ÌâÄ¿ÄÚÈÝ

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____¡£

£¨14·Ö£¬Ã¿¿Õ2·Ö£©
£¨1£©
£¨2£©ÏÈÓа×É«³Áµí£¬Ñ¸ËÙ±äΪ»ÒÂÌÉ«£¬×îºó±ä³ÉºìºÖÉ«
£¨3£©NH3+H2ONH4++OH-
£¨4£©CO+PdCl2+H2O=Pd+CO2+2H++2Cl-
£¨5£©b
£¨6£©´óÓÚ£»H2(g)+ O2(g)=H2O(l)  ¦¤H=-124.6KJ/mol

½âÎöÊÔÌâ·ÖÎö£º¸ù¾ÝÌâÒâ¿ÉÒÔÍÆÖªA¡¢B¡¢C¡¢D¡¢E·Ö±ðΪH¡¢C¡¢N¡¢O¡¢Na£»
£¨1£©NaÓëHÐγÉÀë×ÓÐÍ»¯ºÏÎï
£¨2£©¶þ¼ÛÌúÒ×Ñõ»¯£¬ÓëNaoH½áºÏºó£¬Éú³É°×É«ÇâÑõ»¯ÑÇÌú£¬Öð½¥Ñõ»¯³ÉÇâÑõ»¯Ìú£»
£¨3£©Ò»Ë®ºÏ°±µçÀë·½³Ìʽ
£¨4£©Ñõ»¯»¹Ô­·´Ó¦·½³ÌʽÊéд£¬ÏÈÅжϷ´Ó¦ÎïºÍÉú³ÉÎȻºóÅäƽ£»
£¨5£©ÔªËØÖÜÆÚÂÉ¿ÉÖª£¬·Ç½ðÊôÐÔԽǿ£¬Æø̬Ç⻯ÎïζÈÐÔԽǿ£¬aÕýÈ·£»µç×Ó²ãÊýÔ½¶à£¬°ë¾¶Ô½´ó£¬¹ÊS2->Na+£¬b´íÎó£»NΪ°ë³äÂú£¬µÚÒ»µçÀëÄÜ´óÓÚO£¬cÕýÈ·£»·Ç½ðÊôÐÔԽǿ£¬µç¸ºÐÔÔ½´ó£¬dÕýÈ·£»
£¨6£©ÓÉ¢Ú·´Ó¦ºÍË®´ÓÆø̬µ½ÒºÌ¬·ÅÈÈ¿ÉÖª£¬CH3OHµÄȼÉÕÈÈ´óÓÚ192.9KJ£»ÓɸÇ˹¶¨ÂÉ¿ÉÖª£¬ÇâÆøµÄȼÉÕÈÈ£¬²¢Ð´³öÈÈ»¯Ñ§·½³Ìʽ¡£
¿¼µã£º¿¼²éÎïÖʽṹÏà¹Ø֪ʶ£¬Éæ¼°Ô­×ӽṹ¡¢ÔªËØÖÜÆÚÂÉ¡¢ÔªËؼ°»¯ºÏÎïÐÔÖÊ£¬ÒÔ¼°È¼ÉÕÈȵļÆË㣬ÈÈ»¯Ñ§·½³ÌʽµÄÊéдµÈ¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¶ÌÖÜÆÚÔªËØA¡¢B¡¢C¡¢DÔÚÔªËØÖÜÆÚ±íÖеÄÏà¶ÔλÖÃÈçͼËùʾ£¬ÆäÖÐBËù´¦µÄÖÜÆÚÐòÊýÓë×åÐòÊýÏàµÈ¡£ÌîдÏÂÁпհס£

 
 
