¡¾ÌâÄ¿¡¿³£ÎÂÏ£¬½«Ä³Ò»ÔªËáHAºÍNaOHÈÜÒºµÈÌå»ý»ìºÏ£¬ÊµÑéÐÅÏ¢ÈçÏ£º

ʵÑé±àºÅ

c(HA)/ mol¡¤L£­1

c(NaOH)/ mol¡¤L£­1

·´Ó¦ºóÈÜÒºpH

¼×

0.1

0.1

pH£½9

ÒÒ

c1

0.2

pH£½7

ÏÂÁÐÅжϲ»ÕýÈ·µÄÊÇ

A.c1Ò»¶¨´óÓÚ0.2 mol¡¤L£­1

B.HAµÄµçÀë·½³ÌʽÊÇHAH£«£«A£­

C.¼×·´Ó¦ºóÈÜÒºÖУºc(Na£«) £¾ c(OH£­)£¾ c(A£­) £¾ c(H£«)

D.ÒÒ·´Ó¦ºóÈÜÒºÖУºc(Na£«) £¼ c(HA)£«c(A£­)

¡¾´ð°¸¡¿C

¡¾½âÎö¡¿

ÓÉʵÑé¼×¿ÉÖª£¬ËáºÍ¼îÇ¡ºÃ·´Ó¦£¬·´Ó¦ºóÉú³ÉpH£½9µÄNaAÈÜÒº£¬ËµÃ÷A-Àë×ÓË®½âʹÈÜÒºÏÔ¼îÐÔ£¬¹ÊHAΪÈõËá¡£

A¡¢ÈÜÒºÖÐÒ»¶¨´æÔÚc£¨OH-£©+c£¨A-£©=c£¨Na+£©+c£¨H+£©£¬»ìºÏÈÜÒºpH=7˵Ã÷ÈÜÒº³ÊÖÐÐÔ£¬c£¨OH-£©=c£¨H+£©£¬ËùÒÔc£¨A-£©=c£¨Na+£©£¬·´Ó¦ºóÉú³ÉNaAÈÜÒº£¬A-Àë×ÓË®½âʹÈÜÒºÏÔ¼îÐÔ£¬¹ÊHA¹ýÁ¿²ÅÄÜʹÈÜÒºÏÔÖÐÐÔ£¬Òò´Ë£¬Ô­ÈÜҺŨ¶Èc1£¾0.2 molL-1£¬AÕýÈ·£»

B¡¢HAÊÇÈõËá´æÔÚµçÀëƽºâ£¬HAµÄµçÀë·½³ÌʽÊÇHAH++A-£¬BÕýÈ·£»

C¡¢·´Ó¦ºóÉú³ÉNaAÈÜÒº£¬A-Àë×ÓË®½âÏÔ¼îÐÔ£¬ÈÜÒºÖÐÀë×ÓŨ¶È´óСc£¨Na+£©£¾c£¨A-£©£¾c£¨OH-£©£¾c£¨H+£©£¬C´íÎó£»

D¡¢ÒÒÈÜÒºÖÐÒ»¶¨´æÔÚc£¨OH-£©+c£¨A-£©=c£¨Na+£©+c£¨H+£©£¬»ìºÏÈÜÒºpH=7˵Ã÷ÈÜÒº³ÊÖÐÐÔ£¬c£¨OH-£©=c£¨H+£©£¬ËùÒÔc£¨A-£©=c£¨Na+£©£¬c£¨Na+£©£¼c£¨HA£©+c£¨A-£©£¬DÕýÈ·¡£

¹ÊÑ¡C¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÓÃ1.0mol/LµÄNaOHÈÜÒºÖкÍijŨ¶ÈµÄH2SO4ÈÜÒº£¬ÆäË®ÈÜÒºµÄpHºÍËùÓÃNaOHÈÜÒºµÄÌå»ý±ä»¯¹ØϵÈçͼËùʾ£¬ÔòÔ­H2SO4ÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈºÍÍêÈ«·´Ó¦ºóÈÜÒºµÄ´óÖÂÌå»ýÊÇ£¨ £©

A£®1.0 mol/L£¬20 mL B£®0.5 mol/L£¬40 mL

C£®0.5 mol/L£¬80 mL D£®1.0 mol/L£¬80 mL

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿´ÓÎïÖÊA(ijÕýÑÎ)µÄË®ÈÜÒº³ö·¢ÓÐÏÂÃæËùʾµÄһϵÁб仯£º

