ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿¸ß´¿¹è¾§ÌåÊÇÐÅÏ¢¼¼ÊõµÄÖØÒª²ÄÁÏ¡£

£¨1£©¹¤ÒµÉÏÓÃʯӢºÍ½¹Ì¿¿ÉÒÔÖƵôֹ衣ÒÑÖª·´Ó¦¹ý³ÌµÄÄÜÁ¿±ä»¯ÈçÏÂͼ

д³öÓÃʯӢºÍ½¹Ì¿ÖÆÈ¡´Ö¹èµÄÈÈ»¯Ñ§·½³Ìʽ______________________________¡£

£¨2£©Ä³Í¬Ñ§Éè¼ÆÏÂÁÐÁ÷³ÌÖƱ¸¸ß´¿¹è£º

¢ÙYµÄ»¯Ñ§Ê½Îª____________________¡£

¢Úд³ö·´Ó¦IµÄÀë×Ó·½³Ìʽ________________________________________¡£

¢Ûд³ö·´Ó¦IVµÄ»¯Ñ§·½³Ìʽ________________________________________¡£

¢Ü¼×Íé·Ö½âµÄζÈÔ¶Ô¶¸ßÓÚ¹èÍ飨SiH4£©£¬ÓÃÔ­×ӽṹ½âÊÍÆäÔ­Òò______________________¡£

£¨3£©½«´Ö¹èת»¯³ÉÈýÂȹèÍ飨SiHCl3£©£¬½øÒ»²½·´Ó¦Ò²¿ÉÒÔÖƵôֹ衣Æä·´Ó¦£ºSiHCl3(g)+H2(g)Si(s)+3HCl(g)£¬²»Í¬Î¶ÈÏ£¬SiHCl3µÄƽºâת»¯ÂÊËæ·´Ó¦ÎïµÄͶÁϱȵı仯¹ØϵÈçͼËùʾ¡£ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ__________£¨Ìî×Öĸ£©¡£

A£®¸Ã·´Ó¦ÊÇ·ÅÈÈ·´Ó¦

B£®ºá×ø±ê±íʾµÄͶÁϱȿÉÒÔÊÇ

C£®¸Ã·´Ó¦µÄƽºâ³£ÊýËæζÈÉý¸ß¶øÔö´ó

D£®Êµ¼ÊÉú²úÖÐΪÌá¸ßSiHCl3µÄÀûÓÃÂÊ£¬¿ÉÒÔÊʵ±Ôö´óѹǿ

¡¾´ð°¸¡¿ SiO2(s)+2C(s)=Si(s)+2CO(g) H=+638.4kJ/mol H2SiO3»òH4SiO4 SiO2+2OH-=SiO32-+H2O SiO2+4MgMg2Si+2MgO ÖÜÆÚ±íÖУ¬¹èÖÐ̼ÊôÓÚͬÖ÷×壬ԭ×Ӱ뾶Si´óÓÚC£¬¹èÔªËصķǽðÊôÐÔÈõÓÚ̼ԪËØ£¬¹èÍéµÄÈÈÎȶ¨ÐÔÈõÓÚ¼×Íé BC

