ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿ÖÜÆÚ±íÇ°ËÄÖÜÆÚµÄÔªËØ A¡¢B¡¢C¡¢D¡¢E£¬Ô­×ÓÐòÊýÒÀ´ÎÔö´ó¡£AµÄºËÍâµç×Ó×ÜÊýÓëÆäÆÚÐòÊýÏàͬ£¬BºÍDλÓÚͬһÖÜÆÚÇÒδ³É¶Ôµç×ÓÊýµÈÓÚÆäÖÜÆÚÐòÊý£¬EΪµÚËÄÖÜÆÚÔªËØ£¬×îÍâ²ãÖ»ÓÐÒ»¸öµç×Ó£¬´ÎÍâ²ãµÄËùÓйìµÀ¾ù³äÂúµç×Ó¡£

(1)B¡¢C¡¢DÈýÖÖÔªËصÚÒ»µçÀëÄÜÓÉ´óµ½Ð¡µÄ˳ÐòΪ___(ÌîÔªËØ·ûºÅ)£¬E»ù̬ԭ×Ó¼Û²ãµç×ÓÅŲ¼Í¼Îª_____¡£

(2)д³öÓÉÒÔÉÏÔªËØ×é³ÉµÄBD2µÄµÈµç×ÓÌåµÄ·Ö×Ó _________¡£

(3)ÒÑÖªD¿ÉÐγÉD3+Àë×Ó£¬¸ÃÀë×ÓÖÐÐÄÔ­×ÓÔÓ»¯·½Ê½Îª___£¬Á¢Ìå¹¹ÐÍΪ__¡£

(4)ζȽӽü·Ðµãʱ£¬DµÄ¼òµ¥Ç⻯ÎïµÄʵ²â·Ö×ÓÁ¿Ã÷ÏÔ¸ßÓÚÓÃÔ­×ÓÁ¿ºÍ»¯Ñ§Ê½¼ÆËã³öÀ´µÄ·Ö×ÓÁ¿£¬Ô­ÒòÊÇ _______¡£

(5)ÎÞÉ«µÄ[E(CA3)2]+ÔÚ¿ÕÆøÖв»Îȶ¨¡¢Á¢¼´±»Ñõ»¯³ÉÉîÀ¶É«µÄ[E (CA3)4]2+£¬ÀûÓÃÕâ¸öÐÔÖʿɳýÈ¥ÆøÌåÖеÄÑõÆø£¬¸Ã·´Ó¦µÄÀë×Ó·½³ÌΪ________¡£

(6)ÒÑÖªEºÍDÐγɵÄÒ»ÖÖ¾§Ìå°û½á¹¹ÈçͼËùʾ£¬ÒÑÖª¾§°û±ß³¤Îªanm£¬°¢·ü¼ÓµÂÂÞ³£ÊýΪNA£¬Ôò¸Ã¾§ÌåµÄÃܶÈΪ_________ g/cm3(Áгö¼ÆËã±í´ïʽ¼´¿É)¡£

¡¾´ð°¸¡¿N£¾O£¾C N2O sp2 VÐÎ Ë®ÕôÆøÖд󲿷ֵÄË®·Ö×ÓÒòΪÇâ¼ü¶øÏ໥µÞºÏ£¬Ðγɵ޺ϷÖ×Ó 4[Cu(NH3)2]++O2+8NH3H2O=4[Cu(NH3)4]2++4OH-+6H2O

¡¾½âÎö¡¿

AµÄºËÍâµç×Ó×ÜÊýÓëÆäÖÜÆÚÊýÏàͬ£¬ÔòAÊÇHÔªËØ£»BºÍDλÓÚͬһÖÜÆÚÇÒδ³É¶Ôµç×ÓÊýµÈÓÚÆäÖÜÆÚÐòÊý£¬ÈôÁ½Õ߶¼ÎªµÚÈýÖÜÆÚ£¬Î´³É¶Ôµç×ÓÊýΪ3£¬·ûºÏÌõ¼þµÄÔªËصÚÈýÖÜÆÚÖ»ÓÐÒ»¸ö£¬²»·ûºÏÌâÒ⣬ÈôB¡¢DΪµÚ¶þÖÜÆÚ£¬ÔòºËÍâÓÐ2¸öδ³É¶Ôµç×Ó£¬¼´2p2ºÍ2p4£¬ËùÒÔBΪCÔªËØ£¬DΪOÔªËØ£¬ÔòCΪNÔªËØ£»EΪµÚËÄÖÜÆÚÔªËØ£¬×îÍâ²ãÖ»ÓÐÒ»¸öµç×Ó£¬´ÎÍâ²ãµÄËùÓйìµÀ¾ù³äÂúµç×Ó£¬ÔòEÊÇCuÔªËØ¡£