A
 
B
C
 
D
 
£¨1£©Ð´³öCµÄÑõ»¯ÎïµÄÒ»ÖÖÓÃ;£º        
£¨2£©Bµ¥ÖÊÓëFe2O3·´Ó¦Ê±£¬Ã¿ÏûºÄ13.5g Bʱ·ÅÈÈ213kJ£¬¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽÊÇ       
£¨3£©Îª·ÀÖ¹AÔªËصÄÑõ»¯ÎïAO2ÎÛȾ¿ÕÆø£¬¿Æѧ¼ÒÑ°ÇóºÏÊʵĻ¯ºÏÎïGºÍ´ß»¯¼Á£¬ÒÔʵÏÖ·´Ó¦£º
AO2+X´ß»¯¼ÁA2 +H2O+n Z(δÅäƽ£¬n¿ÉÒÔΪ0)¡£ÉÏÊö·´Ó¦Ê½ÖеÄX²»¿ÉÄÜÊÇ        (Ìî±êºÅ)¡£
a£®NH3      b£®CO      c£®CH3CH2OH       d£®H2O2
£¨4£©AÔªËصÄ×î¼òµ¥Ç⻯Îï¼×¡¢DÔªËصÄ×î¸ß¼ÛÑõ»¯Îï¶ÔÓ¦µÄË®»¯ÎïÒÒ¶¼ÊǺÜÖØÒªµÄ»ù´¡»¯¹¤Ô­ÁÏ¡£
¢ÙÒ»¶¨Ìõ¼þÏ£¬¼×Ôڹ̶¨Ìå»ýµÄÃܱÕÈÝÆ÷Öз¢Éú·Ö½â·´Ó¦£¨¡÷H>0£©²¢´ïƽºâºó£¬½ö¸Ä±äϱíÖз´Ó¦Ìõ¼þx£¬¸ÃƽºâÌåϵÖÐËæxµÝÔöyµÝ¼õµÄÊÇ      £¨Ñ¡ÌîÐòºÅ£©¡£
Ñ¡Ïî
a
b
c
d
x
ζÈ
ζÈ
¼ÓÈëH2µÄÎïÖʵÄÁ¿
¼ÓÈë¼×µÄÎïÖʵÄÁ¿
y
¼×µÄÎïÖʵÄÁ¿
ƽºâ³£ÊýK
¼×µÄת»¯ÂÊ
Éú³ÉÎïÎïÖʵÄÁ¿×ܺÍ
 
¢Ú25¡æʱ£¬Íùa mol¡¤L£­1µÄ¼×µÄË®ÈÜÒºÖеμÓ0.01 mol¡¤L£­1ÒÒÈÜÒº£¬µ±Á½ÖÖÈÜÒºµÈÌå»ý»ìºÏʱ£¬ÈÜÒº³ÊÖÐÐÔ(ÉèζȲ»±ä)¡£µÎ¼Ó¹ý³ÌÖÐÈÜÒºµÄµ¼µçÄÜÁ¦      (Ìî¡°ÔöÇ¿¡±¡¢¡°¼õÈõ¡±»ò¡°²»±ä¡±)£»
ËùµÃ»ìºÏÈÜÒºÖÐA¡¢DÁ½ÖÖÔªËصÄÎïÖʵÄÁ¿µÄ¹ØϵΪ£ºA     2D£¨Ìî¡°´óÓÚ¡±¡¢¡°µÈÓÚ¡±»ò¡°Ð¡ÓÚ¡±£©£»¼×ÖÐÈÜÖʵĵçÀëƽºâ³£ÊýKb£½      (Óú¬aµÄ´úÊýʽ±íʾ)¡£

ϱí¸ø³öÁËÎåÖÖÔªËصÄÏà¹ØÐÅÏ¢£¬ÆäÖÐw¡¢X¡¢Y¡¢ZΪ¶ÌÖÜÆÚÔªËØ£¬Ô­×ÓÐòÊýÒÀ´ÎµÝÔö¡£

ÔªËØ
Ïà¹ØÐÅÏ¢
W
µ¥ÖÊΪÃܶÈ×îСµÄÆøÌå
X
ÔªËØ×î¸ßÕý¼ÛÓë×îµÍ¸º¼ÛÖ®ºÍΪ0
Y
¹¤ÒµÉÏͨ¹ý·ÖÀëҺ̬¿ÕÆø»ñµÃÆäµ¥ÖÊ£¬¸Ãµ¥ÖʵÄijÖÖͬËØÒìÐÎÌåÊDZ£»¤µØÇòµØ±í»·¾³µÄÖØÒªÆÁÕÏ
Z
´æÔÚÖÊÁ¿ÊýΪ23£¬ÖÐ×ÓÊýΪ12µÄºËËØ
T
½ðÊôµ¥ÖÊΪºìÉ«£¬µ¼µçÐԺã¬ÊÇÈËÌå²»¿ÉȱÉٵĻÕÁ¿ÔªËØ£¬ÑæÉ«·´Ó¦Ê±»ðÑæΪÂÌÉ«
 
