᣬΪʲôȴ²»ÄÜʹCuÍêÈ«ÈܽâµÄÔ­ÒòÊÇËæ×Å·´Ó¦½øÐУ¬ÁòËá±»ÏûºÄ£¬²úÎïÓÐË®Éú³É£¬ËùÒÔŨÁòËá±ä³ÉÏ¡ÁòËᣬ·´Ó¦»áÍ£Ö¹£®Ê¹ÓÃ×ãÁ¿µÄÏÂÁÐÒ©Æ·²»ÄÜÓÃÀ´Ö¤Ã÷·´Ó¦½áÊøºóµÄÉÕÆ¿ÖеÄÈ·ÓÐÓàËáµÄÊÇb£¨ÌîÐòºÅ£©£®
a£®Fe·Û           b£®BaCl2ÈÜÒº      c£®CuO     d£®Na2CO3ÈÜÒº
£¨6£©ÊµÑéÖÐijѧÉúÏòAÖз´Ó¦ºóÈÜÒºÖÐͨÈëÒ»ÖÖ³£¼ûÆøÌåµ¥ÖÊ£¬Ê¹Í­Æ¬È«²¿ÈܽâÇÒ½öÉú³ÉÁòËáÍ­ÈÜÒº£¬ÇëÎÊ¸ÃÆøÌåµ¥ÖÊÊÇO2£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ2Cu+O2+2H2SO4¨T2CuSO4+2H2O£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

5£®ÏÂÁÐÎïÖÊÖÐͬ·ÖÒì¹¹ÌåÊýÄ¿×î¶àµÄÊÇ£¨¡¡¡¡£©
A£®C3H8OB£®C4H10OC£®C4H8Cl2D£®C5H11Cl

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

2£®ÓÃÈçͼװÖýøÐÐÏàӦʵÑ飬²»ÄܴﵽʵÑéÄ¿µÄÊÇ£¨¡¡¡¡£©
A£®Í¼1ËùʾװÖÿÉÖÆ±¸ÇâÑõ»¯ÑÇÌú
B£®Í¼2ËùʾװÖÿɵç½âʳÑÎË®ÖÆÇâÆøºÍCu£¨OH£©2
C£®Í¼3ËùʾװÖÿÉÑéÖ¤°±Æø¼«Ò×ÈÜÓÚË®
D£®Í¼4ËùʾװÖÃÓÃÓÚCuºÍŨÁòËá·´Ó¦ÖÆÈ¡ÉÙÁ¿µÄSO2ÆøÌå

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

4£®ÏÂÁÐʵÑé²Ù×÷ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®·ÖÒº²Ù×÷ʱ·ÖҺ©¶·ÖÐϲãÒºÌå´ÓÏ¿ڷųö£¬ÉϲãÒºÌå´ÓÉϿڵ¹³ö
B£®ÓÃÕô·¢µÄ·½·¨Ê¹NaCl´ÓÈÜÒºÖÐÎö³öʱ£¬Ó¦½«Õô·¢ÃóÖÐNaClÈÜҺС»ð¼ÓÈÈÕô¸É
C£®ÀûÓÃÕôÁóµÄ·½·¨½øÐÐÒÒ´¼ÓëË®µÄ·ÖÀëʱ£¬Î¶ȼÆË®ÒøÇòÓ¦·ÅÔÚÒºÃæÒÔÏÂÀ´×¼È·²â¶¨Î¶È
D£®ÝÍÈ¡²Ù×÷ʱ£¬Ó¦Ñ¡ÔñÓлúÝÍÈ¡¼Á£¬ÇÒÝÍÈ¡¼ÁµÄÃܶȱØÐë±ÈË®´ó