¡¾½âÎö¡¿£¨1£©¢ÙSi(s)+O2(g)==SiO2(g) ¡÷H=-859.4kl/mol£»¢Ú2C(s)+ O2(g)==2CO(g) ¡÷H=-221.0kl/mol£»ÓÉ¢Ú-¢Ù¿ÉµÃʯӢºÍ½¹Ì¿ÖÆÈ¡´Ö¹èµÄÈÈ»¯Ñ§·½³Ìʽ£ºSiO2(s)+2C(s)=Si(s)+2CO(g) H=+638.4kJ/mol£»£¨2£©¢ÙXΪ¹èËáÄÆ£¬ÔòYΪ¹èËᣬ¿Éд³ÉH2SiO3»òH4SiO4£»¢Ú·´Ó¦IµÄÀë×Ó·½³ÌʽΪ£ºSiO2+2OH-=SiO32-+H2O£»¢ÛþÓë¶þÑõ»¯¹èÔÚ¸ßÎÂÌõ¼þÏ·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£ºSiO2+4MgMg2Si+2MgO£»¢Ü¼×Íé·Ö½âµÄζÈÔ¶Ô¶¸ßÓÚ¹èÍ飨SiH4£©Ô´ÓÚ̼Çâ¼üÓë¹èÇâ¼üµÄ¼üÄÜ´óС£¬ÖÜÆÚ±íÖУ¬¹è¡¢Ì¼ÊôÓÚͬÖ÷×壬ԭ×Ӱ뾶Si´óÓÚC£¬¹èÔªËصķǽðÊôÐÔÈõÓÚ̼ԪËØ£¬Ì¼Çâ¼üµÄ¼üÄÜ´óÓÚ¹èÇâ¼üµÄ¼üÄÜ£¬¹èÍéµÄÈÈÎȶ¨ÐÔÈõÓÚ¼×Íé¡££¨3£©µ±Í¶ÁϱÈÒ»¶¨Ê±Ê±£¨»­µÈͶÁϱÈÀýÏߣ©£¬ËæζÈÉý¸ß£¨T1´óÓÚT2´óÓÚT3£©£¬SiHCl3µÄƽºâת»¯ÂÊÔö´ó£¬ËµÃ÷Õý·´Ó¦ÊÇÎüÈÈ·´Ó¦£¬¹ÊA´íÎó£»Ôö´óÇâÆøµÄŨ¶È¿ÉÌá¸ßSiHCl3µÄƽºâת»¯ÂÊ£¬BÕýÈ·£»ÒòÕý·´Ó¦ÊÇÎüÈÈ·´Ó¦£¬Éý¸ßζȣ¬Æ½ºâÕýÏòÒƶ¯£¬·´Ó¦µÄƽºâ³£ÊýÔö´ó£¬CÕýÈ·£»ÓÉÓÚ·´Ó¦Ç°ÆøÌå·Ö×ÓÊýСÓÚ·´Ó¦ºóÆøÌå·Ö×ÓÊý£¬Ôö´óѹǿ£¬Æ½ºâÄæÏòÒƶ¯£¬SiHCl3µÄƽºâת»¯ÂʽµµÍ£¬D´íÎó¡£´ð°¸Ñ¡BC¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿ÓÉN¡¢BµÈÔªËØ×é³ÉµÄÐÂÐͲÄÁÏÓÐ׏㷺ÓÃ;¡£

£¨1£©B2H6ÊÇÒ»ÖÖ¸ßÄÜȼÁÏ£¬ËüÓëCl2·´Ó¦Éú³ÉµÄBCl3¿ÉÓÃÓÚ°ëµ¼Ìå²ôÔÓ¹¤ÒÕ¼°¸ß´¿¹èµÄÖÆÔ죻ÓɵڶþÖÜÆÚÔªËØ×é³ÉµÄÓëBCl3»¥ÎªµÈµç×ÓÌåµÄÒõÀë×ÓΪ_______£¨ÌîÀë×Ó·ûºÅ£¬ÌîÒ»¸ö£©¡£

£¨2£©°±ÅðÍ飨H3N¡úBH3£©ºÍTi£¨BH4£©3¾ùΪ¹ãÊܹØ×¢µÄÐÂÐÍ»¯Ñ§Ç⻯Îï´¢Çâ²ÄÁÏ£®

¢ÙH3N¡úBH3ÖÐBÔ­×ÓµÄÍâΧµç×ÓÅŲ¼Í¼_________¡£

¢ÚTi£¨BH4£©3ÓÉTiCl3ºÍLiBH4·´Ó¦ÖƵã¬Ð´³ö¸ÃÖƱ¸·´Ó¦µÄ»¯Ñ§·½³Ìʽ____£»»ù̬Ti3+µÄ³É¶Ôµç×ÓÓÐ___¶Ô£¬BH4-µÄÁ¢Ìå¹¹ÐÍÊÇ____£»Ti£¨BH4£©3Ëùº¬»¯Ñ§¼üµÄÀàÐÍÓÐ____£»

¢Û°±ÅðÍé¿ÉÓÉÁùÔª»·×´»¯ºÏÎHB=NH£©3ͨ¹ýÈçÏ·´Ó¦ÖƵãº3CH4+2£¨HB=NH£©3+6H2O¡ú3CO2+6H3BNH3£»ÓëÉÏÊö»¯Ñ§·½³ÌʽÓйصÄÐðÊö²»ÕýÈ·µÄÊÇ _____________

A£®°±ÅðÍéÖдæÔÚÅäλ¼ü

B£®µÚÒ»µçÀëÄÜ£ºN£¾O£¾C£¾B

C£®·´Ó¦Ç°ºó̼ԭ×ӵĹìµÀÔÓ»¯ÀàÐͲ»±ä

D£®CH4¡¢H2O¡¢CO2¶¼ÊǷǼ«ÐÔ·Ö×Ó

£¨3£©Á×»¯Åð£¨BP£©ÊÇÊܵ½¸ß¶È¹Ø×¢µÄÄÍÄ¥²ÄÁÏ£¬Èçͼ1ΪÁ×»¯Å𾧰û£»