¸ù¾ÝÉÏÊöÐÅÏ¢Ìî¿Õ£º
£¨1£©ÔªËØYÔÚÔªËØÖÜÆÚ±íÖеÄλÖÃÊÇ              ¡£XY2ÓɹÌ̬±äΪÆø̬ËùÐè¿Ë·þµÄ΢Á£¼ä×÷ÓÃÁ¦ÊÇ                       ¡£
£¨2£©»¯ºÏÎï¼×ÊÇÒ»ÖÖÇ¿Ñõ»¯¼Á£¬ÓÉÔªËØYºÍZ×é³É£¬Ð´³ö¼×µÄ»¯Ñ§Ê½£º              £¬
£¨3£©»¯ºÏÎïÒÒÓÉÔªËØWºÍX×é³É£¬ÒÒÊÇͬʱº¬¼«ÐÔ¹²¼Û¼üºÍ·Ç¼«ÐÔ¹²¼Û¼üµÄÏà¶Ô·Ö×ÓÖÊÁ¿×îСµÄ·Ö×Ó¡£ÔÚ25¡æ¡¢101kpaÏ£¬ÒÑÖª2gµÄÆøÌåÒÒÔÚY2ÆøÌåÖÐÍêȫȼÉÕºó»Ö¸´ÖÁԭ״̬£¬·ÅÈÈQkJ£¬¸ÃȼÉÕ·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽÊÇ                             ¡£
£¨4£©»¯ºÏÎï±û½öÓÉW¡¢X¡¢Y¡¢Z¡¢TÖеÄËÄÖÖÔªËØ×é³É£¬ËÄÖÖÔªËصÄÖÊÁ¿±ÈΪ1£º6£º40£º64£¬»¯ºÏÎï±ûÖк¬ÓÐÁ½ÖÖÒõÀë×Ó£¬Ð´³öȼÉÕ±ûµÄ»¯Ñ§·½³Ìʽ                         ¡£

ϱíΪԪËØÖÜÆÚ±íµÄÒ»²¿·Ö¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺

¸ù¾ÝÔªËØÔÚÖÜÆÚ±íÖеÄÏà¶ÔλÖÿÉÖª¢Ù¡«¢â·Ö±ðÊÇH¡¢C¡¢N¡¢O¡¢Mg¡¢Al¡¢Cl¡¢Ca¡¢Mn¡¢Fe¡£
£¨1£©ÉÏÊöÔªËØÖУ¬ÊôÓÚsÇøµÄÊÇ____________(ÌîÔªËØ·ûºÅ)¡£
£¨2£©Ð´³öÔªËآܵĻù̬ԭ×ӵļ۵ç×ÓÅŲ¼Í¼____________________¡£
£¨3£©ÔªËصÚÒ»µçÀëÄÜΪ¢á________¢â (Ìî¡°´óÓÚ¡±»ò¡°Ð¡ÓÚ¡±)¡£
£¨4£©ÔªËØ¢ÛÆø̬Ç⻯ÎïµÄVSEPRÄ£ÐÍΪ________£»¸Ã·Ö×ÓΪ________·Ö×Ó(Ìî¡°¼«ÐÔ¡±»ò¡°·Ç¼«ÐÔ¡±)¡£ÏòÁòËáÍ­ÈÜÒºÖÐÖðµÎ¼ÓÈëÆäË®ÈÜÒº£¬¿É¹Û²ìµ½µÄÏÖÏóΪ_____________________________¡£
£¨5£©¢Þµ¥Öʵľ§ÌåÖÐÔ­×ӵĶѻý·½Ê½ÈçÏÂͼ¼×Ëùʾ£¬Æ侧°ûÌØÕ÷ÈçÏÂͼÒÒËùʾ£¬Ô­×ÓÖ®¼äÏ໥λÖùØϵµÄƽÃæͼÈçÏÂͼ±ûËùʾ¡£

