¡¾ÌâÄ¿¡¿Ä³Ð£Ñо¿ÐÔѧϰС×éµÄͬѧ½øÐÐÁËÒÔÏ»¯Ñ§ÊµÑ飺½«½ðÊô¸ÆÖÃÓÚ¿ÕÆøÖÐȼÉÕ£¬È»ºóÏòËùµÃ¹ÌÌå²úÎïÖмÓÈëÒ»¶¨Á¿ÕôÁóË®£¬´Ë¹ý³ÌÖз´Ó¦·Å³ö´óÁ¿µÄÈÈ£¬²¢ÇҷųöÓгôζµÄÆøÌå¡£

£¨1£©¼×ͬѧÌá³ö£ºÔËÓÃÀà±ÈѧϰµÄ˼Ï룬CaÓëMgλÓÚͬһÖ÷×壬»¯Ñ§ÐÔÖʾßÓÐÒ»¶¨µÄÏàËÆÐÔ¡£Çëд³öCaÔÚ¿ÕÆøÖÐȼÉÕ·¢Éú·´Ó¦µÄ»¯Ñ§·½³Ìʽ_______________________________¡£

£¨2£©ÒÒͬѧÌá³ö£ºCaµÄÐÔÖʱÈNa»îÆã¬ÔÚ¿ÕÆøÖÐȼÉÕ»¹Ó¦ÓÐCaO2Éú³É£¬Çëд³öȼÉÕºó¹ÌÌå²úÎïÓëË®·´Ó¦·Å³öÆøÌåµÄ»¯Ñ§·½³Ìʽ______________________________¡£±ûͬѧÌá³öÓÃʵÑéµÄ·½·¨Ì½¾¿·Å³ö³ôζÆøÌåµÄ³É·Ö£º

£¨²éÔÄ×ÊÁÏ£©

1£®CaO2ÓöË®·´Ó¦Éú³ÉH2O2£¬H2O2¿ÉÄÜ»á·Ö½â²úÉúÒ»¶¨Á¿µÄO3¡£

2£®µâÁ¿·¨ÊÇ×î³£ÓõijôÑõ²â¶¨·½·¨£¬ÆäÔ­ÀíΪǿÑõ»¯¼Á³ôÑõ(O3)Óëµâ»¯¼Ø(KI)Ë®ÈÜÒº·´Ó¦Éú³ÉÓÎÀëµâ(I2)£¬³ôÑõת»¯ÎªÑõÆø¡£·´Ó¦Ê½ÎªO3£«2KI£«H2O===O2£«I2£«2KOH¡£

£¨Ìá³ö¼ÙÉ裩

¼ÙÉè1£º¸Ã³ôζÆøÌåÖ»ÓÐNH3£»

¼ÙÉè2£º¸Ã³ôζÆøÌåÖ»ÓÐ________£»

¼ÙÉè3£º¸Ã³ôζÆøÌ庬ÓÐ________¡£

£¨Éè¼Æ·½°¸¡¡½øÐÐʵÑé̽¾¿£©

£¨3£©¸ÃС×éͬѧÉè¼ÆÈçÏÂʵÑé·½°¸£¬²¢½øÐÐʵÑ飬ÑéÖ¤ÉÏÊö¼ÙÉè¡£ÇëÍê³ÉÏà¹ØµÄʵÑé²Ù×÷²½Öè¡¢Ô¤ÆÚÏÖÏó¼°½áÂÛ(ÒÇÆ÷×ÔÑ¡)¡£

