B.ÅäºÏÎﶼÓÐÄÚ½çºÍÍâ½ç
C.Åäλ¼üµÄÇ¿¶ÈµÍÓÚÀë×Ó¼ü»ò¹²¼Û¼ü
D.ÅäºÏÎïÖУ¬ÐγÉÌåÓëÅäλÔ×Ó¼äÒÔÅäλ¼ü½áºÏ 12.ÏÂÁз½·¨ÖпÉÓÃÓÚ¼õÃâ·ÖÎö²âÊÔÖеÄżȻÎó²îµÄÊÇ A.Ôö¼ÓƽÐвâÑéµÄ´ÎÊý B.½øÐжÔÕÕÊÔÑé
C.½øÐÐÒÇÆ÷У׼ D.½øÐпհ×ÊÔÑé
2.240?20.113.¼ÆËãÆ÷ËãµÃµÄ½á¹ûΪ2.2512£¬°´ÓÐЧÊý×ÖÔËËã¹æÔò£¬Ó¦ÐÞԼΪ
100?0.20A.2.2 B.2.3 C.2.25 D.2.251
14.0.10mol.L-1 NH3.H2OÓë0.10mol.L-1 NH4ClµÈÌå»ý»ìºÏÐγɻº³åÈÜÒº£¬¸ÃÈÜÒºµÄpHÖµµÈÓÚ£¨ÒÑÖªNH3.H2OµÄpKb=4.74£© A.10.83 B.4.44 C.9.26 D.5.04
15.ÔÚËá¼îµÎ¶¨ÖÐÑ¡Ôñָʾ¼Áʱ¿É²»¿¼ÂÇÄĸöÒòËØ A.µÎ¶¨Í»Ô¾µÄ·¶Î§ B.ָʾ¼ÁµÄ±äÉ«·¶Î§ C.ָʾ¼ÁµÄÑÕÉ«±ä»¯ D.ָʾ¼Á·Ö×ÓÁ¿µÄ´óС ¶þ.Ìî¿ÕÌâ
1.ijζÈÏ£¬Ò»ÈÝÆ÷º¬ÓÐ3.0 mol O2£¬2.0 mol N2ºÍ1.0 mol Ar¡£Èç¹û»ìºÏÆøÌåµÄ×ÜѹÁ¦Îª1kPa,ÔòÑõÆøµÄ·ÖѹP(O2)=£¨ £©kPa¡£
2.·´Ó¦C(s)+H2O(g)=CO2(g)+H2(g),¦¤H£¾0,µ±Éý¸ßζÈʱ£¬¸Ã·´Ó¦µÄƽºâ³£Êý½«£¨ £©£»Ôö´óϵͳѹÁ¦»áʹƽºâ£¨ £©Òƶ¯¡£
3.ij·â±ÕϵͳÔÚÑ»·¹ý³ÌÖÐÎüÊÕÈÈÁ¿7.50kJ£¬Ôò¹ý³ÌÖвúÉúµÄ¹¦ W=£¨ £©kJ¡£
4.·´Ó¦C(s)+H2O(g)=CO2(g)+H2(g)µÄ±ê׼ƽºâ³£Êý±í´ïʽΪ£¨ £©¡£
5.Èôµç¼«µçÊÆ´úÊýÖµÔ½´ó£¬Ôòµç¶ÔËù¶ÔÓ¦µÄ»¹ÔÐÍÎïÖʵĻ¹ÔÄÜÁ¦Ô½£¨ £©£¬Ñõ»¯ÐÍÎïÖʵÄÑõ»¯ÄÜÁ¦Ô½£¨ £©¡£
6.ÓÃ0.1000 mol.L-1 NaOHµÎ¶¨20.00 mL0.1000 mol.L-1 HClÈÜÒº£¬Ëæ×ŵζ¨µÄ½øÐУ¬pHÖð½¥£¨ £©£¬µÎ¶¨ÖÁ»¯Ñ§¼ÆÁ¿µãʱÈÜÒºpH=£¨ £©¡£¸ÃµÎ¶¨µÄpHͻԾ·¶Î§Îª£¨ £©£¬Ëá¼îÈÜҺŨ¶ÈÔö´óʱ£¬pHͻԾ·¶Î§£¨ £©¡£
7.º¬ÓÐ0.100 mol.LAg,0.100 mol.LFeºÍ0.010 mol.LFeµÄÈÜÒºÖУ¬¿É·¢ÉúÈçÏ·´Ó¦£ºFe2+£¨aq£©+ Ag+£¨aq£©= Fe3+£¨aq£©+ Ag£¨s£©,ÇóµÃ·´Ó¦ÉÌQ=£¨ £©£¬ÒÑÖª25¡æʱ£¬K¦È=5.0£¬Ôò·´Ó¦×Ô·¢½øÐеķ½ÏòÊÇ£¨ £©¡£
