µ±Ç°Î»ÖãºÊ×Ò³ > дóѧ»¯Ñ§1--3Õ´ð°¸
3£®ÈÜÖÊÊÇÇ¿µç½âÖÊ»òÆäŨ¶È½Ï´óʱ£¬ÈÜÒºµÄÕôÆøÑ¹Ï½µ²»·ûºÏÀÎÚ¶û¶¨Âɵ͍Á¿¹ØÏµ¡£
( ¡Ì )
4£®ÒºÌåµÄÄý¹Ìµã¾ÍÊÇÒºÌåÕô·¢ºÍÄý¾ÛËÙÂÊÏàµÈʱµÄζȡ£ ( ¡Á )
5£®ÖÊÁ¿ÏàµÈµÄ¶¡¶þ°·£¨H2N(CH2)4NH2£©ºÍÄòËØ£¨CO(NH2)2£©·Ö±ðÈÜÓÚ1000 gË®ÖУ¬ËùµÃ Á½ÈÜÒºµÄÄý¹ÌµãÏàͬ¡£ ( ¡Á ) 6£®³£ÀûÓÃÏ¡ÈÜÒºµÄÉøÍ¸Ñ¹À´²â¶¨ÈÜÖʵÄÏà¶Ô·Ö×ÓÖÊÁ¿¡£ ( ¡Ì ) 7£® ÔÚ100gË®ÖÐÈܽâ5.2gij·Çµç½âÖÊ£¬¸Ã·Çµç½âÖʵÄĦ¶ûÖÊÁ¿Îª60£¬´ËÈÜÒº ÔÚ±ê׼ѹÁ¦ÏµķеãΪ373.60K¡£ ( ¡Ì ) 8£®ÈõËá»òÈõ¼îµÄŨ¶ÈԽС£¬Æä½âÀë¶ÈҲԽС£¬ËáÐÔ»ò¼îÐÔÔ½Èõ¡£ ( ¡Á ) 9£®ÔÚÒ»¶¨Î¶ÈÏ£¬Ä³Á½ÖÖËáµÄŨ¶ÈÏàµÈ£¬ÆäË®ÈÜÒºµÄpHÖµÒ²±ØÈ»ÏàµÈ¡£( ¡Á )
10£®µ±Èõµç½âÖʽâÀë´ïƽºâʱ£¬Àë×ÓŨ¶ÈԽС£¬½âÀë³£ÊýԽС£¬Èõµç½âÖʵĽâÀëÔ½Èõ¡£
( ¡Á )
11£®ÔÚ»º³åÈÜÒºÖУ¬Ö»ÊÇÿ´Î¼ÓÉÙÁ¿Ç¿Ëá»òÇ¿¼î£¬ÎÞÂÛÌí¼Ó¶àÉٴΣ¬»º³åÈÜҺʼÖÕ¾ßÓлº³å ÄÜÁ¦¡£ ( ¡Á )
12£®ÒÑÖªKs¦È (Ag2CrO4) =1.11¡Á10-12£¬Ks¦È(AgCl)=1.76¡Á10-10£¬ÔÚ0.0100mol¡¤kg-1K2CrO4ºÍ
2?0.1000mol¡¤kg-1KClµÄ»ìºÏÈÜÒºÖУ¬ÖðµÎ¼ÓÈëAgNO3ÈÜÒº£¬ÔòCrO4ÏȳÁµí¡£
( ¡Á )
13. ÓÃEDTA×öÖØ½ðÊôµÄ½â¶¾¼ÁÊÇÒòΪÆä¿ÉÒÔ½µµÍ½ðÊôÀë×ÓµÄŨ¶È¡£ ( ¡Ì )
14£®ÓÉÓÚKa(HAc)>Ka(HCN),¹ÊÏàͬŨ¶ÈµÄNaAcÈÜÒºµÄpH±ÈNaCNÈÜÒºµÄpH´ó¡£( ¡Á)
¶þ£®Ñ¡ÔñÌâ
1£®ÔÚÖÊÁ¿Ä¦¶ûŨ¶ÈΪ1.00mol¡¤kg-1µÄNaCl
Ë®ÈÜÒºÖУ¬ÈÜÖʵÄĦ¶û·ÖÊý
¦ÖB
ºÍÖÊÁ¿·ÖÊý¦ØB
( C )
Ϊ
A£® 1.00, 18.09% B£®0.055, 17.0% C£®0.0177, 5.53% D£®0.180, 5.85%
2. 30%µÄÑÎËáÈÜÒº£¬ÃܶÈΪ1.15g¡¤cm-3£¬ÆäÎïÖʵÄÁ¿Å¨¶ÈCBºÍÖÊÁ¿Ä¦¶ûŨ¶ÈbB·Ö±ðΪ
( A )
-3-1-3-1
A£®9.452mol¡¤dm, 11.74mol¡¤kg B£® 94.52mol¡¤dm, 27.39mol¡¤kg