ÏÞѡʵÑéÊÔ¼Á£ººìɫʯÈïÊÔÖ½¡¢À¶É«Ê¯ÈïÊÔÖ½¡¢pHÊÔÖ½¡¢µí·Û£­KIÈÜÒº¡¢ÕôÁóË®¡£

ʵÑé²Ù×÷

Ô¤ÆÚÏÖÏóÓë½áÂÛ

È¡ÉÙÁ¿·´Ó¦ºó¹ÌÌåÓÚÊÔ¹ÜÖУ¬_____________

______________________________________

¡¾´ð°¸¡¿2Ca£«O22CaO¡¢3Ca£«N2Ca3N2¡¢2Ca£«CO22CaO£«C2CaO2£«2H2O=2Ca(OH)2£«O2¡ü¡¢Ca3N2£«6H2O=3Ca(OH)2£«2NH3¡üO3O3ºÍNH3ÏòÊÔ¹ÜÖмÓÈëÉÙÁ¿Ë®£¬½«ÊªÈóµÄºìɫʯÈïÊÔÖ½ÖÃÓÚÊԹܿڣ»ÁíÈ¡ÉÙÁ¿·´Ó¦ºó¹ÌÌåÓÚÊÔ¹ÜÖУ¬ÏòÊÔ¹ÜÖмÓÈëÉÙÁ¿Ë®£¬½«²úÉúµÄÆøÌåͨÈëµí·Û£­KIÈÜÒºÖÐÈôʪÈóµÄºìɫʯÈïÊÔÖ½ÏÔÀ¶É«£¬ÇÒµí·Û£­KIÈÜÒº²»±äÉ«£¬Ôò¼ÙÉè1³ÉÁ¢£»ÈôʪÈóµÄºìɫʯÈïÊÔÖ½²»ÏÔÀ¶É«£¬ÇÒµí·Û£­KIÈÜÒº±äÀ¶É«£¬Ôò¼ÙÉè2³ÉÁ¢£»ÈôʪÈóµÄºìɫʯÈïÊÔÖ½ÏÔÀ¶É«£¬ÇÒµí·Û£­KIÈÜÒº±äÀ¶É«£¬Ôò¼ÙÉè3³ÉÁ¢

¡¾½âÎö¡¿

£¨1£©ÓÉÓÚþ¿ÉÒԺͿÕÆøÖеÄÑõÆø¡¢¶þÑõ»¯Ì¼¡¢µªÆø·´Ó¦£¬¸ÆÓëþµÄ»¯Ñ§ÐÔÖʾßÓÐÒ»¶¨µÄÏàËÆÐÔ£¬ËùÒÔ¸ÆÒ²¿ÉÒÔºÍÕâÈýÖÖÆøÌå·¢Éú·´Ó¦£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ2Ca+O22CaO¡¢3Ca+N2Ca3N2¡¢2Ca+CO22CaO+C£»£¨2£©ÄÜÓëË®·´Ó¦·Å³öÆøÌåµÄÎïÖÊΪCaO2ºÍCa3N2£¬ÓëË®·¢Éú·´Ó¦µÄ»¯Ñ§·½³Ìʽ·Ö±ðΪ2CaO2+2H2O 2Ca(OH)2+ O2¡ü¡¢Ca3N2+6H2O3Ca(OH)2+2NH3¡ü£»£¨3£©¸ù¾ÝÌâÖÐËù¸øÐÅÏ¢Öª£¬ÓÐÆøζµÄÆøÌå¿ÉÒÔÊÇNH3£¬Ò²¿ÉÒÔÊÇO3£¬»¹¿ÉÒÔÊǶþÕߵĻìºÏÎï¡£ÑéÖ¤ÊÇ·ñº¬ÓÐNH3£¬¿ÉÓÃʪÈóµÄºìɫʯÈïÊÔÖ½£¬ÈôÊÔÖ½±äÀ¶£¬ÔòÖ¤Ã÷ÓÐNH3£»ÑéÖ¤ÊÇ·ñº¬ÓгôÑõ£¬¿ÉÓõí·ÛKIÈÜÒº£¬ÓÉÌâ¸øÐÅÏ¢Öª³ôÑõÄÜʹµí·ÛKIÈÜÒº±äÀ¶£¬ÏòÊÔ¹ÜÖмÓÈëÉÙÁ¿Ë®£¬½«ÊªÈóµÄºìɫʯÈïÊÔÖ½ÖÃÓÚÊԹܿڣ»ÁíÈ¡ÉÙÁ¿·´Ó¦ºó¹ÌÌåÓÚÊÔ¹ÜÖУ¬ÏòÊÔ¹ÜÖмÓÈëÉÙÁ¿Ë®£¬½«²úÉúµÄÆøÌåͨÈëµí·ÛKIÈÜÒºÖÐ £»ÈôʪÈóµÄºìɫʯÈïÊÔÖ½ÏÔÀ¶É«£¬ÇÒµí·ÛKIÈÜÒº²»±äÉ«£¬Ôò¼ÙÉè1³ÉÁ¢£»ÈôʪÈóµÄºìɫʯÈïÊÔÖ½²»ÏÔÀ¶É«£¬ÇÒµí·ÛKIÈÜÒº±äÀ¶É«£¬Ôò¼ÙÉè2³ÉÁ¢£»ÈôʪÈóµÄºìɫʯÈïÊÔÖ½ÏÔÀ¶É«£¬ÇÒµí·ÛKIÈÜÒº±äÀ¶É«£¬Ôò¼ÙÉè3³ÉÁ¢