¢Ù¾§ÌåÖÐPÔ­×ÓÌîÔÚBÔ­×ÓËùΧ³ÉµÄ____¿Õ϶ÖС£

¢Ú¾§ÌåÖÐBÔ­×ÓÖÜΧ×î½üÇÒÏàµÈµÄBÔ­×ÓÓÐ____¸ö¡£

£¨4£©Á¢·½µª»¯ÅðÊÇÒ»ÖÖÐÂÐ͵ij¬Ó²¡¢ÄÍÄ¥¡¢Ä͸ßεĽṹ²ÄÁÏ£¬Æä½á¹¹ºÍÓ²¶È¶¼Óë½ð¸ÕʯÏàËÆ£¬µ«ÈÛµã±È½ð¸ÕʯµÍ£¬Ô­ÒòÊÇ__________¡£Í¼2ÊÇÁ¢·½µª»¯Å𾧰ûÑØzÖáµÄͶӰͼ£¬ÇëÔÚͼÖÐÔ²ÇòÉÏÍ¿¡°¡ñ¡±ºÍ»­¡°¡Á¡±·Ö±ð±êÃ÷BÓëNµÄÏà¶ÔλÖÃ______¡£ÆäÖС°¡ñ¡±´ú±íBÔ­×Ó£¬¡°¡Á¡±´ú±íNÔ­×Ó¡£

¡¾ÌâÄ¿¡¿£¨14·Ö£©Ì¼¡¢µª¡¢ÁòÊÇÖÐѧ»¯Ñ§ÖØÒªµÄ·Ç½ðÊôÔªËØ£¬ÔÚ¹¤Å©ÒµÉú²úÖÐÓй㷺µÄÓ¦Óá£

(1£©ÓÃÓÚ·¢Éä¡°Ì칬һºÅ¡±µÄ³¤Õ÷¶þºÅ»ð¼ýµÄȼÁÏÊÇҺ̬ƫ¶þ¼×루CH3£©2N£­NH2£¬Ñõ»¯¼ÁÊÇҺ̬ËÄÑõ»¯¶þµª¡£¶þÕßÔÚ·´Ó¦¹ý³ÌÖзųö´óÁ¿ÄÜÁ¿£¬Í¬Ê±Éú³ÉÎÞ¶¾¡¢ÎÞÎÛȾµÄÆøÌå¡£ÒÑÖªÊÒÎÂÏ£¬1 gȼÁÏÍêȫȼÉÕÊͷųöµÄÄÜÁ¿Îª42.5kJ£¬Çëд³ö¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ________________________________________¡£

£¨2£©298 Kʱ£¬ÔÚ2LµÄÃܱÕÈÝÆ÷ÖУ¬·¢Éú¿ÉÄæ·´Ó¦£º

2NO2(g)N2O4(g) ¦¤H£½£­a kJ¡¤mol£­1(a£¾0)

N2O4µÄÎïÖʵÄÁ¿Å¨¶ÈËæʱ¼ä±ä»¯Èçͼ¡£´ïƽºâʱ£¬N2O4µÄŨ¶ÈΪNO2µÄ2±¶£¬»Ø´ðÏÂÁÐÎÊÌâ¡£

¢Ù298kʱ£¬¸Ã·´Ó¦µÄƽºâ³£ÊýΪ________¡£

¢ÚÔÚζÈΪT1¡¢T2ʱ£¬Æ½ºâÌåϵÖÐNO2µÄÌå»ý·ÖÊýËæѹǿ±ä»¯ÇúÏßÈçͼËùʾ¡£ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ

a£®A¡¢CÁ½µãµÄ·´Ó¦ËÙÂÊ£ºA£¾C

b£®B¡¢CÁ½µãµÄÆøÌåµÄƽ¾ùÏà¶Ô·Ö×ÓÖÊÁ¿£ºB£¼C

c£®A¡¢CÁ½µãÆøÌåµÄÑÕÉ«£ºAÉCdz

d£®ÓÉ״̬Bµ½×´Ì¬A£¬¿ÉÒÔÓüÓÈȵķ½·¨

¢ÛÈô·´Ó¦ÔÚ398K½øÐУ¬Ä³Ê±¿Ì²âµÃn(NO2£©="0.6" mol n(N2O4£©=1.2mol£¬Ôò´ËʱV£¨Õý£© V£¨Ä棩£¨Ìî¡°>¡±¡¢¡°<¡±»ò¡°=¡±£©¡£