(1)ͬһÖÜÆÚÔªËØ£¬ÔªËصĵÚÒ»µçÀëÄÜËæ×ÅÔ­×ÓÐòÊýµÄÔö´ó¶ø³ÊÔö´óÇ÷ÊÆ£¬µ«µÚIIA×å¡¢µÚVA×åÔªËصÚÒ»µçÀëÄÜ´óÓÚÏàÁÚÔªËØ£¬¹ÊC¡¢N¡¢OµÄµÚÒ»µçÀëÄÜ´óС¹ØϵΪ£ºN£¾O£¾C£»CuΪ29ºÅÔ­×Ó£¬ºËÍâµç×ÓÅŲ¼Ê½Îª[Ar]3d104s1£¬Æä¼Ûµç×ÓÅŲ¼Í¼Îª£»

(2)BD2ΪCO2£¬º¬ÓÐ3¸öÔ­×Ó£¬¼Ûµç×ÓÊýΪ16£¬µÈµç×ÓÌåÊÇÖ¸Ô­×Ó×ÜÊýÏàµÈ£¬¼Ûµç×Ó×ÜÊýÏàµÈµÄ΢Á££¬ËùÒÔÓÉÒÔÉÏÔªËØ×é³ÉµÄÓëCO2»¥ÎªµÈµç×ÓÌåµÄ·Ö×ÓΪN2O£»

(3)DΪOÔªËØ£¬ËùÒÔD3+Àë×ÓΪO3+£¬ÖÐÐÄÑõÔ­×ӵļ۲ãµç×Ó¶ÔÊýΪ2+=2.5£¬°´×÷3¼ÆË㣬ËùÒÔΪsp2ÔÓ»¯£¬¹Âµç×Ó¶ÔÊýΪ1£¬ËùÒÔÁ¢Ìå¹¹ÐÍΪVÐΣ»

(4)ζȽӽüË®µÄ·ÐµãµÄË®ÕôÆøÖдæÔڴ󲿷ֵÄË®·Ö×ÓÒòΪÇâ¼ü¶øÏ໥µÞºÏ£¬Ðγɵ޺ϷÖ×Ó£¬µ¼ÖÂÆä²â¶¨ÖµÆ«´ó£»

(5)ÎÞÉ«µÄ[Cu(NH3)2]+ÔÚ¿ÕÆøÖв»Îȶ¨£¬Á¢¼´±»Ñõ»¯ÎªÉîÀ¶É«µÄ[Cu(NH3)4]2+£¬ÑõÆø×÷Ñõ»¯¼Á£¬¸Ã·´Ó¦Ó¦ÔÚ°±Ë®ÖнøÐУ¬½áºÏÔªËØÊغã¿ÉÖª¸Ã¹ý³ÌÖл¹ÓÐÇâÑõ¸ùºÍË®Éú³É£¬Àë×Ó·½³ÌʽΪ4[Cu(NH3)2]++O2+8NH3H2O=4[Cu(NH3)4]2++4OH-+6H2O£»