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿±½±û´¼Ò춡Ëáõ¥£¨)ÊÇÒ»ÖֺϳÉÏãÁÏ£¬ÆäÖÐÒ»ÖֺϳÉ·ÏßÈçͼ:

ÒÑÖª£ºR-CH=CH2R-CH2CH2OH¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©AµÄ»¯Ñ§Ãû³ÆÊÇ______________£¬BÖк¬ÓеĹÙÄÜÍÅÃû³ÆÊÇ________________¡£

£¨2£©ÔÚÒ»¶¨Ìõ¼þÏ£¬AÓëË®·´Ó¦Ò²¿ÉÒÔÉú³ÉB»òÁíÍâÒ»ÖÖÉú³ÉÎ¸Ã·´Ó¦µÄ·´Ó¦ÀàÐÍΪ___________£¬¡°ÁíÍâÒ»ÖÖÉú³ÉÎµÄ½á¹¹¼òʽΪ_____________________________¡£

£¨3£©C¿É±»ÐÂÖƵÄCu(OH)2Ðü×ÇÒºÑõ»¯£¬Ò²¿ÉÒÔ±»ÆäËûÑõ»¯¼ÁËùÑõ»¯¡£Ð´³öCÓëÒø°±ÈÜÒº·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º_____________________________¡£

£¨4£©DµÄºË´Å¹²ÕñÇâÆ×ÏÔʾΪ6×é·å£¬ÇÒ·åÃæ»ýÖ®±ÈΪ1¡Ã1¡Ã1¡Ã1¡Ã2¡Ã2£¬ÔòDµÄ½á¹¹¼òʽΪ_______________£»DµÄͬ·ÖÒì¹¹ÌåÓжàÖÖ£¬ÆäÖб½»·ÉÏÓÐÁ½¸ö²»º¬»·×´µÄÈ¡´ú»ù£¬²¢ÄÜÓëNa·´Ó¦µÄͬ·ÖÒì¹¹ÌåÓÐ_______ÖÖ£¨²»º¬Á¢ÌåÒì¹¹£©¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÏÂÁÐ˵·¨ÖÐÕýÈ·µÄÊÇ

