ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿´×ÂÈ·ÒËá(H)ÊÇÒ»ÖÖÐÂÐÍ¡¢Ç¿Ð§½âÈÈ¡¢ÕòÍ´¡¢¿¹¹Ø½ÚÑ×Ò©ÎÆäºÏ³É·ÏßÈçÏ£¨²¿·Ö·´Ó¦Ìõ¼þÊ¡ÂÔ£©£º

ÒÑÖª£º

Çë»Ø´ð£º

£¨1£©»¯ºÏÎïFµÄ½á¹¹¼òʽÊÇ__________________£»

£¨2£©Ð´³öC+D¡úEµÄ»¯Ñ§·½³Ìʽ____________________________________________£»

£¨3£©ÏÂÁÐ˵·¨²»ÕýÈ·µÄÊÇ__________¡£

A£®»¯ºÏÎïAÖÐËùÓÐÔ­×ӿ϶¨´¦ÓÚͬһƽÃæ

B£®»¯ºÏÎïB¾ßÓмîÐÔ

C£®»¯ºÏÎïEÓëNaOHÈÜÒº·´Ó¦×î¶àÏûºÄ4molNaOH

D£®´×ÂÈ·ÒËá·Ö×ÓʽΪC16H12Cl2NO4

£¨4£©ÒÑÖªCµÄͬ·ÖÒì¹¹ÌåÓжàÖÖ£¬Ð´³öͬʱÂú×ãÏÂÁÐÌõ¼þµÄËùÓÐͬ·ÖÒì¹¹ÌåµÄ½á¹¹¼òʽ_______¡£

¢ÙIRÆ×±íÃ÷·Ö×ÓÖк¬ÓÐÁ½¸ö±½»·½á¹¹£¬ÇÒ2¸öÂÈÔ­×Ó²»ÔÚͬһ¸ö±½»·ÉÏ£»

¢Ú1H-NMRÆ×ÏÔʾֻÓÐ5ÖÖ²»Í¬»·¾³µÄÇâÔ­×Ó

£¨5£©Éè¼ÆÒÔ±½ºÍÒÒϩΪԭÁÏºÏ³É µÄ·Ïß(ÓÃÁ÷³Ìͼ±íʾ£¬ÎÞ»úÊÔ¼ÁÈÎÑ¡)¡£

______________________________________________________________________________

¡¾´ð°¸¡¿ CD

¡¾½âÎö¡¿

¸ù¾ÝEµÄ½á¹¹¼òʽ£¬¼°·´Ó¦Á÷³Ì¿ÉÖª£¬Aº¬Óб½»·£¬ÇÒÁ½¸öÂÈÔ­×ÓÔÚ¼ä룬Æä½á¹¹¼òʽΪ£º£¬Í¬Àí£¬BµÄ½á¹¹¼òʽΪ£º£¬AÓëB·´Ó¦Éú³ÉC£¬ÔòCµÄ½á¹¹¼òʽΪ£º£»CÓëD·´Ó¦Éú³ÉE£¬¿ÉÍƲâDµÄ½á¹¹¼òʽΪ£ºClCOCH2Cl£»Í¨¹ýGÓëEµÄ½á¹¹¼òʽµÄ¶Ô±È£¬½áºÏ·´Ó¦¿ÉÖªFµÄ½á¹¹¼òʽΪ£º£»

£¨1£©¸ù¾Ý·ÖÎö¿ÉÖª£¬FµÄ½á¹¹¼òʽΪ£º£»

£¨2£©CµÄ½á¹¹¼òʽΪ£º£¬ DµÄ½á¹¹¼òʽΪ£ºClCOCH2Cl£¬CÓëD·´Ó¦Éú³ÉE£¬¸ù¾Ý·´Ó¦¢ÚµÄÌص㣬Ôò·´Ó¦µÄ·½³ÌʽΪ£º£»

£¨3£©A.»¯ºÏÎïAµÄ½á¹¹¼òʽΪ£º£¬ClÔ­×ÓÈ¡´ú±½»·ÉϵÄÇ⣬ÔòËùÓÐÔ­×Ó¹«Æ½Ã棬AÕýÈ·£»

B.»¯ºÏÎïBµÄ½á¹¹¼òʽ£º£¬º¬Óа±»ù£¬°±»ùÏÔ¼îÐÔ£¬BÕýÈ·£»

C.»¯ºÏÎïEΪ£º£¬ëļüÔÚ¼îÐÔÌõ¼þÏÂË®½â£¬ClÔ­×ÓË®½â£¬±½»·ÉϵÄClÔ­×ÓË®½âºóÉú³É·ÓôÇ»ù£¬Óë¼î·´Ó¦£¬Ôò¹²¼Æ6molNaOH£¬C´íÎó£»