-3-1-3
C£®31.51mol¡¤dm, 1.74mol¡¤kg D£® 0.945mol¡¤dm, 2.739mol¡¤kg-1
3.ÏÂÃæÏ¡ÈÜÒºµÄŨ¶ÈÏàͬ£¬ÆäÕôÆøÑ¹×î¸ßµÄÊÇ ( C ) A£®NaClÈÜÒº B£®H3PO4ÈÜÒº C£®C6H12O6ÈÜÒº D£®NH3-H2OÈÜÒº 4.ÏÂÁÐÎïÖÊË®ÈÜÒºÖУ¬Äý¹Ìµã×îµÍµÄÊÇ ( C )
A£® 0.2mol¡¤kg-1 C12H22O11 B£® 0.2mol¡¤kg-1 HAc C£® 0.2mol¡¤kg-1 NaCl D£® 0.1mol¡¤kg-1 HAc
5.ÏàͬŨ¶ÈµÄÏÂÁÐÈÜÒºÖзеã×î¸ßµÄÊÇ ( C )
A£®ÆÏÌÑÌÇ B£®NaCl C£®CaCl2 D£®[Cu(NH3)4]SO4 6£®0.1mol¡¤kg-1 µÄÏÂÁÐÈÜÒºÖÐpH×îСµÄÊÇ ( B ) A£®HAc B£®H2C2O4 C£®NH4Ac D£®H2S
7. ÏÂÁлìºÏÈÜÒº£¬ÊôÓÚ»º³åÈÜÒºµÄÊÇ ( A )A£® 50g 0.2mol¡¤kg-1 HAcÓë 50g 0.1mol¡¤kg-1 NaOH B£® 50g 0.1mol¡¤kg-1 HAcÓë 50g 0.1mol¡¤kg-1 NaOH C£® 50g 0.1mol¡¤kg-1 HAcÓë 50g 0.2mol¡¤kg-1 NaOH D£® 50g 0.2mol¡¤kg-1 HClÓë 50g 0.1mol¡¤kg-1 NH3¡¤H2O
8£®ÈôÓÃHAcºÍNaAcÈÜÒºÅäÖÆpH = 4.5µÄ»º³åÈÜÒº£¬Ôò¶þÕßŨ¶ÈÖ®±ÈΪ ( C )
13.21.88A£® 1.8 B£® 36 C£®1 D£®9
9. ÅäÖÆpH ¡Ö 7µÄ»º³åÈÜÒº,ӦѡÔñ ( D ) A£® K¦È(HAc)=1.8¡Á10-5 B£®K¦È(HCOOH)=1.77¡Á10-4
C£® K¦È(H2CO3)=4.3¡Á10-7 D£®K¦È(H2PO4-)=6.23¡Á10-8
10. AgClÔÚÏÂÁÐÎïÖÊÖÐÈܽâ¶È×î´óµÄÊÇ ( B ) A£® ´¿Ë® B£® 6mol¡¤kg-1 NH3¡¤H2O
-1-1
C£® 0.1mol¡¤kg NaCl D£® 0.1mol¡¤kg BaCl2
11.ÔÚPbI2³ÁµíÖмÓÈë¹ýÁ¿µÄKIÈÜÒº£¬Ê¹³ÁµíÈܽâµÄÔÒòÊÇ ( B ) A£®Í¬Àë×ÓЧӦ B£®Éú³ÉÅäλ»¯ºÏÎï C£®Ñõ»¯»¹Ô×÷Óà D£®ÈÜÒº¼îÐÔÔöÇ¿
12£®ÏÂÁÐ˵·¨ÖÐÕýÈ·µÄÊÇ ( A ) A£®ÔÚH2SµÄ±¥ºÍÈÜÒºÖмÓÈëCu2+£¬ÈÜÒºµÄpHÖµ½«±äС¡£