8.ÒÑÖªÅäÀë×Ó[FeF6]3-,[Fe(CN)6]3-µÄKf¦È·Ö±ðΪ2.04¡Á1014.1.00¡Á1042£¬ÔòÕâÁ½ÖÖÅäÀë×ÓÖУ¨ £©¸üÎȶ¨¡£
9.Óûù×¼ÎïÖÊNa2C2O4ÔÚÇ¿ËáÌõ¼þϱ궨KMnO4ÈÜҺʱ£¬¿ÉÓ㨠£©×÷×ÔÉíָʾ¼Á¡£
13
-1
+
-1
2+
-1
3+
¦È
10.ÅäºÏÎïNa3[AlF6]µÄÍâ½çÊÇ£¨ £©£¬ÄÚ½çÊÇ£¨ £©£¬ÖÐÐÄÀë×ÓÊÇ£¨ £©£¬ÅäλÌ壨 £©£¬ÅäλÊýÊÇ£¨ £©£¬ÅäºÏÎïÃû³ÆÊÇ£¨ £©¡£ 11.K2Cr2O7ÐÔÖÊÎȶ¨¡¢Ò×Ìá´¿£¬Æä±ê×¼ÈÜÒº¿É²ÉÓ㨠£©·¨ÅäÖÆ¡£
12.È·¶¨ÏÂÁÐÓÐЧÊý×ÖµÄλÊý£º£¨1£©lgX=2.65£¨ £©£¬£¨2£©0.01010£¨ £©¡£ 13.Ö±½ÓµâÁ¿·¨ÖУ¬Óõí·Û×÷ָʾ¼Á£¬µ±ÈÜÒº³ÊÏÖ£¨ £©É«Ê±£¬¼´ÎªÖյ㡣 Èý.ÅжÏÌâ
1.ͬһϵͳͬһ״̬¿ÉÄÜÓжà¸öÈÈÁ¦Ñ§ÄÜ¡£ 2.µ±[H+]£¾[OH-]ʱ£¬ÈÜÒº³ÊËáÐÔ¡£ 3.»¯Ñ§¼ÆÁ¿µã¾ÍÊǵζ¨Öյ㡣
4.Ñõ»¯Êý¿ÉÒÔÊÇÕûÊý£¬Ò²¿ÉÒÔÊÇСÊý»ò·ÖÊý¡£ 5.EDTAÓëMÐγÉ1:1µÄòüºÏÎËùÒÔÆäÅäλÊýÊÇ1¡£ 6.ϵͳµÄìÊÖµµÈÓÚºãѹ·´Ó¦ÈÈ¡£ 7.±ê×¼ÈÜÒºÊÇÖªµÀ׼ȷŨ¶ÈµÄÊÔ¼ÁÈÜÒº¡£ 8.¶¨Á¿·ÖÎö¹¤×÷ÒªÇóÔñ¶¨½á¹ûµÄÎó²îµÈÓÚÁã¡£
9.ÔÚ³Áµíת»¯ÖУ¬Èܽâ¶È½ÏСµÄ³ÁµíÈÝÒ×ת»¯ÎªÈܽâ¶È½Ï´óµÄ³Áµí¡£ 10.»º³åÈÜÒº¿ÉÒÔ½øÐÐÈÎÒâ±ÈÀýµÄÏ¡ÊÍ¡£ ËÄ.¼ò´ðÌâ
1.ºÎν·Ö²½³Áµí£¿ÈçºÎÈ·¶¨·Ö²½³ÁµíµÄ˳Ðò£¿ 2.ÅäƽÏÂÁз´Ó¦£º
£¨1£©Ç¿ËáÐÔÌõ¼þÏÂÓÃNa2C2O4×÷»ù×¼ÎïÖʱ궨KMnO4ÈÜÒºµÄÀë×Ó·´Ó¦·½³Ìʽ¡£ £¨2£©Cu + HNO3(Ï¡) ¡úCu£¨NO3£©2 + NO + H2O
3.Åж϶þÔªÈõËáH2C2O4µÄÁ½¼¶H+ÄÜ·ñ±»Ç¿¼îÖ±½ÓµÎ¶¨ºÍ·Ö±ðµÎ¶¨£¨Ð´³öÅжÏÒÀ¾Ý£©ÉèH2C2O4Ũ¶ÈΪ0.1 mol.L-1£¨ÒÑÖªH2C2O4µÄKa1¦È=5.4¡Á10-2, Ka2¦È=5.4¡Á10-5£© 4.¼òÊö½ðÊôָʾ¼ÁµÄ±äÉ«ÔÀí¡£ 5.д³öQ¼ìÑ鷨ȡÉá¿ÉÒÉÖµµÄ»ù±¾²½Öè¡£ Îå.¼ÆËãÌâ