ÈôÒÑÖª¢ÞµÄÔ­×Ӱ뾶Ϊdcm£¬NA´ú±í°¢·ü¼ÓµÂÂÞ³£Êý£¬ÔªËØ¢ÞµÄÏà¶ÔÔ­×ÓÖÊÁ¿ÎªM£¬Çë»Ø´ð£º¾§°ûÖТÞÔ­×ÓµÄÅäλÊýΪ            ,¸Ã¾§ÌåµÄÃܶÈΪ            £¨ÓÃ×Öĸ±íʾ£©
£¨6£©ÊµÑéÖ¤Ã÷£º¢ÝºÍ¢àµÄÑõ»¯Îï¡¢KCl¡¢TiNÕâ4ÖÖ¾§ÌåµÄ½á¹¹ÓëNaCl¾§Ìå½á¹¹ÏàËÆ£¨ÈçÏÂͼËùʾ£©£¬ÒÑÖª3ÖÖÀë×Ó¾§ÌåµÄ¾§¸ñÄÜÊý¾ÝÈçÏÂ±í£º

Àë×Ó¾§Ìå
NaCl
KCl
CaO
¾§¸ñÄÜ/kJ¡¤mol£­1
786
715
3401
 
Ôò¸Ã4ÖÖÀë×Ó¾§Ì壨²»°üÀ¨NaCl£©ÈÛµã´Ó¸ßµ½µÍµÄ˳ÐòÊÇ£º              £¨Óû¯Ñ§Ê½Ìîд£©¡£
ÆäÖТàµÄÑõ»¯ÎᄃÌåÖÐÒ»¸öÑôÀë×ÓÖÜΧºÍËü×îÁÚ½üÇҵȾàÀëµÄÑôÀë×ÓÓР      ¸ö¡£

£¨16·Ö£©ÒÑÖªA¡¢B¡¢C¡¢D¡¢E¡¢F¡¢GºÍH¶¼ÊÇÔªËØÖÜÆÚ±íÖÐÇ°36ºÅµÄÔªËØ£¬ËüÃǵÄÔ­×ÓÐòÊýÒÀ´ÎÔö´ó¡£AÊÇÖÜÆÚ±íÖÐÔ­×Ӱ뾶×îСµÄÔªËØ£¬BÔ­×Ó×îÍâ²ãµç×ÓÊýÊÇÄÚ²ãµç×ÓÊýµÄÁ½±¶£¬CµÄ»ù̬ԭ×ÓºËÍâÓÐ7ÖÖ²»Í¬Ô˶¯×´Ì¬µÄµç×Ó£¬EÊǵ縺ÐÔ×î´óµÄÔªËØ£¬FµÄÇ⻯ÎïºÍ×î¸ß¼ÛÑõ»¯Îï¶ÔÓ¦µÄË®»¯Îï¾ùΪǿËᣬG¡¢H·Ö±ðÊÇÖÜÆÚ±íÖÐ1¡ª18×ÝÁÐÖеĵÚ10¡¢12×ÝÁÐÔªËØ¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©DÔªËØÔÚÖÜÆÚ±íλÖÃÊÇ        £¬GµÄºËÍâ¼Û²ãµç×ÓÅŲ¼Ê½Îª      ¡£
£¨2£©EÓëAÐγɵĻ¯ºÏÎï±ÈFÓëAÐγɵĻ¯ºÏÎïµÄ·Ðµã   £¨Ìî¡°¸ß¡±»ò¡°µÍ¡±£©£¬ÆäÔ­ÒòÊÇ   ¡£
£¨3£©B¡¢CÔ­×ӵĵÚÒ»µçÀëÄܽϴóµÄÊÇ     £¨ÌîÔªËØ·ûºÅ£©£¬ÆäÔ­ÒòÊÇ       ¡£
£¨4£©BD32£­Àë×ÓÖÐBÔ­×Ó²ÉÈ¡   ÔÓ»¯£¬ÈÎдһÖÖÓëBD»¥ÎªµÈµç×ÓÌåµÄ·Ö×ӵĵç×Óʽ   ¡£
£¨5£©ÔªËØHµÄÒ»ÖÖÁò»¯ÎᄃÌåµÄ¾§°û½á¹¹ÈçͼËùʾ£¬¸ÃÁò»¯ÎïµÄ»¯Ñ§Ê½ÊÇ         ¡£ÔªËØHµÄÇâÑõ»¯Îï¿ÉÈÜÓÚ°±Ë®ÖУ¬Éú³ÉºÍÍ­°±ÅäÀë×ÓÏàͬÅäλÊýµÄÀë×Ó£¬Ð´³ö¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽΪ              ¡£

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

¾«Ó¢¼Ò½ÌÍø