£¨1£©Ð´³öA¡«FÎïÖʵĻ¯Ñ§Ê½£º

A__________£»B__________£»C__________£»D__________£»E.__________£»F__________¡£

£¨2£©Ð´³öE¡úFµÄ»¯Ñ§·½³Ìʽ______________________________¡£

£¨3£©¼ø±ðÎïÖÊFÖÐÒõÀë×ӵķ½·¨ÊÇ________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿³£ÎÂÏÂÓÃNaOHÈÜÒºµÎ¶¨H2C2O4ÈÜÒºµÄ¹ý³ÌÖУ¬ÈÜÒºÖУ­lgºÍ£­lg c(HC2O4-)»ò£­lgºÍ£­lg c(C2O42-)µÄ¹ØϵÈçͼËùʾ¡£ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨¡¡¡¡£©

A.Ka1(H2C2O4)ÊýÁ¿¼¶Îª10£­1

B.ÇúÏßN±íʾ£­lgºÍ£­lg c(HC2O4-)µÄ¹Øϵ

C.ÏòNaHC2O4ÈÜÒºÖмÓNaOHÖÁc(HC2O4-)ºÍc(C2O4-)ÏàµÈ£¬´ËʱÈÜÒºpHԼΪ5

D.ÔÚNaHC2O4ÈÜÒºÖÐc(Na£«)£¾c(HC2O4-)£¾c(H2C2O4)£¾c(C2O42-)

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÇúÃÀÌæÄáÊÇÒ»ÖÖÒÖÖƺÚÉ«ËØÁöµÄÐÂÐÍ¿¹°©Ò©ÎÏÂÃæÊǺϳÉÇúÃÀÌæÄáÖмäÌåGµÄ·´Ó¦Â·Ïߣº

ÒÑÖª£º¢ÙD·Ö×ÓÖÐÓÐ2¸ö6Ôª»·£»

Çë»Ø´ð£º

£¨1£©»¯ºÏÎïAµÄ½á¹¹¼òʽ___________¡£AÉú³ÉBµÄ·´Ó¦ÀàÐÍ___________¡£

£¨2£©ÏÂÁÐ˵·¨²»ÕýÈ·µÄÊÇ___________¡£

A.B¼ÈÄܱíÏÖ¼îÐÔÓÖÄܱíÏÖËáÐÔ

B.1moCÔÚ¼îÈÜÒºÖÐÍêÈ«Ë®½â×î¶à¿ÉÒÔÏûºÄ4 molOH£­

C.DÓëPOCl3µÄ·´Ó¦»¹»áÉú³ÉEµÄÒ»ÖÖͬ·ÖÒì¹¹Ìå

D.GµÄ·Ö×ÓʽΪC16H18O3N4

£¨3£©Ð´³öC¡úDµÄ»¯Ñ§·½³Ìʽ____________________________________________¡£

£¨4£©XÊDZÈA¶à2¸ö̼ԭ×ÓµÄAµÄͬϵÎд³ö·ûºÏÏÂÁÐÌõ¼þµÄX¿ÉÄܵĽṹ¼òʽ£º_______________________________________________________¡£

¢Ù1H-NMRÆ×ÏÔʾ·Ö×ÓÖÐÓÐ3ÖÖÇâÔ­×Ó£¬¢ÚIRÆ×ÏÔʾ·Ö×ÓÖÐÓб½»·Ó룭NH2ÏàÁ¬½á¹¹

£¨5£©Á÷³ÌÖÐʹÓõÄDMF¼´N£¬N-¶þ¼×»ù¼×õ£°·½á¹¹¼òʽΪ£¬Êdz£ÓõÄÓлúÈܼÁ¡£Éè¼ÆÒÔ¼×´¼ºÍ°±ÎªÖ÷ÒªÔ­ÁÏÖÆÈ¡DMFµÄºÏ³É·Ïß(ÓÃÁ÷³Ìͼ±íʾ£¬ÆäËûÎÞ»úÊÔ¼ÁÈÎÑ¡)¡£_____________

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿Ä³Ç¦ËáÐîµç³ØµÄ¹¤×÷Ô­ÀíÈçͼËùʾ£¬Æä×Ü·´Ó¦Ê½Îª¡£ÏÂÁÐÅжϲ»ÕýÈ·µÄÊÇ

A.±ÕºÏKʱ£¬dµç¼«µÄ·´Ó¦Ê½Îª

B.µ±µç·ÖÐתÒƵç×Óʱ£¬¢ñÖÐÏûºÄµÄΪ

C.±ÕºÏKʱ£¬¢òÖÐÏòcµç¼«Ç¨ÒÆ

D.±ÕºÏKÒ»¶Îʱ¼äºó£¬¢ò¿Éµ¥¶À×÷Ϊԭµç³Ø£¬dµç¼«ÎªÕý¼«

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÓûîÐÔÌ¿»¹Ô­´¦ÀíµªÑõ»¯ÎÓйط´Ó¦ÎªC(s)+2NO(g) N2(g)+CO2(g)¡£

£¨1£©Ð´³öÉÏÊö·´Ó¦µÄƽºâ³£Êý±í´ïʽ_______________¡£

£¨2£©ÔÚ2LºãÈÝÃܱÕÆ÷ÖмÓÈë×ãÁ¿CÓëNO·¢Éú·´Ó¦£¬ËùµÃÊý¾ÝÈç±í£¬»Ø´ðÏÂÁÐÎÊÌâ¡£

ʵÑé±àºÅ

ζÈ/¡æ

ÆðʼʱNOµÄÎïÖʵÄÁ¿/mol

ƽºâʱN2µÄÎïÖʵÄÁ¿/mol

1

700

0.40

0.09

2

800

0.24

0.08

¢Ù½áºÏ±íÖÐÊý¾Ý£¬Åжϸ÷´Ó¦µÄ¡÷H____0(Ìî¡°£¾¡±»ò¡°£¼¡±)£¬ÀíÓÉÊÇ_________¡£

¢ÚÅжϸ÷´Ó¦´ïµ½Æ½ºâµÄÒÀ¾ÝÊÇ_______¡£

A.ÈÝÆ÷ÄÚÆøÌåÃܶȺ㶨 B.ÈÝÆ÷ÄÚ¸÷ÆøÌåŨ¶Èºã¶¨

C.ÈÝÆ÷ÄÚѹǿºã¶¨ D.2vÕý£¨NO£©= vÄ棨N2£©

£¨3£©700¡æʱ£¬ÈôÏò2LÌå»ýºã¶¨µÄÃܱÕÈÝÆ÷ÖгäÈëÒ»¶¨Á¿N2ºÍCO2·¢Éú·´Ó¦£ºN2(g)+CO2(g)C(s)+2NO(g) £»ÆäÖÐN2¡¢NOÎïÖʵÄÁ¿Ëæʱ¼ä±ä»¯µÄÇúÏßÈçÏÂͼËùʾ¡£Çë»Ø´ðÏÂÁÐÎÊÌâ¡£

¢Ù0¡«10 minÄÚµÄCO2ƽ¾ù·´Ó¦ËÙÂÊv£½____________¡£

¢ÚͼÖÐAµãv(Õý)___v(Äæ)£¨Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°£½¡±£©¡£

¢ÛµÚ10 minʱ£¬Íâ½ç¸Ä±äµÄÌõ¼þ¿ÉÄÜÊÇ_____________¡£

A£®¼Ó´ß»¯¼Á B£®Ôö´óCµÄÎïÖʵÄÁ¿

C£®¼õСCO2µÄÎïÖʵÄÁ¿ D£®ÉýΠE£®½µÎÂ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÃÌËáﮣ¨LiMn2O4)ÊÇÐÂÐÍï®Àë×Óµç³Ø³£ÓõÄÕý¼«²ÄÁÏ¡£¹¤ÒµÉÏÒÔÈíÃÌ¿ó½¬ÎªÔ­ÁÏ¿ÉÖƱ¸ÃÌËáﮣ¬Í¬Ê±ÖƵø±²úÆ·MnSO4¡¤H2O¾§Ì壬ÆäÁ÷³ÌÈçͼËùʾ¡£

ÒÑÖª£º£¨1£©ÈíÃ̿󽬵ÄÖ÷Òª³É·ÖΪMnO2£¬»¹º¬ÓÐFe2O3£¬MgO¡¢Al2O3£¬CaO£¬SiO2µÈÔÓÖÊ¡£

£¨2£©Î¶ȸßÓÚ27¡æʱ£¬MnSO4¾§ÌåµÄÈܽâ¶ÈËæζÈÉý¸ß¶øÖð½¥½µµÍ¡£

£¨3£©ÓйØÎïÖʵÄÈܶȻý³£ÊýÈçÏÂ±í£º

£¨1£©¡°½þ³ö¡±¹ý³ÌÖÐMnO2ת»¯ÎªMn2+µÄÀë×Ó·½³ÌʽΪ_____¡£¸Ã¹ý³ÌÖУ¬ÎªÌá¸ßÈíÃÌ¿óÖÐMnO2µÄ½þ³öÂÊ£¬ÏÂÁдëÊ©¿ÉÐеÄÓÐ_____£¨Ìî×Öĸ£©¡£

A£®²»¶Ï½Á°è£¬Ê¹SO2ºÍÈíÃ̿󽬳ä·Ö½Ó´¥

B£®Ôö´óͨÈëSO2µÄÁ÷ËÙ

C£®Êʵ±ÉýÎÂ

D£®¼õÉÙÈíÃ̿󽬵ĽøÈëÁ¿

£¨2£©µÚ1²½³ýÔÓÖмÓÈëH2O2µÄÄ¿µÄÊÇ_____¡£

£¨3£©µÚ2²½³ýÔÓ£¬Ö÷ÒªÊǽ«Ca2+£¬Mg2+ת»¯ÎªÏàÓ¦µÄ·ú»¯Îï³Áµí³ýÈ¥£¬ÆäÖÐMnF2³ýÈ¥Mg2+·´Ó¦µÄÀë×Ó·½³ÌʽΪMnF2(s)+Mg2+(aq)=Mn2+(aq)+MgF2(s)£¬¸Ã·´Ó¦µÄƽºâ³£ÊýΪ_____¡£

£¨4£©Í¼ÖеÄһϵÁвÙ×÷Ö¸µÄÊÇ_____¡£

£¨5£©½«MnO2ºÍLi2CO3°´4£º1µÄÎïÖʵÄÁ¿Ö®±ÈÅäÁÏ£¬»ìºÏ½Á°è£¬È»ºó¸ßÎÂìÑÉÕ600~750¡æ£¬ÖÆÈ¡²úÆ·LiMn2O4¡£Ð´³ö¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º_____¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÂÈ»¯ï§Ë׳Ʊɰ£¬Ö÷ÒªÓÃÓڸɵç³Ø¡¢»¯·ÊµÈ¡£Ä³»¯Ñ§Ñо¿Ð¡×éÉè¼ÆÈçͼʵÑéÖƱ¸Â±É°²¢½øÐÐÔªËزⶨ¡£

¢ñ.ʵÑéÊÒÖƱ¸Â±É°ËùÐèµÄ×°ÖÃÈçͼËùʾ£¬×°ÖÿÉÖظ´Ñ¡Óá£

£¨1£©×°ÖýӿÚÁ¬½Ó˳ÐòÊÇ___¡úa£»b¡û___¡£

£¨2£©C×°ÖõÄ×÷ÓÃÊÇ___£¬D×°ÖÃÊ¢×°µÄÎïÖÊÊÇ___¡£

£¨3£©Ð´³öÓÃÉÏÊö×°ÖÃÖƱ¸°±ÆøµÄÒ»×éÊÔ¼Á£º___¡£

¢ò.²â¶¨Â±É°ÖÐClÔªËغÍNÔªËصÄÖÊÁ¿Ö®±È¡£

¸ÃÑо¿Ð¡×é׼ȷ³ÆÈ¡ag±ɰ£¬Óë×ãÁ¿Ñõ»¯Í­»ìºÏ¼ÓÈÈ£¬³ä·Ö·´Ó¦ºó°ÑÆøÌå²úÎï°´ÈçͼװÖýøÐÐʵÑé¡£ÊÕ¼¯×°ÖÃÊÕ¼¯µ½µÄÆøÌåΪ¿ÕÆøÖк¬Á¿×î¶àµÄÆøÌ壬ÆäÌå»ý»»Ëã³É±ê×¼×´¿öϵÄÌå»ýΪVL£¬¼îʯ»ÒÔöÖØbg¡£

£¨4£©E×°ÖÃÄÚµÄÊÔ¼ÁΪ___£¬Â±É°ÓëÑõ»¯Í­»ìºÏ¼ÓÈÈ·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ___¡£

£¨5£©Â±É°ÖÐClÔªËغÍNÔªËصÄÖÊÁ¿Ö®±ÈΪ___(Óú¬b¡¢VµÄʽ×Ó±íʾ)¡£

£¨6£©ÎªÁ˲ⶨ±ɰÖÐÂÈÔªËصÄÖÊÁ¿£¬ËûÃÇÉè¼ÆµÄʵÑé·½°¸Êǽ«ag±ɰÍêÈ«ÈܽâÓÚË®£¬¼ÓÈë¹ýÁ¿AgNO3ÈÜÒº£¬È»ºó²â¶¨Éú³É³ÁµíµÄÖÊÁ¿¡£ÇëÄãÆÀ¼Û¸Ã·½°¸ÊÇ·ñºÏÀí£¬²¢ËµÃ÷ÀíÓÉ£º___¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