1.¶ÔÓÚ·´Ó¦£ºCCl4(l) + H2(g) = HCL(g) + CHCl3(l),ÒÑÖª¦¤rHm¦È£¨298K£©=-90.34KJ.mol, ¦¤rSm£¨298K£©=4.15¡Á10KJ.mol,Çó298Kʱ±ê׼״̬ϸ÷´Ó¦µÄ¦¤rGm¦È£¨298K£©£¬²¢ÅжÏÔÚ¸ÃÌõ¼þÏ·´Ó¦ÄÜ·ñÕýÏò×Ô·¢½øÐУ¿ 2.¼ÆËã²¢ÅжϳÁµíµÄÉú³É£º
£¨1£©½«2.0¡Á10-3mol.L-1BaCl2ÈÜÒººÍ2.0¡Á10-3mol.L-1Na2SO4ÈÜÒºµÈÌå»ý»ìºÏ£¬ÎÊÓÐÎÞBaSO4³ÁµíÉú³É£¿£¨ÒÑÖªKsp¦È=1.08¡Á10-10£©
(2)ÔÚ0.20LµÄ0.50 mol.L-1MgCl2ÈÜÒºÖмÓÈëµÈÌå»ýµÄ0.10 mol.L-1 NH3.H2OÈÜÒº£¬ÎÊÓÐÎÞMg(OH)2³ÁµíÉú³É£¿£¨ÒÑÖªMg(OH)2µÄKsp¦È=5.6¡Á10-12£¬NH3.H2OµÄKb¦È=1.8
14
-1
¦È
-2
-1
¡Á10-5£©(×¢£ºÉæ¼°¿ª·½ÔËËãʱ£¬Áгö¹«Ê½´úÈëÊý¾Ý¼´¿É¡£)
3.³ÆÈ¡´¿½ðÊôп0.2850g£¬ÈÜÓÚÑÎËáºó£¬ÒÔÕôÁóË®ÔÚ250mlÈÝÁ¿Æ¿Öж¨ÈÝ¡£ £¨1£©¼ÆËã¸ÃZn2+±ê×¼ÈÜÒºµÄŨ¶È£¿
£¨2£©È¡25.00mlÉÏÊöZn2+±ê×¼ÈÜÒº±ê¶¨EDTAÈÜÒº£¬ÏûºÄEDTAÈÜÒº22.55ml£¬Çó¸ÃEDTA±ê×¼ÈÜÒºµÄŨ¶È¡££¨ÒÑÖªM£¨Zn£©=65.38g.mol-1£©
4.±ê¶¨C(NaOH)=0.10 mol.L-1µÄNaOHÈÜÒº¡£ÓûÏûºÄNaOHÈÜÒº25.00ml£¬Ó¦³ÆÈ¡H2C2O4.2H2O»ù×¼ÎïÖʶàÉÙ¿Ë£¿ ÒÑÖªM(H2C2O4.2H2O)=126.1g.mol-1¡£ 5.
ÔÆÄÏÊ¡2013ÄêÆÕͨ¸ßУ¡°×¨Éý±¾¡±ÕÐÉú¿¼ÊÔ¹«¹²»¯Ñ§ÊÔ¾í
Ò».Ñ¡ÔñÌâ
1.25¡æʱÒÔÅÅË®·¨ÊÕ¼¯ÑõÆøÓÚ¸ÖÆ¿ÖУ¬²âµÃ¸ÖÆ¿µÄѹÁ¦Îª150.5kPa,ÒÑÖª25¡æʱˮµÄ±¥ºÍÕôÆøѹΪ3.2kPa£¬Ôò¸ÖÆ¿ÖÐÑõÆøµÄѹÁ¦Îª
A.147.3kPa B.153.7kPa C.150.5 kPa D.101.325 kPa
2.Óñê×¼NaOHÈÜÒºµÎ¶¨Í¬Å¨¶ÈµÄHCl,ÈôÁ½ÕßµÄŨ¶È¾ùÔö´ó10±¶£¬ÒÔÏÂÐðÊöµÎ¶¨ÇúÏßpHͻԾ´óС£¬ÕýÈ·µÄÊÇ
A.»¯Ñ§¼ÆÁ¿µãÇ°0.1%µÄpH¼õС£¬ºó0.1%µÄpHÔö´ó B.»¯Ñ§¼ÆÁ¿µãÇ°ºó0.1%µÄpH¾ùÔö´ó C.»¯Ñ§¼ÆÁ¿µãÇ°ºó0.1%µÄpH¾ù¼õС
D.»¯Ñ§¼ÆÁ¿µãÇ°0.1%µÄpH²»±ä£¬ºó0.1%µÄpHÔö´ó
3.ÒÑÖª¼¸¸öµç¶ÔµÄ±ê×¼µç¼«µçλ¦Õ¦È(ClO4-/ClO3-)=1.19V£¬¦Õ¦È(O3/O2)=2.07V, ¦Õ¦È
