Lop 9 - 2011-Hoc Ki 1-Chinh Thuc
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Transcript of Lop 9 - 2011-Hoc Ki 1-Chinh Thuc
S GIO DC V O TO
S GIO DC V O TO
THNH PH NNG
KIM TRA HC K I - NM HC 2010 - 2011MN: HA HC LP 9
Thi gian: 45 pht (khng k thi gian giao )
CHNH THC
(a) Vit phng trnh ha hc thc hin chuyn ha, ghi r iu kin nu c:
(b) Nu ng dng ca KNO3 trong nng nghip.
(a) Vit phng trnh ha hc nu c xy ra khi cho Fe ln lt tc dng vi Cl2, dung dch HCl, dung dch Cu(NO3)2.
(b) Lm th no tinh ch dung dch mui Cu(NO3)2 c ln AgNO3? Vit phng trnh ha hc minh ha.
Cu 3 (2 im)
(a) Trnh by phng php phn bit hai dung dch NaOH v Ba(OH)2. Vit phng trnh ha hc minh ha.
(b) Th no l gang? Vit cc phng trnh ha hc chnh xy ra trong qu trnh luyn gang (khng vit qu trnh to x)
Cu 4 (2 im)
(a) Tnh khi lng NaOH v khi lng nc cn ly pha c 200 gam dung dch NaOH 10%.
(b) Dung dch NaOH trn trung ha va vi 200 ml dung dch H2SO4. Tnh nng mol ca dung dch H2SO4. Cu 5(2 im) Ha tan 8 gam Cu trong mt lng va V ml dung dch H2SO4 98% (m c) thu c V1 lt SO2 iu kin tiu chun. Tnh V1 v V, bit khi lng ring dung dch H2SO4 98% l 1,84 g/ml. Cho Na = 23, O = 16, H = 1, Cu = 64, S = 32
--------HT--------Ch : Hc sinh c dng bng H thng tun hon, bng tnh tan v my tnh c nhn theo quy nh ca B Gio dc v o to. S GIO DC V O TO
THNH PH NNG
KIM TRA HC K I - NM HC 2010 - 2011HNG DN CHM MN HA HC 9
P NIM
Cu 1 (2 im) (a)Vit ng cc phn ng (1-3) , mi phn ng 0,5 , phn ng (4) : 0,25
(b) Lm phn bn, cung cp nguyn t nit v kali cho cy trng: 0,25 1,75 0,25
Cu 2 (2 im)(a)Vit ng mi phn ng 0,25
(b) Ngm 1 thanh Cu d vo dung dch hn hp, AgNO3 s b loi Cu + 2AgNO3 ( Cu(NO3)2 + 2Ag 1,5
0,25
0,25
Cu 3 (2 im) (a)Trch mu th, cho hai mu th tc dng vi dung dch H2SO4, mu th tc dng sinh kt ta trng cha Ba(OH)2, mu th cn li cha NaOH Ba(OH)2 + H2SO4 ( BaSO4 + 2H2O
Hc sinh c th dng CO2 hoc mui sunfat thay cho H2SO4 (b)Nu ng nh ngha Vit ng cc phng trnh ha hc : C ( CO2 , CO2 ( CO , CO kh Fe2O3 0,5
0,5 0,25 0,75
Cu 4 (2 im); (a) Khi lng NaOH cn ly : mNaOH = = 20 (g) Khi lng nc = 200 20 = 180 (g)
(b) PTHH (1) 2NaOH + H2SO4 ( Na2SO4 + 2H2O
T (1) (mol)
CM ( H2SO4) = = 1,25 (M) 0,5
0,25
0,50,5 0,25
Cu 5 (2 im):
nCu = 8 : 64 = 0,125 (mol) PTHH : (1) Cu + 2H2SO4 CuSO4 + SO2 + 2H2O T (1) ( s mol SO2 sinh ra = s mol Cu = 0,125 (mol)
V1 = 0,125 ( 22,4 = 2,8 (lt)T (1) ( s mol H2SO4 cn dng = 2 s mol Cu = 2(0,125 = 0,25 (mol)
Th tch dung dch H2SO4 cn dng = V = = 13,59 (ml) PTHH: 0,5Tnh ra V1 0,5
Tnh ra V
1,0
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