D. ´×ÂÈ·ÒËá·Ö×ÓʽΪC16H13Cl2NO4£¬D´íÎó£»

´ð°¸ÎªCD

£¨4£©CµÄ·Ö×ÓʽΪ£ºC12H9NCl2£¬¢ÙIRÆ×±íÃ÷·Ö×ÓÖк¬ÓÐÁ½¸ö±½»·½á¹¹£¬ÇÒ2¸öÂÈÔ­×Ó²»ÔÚͬһ¸ö±½»·ÉÏ¢Ú1H-NMRÆ×ÏÔʾֻÓÐ5ÖÖ²»Í¬»·¾³µÄÇâÔ­×Ó£¬ÔòÁ½¸ö±½»·Îª¶Ô³Æ½á¹¹£¬¿ÉÄܵĽṹ¼òʽΪ£º£»

£¨5£©¸ù¾Ý·´Ó¦¢Ù¿ÉÖª£¬±½»·ÉÏÌí¼ÓÖ§Á´ÔòÐ裬Éú³É£¬ÔòÐèÒªÂÈ´úÌþºÍÇ軯ÄÆÔÚËáÐÔÌõ¼þÏ·´Ó¦ÖÆÈ¡£¬¹Ê¿ÉÏÈÓÃÒÒÏ©ÖÆÈ¡ÂÈ´úÌþ£¬ÔÙÓëÇ軯ÄÆ·´Ó¦¼´¿É£¬Á÷³ÌΪ£º£»

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿µªµÄ»¯ºÏÎïÓÃ;¹ã·º¡£»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©ÔÚÒ»¶¨Ìõ¼þÏ£¬µªÆøÄܺÍË®ÕôÆø·´Ó¦Éú³É°±ÆøºÍÑõÆø2N2£¨g£©£«6H2O£¨g£©=4NH3£¨g£©£«3O2£¨g£©¡÷H£¬Óë¸Ã·´Ó¦Ïà¹ØµÄ»¯Ñ§¼ü¼üÄÜÊý¾ÝÈçÏ£º

»¯Ñ§¼ü

N¡ÔN

H¡ªO

N¡ªH

O=O

E£¨kJ/mol£©

946

463

391

496

Ôò¸Ã·´Ó¦µÄ¡÷H£½________kJ¡¤mol-1¡£

£¨2£©ÔÚºãÈÝÃܱÕÈÝÆ÷ÖгäÈë2 mol N2O5Óë1molO2·¢Éú·´Ó¦4NO2 (g) + O2 (g) 2N2O5 (g) ¡÷H¡£

¢ÙÒÑÖªÔÚ²»Í¬Î¶ÈϲâµÃN2O5µÄÎïÖʵÄÁ¿Ëæʱ¼äµÄ±ä»¯ÈçͼËùʾ£¬¸Ã·´Ó¦µÄ¡÷H_____0£¨Ìî¡°£¾¡±¡°£¼¡±»ò¡°£½¡±£©¡£¸ßÎÂϸ÷´Ó¦ÄÜÄæÏò×Ô·¢½øÐУ¬ÆäÔ­ÒòÊÇ___________________¡£

¢ÚÏÂÁÐÓйظ÷´Ó¦µÄ˵·¨ÕýÈ·µÄÊÇ_______£¨Ìî±êºÅ£©¡£

A£®À©´óÈÝÆ÷Ìå»ý£¬Æ½ºâÏòÄæ·´Ó¦·½ÏòÒƶ¯£¬»ìºÏÆøÌåÑÕÉ«±äÉî

B£®ºãκãÈÝ£¬ÔÙ³äÈë2 mol NO2ºÍ1molO2£¬Ôٴδﵽƽºâʱ£¬NO2µÄת»¯ÂÊÔö´ó

C£®ºãκãÈÝ£¬µ±ÈÝÆ÷ÄÚµÄÃܶȱ£³Ö²»±äʱ£¬·´Ó¦´ïµ½ÁËƽºâ״̬

D£®Èô¸Ã·´Ó¦µÄƽºâ³£ÊýÔö´ó£¬ÔòÒ»¶¨ÊǽµµÍÁËζÈ

£¨3£©N2O5ÊÇÒ»ÖÖÐÂÐÍÂÌÉ«Ïõ»¯¼Á£¬ÆäÖƱ¸¿ÉÒÔÓÃÅðÇ⻯ÄÆȼÁϵç³Ø×÷µçÔ´£¬²ÉÓõç½â·¨ÖƱ¸µÃµ½N2O5£¬¹¤×÷Ô­ÀíÈçͼËùʾ¡£ÔòÅðÇ⻯ÄÆȼÁϵç³ØµÄ¸º¼«·´Ó¦Ê½Îª_________¡£