15
(Cr3+/Cr2+)=-0.410V,(Cu2+/Cu+)=0.158V,Ôò¸÷µç¶ÔÖÐ×îÇ¿µÄÑõ»¯¼ÁºÍ×îÇ¿µÄ»¹Ô¼ÁÊÇ
A.ClO4-ºÍCr2+ B.Cr3+ºÍO2 C.O3ºÍCr2+ D.O3ºÍCu+ 4.ÏÂÁз´Ó¦ÖеĦ¤rHm¦ÈµÈÓÚAgBr(s)µÄ¦¤fHm¦ÈµÄÊÇ
A.Ag+(ag)+Br-(ag)£½AgBr(s) B.2Ag (s)+Br2(g)£½2AgBr(s) C.Ag (s)+1/2Br2(l)£½AgBr(s) D.Ag (s)+1/2Br2(g)£½2AgBr(s) 5.ijÌåϵ¾Ñ»·¹ý³Ì»Øµ½Æðʼ״̬£¬ÏÂÁÐÁ¿Öв»Ò»¶¨ÎªÁãµÄÊÇ A.U B.H C.S D.Q
6.384Kʱ·´Ó¦2NO2(g)=N2O4(g), K¦È=3.9¡Á10-2,ͬζÈÏ·´Ó¦NO2(g)=1/2N2O4(g)µÄ KӦΪ
A.K¦È =1/3.9¡Á10-2 B.K¦È =3.9¡Á10-2 C.K¦È =1.95¡Á10-2 D.K¦È =¡Ì3.9¡Á10-2
7.ijζÈʱ£¬·´Ó¦¢Ù¡¢¢ÚºÍ¢ÛµÄ±ê׼ƽºâ³£Êý·Ö±ðΪK1¦È¡¢K2¦ÈºÍK3¦È£¬Ôò·´Ó¦¢ÜµÄK4¦ÈµÈÓÚ
¢ÙCoO(s) + CO(g) = Co(s) + CO2(g) ¢ÚCO2(g) + H2(g) = CO(g) + H2O(l) ¢ÛH2O(l) = H2O(g)
¢ÜCoO(s) + H2(g) = Co(s) + H2O(g) A.K1+ K2+ K3
¦È
¦È
¦È
¦È
B.K1- K2-K3
¦È¦È¦È
C.K1¦È¡¨ K2¦È¡¨K3¦È D.K1¦È¡¨ K2¦È/K3¦È
-8.É谱ˮµÄŨ¶ÈΪC,Èô½«ÆäÏ¡ÊÍÒ»±¶£¬ÈÜÒºÖеÄOHŨ¶ÈΪ ??CKbCKbCA. B.2c C. D.
2229.ÒÑÖªH3PO4µÄpKa1¦È¡¢pKa2¦È¡¢pKa3¦È·Ö±ðΪ2.12¡¢7.20¡¢12.36£¬ÔòPO43-µÄpKb1¦È
Ϊ
A.11.88 B.1.64 C.6.80 D.2.12
¦È
10.Ca3(PO4)2µÄÈܽâ¶ÈSÓëKspÖ®¼äµÄ¹ØϵΪSµÈÓÚ ??KK??spspA.5 B.Ksp C.5Ksp D.
108108¦È
11.ÏÂÁе缫ÖУ¬¦ÕÖµ×î¸ßµÄÊÇ
A.[Ag(NH3)2+]/Ag B.Ag+/Ag C.[Ag(CN)2+]/Ag D.AgCl/Ag
12.·´Ó¦3A2++2B=3A+2B3+ÔÚ±ê×¼×´¿öϵç³Øµç¶¯ÊÆΪ1.8V,ÔÚijŨ¶Èʱ²âµÃµç³Øµç¶¯ÊÆΪ1.6V£¬Ôò´Ë·´Ó¦µÄlgK¦ÈÖµ¿ÉÒÔ±íʾΪ A.3¡Á1.8/0.0592 B.6¡Á1.8/0.0592 C.6¡Á1.6/0.0592 D.3¡Á1.6/0.0592 13.ÏÂÁÐÇé¿öÖÐÒýÆðżȻÎó²îµÄÊÇ
16