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

14£®¸ù¾ÝÏÂÁÐÊÂʵµÃ³öµÄ½áÂÛÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®ÊÔÑù$¡ú_{ˮԡ¼ÓÈÈ}^{Òø°±ÈÜÒº}$²úÉúÒø¾µ½áÂÛ£º¸ÃÊÔÑùÊÇÈ©Àà
B£®Ä³Â±´úÌþÊÔÑù$¡ú_{¼ÓÈÈ}^{ÇâÑõ»¯ÄÆÈÜÒº}$$\stackrel{µÎÈëÏõËáÒøÈÜÒº}{¡ú}$×îÖյijÁµí²»Êǰ×É«£®½áÂÛ£º¸Ã±´úÌþÖв»º¬ÂÈÔ­×Ó
C£®Ä³ÈÜÒº$\stackrel{´ò¿ªÆ¿¸Ç}{¡ú}$ð×Ű×Îí$\stackrel{ÓÃÕºÓÐŨ°±Ë®²£Á§°ô}{¡ú}$²úÉú´óÁ¿°×ÑÌ£¬½áÂÛ£º´ËÈÜҺΪŨÑÎËá
D£®ÎÞÉ«ÈÜÒº$\stackrel{ÑæÉ«·´Ó¦}{¡ú}$³Ê»ÆÉ«£º½áÂÛ£º´ËÈÜÒºÒ»¶¨º¬ÓÐÄÆÔªËØ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

1£®²ÄÁÏÊÇΪÈËÀàÉç»áËùÐèÒª²¢ÄÜÓÃÓÚÖÆÔìÓÐÓÃÆ÷ÎïµÄÎïÖÊ£®°´ÓÃ;¿É·ÖΪ½á¹¹²ÄÁÏ£¬¹¦ÄܲÄÁϵȣ»°´»¯Ñ§×é³ÉºÍÌØÐÔÓÖ¿É·Ö³ÉËÄÀ࣬Ç뽫ÏÂÁÐÎïÖʵıêºÅÌîÔÚÏàÓ¦µÄ¿Õ¸ñÖУº
A£®Ë®Äà   B£®°ëµ¼Ìå²ÄÁÏ    C£®ËÜÁÏ   D£®³¬Ó²Ä͸ßβÄÁÏ   E£®ÌÕ´É  F£®ÆÕͨºÏ½ð   G£®ºÏ³ÉÏ𽺺ϳÉÏËά    H£®²£Á§
£¨1£©ÊôÓÚ´«Í³ÎÞ»ú·Ç½ðÊô²ÄÁϵÄÓÐAEH£»
£¨2£©ÊôÓÚÐÂÐÍÎÞ»ú·Ç½ðÊô²ÄÁϵÄÓÐBD£»
£¨3£©ÊôÓÚ½ðÊô²ÄÁϵÄÓÐF£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

18£®ÏÂÁÐÓлúÎïÓëÆäÃû³ÆÏà·ûµÄÊÇ£¨¡¡¡¡£©
A£®3£¬3-¶þ¼×»ù¶¡Íé   B£®l-¶¡Ï©£ºCH2=CH-CH2-CH3
C£®¼ä¶þ¼×±½£ºD£®±½¼×´¼£º

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

19£®X¡¢Y¡¢Z¡¢Q¡¢EÎåÖÖÔªËØÖУ¬XÔ­×ÓºËÍâµÄM²ãÖÐÖ»ÓÐÁ½¶Ô³É¶Ôµç×Ó£¬YÔ­×ÓºËÍâµÄL²ãµç×ÓÊýÊÇK²ãµÄÁ½±¶£¬ZÊǵؿÇÄÚº¬Á¿£¨ÖÊÁ¿·ÖÊý£©×î¸ßµÄÔªËØ£¬QµÄºËµçºÉÊýÊÇXÓëZµÄºËµçºÉÊýÖ®ºÍ£¬EÔÚÔªËØÖÜÆÚ±íµÄ¸÷ÔªËØÖе縺ÐÔ×î´ó£®Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©X¡¢YµÄÔªËØ·ûºÅÒÀ´ÎΪS¡¢C£»
£¨2£©QµÄÔªËØ·ûºÅÊÇCr£¬ËüÊôÓÚµÚËÄÖÜÆÚ£¬ËüµÄºËÍâµç×ÓÅŲ¼Ê½Îª1s22s22p63s23p63d54s1£¬ÔÚÐγɻ¯ºÏÎïʱËüµÄ×î¸ß»¯ºÏ¼ÛΪ+6£®

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