B£®·Ö²½³ÁµíµÄ½á¹û×ÜÄÜʹÁ½ÖÖÈܶȻý²»Í¬µÄÀë×Óͨ¹ý³Áµí·´Ó¦ÍêÈ«·ÖÀ뿪¡£ C£®Ëùν³ÁµíÍêÈ«ÊÇÖ¸³Áµí¼Á½«ÈÜÒºÖÐijһÀë×Ó³ý¾»ÁË¡£
D£®ÈôijϵͳµÄÈÜÒºÖÐÀë×Ó»ýµÈÓÚÈܶȻý£¬Ôò¸Ãϵͳ±ØÈ»´æÔÚ¹ÌÏà¡£
13£®ÏÂÁÐÅäºÏÎïµÄÖÐÐÄÀë×ÓµÄÅäλÊý¶¼ÊÇ6£¬ÏàͬŨ¶ÈµÄË®ÈÜÒºµ¼µçÄÜÁ¦×îÇ¿µÄÊÇ(D) A£® K2[MnF6] B£® [Co(NH3)6]Cl3 C£® [Cr(NH3)4]Cl3 D£® K4[Fe(CN)6]
Èý£®Ìî¿ÕÌâ
1£®Ï¡ÈÜÒºµÄÒÀÊýÐÔÊÇÖ¸ÈÜÒºµÄ_ÕôÆøÑ¹Ï½µ_¡¢___·ÐµãÉý¸ß_____¡¢__Äý¹ÌµãϽµ___ºÍ_ÉøÍ¸Ñ¹_¡£ËüÃǵÄÊýÖµÖ»ÓëÈÜÖʵÄ__Á£×ÓÊýÄ¿£¨Ò»¶¨Á¿ÈܼÁÖÐÈÜÖʵÄÎïÖʵÄÁ¿£©_³ÉÕý±È¡£ 2£®ÏÂÁÐË®ÈÜÒº£¬°´Äý¹ÌµãÓɸߵ½µÍµÄ˳ÐòÅÅÁÐ(ÓÃ×Öĸ±íʾ) __ D > C > A > B ¡£ A.1mol¡¤kg-1 KCl B.1mol¡¤kg-1 Na2SO4
-1-1
C.1mol¡¤kgÕáÌÇ D.0.1mol¡¤kgÕáÌÇ
3£®HAcÈÜÒºÖеÎÈë2µÎ¼×»ù³Èָʾ¼Á£¬ÈÜÒºÏÔ _ºì__ É«£¬ÈôÔÙ¼ÓÈëÉÙÁ¿NaAc(s)£¬ÈÜÒºÓÉ _ºì__ É«±äΪ __»Æ__ É«£¬ÆäÔÒòÊÇ ___ͬÀë×ÓЧӦ__¡£
???K?bAgbs4£®Ag2CrO4µÄÈܶȻý³£Êý±í´ïʽΪ
3????22??bCrO4b?,ÆäÈܽâ¶ÈSÓëKsµÄ¹Ø
???S?ϵΪ£º
5£®Ìî±í£º »¯Ñ§Ê½ ?Ks4¡£
Ãû³Æ ÖÐÐÄÅäλÌå Àë×Ó Pt4+ ÅäλÔ×Ó ÅäλÊý ÅäÀë×ÓµçºÉ +2 [Pt(NH3)4(NO2)Cl]SO4 ÁòËáÒ»ÂÈ?Ò»Ïõ»ù?ËݱºÏ²¬(¢ô) ¶þÂÈ»¯ÈýÒÒ¶þ°·ºÏÄø(¢ò) ÒÒ¶þ°·ËÄÒÒËá¸ùºÏÌú(¢ò)ÅäÀë×Ó ËÄÒìÁòÇè¸ù?¶þ°±ºÏîÜ(¢ó)Ëáï§ ¶þÂÈ»¯Ò»ÑÇÏõ NH3¡¢N¡¢O¡¢6 Cl NO2-¡¢Cl- [Ni(en)3]Cl2 [Fe(EDTA] 2-Ni2+ Fe2+ en EDTA4- N 6 +2 -2 -1 N¡¢O 6 NH4[Co(NCS)4(NH3)2] Co3+ NCS-¡¢NH3 Co3+ ONO-¡¢N¡¢N 6 [Co(ONO)(NH3)3(H2O)2]Cl2 O¡¢N¡¢6 +2 Ëá¸ù?Èý°±?¶þË®ºÏîÜ(¢ó) [Co(C2O4)3]3- Èý²ÝËá¸ùºÏîÜ(¢ó)ÅäÀë×Ó NH3¡¢H2O Co3+ C2O42- O O 6 -3 6£®ÊÔÈ·¶¨ÏÂÁз´Ó¦Ïò _ÓÒ____ ·½½øÐУº