(6))CuºÍOÐγÉÒ»ÖÖ¾§Ì壬¸Ã¾§°ûÖÐCuÔ­×Ó¸öÊý=4¡¢OÔ­×Ó¸öÊý=8¡Á+6¡Á=4£¬ËùÒÔ¾§°ûµÄÖÊÁ¿Îªg£¬¸Ã¾§°ûÌå»ýV=(a¡Á10-7 cm)3£¬Ôò¸Ã¾§ÌåÃܶȡ£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿ÔÚ¹¤Òµ·ÏÆøºÍÆû³µÎ²ÆøÖк¬ÓжàÖÖµªÑõ»¯ÎÖ÷ÒªÊÇNO ºÍNO2£¬¶¼ÒÔNOxÀ´±íʾ¡£NOxÄÜÆÆ»µ³ôÑõ²ã£¬²úÉú¹â»¯Ñ§ÑÌÎí£¬ÊÇÔì³É´óÆøÎÛȾµÄÖ÷ÒªÀ´Ô´Ö®Ò»¡£»Ø´ðÏÂÁÐÎÊÌ⣺

(1)ÒÑÖª 1mol ·Ö×Ó·Ö½âΪµ¥¸öÔ­×ÓËùÐèÒªµÄÄÜÁ¿Îª½âÀëìÊ¡£ N2(g)¡¢NO(g)¡¢O2(g)µÄ½âÀëìÊ·Ö±ðΪ941.7¡¢631.8¡¢493.7(µ¥Î»kJ/mol)£¬¼ÆËã·´Ó¦ 2NO(g) = N2(g) + O2(g)µÄ¡÷H=_______kJ/mol£¬ÊÔÅжÏNO(g)³£Î¡¢³£Ñ¹ÏÂÄÜ·ñ×Ô·¢·Ö½â ________ÌîÄÜ»ò²»ÄÜ)¡£

(2)Ϊ·ÀÖ¹¹â»¯Ñ§ÑÌÎí£¬³ý´Ó¹¤³§¡¢Æû³µµÄÉè¼Æ½øÐиĽøÍ⣬ҲҪ²ÉÓÃijЩ»¯Ñ§·½·¨¡£Óý¹Ì¿»¹Ô­NOx µÄ·´Ó¦Îª2NOx(g) + xC(s)N2(g) + xCO2(g)

¢ñ.ÔÚºãÎÂÌõ¼þÏ£¬2 molNO2(g)ºÍ×ãÁ¿ C(s)·´Ó¦£¬²âµÃƽºâʱ NO2(g)ºÍ CO2(g)µÄÎïÖʵÄÁ¿Å¨¶ÈÓëƽºâ×ÜѹµÄ¹ØϵÈçͼËùʾ£º

¢ÙA¡¢BÁ½µãNO2ƽºâת»¯ÂʵĹØϵ¦Á(A)____¦Á(B)£¬Æ½ºâ³£Êý¹ØϵK(A)_____K(B)(Ì¡¢£¼»ò£½)¡£

¢Ú¼ÆËãCµãʱ¸Ã·´Ó¦µÄѹǿƽºâ³£ÊýKp=_____MPa(KpÊÇÓÃƽºâ·Öѹ´úÌæƽºâŨ¶È)¼ÆË㣬·Öѹ=×Üѹ¡ÁÎïÖʵÄÁ¿·ÖÊý)¡£

¢ò.ÏÖÔÚÏòÈÝ»ý¾ùΪ 2L µÄÁ½¸öÃܱÕÈÝÆ÷A¡¢BÖмÓÈëÒ»¶¨Á¿µÄ NO(g)ºÍ×ãÁ¿µÄC(s)£¬ÏàͬζÈϲâµÃÁ½ÈÝÆ÷ÖÐn(NO)Ëæʱ¼ä±ä»¯Çé¿öÈç±íËùʾ£º

0

20

40

60

80

n(NO)/mol(A)

2

1.5

1.1

0.8

0.8

n(NO)/mol(B)

1

0.8

0.65

0.53

0.45

BÈÝÆ÷ÄÚ·´Ó¦ÔÚ 100sʱ´ïµ½Æ½ºâ״̬£¬Ôò0~100sÄÚÓà NO ±íʾµÄƽ¾ù·´Ó¦ËÙÂÊΪv(NO)= ____________¡£

(3)½üÄêÀ´µç»¯Ñ§·½·¨ÔÚ´¦ÀíµªÑõ»¯Îï·½ÃæÒ²Æðµ½ÁËÒ»¶¨µÄ×÷Óã¬ÈçͼÊÇÒ»ÖÖ°±ÆøÒ»¶þÑõ»¯µªÈ¼Áϵç³Ø£¬³£ÎÂÏ¿ɽ«¶þÑõ»¯µª×ª»¯ÎªµªÆø¡£