A. ±ê×¼×´¿öÏ£¬22.4 LË®ÖÐËùº¬µÄ·Ö×ÓÊýԼΪ6.02¡Á1023¸ö

B. 1 mol Cl2Öк¬ÓеÄÔ­×ÓÊýΪNA

C. ±ê×¼×´¿öÏ£¬a LÑõÆøºÍµªÆøµÄ»ìºÏÎﺬÓеķÖ×ÓÊýԼΪ¡Á6.02¡Á1023¸ö

D. ´Ó1 L0.5 mol¡¤L£­1NaClÈÜÒºÖÐÈ¡³ö100 mL£¬Ê£ÓàÈÜÒºÖÐNaClÎïÖʵÄÁ¿Å¨¶ÈΪ0.45 mol¡¤L£­1

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿

(1)ÓªÑøƽºâ¡¢ºÏÀíÓÃÒ©ÊDZ£Ö¤ÈËÌ彡¿µºÍÉú»îÖÊÁ¿µÄÖØҪ;¾¶¡£

¢ÙÏÂÁÐÎïÖÊÖУ¬²»ÄÜΪÈËÌåÌṩÄÜÁ¿µÄÊÇ_______(Ìî×Öĸ)¡£

a.µ°°×ÖÊ b.ÏËάËØ c.ÓÍÖ¬

¢Úµ±³öÏÖÍâÉ˸ÐȾʱ£¬Ò½Éú»á½¨ÒéʹÓÃÒÔϳ£ÓÃÒ©ÎïÖеÄ______(Ìî×Öĸ)¡£

a.¸´·½ÇâÑõ»¯ÂÁƬ b.°¢Ë¾Æ¥ÁÖ c.ÅÌÄáÎ÷ÁÖ

¢ÛʳƷÌí¼Ó¼ÁÑÇÏõËáÄƵÄÍâ¹ÛÏñʳÑβ¢ÓÐÏÌ棬Ëü²»µ«ÊÇ·À¸¯¼Á£¬»¹¾ßÓп¹Ñõ»¯×÷Óá£ÑÇÏõËáÄÆÊôÓÚ____(Ìî×Öĸ)¡£µ«ÑÇÏõËáÄÆ»áÓëÈâÀàµÄµ°°×ÖÊ·´Ó¦£¬Éú³ÉÒ»ÖÖÖ°©»¯ºÏÎ¡ªÑÇÏõ°·¡£ËùÒÔ²»¿É³¤ÆÚ»ò´óÁ¿½øʳëçÖÆÀàÈâÀà¡£

a.µ÷ζ¼Á b.×ÅÉ«¼Á c.·¢É«¼Á

(2)»ý¼«±£»¤Éú̬»·¾³¿ÉʵÏÖÈËÓë×ÔÈ»µÄºÍг¹²´¦¡£

¢ÙÏòúÖмÓÈëʯ»Òʯ£¬¿ÉÓÐЧ¼õÉÙ_______µÄÅÅ·Å¡£

¢ÚÏòº¬ÓÐHg2+µÄ·ÏË®ÖмÓÈë____£¬¿ÉÓÐЧ³ýÈ¥¸ÃÖؽðÊôÀë×Ó¡£

¢ÛÏÂÁÐÎïÖÊÄܸøË®Ìåɱ¾úÏû¶¾£¬ÓÖÄÜʹˮÌå¾»»¯µÄÊÇ______(Ìî×Öĸ)¡£

A.Na2FeO4(aq) B.KAl(SO4)2¡¤12H2O C.Ca(ClO)2(aq) D.NaClO(aq)

¢ÜÓöþÑõ»¯Ì¼Éú²ú»¯¹¤²úÆ·£¬ÓÐÀûÓÚ¶þÑõ»¯Ì¼µÄ´óÁ¿»ØÊÕ¡£Ä³¹¤ÒµÉú²úÖÐÀûÓÃCO2ºÍC2H4¼°Ë®ÕôÆøÔÚ´ß»¯¼ÁÌõ¼þϺϳÉÒÒËá(Ô­×ÓÀûÓÃÂÊ100%)£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ____________¡£

(3)²ÄÁÏÊÇÈËÀàÉú´æºÍ·¢Õ¹µÄÎïÖÊ»ù´¡¡£

¢ÙÆû³µÐÐÊ»ÔÚ¡°³¬¼¶¸ßËÙ¹«Â·¡±ÉÏÐÐÊ»£¬×Ô¶¯ÊÕ·Ñϵͳ»áͨ¹ý³µÔØоƬ¶Ô³µÁ¾½øÐÐ×Ô¶¯ÊÕ·Ñ¡£ÖÆȡоƬµÄÖ÷ÒªÔ­ÁÏÊÇ_____ (Ìî×Öĸ)¡£

a.¹è b.ʯī c.¶þÑõ»¯¹è

¢ÚÔÚÏÂÁвÄÁÏÖУ¬ÊôÓÚ¸´ºÏ²ÄÁϵÄÊÇ____ (Ìî×Öĸ)¡£

a.¸Ö»¯²£Á§ b.¶¡±½Ï𽺠c.µª»¯¹èÌÕ´É d.¸Ö½î»ìÄýÍÁ

¢Û»ù´¡½¨ÉèÐèÒª´óÁ¿µÄË®Ä࣬ˮÄàÊôÓÚ______(Ìî×Öĸ)£¬Ë®ÄàµÄ±£ÖÊÆÚͨ³£Ö»ÓÐÈý¸öÔ£¬²»Äܳ¤ÆÚ±£´æµÄÔ­ÒòÊÇ_______________¡£

a.½ðÊô²ÄÁÏ b.ÎÞ»ú·Ç½ðÊô²ÄÁÏ c.Óлú¸ß·Ö×Ó²ÄÁÏ

¢ÜÓлú²£Á§ÊÇÓÉÓлúÎïX¼Ó¾ÛÖƵõÄÈÈËÜÐÔËÜÁÏ£¬ÎªÍ¸Ã÷Èç²£Á§×´µÄÎÞÉ«¹ÌÌ壬¿ÉÓÃÒÔÖÆÔ캽¿Õ´°²£Á§¡¢ÒDZíÅÌ¡¢Íâ¿ÆÕÕÃ÷µÆ¡¢×°ÊÎÆ÷ºÍÉú»îÓÃÆ·µÈ£¬Æä½á¹¹¼òʽÈçͼËùʾ¡£Çëд³öX µÄ½á¹¹¼òʽ__________¡£