ZnCO3(s) + 4CN- [Zn(CN)4]2-+ CO32-
7£®ÐγÉÅäλ¼üʱ£¬ÖÐÐÄÔ×Ó±ØÐë¾ßÓÐ _¿Õ¹ìµÀ_______£¬ÅäλÌ屨Ðë¾ßÓÐ _¹Â¶Ôµç×Ó_¡£
?8£®¸ù¾ÝËá¼îÖÊ×ÓÀíÂÛ£¬H2PO4?£»H2[PtCl6]£»HSO4£»[Fe(H2O)6]3+ µÄ¹²éî¼îµÄ
2?2?HPO£
4»¯Ñ§Ê½·Ö±ðÊÇ, H[PtCl6]- £¬SO4 ºÍ [Fe(H2O)5OH]2+ ¡£
9£®ÒÑÖªNH3µÄKbΪ1.76¡Á10-5£¬NH4Àë×ÓµÄKaֵΪ 5.68¡Á10 ¡£
-10
???10£®¸ù¾ÝËá¼îÖÊ×ÓÀíÂÛ£¬ÏÂÁÐÎïÖÊÖÐ
£
£
?NH4¡¢H3PO4 ¡¢H2S 2+
?3¡¢HSO3? ÊÇË᣻ PO4£
2?3¡¢CO¡¢
£¡¢
CN¡¢OH¡¢NO ÊǼ [Fe(H2O)5OH] Á½ÐÔÎïÖÊ¡£
?2¡¢HS?¡¢H2PO42?¡¢HPO4¡¢H2O ÊÇËÄ£®ÎÊ´ðÌâ
1£®ÈÜÒºµÄ·ÐµãÉý¸ßºÍÄý¹Ìµã½µµÍÓëÈÜÒºµÄ×é³ÉÓкιØÏµ£¿ 2£® ÔõÑùºâÁ¿»º³åÈÜÒº»º³åÄÜÁ¦µÄ´óС£¿ 3£®ÊÔÌÖÂÛÔõÑù²ÅÄÜʹÄÑÈܳÁµíÈܽ⡣
4£®ÊÔÓÃÆ½ºâÒÆ¶¯µÄ¹Ûµã˵Ã÷ÏÂÁÐÊÂʵ½«²úÉúʲôÏÖÏó¡£
(1)Ïòº¬ÓÐAg2CO3³ÁµíµÄÈÜÒºÖмÓÈëNa2CO3¡£ (2)Ïòº¬ÓÐAg2CO3³ÁµíµÄÈÜÒºÖмÓÈëÂÈË®¡£ (3)Ïòº¬ÓÐAg2CO3³ÁµíµÄÈÜÒºÖмÓÈëHNO3 5£®ÊÔ˵Ã÷ʲô½ÐòüºÏÎï¡£
6£®Ëá¼îÖÊ×ÓÀíÂÛÓëµçÀëÀíÂÛÓÐÄÄÐ©Çø±ð£¿
Î壮¼ÆËã
1£®ÔÚ100cm3Ë®(ÃܶÈΪ1.0g¡¤cm-3)ÖÐÈܽâ17.1gÕáÌÇ(C12H22O11)£¬ÈÜÒºµÄÃܶÈΪ1.0638g¡¤cm-3£¬ ÊÔ¼ÆË㣺
(1) ÈÜÒºµÄÖÊÁ¿·ÖÊý;
(2) ÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶È£» (3) ÈÜÒºµÄÖÊÁ¿Ä¦¶ûŨ¶È¡£ (4) ÕáÌǺÍË®µÄĦ¶û·ÖÊý¡£
wB? ½â£º(1) ÈÜÒºµÄÖÊÁ¿·ÖÊý£º
(2) ÒÑÖªÕáÌÇĦ¶ûÖÊÁ¿Îª£º342g¡¤mol-1£¬Ôò£º ÎïÖʵÄÁ¿Å¨¶È£º cB = 0.454mol¡¤dm-3
17.1g?100%?14.67.1g
17.1g/342g?mol?1?1bB??1000g?0.5mol?kg3?3100cm?1.0g?cm(3) ÖÊÁ¿Ä¦¶ûŨ¶È£º
100cm3?1.0g?cm?3n??5.56mol?118g?mol(4) Ħ¶û·ÖÊý£ºH2OµÄÎïÖʵÄÁ¿
n?