¢Ùc¿ÚͨÈëµÄÆøÌåΪ______ £¬Ð´³ö¸º¼«·´Ó¦µÄ·½³Ìʽ ________¡£

¢ÚÈôa¡¢d¿Ú²úÉúµÄÆøÌåÌå»ý¹²Îª1.568L(±ê¿öÏÂ)£¬µç·ÖÐͨ¹ýµÄµç×ÓÊýΪ____¡£

¡¾ÌâÄ¿¡¿Áò´úÁòËáÄÆ (Na2S2O3)ÔÚÉú²úÉú»îÖоßÓй㷺ӦÓá£Áò»¯¼î·¨Êǹ¤ÒµÉÏÖÆÈ¡Áò´úÁòËáÄƵķ½·¨Ö®Ò»¡£ÊµÑéÊÒÄ£Ä⹤ҵÉú²ú×°ÖÃÈçͼËùʾ£º

(1)ÀûÓÃÈçͼװÖýøÐÐʵÑ飬Ϊ±£Ö¤ÁòËá˳ÀûµÎϵIJÙ×÷ÊÇ_______¡£

(2)×°ÖÃBÖÐÉú³ÉµÄNa2S2O3ͬʱ»¹Éú³ÉCO2£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪ_______£»ÔÚ¸Ã×°ÖÃÖÐʹÓöà¿×ÇòÅݵÄÄ¿µÄÊÇ_____¡£

(3)×°ÖÃCµÄ×÷ÓÃÊǼìÑé×°ÖÃBÖÐSO2µÄÎüÊÕЧ¹û£¬CÖпÉÑ¡ÔñµÄÊÔ¼ÁÊÇ__(Ìî×Öĸ)¡£

a.H2O2ÈÜÒº b.äåË® c.KMnO4ÈÜÒº d.BaCl2ÈÜÒº

(4)Na2S2O3ÈÜÒº³£ÓÃÓڲⶨ·ÏË®ÖÐBa2+Ũ¶È¡£

¢ÙÈ¡·ÏË®20.00mL£¬¿ØÖÆÊʵ±µÄËá¶È£¬¼ÓÈë×ãÑÎK2Cr2O7ÈÜÒº£¬µÃµ½ BaCrO4 ³Áµí£¬¹ýÂËÏ´µÓºóÓÃÊÊÁ¿Ï¡ËáÈܽ⣬´Ëʱ CrO42-È«²¿×ª»¯ÎªCr2O72-£»ÔÙ¼Ó¹ýÁ¿ KIÈÜÒº£¬½«Cr2O72- ³ä·Ö·´Ó¦£»È»ºó¼ÓÈëµí·ÛÈÜÒº×÷ָʾ¼Á£¬ÓÃ0.100 mol/LµÄNa2S2O3 ÈÜÒº½øÐе樣º(I2 +2 S2O32-= S4O62-+ 2I-)£¬µÎ¶¨ÖÕµãµÄÏÖÏóΪ__________¡£Æ½Ðеζ¨3´Î£¬ÏûºÄNa2S2O3 ÈÜÒºµÄƽ¾ùÓÃÁ¿Îª18.00mL¡£Ôò¸Ã·ÏË®ÖÐBa2+ µÄÎïÖʵÄÁ¿Å¨¶ÈΪ____mol/L£¬

¢ÚÔڵζ¨¹ý³ÌÖУ¬ÏÂÁÐʵÑé²Ù×÷»áÔì³ÉʵÑé½á¹ûÆ«¸ßµÄÊÇ______(Ìî×Öĸ)¡£

a.µÎ¶¨¹ÜδÓÃNa2S2O3ÈÜÒºÈóÏ´

b.µÎ¶¨ÖÕµãʱ¸©ÊÓ¶ÁÊý

c.׶ÐÎÆ¿ÓÃÕôÁóˮϴµÓºóδ½øÐиÉÔï´¦Àí

d.µÎ¶¨¹Ü¼â×ì´¦µÎ¶¨Ç°ÎÞÆøÅÝ£¬µÎ¶¨Öյ㷢ÏÖÓÐÆøÅÝ

¡¾ÌâÄ¿¡¿Á×ÊÇÖØÒªµÄÔªËØ£¬ÄÜÐγɶàÖÖº¬ÑõËá¡£»Ø´ðÏÂÁÐÎÊÌâ:

(1)´ÎÁ×Ëá(H3PO2)ÊÇÒ»ÔªËᣬÆäµçÀë³£ÊýµÄÖµK=9¡Á103¡£Ïò10mL0.1 molL-1H3PO2ÈÜÒºÖмÓÈë30mLµÈÎïÖʵÄÁ¿Å¨¶ÈµÄNaOHÈÜÒº£¬Ð´³ö·´Ó¦µÄÀë×Ó·½³Ìʽ_________£¬c(Na+)+(H2PO2-)+c(H3PO2)=______(ºöÂÔ»ìºÏºóÈܲ¨Ìå»ýµÄ±ä»¯)¡£

(2)ÑÇÁ×Ëá(H3PO3)ÊǶþÔªÈõËᣬ 25¡æʱÑÇÁ×ËáµÄµçÀë³£ÊýµÄֵΪK1=1¡Á10-2¡¢k2=2.6¡Á10£­7,ÔòNaH2PO3ÈÜÒºÏÔÐÔ_____(Ìî¡°Ëᡱ¡°¼î¡±»ò¡°ÖС±)£¬Ô­ÒòÊÇ____(½áºÏ»¯Ñ§ÓÃÓï¼°Êý¾Ý¼ÆËã½øÐнâÊÍ)

(3)ÒÑÖªHFµÄµçÀë³£ÊýµÄֵΪK=3.6¡Á10-4,½«×ãÁ¿HFÈÜÒººÍNa2HPO3ÈÜÒº·´Ó¦£¬ÆäÀë×Ó·½³ÌʽΪ______¡£

(4)ÑÇÁ×Ëá¾ßÓÐÇ¿»¹Ô­ÐÔ¡£»¯Ñ§ÊµÑéС×éÀûÓõζ¨·¨²â¶¨Ä³ÑÇÁ×ËáÈÜÒºµÄŨ¶È£¬È¡25.00mLµÄÑÇÁ×ËáÈÜÒº·ÅÈë׶ÐÎÆ¿ÖУ¬ÓÃ0.10 molL-1µÄ¸ßÃÌËá¼ØÈÜÒº½øÐе樣¬·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ5H3PO3+ 2MnO4-+6H+ = 5H3PO4+ 2Mn2+ +3H2O¡£

Èý´ÎµÎ¶¨ÊµÑéµÄÊý¾Ý·Ö±ðÈçÏÂ:

ʵÑé±àºÅ

µÎ¶¨Ç°¶ÁÊý

µÎ¶¨ºó¶ÁÊý

1

0.50

22.50

2

1.50

24.50

3

1.00

22.00

¢ÙÑÇÁ×ËáÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈΪ______¡£

¢Ú¹ØÓÚ¸ÃʵÑéÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ______(ÌîдÐòºÅ)¡£

a È¡ÑÇÁ×ËáÈÜÒºµÄµÎ¶¨¹Ü£¬Ï´µÓºóδÈóÏ´£¬µ¼Ö½á¹ûÆ«µÍ

b Ê¢¸ßÃÌËá¼ØÈÜÒºµÄµÎ¶¨¹ÜµÎ¶¨Ç°ÓÐÆøÅÝ£¬µÎ¶¨ºóÎÞÆøÅÝ£¬µ¼Ö½á¹ûÆ«¸ß

c µÎ¶¨¹ý³ÌÖÐÑÛ¾¦Ö»×¢Êӵζ¨¹ÜÖÐÒºÃæ±ä»¯£¬²¢×öºÃ¼Ç¼

d ׶ÐÎƿδ¸ÉÔïµ×²¿ÓÐË®£¬»áµ¼Ö½á¹ûÆ«µÍ

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

¾«Ó¢¼Ò½ÌÍø