¢ÝÌ«ÑôÄܵç³ØÐèÒªÓõ½¸ß´¿¹èΪԭÁÏ¡£¸ßÎÂϽ¹Ì¿ºÍʯӢ·´Ó¦¿ÉÒÔÖƵôֹ裬¸Ã·´Ó¦µÄ·½³ÌʽΪ__________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÏÂͼÊÇij»¯Ñ§ÐËȤС×éÉè¼ÆµÄÀûÓõç×ÓÀ¬»ø(º¬70% Cu¡¢25% Al¡¢4% Fe¼°ÉÙÁ¿Au¡¢Pt)ÖƱ¸ÁòËáÍ­ºÍÁòËáÂÁ¾§ÌåµÄ·Ïߣº

ÒÑÖªÏÂÁÐÐÅÏ¢£ºCu¿ÉÓëÏ¡ÁòËáºÍH2O2µÄ»ìºÏÒº·´Ó¦Éú³ÉÁòËáÍ­£»Ìú¡¢ÂÁ¡¢Í­µÈÀë×ÓÒÔÇâÑõ»¯ÎïÐÎʽ³ÁµíʱÈÜÒºµÄpHÈçÏÂ±í£º

³ÁµíÎï

Fe(OH)3

Al(OH)3

Cu(OH)2

¿ªÊ¼³Áµí

1.1

4.0

5.4

ÍêÈ«³Áµí

3.2

5.2

6.7

£¨1£©Ð´³öCuÓëÏ¡ÁòËáºÍH2O2µÄ»ìºÏÒº·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º_________________________________¡£

£¨2£©ÔÚ²Ù×÷¢òÖУ¬xµÄÈ¡Öµ·¶Î§ÊÇ____________¡£

£¨3£©ÔÚ²Ù×÷¢óÖУ¬Õô·¢Å¨ËõÐèÒªµÄ²£Á§ÒÇÆ÷ÓÐ________________¡£

£¨4£©ÓÉÂËÔüaÖÆÈ¡Al2(SO4)3¡¤18H2O£¬Ì½¾¿Ð¡×éÉè¼ÆÁËÈýÖÖ·½°¸£º

¼×£ºÂËÔüa¨D¡úH2SO4¨D¡ú²Ù×÷¢ó¨D¡úAl2(SO4)3¡¤18H2O

ÒÒ£ºÂËÔüa¨D¡úH2SO4¨D¡úÊÊÁ¿Al·Û£¬¹ýÂ˨D¡ú²Ù×÷¢ó¨D¡úAl2(SO4)3¡¤18H2O

±û£ºÂËÔüa¨D¡úNaOHÈÜÒº£¬¹ýÂ˨D¡úH2SO4¨D¡ú²Ù×÷¢ó¨D¡úAl2(SO4)3¡¤18H2O

×ۺϿ¼ÂÇÉÏÊöÈýÖÖ·½°¸£¬×î¾ß¿ÉÐÐÐÔµÄÊÇ______(ÌîÐòºÅ)¡£

£¨5£©Îª²â¶¨CuSO4¡¤5H2O¾§ÌåµÄ´¿¶È£¬½øÐÐÏÂÁÐʵÑ飺ȡa g ÊÔÑùÅä³É100 mLÈÜÒº£¬Ã¿´ÎÈ¡20.00 mL£¬Ïû³ý¸ÉÈÅÀë×Óºó£¬ÓÃb mol¡¤L£­1 EDTA(Na2H2Y)±ê×¼ÈÜÒºµÎ¶¨ÆäÖеÄCu2£«(Àë×Ó·½³ÌʽΪCu2£«£«H2Y2£­===CuY2£­£«2H£«)£¬µÎ¶¨ÖÁÖյ㣬ƽ¾ùÏûºÄEDTAÈÜÒº12.00 mL(³£ÎÂʱ£¬5%µÄNa2H2YË®ÈÜÒº£¬ÆäpHΪ4¡«6)£¬ÔòCuSO4¡¤5H2O¾§ÌåµÄ´¿¶ÈÊÇ________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿»ìºÏÎï·ÖÀëºÍÌá´¿³£ÓÃÏÂͼװÖýøÐУ¬°´ÒªÇó»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©ÒÇÆ÷¢ÙµÄÃû³Æ________________________¡£

