ÌâÄ¿ÄÚÈÝ

ÎåÖÖ¶ÌÖÜÆÚÔªËØX¡¢Y¡¢Z¡¢M¡¢NµÄÔ­×ÓÐòÊýÒÀ´ÎÔö´ó£¬X¡¢Y¡¢Z¡¢MÊÇ×é³Éµ°°×ÖʵĻù´¡ÔªËØ£¬MÔ­×ÓµÄ×îÍâ²ãµç×ÓÊýÊǵç×Ó²ãÊýµÄ3±¶£¬N2M2¿ÉÓÃÓÚDZˮͧÖÐ×÷ΪÑõÆøµÄÀ´Ô´£¬ÏÂÁÐ˵·¨´íÎóµÄÊÇ
A£®ÓëYͬÖ÷×åµÄ¶ÌÖÜÆÚÔªËØ¿ÉÓëÇ¿¼î·´Ó¦Éú³ÉÇâÆø
B£®XºÍMÐγɵĻ¯ºÏÎï·Ö×ÓÖпÉÄܺ¬ÓзǼ«ÐÔ¹²¼Û¼ü
C£®ÓÉX¡¢Z¡¢MÐγɵĻ¯ºÏÎïÒ»¶¨Êǹ²¼Û»¯ºÏÎï
D£®µÈÎïÖʵÄÁ¿µÄN2YM3·ÛÄ©ºÍNXYM3·ÛÄ©·Ö±ðÓëµÈŨ¶ÈµÄÑÎËá·´Ó¦£¬ºóÕß·´Ó¦ËÙÂÊÒª¿ì
C

ÊÔÌâ·ÖÎö£ºÓÉÌâÒâ¿ÉÖªX¡¢Y¡¢Z¡¢M¡¢N·Ö±ðΪH¡¢C¡¢N¡¢O¡¢Na¡£
A¡¢ÓëYͬÖ÷×å¶ÌÖÜÆÚÔªËØΪSi£¬¿ÉÒÔÓëÇ¿¼îÉú³ÉÇâÆø£¬ÕýÈ·£»B¡¢XºÍM¿ÉÐγÉH2O2£¬ÆäÖÐÒõÀë×ÓÖк¬ÓзǼ«ÐÔ¼ü£¬ÕýÈ·£»C£¬X¡¢Z¡¢M¿ÉÐγÉNH4NO3£¬ÊÇÀë×Ó»¯ºÏÎ´íÎó£»D¡¢µÈÎïÖʵÄÁ¿µÄNaCO3ºÍNaHCO3£¬ÆäÖÐHCO3-¸üÒ×½áºÏH+£¬Òò´Ë·´Ó¦ËÙÂʸü¿ì£¬ÕýÈ·¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
A¡¢B¡¢C¡¢DÊÇÔªËØÖÜÆÚ±íÇ°ËÄÖÜÆÚ³£¼ûµÄËÄÖÖÔªËØ£¬Ô­×ÓÐòÊýÒÀ´ÎÔö´ó£¬ ËüÃǵÄÏà¹Ø½á¹¹ºÍÐÔÖÊÐÅÏ¢ÈçϱíËùʾ£¬Çë½áºÏÏà¹ØÐÅÏ¢£¬Íê³ÉÏà¹ØµÄÎÊÌ⣺
ÔªËØ
Ïà¹Ø½á¹¹ºÍÐÔÖÊ
A
AΪ·Ç½ðÊôÔªËØ£¬ÆäÒ»ÖÖµ¥ÖÊÊǵçµÄÁ¼µ¼Ì壬ÄÑÈÛ¡¢ÖÊÈí²¢ÓÐÈó»¬ÐÔ£¬¿ÉÓÃ×÷µç¼«²ÄÁÏ
B
Êǵ°°×ÖʵÄ×é³ÉÔªËØÖ®Ò»£¬Ô­×ÓºËÍâÓÐÈýÖÖ²»Í¬ÄÜÁ¿µÄµç×Ó£¬ÇÒδ³É¶Ôµç×ÓÊý×î¶à
C
Æäµ¥ÖʼÈÄÜÓëÇ¿ËáÈÜÒºÓ¦ÓÖÄÜÓëÇ¿¼îÈÜÒº·´Ó¦£¬Æä¼òµ¥Àë×ÓÔÚµÚÈýÖÜÆÚµ¥ºËÀë×ÓÖа뾶×îС
D
DµÄ¸ß¼ÛÂÈ»¯ÎïµÄÑÎÈÜÒº³£ÓÃÓÚÓ¡Ë¢µç·¼«µç·µÄ¡°¿ÌÊ´¡±
 
£¨1£©DλÓÚÔªËØÖÜÆÚ±íµÚ  ÖÜÆÚµÚ  ×壬Æä»ù̬ԭ×ÓºËÍâ¼Ûµç×ÓÅŲ¼Ê½Îª               £»ÔªËØDµÄ¸ß¼ÛÂÈ»¯ÎïÐγɵÄÑÎÈÜÒº¿ÌÊ´Ó¡Ë¢µç·°åµç·Ëù·¢Éú·´Ó¦ µÄÀë×Ó·½³ÌʽΪ                      £»