ÕáÌǵÄÎïÖʵÄÁ¿
17.1g?0.05mol?1342g?mol
xÌÇ?xH2O0.05mol?8.91?10?3(5.56?0.05)mol
5.56mol??0.991(5.56?0.05)mol
2£®½«0.450gijµç½âÖÊÈÜÓÚ30.0gË®ÖУ¬Ê¹ÈÜÒºÄý¹Ìµã½µµ½£0.150¡æ¡£¼ÆËã¸Ã·Çµç½âÖʵÄÏà
¶Ô·Ö×ÓÖÊÁ¿¡£ M = 186g¡¤mol-1
3£®Ä³Å¨¶ÈµÄÕáÌÇÈÜÒºÔÚ£0.250¡æÊ±½á±ù¡£´ËÈÜÒºÔÚ20¡æÊ±µÄÕôÆøÑ¹Îª¶à´ó£¿ÉøÍ¸Ñ¹Îª¶à
´ó£¿
ÕáÌǵÄÕôÆøÑ¹Îª£ºP(ÕáÌÇ) = 2327.53Pa
¡Ç(ÕáÌÇ)= 326.4kPa
4£®¼ÆËãÏÂÁÐÈÜÒºÖеÄb(H+)¡¢b(Ac-)ºÍ£¨1£©¡¢£¨2£©µÄ½âÀë¶È¦Á£º
-1
£¨1£©0.050mol¡¤kg HAcÈÜÒº£» £¨2£©0.10mol¡¤kg -1 HAcÈÜÒºÖмÓÈëµÈÖÊÁ¿µÄ0.050 mol¡¤kg-1 KAcÈÜÒº£» £¨3£©0.10 mol¡¤kg-1 HAcÈÜÒºÖмÓÈëµÈÖÊÁ¿µÄ0.05 mol¡¤kg-1 HClÈÜÒº£» £¨4£©0.10 mol¡¤kg-1 HAcÈÜÒºÖмÓÈëµÈÖÊÁ¿µÄ0.05 mol¡¤kg-1 NaOHÈÜÒº¡£
£¨1£©
9.4?10?4???100%?1.88%0.05
£¨2£© ¦Á=0.07%
£¨3£© b(Ac-)=3.52¡Á10-5 mol¡¤kg-1
+-5 -1-£¨4£©b(H)£½1.76¡Á10mol¡¤kg b(Ac)=0.025mol¡¤kg-1
5£®0.010mol¡¤kg-1 µÄijһÈõËáÈÜÒº£¬ÔÚ298Kʱ£¬²â¶¨ÆäpHֵΪ5.0£¬
??KK?aaÇ󣺣¨1£©¸ÃËáµÄºÍ¡££¨2£©¼ÓÈë1±¶Ë®Ï¡ÊͺóÈÜÒºµÄpHÖµ¡¢ºÍ?¡£
½â£º£¨1£©
?Ka??10-5?20.010
10?5???100%?0.10%0.01
£¨2£©
?Ka??1.0?10?8?7.071?10??62?6
0.005?7.071?10
?67.071?10???100%?0.14%0.005 )
?1.0?10?86£®¼ÆËã20¡æÊ±£¬ÔÚ0.10 mol¡¤kg-1ÇâÁòËá±¥ºÍÈÜÒºÖУº +2-£¨1£©b(H)¡¢b(S)ºÍpH£»
2-£¨2£©ÈçÓÃHClµ÷½ÚÈÜÒºµÄËá¶ÈΪpH£½2.00ʱ£¬ÈÜÒºÖеÄSŨ¶ÈÊǶàÉÙ£¿¼ÆËã½á¹û˵Ã÷
ʲôÎÊÌ⣿ ½â£º£¨1£©b(H+) = b(HS-) = 9.5¡Á10-5 mol¡¤kg-1 pH = -lg9.5¡Á10-5 = 4.02 £¨2£©b(S2-)= 1.0¡Á10-16 mol¡¤kg-1
7£®ÔÚ18¡æÊ±£¬PbSO4µÄÈܶȻýΪ1.82¡Á10-8£¬ÊÔÇóÔÚÕâ¸öζÈPbSO4ÔÚ0.1mol¡¤kg-1 K2SO4ÈÜ
ÒºÖеÄÈܽâ¶È¡£
b(Pb2+) = 1.82¡Á10-7 mol¡¤kg-1
8£®ÔÚ18¡æÊ±£¬AgBrµÄÈܶȻýΪ5.35¡Á10-13£¬ÔÚ´¿Ë®ÖÐAgBrµÄÈܽâ¶ÈÊǶàÉÙ£¿ÔÚ0.10mol¡¤kg-1
µÄNaBrÈÜÒºÖÐAgBrµÄÈܽâ¶ÈÊǶàÉÙ£¿
?K2b(S) ¡Ö= 1.1¡Á10-12 mol¡¤kg-1 2-
¹²·ÖÏí92ƪÏà¹ØÎĵµ