£¨2£©Ã÷½ºÊÇË®ÈÜÐÔµ°°×ÖÊ»ìºÏÎÈÜÓÚË®ÐγɽºÌå¡£·ÖÀëÃ÷½ºµÄË®ÈÜÒºÓëNa2CO3¡¢Na2SO4µÄ»ìºÏÈÜҺӦѡÓÃ×°ÖõÄΪ£¨ÓÃÉÏͼ×ÖĸÌîд£©____________________¡£ÈçºÎÖ¤Ã÷SO42-ÒÑ·ÖÀë³öÀ´______________________¡£

£¨3£©ÔÚ×°ÖÃDÖмÓÈë10 mLµâË®£¬È»ºóÔÙ×¢Èë4 mL±½£¬¸ÇºÃ²£Á§Èû£¬°´²Ù×÷¹æÔò·´¸´Õñµ´ºó¾²Öù۲쵽µÄÏÖÏóÊÇ£º________________________¡£²Ù×÷Íê±Ïºó£¬ÎªµÃµ½µâ²¢»ØÊÕ±½¿ÉÓÃ______________·¨¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿Ä³Ç¿ËáÐÔÈÜÒºXÖпÉÄܺ¬ÓÐFe2+¡¢A13+¡¢NH4+¡¢CO32-¡¢SO32-¡¢SO42-¡¢Cl-ÖеÄÈô¸ÉÖÖ£¬ÏÖÈ¡XÈÜÒº½øÐÐÁ¬ÐøʵÑ飬ʵÑé¹ý³Ì¼°²úÎïÈçÏ£¬ÏÂÁÐ˵·¨²»ÕýÈ·µÄÊÇ

A. X Öп϶¨´æÔÚ Fe2+¡¢A13+¡¢NH4+¡¢SO42-

B. XÖв»ÄÜÈ·¶¨ÊÇ·ñ´æÔÚµÄÀë×ÓÊÇAl3+ºÍCl-

C. ÈÜÒºEºÍÆøÌåF·¢Éú·´Ó¦£¬Éú³ÉÎïΪÑÎÀà

D. ÆøÌåAÊÇNO

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿Ä³ÐËȤС×éµÄͬѧΪÁËÖƱ¸ÂÈÆø²¢Ì½¾¿ÆäÐÔÖÊ£¬»Ø´ðÏÂÁÐÎÊÌâ¡£

¢ñ£®Ð´³öʵÑéÊÒÖÆÈ¡ÂÈÆøµÄÀë×Ó·½³Ìʽ£º_______________________________

¢ò£®¼×ͬѧÉè¼ÆÈçͼËùʾװÖÃÑо¿ÂÈÆøÄÜ·ñÓëË®·¢Éú·´Ó¦£®ÆøÌåaÊǺ¬ÓÐÉÙÁ¿¿ÕÆøºÍË®ÕôÆøµÄÂÈÆø£®Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©Å¨ÁòËáµÄ×÷ÓÃÊÇ___________________¡£

£¨2£©Ö¤Ã÷ÂÈÆøºÍË®·´Ó¦µÄʵÑéÏÖÏóΪ__________¡£

£¨3£©IClµÄÐÔÖÊÓëCl2ÀàËÆ£¬Ð´³öIClÓëË®·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ______________________¡£

£¨4£©Èô½«ÂÈÆøͨÈëʯ»ÒÈéÖÆȡƯ°×·Û£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ______________________¡£Æ¯°×·ÛÈÜÓÚË®ºó£¬Óöµ½¿ÕÆøÖеÄCO2£¬¼´²úÉúƯ°×¡¢É±¾ú×÷Ó㬷´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ_________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿½µµÍ´óÆøÖÐCO2µÄº¬Á¿¼°ÓÐЧµØ¿ª·¢ÀûÓÃCO2£¬ÒÑÒýÆðÁËÈ«ÊÀ½çµÄÆÕ±éÖØÊÓ