£¨2£©ÄÆÓëBÔªËØÐγɵÄNa3B¾§ÌåÊôÓÚ       ¾§Ì壨Ìî¡°·Ö×Ó¡±¡¢¡°Ô­×Ó¡±¡¢¡°Àë×Ó¡±£©¡£ ·ÖÎöBÔªËصÄÆø̬Ç⻯ÎKÒ×ÈÜÓÚË®µÄÔ­Òò£¬³ýÁËÒòΪËüÃǾùÊǼ«ÐÔ·Ö×ÓÖ®Í⣬»¹ÒòΪ                               £»
£¨3£©ÔªËØAÓëÔªËØBÏà±È£¬·Ç½ðÊôÐÔ½ÏÇ¿µÄÊÇ         £¨ÓÃÔªËØ·ûºÅ±íʾ£©£¬¿ÉÒÔ×÷ΪÅжÏÁ½Õ߷ǽðÊôÐÔÇ¿ÈõµÄÒÀ¾ÝµÄÊÇ        ¡£
a£®³£ÎÂÏÂAµÄµ¥ÖʺÍBµÄµ¥ÖÊ״̬²»Í¬
b£®¸ÃÁ½ÖÖÔ­×ÓÐγɵĹ²¼Û¼üÖй²Óõç×Ó¶ÔµÄÆ«Ïò
c£®×î¸ß¼ÛÑõ»¯Îï¶ÔӦˮ»¯ÎïµÄËáÐÔÇ¿Èõ
d£®µ¥ÖÊÓëËá·´Ó¦µÄÄÑÒ׳̶È
£¨4£©ÒÑÖªš°5.4gÔªËØCµÄµ¥ÖÊ¿ÉÓëÔªËØDµÄµÍ¼ÛÑõ»¯Îï·´Ó¦£¬·Å³ö346.2kJµÄÈÈÁ¿¡£ÊÔд³ö¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ£º                                            ¡£
ÏÖÓÐÎåÖÖ¶ÌÖÜÆÚÖ÷×åÔªËØA¡¢B¡¢C¡¢   D¡¢E£¬ÆäÔ­×ÓÐòÊýÒÀ´ÎÔö´ó¡£AÔ­×ÓÔ¼Õ¼ÓîÖæÖÐÔ­×Ó×ÜÊýµÄ88.6%£¬A+ÓÖ³ÆΪÖÊ×Ó£ºBÊÇÐγɻ¯ºÏÎïÖÖÀà×î¶àµÄÔªËØ£¬CÔªËصÄ×î¼òµ¥µÄÇ⻯ÎïYµÄË®ÈÜÒºÏÔ¼îÐÔ£®EÊǶÌÖÜÆÚÔªËØÖе縺ÐÔ×îСµÄÔªËØ¡£A¡¢B¡¢C¡¢EËÄÖÖÔªËض¼ÄÜÓëDÔªËØÐγÉÔ­×Ó¸öÊý±È²»ÏàͬµÄ³£¼û»¯ºÏÎï¡£ÊԻشðÏÂÁÐÎÊÌ⣺
£¨1£©Ð´³öA¡¢EÁ½ÔªËØÐγɵÄÔ­×Ó¸öÊý±ÈΪ1:1µÄ»¯ºÏÎïµÄµç×Óʽ____¡£
£¨2£©ÏòÂÈ»¯ÑÇÌúÈÜÒºµÎ¼Ó¹ýÁ¿µÄEµÄ×î¸ß¼ÛÑõ»¯Îï¶ÔӦˮ»¯ÎïµÄÈÜÒº£¬ÏÖÏóÊÇ____¡£
£¨3£©YÈÜÒºÏÔ¼îÐÔµÄÔ­ÒòÊÇ(ÓÃÒ»¸öÀë×Ó·½³Ìʽ±íʾ£©____¡£
£¨4£©¼ìÑéÆû³µÎ²ÆøÖк¬ÓеĻ¯ºÏÎïBDµÄ·½·¨ÊÇ£ºÏòËáÐÔPdC12ÈÜÒºÖÐͨAÆû³µÎ²Æø£¬ÈôÉú³ÉºÚÉ«³Áµí£¨Pd£©£¬Ö¤Ã÷Æû³µÎ²ÆøÖк¬ÓÐBD¡£Ð´³ö·´Ó¦µÄÀë×Ó·½³Ìʽ____¡£
£¨5£©ÏÂÁÐÓйØÎïÖÊÐÔÖʵıȽÏÖУ®²»ÕýÈ·µÄÊÇ               ¡£
a£®ÈÈÎȶ¨ÐÔ£ºH2S>SiH4  b£®Àë×Ӱ뾶£ºNa+>S2£­