£¨3£©NH4HSO4ÔÚ·ÖÎöÊÔ¼Á¡¢Ò½Ò©¡¢µç×Ó¹¤ÒµÖÐÓÃ;¹ã·º¡£ÏÖÏò100 mL 0.1 mol¡¤L£­1NH4HSO4ÈÜÒºÖеμÓ0.1 mol¡¤L£­1NaOHÈÜÒº£¬µÃµ½µÄÈÜÒºpHÓëNaOHÈÜÒºÌå»ýµÄ¹ØϵÇúÏßÈçͼËùʾ¡£

ÊÔ·ÖÎöͼÖÐa¡¢b¡¢c¡¢d¡¢eÎå¸öµã£¬

¢ÙË®µÄµçÀë³Ì¶È×î´óµÄÊÇ__________£»

¢ÚÆäÈÜÒºÖÐc(OH-)µÄÊýÖµ×î½Ó½üNH3¡¤H2OµÄµçÀë³£ÊýKÊýÖµµÄÊÇ £»

¢ÛÔÚcµã£¬ÈÜÒºÖи÷Àë×ÓŨ¶ÈÓÉ´óµ½Ð¡µÄÅÅÁÐ˳ÐòÊÇ_______¡£

¡¾ÌâÄ¿¡¿Ä³Ð¡×éÀûÓÃÈçͼװÖã¬Óñ½ÓëäåÔÚFeBr3´ß»¯×÷ÓÃÏÂÖƱ¸äå±½£º

±½

äå

äå±½

ÃܶÈ/g¡¤cm-3

0.88

3.10

1.50

·Ðµã/¡ãC

80

59

156

Ë®ÖÐÈܽâÐÔ

΢ÈÜ

΢ÈÜ

΢ÈÜ

ʵÑé¹ý³Ì£ºÔÚaÖмÓÈë15mLÎÞË®±½ºÍÉÙÁ¿Ìúм£¬ÔÚbÖÐСÐļÓÈë4.0mLҺ̬äå¡£ÏòaÖеÎÈ뼸µÎäå¡£·´Ó¦¾çÁÒ½øÐС£·´Ó¦Í£Ö¹ºó°´ÈçÏÂÁ÷³Ì·ÖÀëÌá´¿²úÆ·£º

¡¡

£¨1£©ÉÕÆ¿ÖÐÓдóÁ¿ºì×ØÉ«ÕôÆø£¬ÊÔ¹ÜdÖеÄÏÖÏóÊÇ£º¢Ù______________£»¢Ú ÕôÁóË®Öð½¥±ä³É»ÆÉ«¡£cµÄ×÷ÓÃÊÇ___________________________________¡£

£¨2£©·ÖÀëÌᴿʱ£¬²Ù×÷¢ñΪ______________£¬²Ù×÷¢òΪ_________________¡£

£¨3£©¡°Ë®Ï´¡±ÐèÒªÓõ½µÄ²£Á§ÒÇÆ÷ÊÇ_________¡¢ÉÕ±­£¬Ïò¡°Ë®Ï´¡±ºóËùµÃË®ÏàÖеμÓKSCNÈÜÒº£¬ÈÜÒº±äºìÉ«¡£ÍƲâˮϴµÄÖ÷ҪĿµÄÊdzýÈ¥__________________¡£

£¨4£©¡°NaOHÈÜҺϴ¡±Ê±·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ________________________¡£

£¨5£©ÒÑÖª±½Óëäå·¢ÉúµÄÊÇÈ¡´ú·´Ó¦£¬ÍƲⷴӦºóÊÔ¹ÜdÖÐÒºÌ庬ÓеÄÁ½ÖÖ´óÁ¿Àë×ÓÊÇH+ºÍBr£­£¬Éè¼ÆʵÑé·½°¸ÑéÖ¤ÍƲ⡣£¨ÏÞÑ¡ÊÔ¼Á£ºMg¡¢CCl4¡¢AgNO3aq¡¢H2O£©

ʵÑé²½Öè

Ô¤ÆÚÏÖÏó

½áÂÛ

²½Öè1£º½«ÊÔ¹ÜdÖÐÒºÌåתÈë·ÖҺ©¶·£¬

__________________________________£¬½«ËùÈ¡ÈÜÒºµÈ·Ö³ÉÁ½·Ý£¬ÖÃÓÚA¡¢BÁ½ÊÔ¹ÜÖУ¬½øÐв½Öè2¡¢3¡£

²½Öè2£º ¡£

Ö¤Ã÷ÓÐ ´æÔÚ

²½Öè3£º ¡£

Ö¤Ã÷ÓÐ ´æÔÚ

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

¾«Ó¢¼Ò½ÌÍø