£¨1£©CO2¼ÓÇâºÏ³ÉDME£¨¶þ¼×ÃÑ£©Êǽâ¾öÄÜԴΣ»úµÄÑо¿·½ÏòÖ®Ò»¡£

¢Ù2CO2(g) + 6H2(g)CH3OCH3(g) + 3H2O(g) ¡÷H= -122.4kJ¡¤mol-1

ijζÈÏ£¬½«2.0 mol CO2(g) ºÍ6.0 mol H2(g)³äÈëÈÝ»ýΪ2 LµÄÃܱÕÈÝÆ÷ÖУ¬·´Ó¦µ½´ïƽºâʱ£¬¸Ä±äѹǿºÍζȣ¬Æ½ºâÌåϵÖÐ CH3OCH3(g) µÄÎïÖÊ·ÖÊý±ä»¯Çé¿öÈçͼËùʾ£¬ÔòP1_______P2£¨Ìî¡°>¡±¡°<¡±»ò¡°=¡±£¬ÏÂͬ£©¡£ÈôT1¡¢P1£¬T3¡¢P3ʱƽºâ³£Êý·Ö±ðΪK1¡¢K3£¬ÔòK1________K3£¬T1¡¢P1ʱH2µÄƽºâת»¯ÂÊΪ______________¡£

¢ÚÔÚºãÈÝÃܱÕÈÝÆ÷Àï°´Ìå»ý±ÈΪ1¡Ã3³äÈë¶þÑõ»¯Ì¼ºÍÇâÆø£¬Ò»¶¨Ìõ¼þÏ·´Ó¦´ïµ½Æ½ºâ״̬¡£µ±¸Ä±ä·´Ó¦µÄijһ¸öÌõ¼þºó£¬ÏÂÁÐÄÜ˵Ã÷ƽºâÒ»¶¨ÏòÄæ·´Ó¦·½ÏòÒƶ¯µÄÊÇ______£¨ÌîÐòºÅ£©¡£

A£®·´Ó¦ÎïµÄŨ¶ÈÔö´ó B£®»ìºÏÆøÌåµÄÃܶȼõС

C£®Õý·´Ó¦ËÙÂÊСÓÚÄæ·´Ó¦ËÙÂÊ D£®ÇâÆøµÄת»¯ÂʼõС

£¨2£©½«Ò»¶¨Á¿µÄCO2ÆøÌåͨÈëÇâÑõ»¯ÄƵÄÈÜÒºÖУ¬ÏòËùµÃÈÜÒºÖбߵμÓÏ¡ÑÎËá±ßÕñµ´ÖÁ¹ýÁ¿¡¢²úÉúµÄÆøÌåÓë¼ÓÈëÑÎËáÎïÖʵÄÁ¿µÄ¹ØϵÈçͼ£¨ºöÂÔÆøÌåµÄÈܽâºÍHClµÄ»Ó·¢£©¡£Çë»Ø´ð£ºµ±¼ÓÈëHCl µÄÎïÖʵÄÁ¿Îª1 molʱ£¬ÈÜÒºÖÐËùº¬ÈÜÖʵĻ¯Ñ§Ê½__________£¬aµãÈÜÒºÖи÷Àë×ÓŨ¶ÈÓÉ´óµ½Ð¡µÄ¹ØϵʽΪ____________________________________¡£

£¨3£©CO2ÔÚ×ÔÈ»½çÑ­»·Ê±¿ÉÓëCaCO3·´Ó¦£¬CaCO3ÊÇÒ»ÖÖÄÑÈÜÎïÖÊ£¬ÆäKsp = 2.8¡Á10£­9¡£CaCl2ÈÜÒºÓëNa2CO3ÈÜÒº»ìºÏ¿ÉÐγÉCaCO3³Áµí£¬ÏÖ½«µÈÌå»ýµÄCaCl2ÈÜÒºÓëNa2CO3ÈÜÒº»ìºÏ£¬Èô»ìºÏÇ°Na2CO3ÈÜÒºµÄŨ¶ÈΪ2¡Á10£­4 mol¡¤L£­1£¬ÔòÉú³É³Áµí¼ÓÈëCaCl2ÈÜÒºµÄ×îСŨ¶ÈΪ___________ mol¡¤L£­1¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