£¨4£©X¡¢Y¡¢Z¡¢W·Ö±ðÊÇHNO3¡¢NH4NO3¡¢NaOH¡¢NaNO2ËÄÖÖÇ¿µç½âÖÊÖеÄÒ»ÖÖ¡£Ï±íÊdz£ÎÂÏÂŨ¶È¾ùΪ0£®01molL¡ª1µÄX¡¢Y¡¢Z¡¢WÈÜÒºµÄpH¡£½«X¡¢Y¡¢Z¸÷1molͬʱÈÜÓÚË®Öеõ½»ìºÏÈÜÒº£¬Ôò»ìºÏÈÜÒºÖи÷Àë×ÓµÄŨ¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòΪ________¡£

0£®01molL¡ª1µÄÈÜÒº

X

Y

Z

W

pH

12

2

8.5

4.5

£¨5£©µªµÄÑõ»¯ÎïÓëÐü¸¡ÔÚ´óÆøÖеĺ£ÑÎÁ£×ÓÏ໥×÷ÓÃʱ£¬Éæ¼°ÈçÏ·´Ó¦£º

I£º2NO2(g)+NaCl(s) NaNO3 (s)+ClNO(g) K1

¢ò£º2NO(g)+Cl2 (g) 2CNO(g) K2

¢Ù4NO2£¨g£©£«2NaCl£¨s£©2NaNO3£¨s£©£«2NO£¨g£©£«Cl2£¨g£©µÄƽºâ³£ÊýK£½____£¨ÓÃK1¡¢K2±íʾ£©¡£

¢ÚÔÚºãÎÂÌõ¼þÏ£¬Ïò2LºãÈÝÃܱÕÈÝÆ÷ÖмÓÈë0£®2 mol NOºÍ0£®1 mol Cl2£¬10minʱ·´Ó¦¢ò´ïµ½Æ½ºâ£¬²âµÃ10minÄÚv(ClNO)£½7.5¡Á10-3molL-1min-1£¬ÔòƽºâʱNOµÄת»¯ÂʦÁ1=____£»ÈôÆäËûÌõ¼þ²»±ä£¬·´Ó¦¢òÔÚºãѹÌõ¼þϽøÐУ¬Æ½ºâʱNOµÄת»¯ÂʦÁ2__¦Á1£¨Ìî¡°£¾¡±¡°£¼¡±»ò¡°£½¡±£©¡£

¡¾ÌâÄ¿¡¿×ÊÔ´µÄ¸ßЧÀûÓöԱ£»¤»·¾³¡¢´Ù½ø¾­¼Ã³ÖÐø½¡¿µ·¢Õ¹¾ßÓÐÖØÒª×÷Óá£Á×β¿óÖ÷Òªº¬Ca5(PO4)3FºÍCaCO3¡¤MgCO3¡£Ä³Ñо¿Ð¡×éÌá³öÁËÁ×β¿ó×ÛºÏÀûÓõÄÑо¿·½°¸£¬ÖƱ¸¾ßÓÐÖØÒª¹¤ÒµÓÃ;µÄCaCO3¡¢Mg(OH)2¡¢P4ºÍH2£¬Æä¼ò»¯Á÷³ÌÈçÏ£º

ÒÑÖª£º¢ÙCa5(PO4)3FÔÚ950¡æ²»·Ö½â£»

¢Ú4Ca5(PO4)3F+18SiO2+30C2CaF2+30CO+18CaSiO3+3P4

Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©950¡æìÑÉÕÁ×β¿óÉú³ÉÆøÌåµÄÖ÷Òª³É·ÖÊÇ___________¡£

£¨2£©ÊµÑéÊÒ¹ýÂËËùÐèµÄ²£Á§ÒÇÆ÷ÊÇ_____________¡£

£¨3£©NH4NO3ÈÜÒºÄÜ´ÓÁ׿ó¢ñÖнþÈ¡³öCa2+µÄÔ­ÒòÊÇ__________¡£

£¨4£©ÔÚ½þÈ¡Òº¢òÖÐͨÈëNH3£¬·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ____________¡£