c£®µÚÒ»µçÀëÄÜN>O   d£®ÔªËص縺ÐÔ£ºC>H
£¨6£©ÒÑÖª£º¢ÙCH3OH(g)+H2O(g)=CO2(g)+3H2(g)  ¡÷H=+49.0kJ/mol
¢ÚCH3OH(g)+3/2O2(g)=CO2(g)+2H2O(g)  ¡÷H=£­192£®9kJ/mol
ÓÉÉÏÊö·½³Ìʽ¿ÉÖª£®CH3OHµÄȼÉÕÈÈ____£¨Ìî¡°´óÓÚ¡±¡¢¡°µÈÓÚ¡±»òСÓÚ¡±£©192.9kJ/mol¡£ÒÑ֪ˮµÄÆø»¯ÈÈΪ44 kJ/mol£®Ôò±íʾÇâÆøȼÉÕÈȵÄÈÈ»¯Ñ§·½³ÌʽΪ____¡£
A¡¢B¡¢C¡¢D¡¢E¡¢FÊǶÌÖÜÆÚÔªËØ£¬ÖÜÆÚ±íÖÐAÓëB¡¢BÓëCÏàÁÚ£¬CÓëEͬÖ÷×壬AÓëC×îÍâ²ãµç×ÓÊýÖ®±ÈΪ2:3£¬BµÄ×îÍâ²ãµç×ÓÊý±ÈCµÄ×îÍâ²ãµç×ÓÊýÉÙ1¸ö£» FÔªËصÄÔ­×ÓÔÚÖÜÆÚ±íÖа뾶×îС£»³£¼û»¯ºÏÎïD2C2ÓëË®·´Ó¦Éú³ÉCµÄµ¥ÖÊ£¬ÇÒÈÜҺʹ·Ó̪ÈÜÒº±äºì¡£
£¨1£©EµÄÃû³ÆΪ______£»DµÄ×î¸ß¼ÛÑõ»¯ÎïµÄË®»¯ÎïµÄµç×Óʽ£º_________¡£FAB·Ö×ӵĽṹʽΪ_____
£¨2£©A¡¢B¡¢CµÄÇ⻯ÎïÎȶ¨ÐÔ˳ÐòΪ£¨Ó÷Ö×Óʽ±íʾ£¬ÓÉ´óµ½Ð¡£©___________________£»BµÄÇ⻯ÎïºÍBµÄ×î¸ß¼ÛÑõ»¯ÎïµÄË®»¯Îï·´Ó¦Éú³ÉZ£¬ÔòZÖеĻ¯Ñ§¼üÀàÐÍΪ____________________¡£
£¨3£©Á½ÖÖ¾ùº¬C¡¢D¡¢E¡¢FËÄÖÖÔªËصĻ¯ºÏÎïÏ໥·´Ó¦·Å³öÆøÌåµÄ·´Ó¦Àë×Ó·½³ÌʽΪ____________¡£
£¨4£©Ò»¶¨Á¿µÄD2C2ÓëAC2·´Ó¦ºóµÄ¹ÌÌåÎïÖÊ£¬Ç¡ºÃÓ뺬0.8mol HClµÄÏ¡ÑÎËáÍêÈ«·´Ó¦£¬²¢ÊÕ¼¯0.25 mol ÆøÌ壬Ôò¹ÌÌåÎïÖʵÄ×é³ÉΪ£¨Ð´Çå³É·ÖºÍÎïÖʵÄÁ¿£©______________________¡£
£¨5£©ÔÚÈÝ»ý²»±äµÄÃܱÕÈÝÆ÷ÖнøÐÐÈçÏ·´Ó¦£º3F2(g)+B2(g)2BF3(g)£¬Èô½«Æ½ºâÌåϵÖи÷ÎïÖʵÄŨ¶È¶¼Ôö¼Óµ½Ô­À´µÄ2±¶£¬Ôò²úÉúµÄ½á¹ûÊÇ______________________£¨ÌîÐòºÅ£©¡£
A£®Æ½ºâ²»·¢ÉúÒƶ¯¡¡¡¡     B£®·´Ó¦ÎïµÄת»¯ÂʼõС
C£®BF3µÄÖÊÁ¿·ÖÊýÔö¼Ó¡¡ ¡¡ D£®ÕýÄæ·´Ó¦ËÙÂʶ¼Ôö´ó

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø