¡¾ÌâÄ¿¡¿³£ÎÂÏ£¬Ïò20 mL 0£®2 mol¡¤L-1¶þÔªËáH2AÈÜÒºÖеμÓ0£®2 mol¡¤L-l NaOHÈÜÒº£¬ÓйØ΢Á£ÎïÖʵÄÁ¿±ä»¯Èçͼ¡£ÏÂÁÐÐðÊöÕýÈ·µÄÊÇ£¨ £©

A. µ±V£¨NaOH£©="20" mLʱ£¬ÈÜÒºÖи÷Àë×ÓŨ¶ÈµÄ´óС˳ÐòΪ£ºc£¨Na+£©>c£¨HA-£©>c£¨ A2-£©>c£¨ OH-£©> £¨£¨H+£©

B. µÈÌå»ýµÈŨ¶ÈµÄNaOHÈÜÒºÓëH2AÈÜÒº»ìºÏºó£¬ÆäÈÜÒºÖÐË®µÄµçÀë³Ì¶È±È´¿Ë®ÖеĴó

C. µÈŨ¶ÈH2AºÍNaHAµÄ»ìºÏÈÜÒºÖÐÎÞÂÛ¼ÓÈëÉÙÁ¿µÄÇ¿Ëá»òÇ¿¼î£¬ÈÜÒºµÄpH±ä»¯¶¼²»´ó

D. µ±V£¨NaOH£© ="40" mLʱ£¬Éý¸ßζȣ¬c£¨Na+£©£¯c£¨A2-£©¼õС

¡¾´ð°¸¡¿C

¡¾½âÎö¡¿

ÊÔÌâA¡¢µ±V£¨NaOH£©="20" mLʱ£¬0£®2 mol¡¤L-lNaOHÓë20 mL 0£®2 mol¡¤L-1H2A·´Ó¦Éú³ÉNaHA£¬Àë×ÓŨ¶È´óСµÄ˳ÐòΪc£¨Na+£©>c£¨HA-£©>c£¨OH-£©> c£¨H+£©> c£¨ A2-£©£¬´íÎó£»B¡¢µÈÌå»ýµÈŨ¶ÈµÄNaOHÈÜÒºÓëH2AÈÜÒº»ìºÏÉú³ÉNaHA£¬ÓÐͼÏñ¿ÉÖª£¬µ±¼ÓÈë20mLNaOHÈÜҺʱ£¬A2-µÄŨ¶È´óÓÚH2A£¬ËµÃ÷HA-µÄµçÀë³Ì¶È´óÓÚË®½â³Ì¶È£¬Ôò»áÒÖÖÆË®µÄµçÀë³Ì¶È£¬¹ÊÆäÈÜÒºÖÐË®µÄµçÀë³Ì¶È±È´¿Ë®ÖеÄС£¬´íÎó£»C¡¢µÈŨ¶ÈH2AºÍNaHAµÄ»ìºÏÈÜÒºÖÐÎÞÂÛ¼ÓÈëÉÙÁ¿µÄÇ¿Ëá»òÇ¿¼î£¬ÈÜÒºµÄpH±ä»¯¶¼²»´ó£¬ÕýÈ·£»D¡¢µ±V£¨NaOH£© ="40" mLʱ£¬Éú³ÉNa2A£¬A2-»á·¢ÉúË®½â£¬Éý¸ßζȣ¬´Ù½øË®½â³Ì¶È£¬Ôòc£¨A2-£©¼õС£¬Ôòc£¨Na+£©/c£¨A2-£©Ôö´ó£¬¹Ê´ð°¸ÎªC¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿¼×´¼×÷Ϊ»ù±¾µÄÓлú»¯¹¤²úÆ·ºÍ»·±£¶¯Á¦È¼ÁϾßÓйãÀ«µÄÓ¦ÓÃÇ°¾°£¬CO2¼ÓÇâºÏ³É¼×´¼ÊǺÏÀíÀûÓÃCO2µÄÓÐЧ;¾¶¡£ÓÉCO2ÖƱ¸¼×´¼¹ý³Ì¿ÉÄÜÉæ¼°·´Ó¦ÈçÏ£º

·´Ó¦¢ñ£ºCO2(g)£«H2(g) CO(g)£«H2O(g) ¦¤H1£½£«41.19 kJ¡¤mol£­1

·´Ó¦¢ò£ºCO(g)£«2H2(g) CH3OH(g) ¦¤H2

·´Ó¦¢ó£ºCO2(g)£«3H2g) CH3OH(g)£«H2O(g) ¦¤H3£½£­49.58 kJ¡¤mol£­1

»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©·´Ó¦¢ñ¡¢¢ò¡¢¢óµÄ»¯Ñ§Æ½ºâ³£Êý·Ö±ðΪK1¡¢K2¡¢K3¡¢ÔòK3=__________£¨ÓÃKµÄ´úÊýʽ±íʾ£©

£¨2£©ÔÚºãѹÃܱÕÈÝÆ÷ÖУ¬³äÈëÒ»¶¨Á¿µÄ H2 ºÍ CO2£¨¼Ù¶¨½ö·¢Éú·´Ó¦¢ó£©£¬ÊµÑé²âµÃ·´Ó¦ÎïÔÚ²»Í¬Î¶ÈÏ£¬·´Ó¦ÌåϵÖÐ CO2 µÄƽºâת»¯ÂÊÓëѹǿµÄ¹ØϵÇúÏßÈçͼ 1 Ëùʾ¡£

¢Ù±È½Ï T1 Óë T2 µÄ´óС¹Øϵ£ºT1____________T2£¨Ìî¡°£¼¡±¡¢¡°£½¡±»ò¡°£¾¡±£©

¢ÚÔÚ T1 ºÍ p6 µÄÌõ¼þÏ£¬ÍùÃܱÕÈÝÆ÷ÖгäÈë 3 mol H2 ºÍ 1 mol CO2£¬¸Ã·´Ó¦ÔÚµÚ 5 min ʱ´ïµ½Æ½ºâ£¬´ËʱÈÝÆ÷µÄÌå»ýΪ 1.8 L£¬Ôò¸Ã·´Ó¦ÔÚ´ËζÈϵÄƽºâ³£ÊýΪ________¡£

a.Èô´ËÌõ¼þÏ·´Ó¦ÖÁ 3 min ʱ¿Ì£¬¸Ä±äÌõ¼þ²¢ÓÚ A µã´¦´ïµ½Æ½ºâ£¬CH3OH µÄŨ¶ÈË淴Ӧʱ¼äµÄ±ä»¯Ç÷ÊÆÈçͼ 2Ëùʾ£¨3¡«4 min µÄŨ¶È±ä»¯Î´±íʾ³öÀ´£©£¬Ôò¸Ä±äµÄÌõ¼þΪ __________£¬ÇëÓà H2 µÄŨ¶È±ä»¯¼ÆËã´Ó4 min¿ªÊ¼µ½ AµãµÄ·´Ó¦ËÙÂÊv(H2)= _________(±£ÁôÁ½Î»Ð¡Êý)¡£

b.ÈôζȲ»±ä£¬Ñ¹Ç¿ºã¶¨ÔÚ p8 µÄÌõ¼þÏÂÖØдﵽƽºâʱ£¬ÈÝÆ÷µÄÌå»ý±äΪ_______L

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÇâÆø×÷ΪÇå½àÄÜÔ´ÓÐ׏㷺µÄÓ¦ÓÃÇ°¾°£¬º¬ÁòÌìÈ»ÆøÖƱ¸ÇâÆøµÄÁ÷³ÌÈçÏ¡£

Çë»Ø´ðÏÂÁÐÎÊÌ⣺

I£®×ª»¯ÍÑÁò£º½«ÌìÈ»ÆøѹÈëÎüÊÕËþ£¬30¡æʱ£¬ÔÚT£®F¾ú×÷ÓÃÏ£¬ËáÐÔ»·¾³ÖÐÍÑÁò¹ý³ÌʾÒâͼÈçÏ¡£

(1)¹ý³ÌiµÄÀë×Ó·´Ó¦·½³ÌʽΪ_________________________________________¡£

(2)ÒÑÖª£º

¢ÙFe3+ÔÚpH=l.9ʱ¿ªÊ¼³Áµí£¬pH=3.2ʱ³ÁµíÍêÈ«¡£

¢Ú30¡æʱ£¬ÔÚT.F¾ú×÷ÓÃÏ£¬²»Í¬pHµÄFeSO4ÈÜÒºÖÐFe2+µÄÑõ»¯ËÙÂÊÈçÏÂ±í¡£

pH

0.7

1.1

1.5

1.9

2.3

2.7

Fe2+µÄÑõ»¯ËÙÂÊ/g¡¤L-1¡¤h-1

4.5

5.3

6.2

6.8

7.0

6.6

ÔÚת»¯ÍÑÁòÖУ¬ÇëÔÚÉϱíÖÐÑ¡Ôñ×î¼ÑpH·¶Î§ÊÇ_______<pH<_______£¬ÕâÑùÑ¡ÔñµÄÔ­ÒòÊÇ£º_______________________________________________¡£

¢ò£®ÕôÆøת»¯£ºÔÚ´ß»¯¼ÁµÄ×÷ÓÃÏ£¬Ë®ÕôÆø½«CH4Ñõ»¯¡£½áºÏÏÂͼ»Ø´ðÎÊÌâ¡£

(3)¢Ù¸Ã¹ý³ÌµÄÈÈ»¯Ñ§·½³ÌʽÊÇ__________________________________________¡£

¢Ú±È½ÏѹǿP1ºÍp2µÄ´óС¹Øϵ£ºP1 _________ P2(Ñ¡Ìî¡°>¡±¡°<¡±»ò¡°=¡±)¡£

¢ÛÔÚÒ»¶¨Î¶ȺÍÒ»¶¨Ñ¹Ç¿ÏµÄÌå»ý¿É±äµÄÃܱÕÈÝÆ÷ÖгäÈë1molCH4ºÍ1molµÄË®ÕôÆø³ä·Ö·´Ó¦´ïƽºâºó£¬²âµÃÆðʼʱ»ìºÏÆøµÄÃܶÈÊÇƽºâʱ»ìºÏÆøÃܶȵÄ1.4±¶£¬Èô´ËʱÈÝÆ÷µÄÌå»ýΪ2L,Ôò¸Ã·´Ó¦µÄƽºâ³£ÊýΪ______________(½á¹û±£Áô2λÓÐЧÊý×Ö)¡£

¢ó£®CO±ä»»£º500¡æʱ£¬CO½øÒ»²½ÓëË®·´Ó¦Éú³ÉCO2ºÍH2¡£

¢ô£®H2Ìá´¿£º½«CO2ºÍH2·ÖÀëµÃµ½H2µÄ¹ý³ÌÈçʾÒâͼ

(4)ÎüÊÕ³ØÖз¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ____________________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÏÂÁл¯ºÏÎïÖÐͬ·ÖÒì¹¹ÌåÊýÄ¿×îÉÙµÄÊÇ

A. ¶¡ÍéB. C5H11Cl

C. ÎìÏ©D. ¶þäå±ûÍé

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÔÚÒ»¶¨Ìõ¼þÏ£¬ÔÚÈÝ»ýΪ2LµÄºãÈÝÃܱÕÈÝÆ÷ÖУ¬½«2molNÆøÌåºÍ3molMÆøÌåÏà»ìºÏ£¬·¢ÉúÈçÏ·´Ó¦£º2N(g)+3M(g)xQ(g)+3R(g)£¬4sºó¸Ã·´Ó¦´ïƽºâʱ£¬Éú³É2.4molR£¬²¢²âµÃQµÄ·´Ó¦ËÙÂÊΪ0.1mol/£¨L¡¤s£©£¬ÏÂÁÐÓйØÐðÊöÕýÈ·µÄÊÇ

A. NµÄת»¯ÂÊΪ80% B. 0¡«4sÄÚ£¬»ìºÏÆøÌåµÄƽ¾ùÏà¶Ô·Ö×ÓÖÊÁ¿²»±ä

C. xֵΪ2 D. ƽºâʱMµÄŨ¶ÈΪ0.6mol/L

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿Ç°ËÄÖÜÆÚµÄA¡¢B¡¢C¡¢DËÄÖÖÔªËØÔÚÖÜÆÚ±íÖоùÓëÔªËØX½ôÃÜÏàÁÚ¡£ÒÑÖªÔªËØX×î¸ß¼ÛÑõ»¯ÎïµÄ»¯Ñ§Ê½ÎªX2O5£¬B¡¢DͬÖ÷×åÇÒBÔªËصÄÔ­×Ӱ뾶ÊÇͬ×åÔªËØÖÐ×îСµÄ£¬CµÄ×î¸ß¼ÛÑõ»¯Îï¶ÔÓ¦µÄË®»¯ÎïÊÇÇ¿Ëá¡£

£¨1£©DÔªËØ»ù̬ԭ×ÓµÄÍâΧµç×ÓÅŲ¼Ê½Îª____________________¡£

£¨2£©A¡¢C¡¢XÈýÖÖÔªËØÔ­×ӵĵÚÒ»µçÀëÄÜÓÉ´óµ½Ð¡µÄ˳ÐòΪ________________£¨ÓÃÏàÓ¦µÄÔªËØ·ûºÅ×÷´ð£©¡£

£¨3£©B¡¢X¡¢DÇ⻯ÎïµÄ·ÐµãÓɸߵ½µÍµÄ˳ÐòΪ_______________£¨ÓÃÏàÓ¦µÄ»¯Ñ§Ê½×÷´ð£©¡£

£¨4£©CÔªËصÄÔ­×Ó¿ÉÐγɶàÖÖÀë×Ó£¬ÊÔÍƲâÏÂÁÐ΢Á£µÄÁ¢Ìå¹¹ÐÍ£¨CΪ×Öĸ£¬²»ÊÇ̼ԪËØ£©£º

΢Á£

CO32-

CO42-

Á¢Ìå¹¹ÐÍÃû³Æ

_______________

_______________

£¨5£©ÔªËØBµÄÒ»ÖÖÇ⻯ÎïB2H4¾ßÓÐÖØÒªµÄÓÃ;¡£ÓйØB2H4µÄ˵·¨ÕýÈ·µÄÊÇ_______¡£

A£®B2H4·Ö×Ó¼ä¿ÉÐγÉÇâ¼ü B£®BÔ­×ÓÊÇsp3ÔÓ»¯

C£®B2H4·Ö×ÓÖк¬ÓÐ5¸ö¦Ò¼üºÍ1¸ö¦Ð¼ü D£®B2H4¾§Ìå±äΪҺ̬ʱÆÆ»µ¹²¼Û¼ü

£¨6£©EÔªËغÍDÔªËØÔÚͬһÖÜÆÚ£¬ÊôÓÚVIII×壬¼Û²ãÓÐÈý¸öµ¥µç×Ó£¬E(OH)2ΪÁ½ÐÔÇâÑõ»¯ÎÔÚŨµÄÇ¿¼îÈÜÒºÖпÉÐγÉE(OH)42-£¬Ð´³öE(OH)2ËáʽµçÀëµÄµçÀë·½³Ìʽ___________________¡£

£¨7£©FÔªËØ»ù̬ԭ×ÓM²ãÉÏÓÐ5¶Ô³É¶Ôµç×Ó£¬FÐγɵĵ¥ÖÊÓЦġ¢¦Ã¡¢¦ÁÈýÖÖͬËØÒìÐÎÌ壬ÈýÖÖ¾§°û£¨ÈçͼËùʾ£©ÖÐFÔ­×ÓµÄÅäλÊýÖ®±ÈΪ___________£¬¦Ä¡¢¦Ã¡¢¦ÁÈýÖÖ¾§°ûµÄ±ß³¤Ö®±ÈΪ_____________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÉèNAΪ°¢·ü¼ÓµÂÂÞ³£Êý£¬ÏÂÁÐÓйØ˵·¨ÕýÈ·µÄÊÇ

A. ³£Î³£Ñ¹Ï£¬1.8g¼×»ù(¡ªCD3)Öк¬ÓеÄÖÐ×ÓÊýΪNA

B. ±ê×¼×´¿öÏ£¬11.2LÒÒÏ©ºÍ»·±ûÍé(C3H6)µÄ»ìºÏÆøÌåÖУ¬¹²Óõç×Ó¶ÔµÄÊýĿΪ3NA

C. ¹ýÁ¿Í­Ó뺬0.4 mol HNO3µÄŨÏõËá·´Ó¦£¬µç×ÓתÒÆÊý´óÓÚ0.2NA

D. ³£ÎÂÏ£¬1L pH=9µÄCH3COONaÈÜÒºÖУ¬·¢ÉúµçÀëµÄË®·Ö×ÓÊýΪ1¡Á10£­9 NA

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿°ÑÖÊÁ¿·ÖÊýΪ98%ÃܶÈΪ1.84g/cm3µÄŨH2SO4ÅäÖƳÉ500ml0.5mol/LµÄÏ¡H2SO4£¬¼ÆË㣺

£¨1£©Å¨H2SO4µÄÎïÖʵÄÁ¿Å¨¶ÈÊǶàÉÙ£¿_____

£¨2£©ËùÐèŨH2SO4µÄÌå»ýÊǶàÉÙ£¿_____(д³öÏà¹Ø¹«Ê½¼°¼ÆËã¹ý³Ì)

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿Îª²â¶¨Ä³²ÝËᾧÌ壨H2C2O42H2O£©ÑùÆ·µÄ´¿¶È£¬ÏÖ³ÆÈ¡Ò»¶¨ÖÊÁ¿µÄ¸ÃÑùÆ·£¬ÅäÖƳÉ100 mLÈÜÒº£¬È¡25.00 mL¸ÃÈÜÒºÓÚ׶ÐÎÆ¿ÖУ¬¼ÓÊÊÁ¿Ï¡ÁòËᣬÓÃ0.100 mol/LµÄKMnO4ÈÜÒºµÎ¶¨(ÔÓÖʲ»²ÎÓë·´Ó¦)¡£ÎªÊ¡È¥¼ÆËã¹ý³Ì£¬µ±³ÆÈ¡µÄÑùÆ·µÄÖÊÁ¿ÎªÄ³Êýֵʱ£¬µÎ¶¨ËùÓÃKMnO4ÈÜÒºµÄºÁÉýÊýÇ¡ºÃµÈÓÚÑùÆ·ÖвÝËᾧÌåµÄÖÊÁ¿·ÖÊýµÄ100±¶¡£ÔòÓ¦³ÆÈ¡ÑùÆ·µÄÖÊÁ¿Îª

A. 2.25 g B. 3.15 g C. 9.00 g D. 12.6 g

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