£¨5£©¹¤ÒµÉϳ£ÓÃÁ×¾«¿ó[Ca5(PO4)3F]ºÍÁòËá·´Ó¦ÖƱ¸Á×Ëá¡£ÒÑÖª25¡æ£¬101kPaʱ£º

CaO(s)+H2SO4(l)=CaSO4(s)+H2O(l) ¦¤H=-271kJ/mol

5 CaO(s)+3H3PO4(l)+HF(g)= Ca5(PO4)3F (s)+5H2O(l) ¦¤H=-937kJ/mol

ÔòCa5(PO4)3FºÍÁòËá·´Ó¦Éú³ÉÁ×ËáµÄÈÈ»¯Ñ§·½³ÌʽÊÇ_________________¡£

£¨6£©ÔÚÒ»¶¨Ìõ¼þÏÂCO(g)+H2O(g)CO2(g)+H2(g)£¬µ±COÓëH2O(g)µÄÆðʼÎïÖʵÄÁ¿Ö®±ÈΪ1:5£¬´ïƽºâʱ£¬COת»¯ÁË¡£Èôa kgº¬Ca5(PO4)3F£¨Ïà¶Ô·Ö×ÓÖÊÁ¿Îª504£©µÄÖÊÁ¿·ÖÊýΪ10%µÄÁ×β¿ó£¬ÔÚÉÏÊö¹ý³ÌÖÐÓÐb%µÄCa5(PO4)3Fת»¯ÎªP4£¬½«²úÉúµÄCOÓëH2O(g)°´ÆðʼÎïÖʵÄÁ¿Ö®±È1:3»ìºÏ£¬ÔòÏàͬÌõ¼þÏ´ïƽºâʱÄܲúÉúH2________kg¡£

¡¾ÌâÄ¿¡¿¼îʽ̼ËáÍ­ [Cu2(OH)2CO3]ÊÇÒ»ÖÖÓÃ;¹ã·ºµÄ»¯¹¤Ô­ÁÏ£¬ÊµÑéÊÒÒÔ·ÏͭмΪԭÁÏÖÆÈ¡¼îʽ̼ËáÍ­µÄ²½ÖèÈçÏ£º

²½ÖèÒ»£º·ÏͭмÖÆÏõËáÍ­

Èçͼ£¬ÓýºÍ·µÎ¹ÜÎüȡŨHNO3»ºÂý¼Óµ½×¶ÐÎÆ¿ÄڵķÏͭмÖÐ(·Ïͭм¹ýÁ¿)£¬³ä·Ö·´Ó¦ºó¹ýÂË£¬µÃµ½ÏõËáÍ­ÈÜÒº¡£

²½Öè¶þ£º¼îʽ̼ËáÍ­µÄÖƱ¸

Ïò´óÊÔ¹ÜÖмÓÈë̼ËáÄƺÍÏõËáÍ­ÈÜÒº£¬Ë®Ô¡¼ÓÈÈÖÁ70¡æ×óÓÒ£¬ÓÃ0.4mol/LµÄNaOHÈÜÒºµ÷½ÚpHÖÁ8.5£¬Õñµ´£¬¾²Ö㬹ýÂË£¬ÓÃÈÈˮϴµÓ£¬ºæ¸É£¬µÃµ½¼îʽ̼ËáÍ­²úÆ·¡£

Íê³ÉÏÂÁÐÌî¿Õ£º

£¨1£©Ð´³öŨÏõËáÓëÍ­·´Ó¦µÄÀë×Ó·½³Ìʽ___________________________________¡£

£¨2£©ÉÏͼװÖÃÖÐNaOHÈÜÒºµÄ×÷ÓÃÊÇ__________________________________________¡£

£¨3£©²½Öè¶þÖУ¬Ë®Ô¡¼ÓÈÈËùÐèÒÇÆ÷ÓÐ_________¡¢________(¼ÓÈÈ¡¢¼Ð³ÖÒÇÆ÷¡¢Ê¯ÃÞÍø³ýÍâ)£»Ï´µÓµÄÄ¿µÄÊÇ______________________________________________¡£²½Öè¶þµÄÂËÒºÖпÉÄܺ¬ÓÐCO32-£¬Ð´³ö¼ìÑéCO32-µÄ·½·¨_______________________________________¡£

£¨4£©Ó°Ïì²úÆ·²úÁ¿µÄÖ÷ÒªÒòËØÓÐ________________________________________¡£

£¨5£©ÈôʵÑéµÃµ½2.42gÑùÆ·(Ö»º¬CuOÔÓÖÊ)£¬È¡´ËÑùÆ·¼ÓÈÈÖÁ·Ö½âÍêÈ«ºó£¬µÃµ½1.80g¹ÌÌ壬´ËÑùÆ·Öмîʽ̼ËáÍ­µÄÖÊÁ¿·ÖÊýÊÇ_________________(±£ÁôÁ½Î»Ð¡Êý)¡£

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

¾«Ó¢¼Ò½ÌÍø