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Gio n ha hc 10 Ban c bn &... Trng THPT BC Nam Sch

HC K ITrang

Mc lc 1Tit 1 n tp u nm 3Tit 2 n tp u nm (tip theo) 7

CHNG I : NGUYN T 10

Tit 3 Thnh phn nguyn t 10Tit 4 Ht nhn nguyn t - Nguyn t ho hc - ng vTit 5 Ht nhn nguyn t - Nguyn t ho hc - ng v (tit theo)Tit 6 Luyn tp: Thnh phn nguyn tTit 7 Cu to v electron ca nguyn tTit 8 Cu to v electron ca nguyn t (tip theo)Tit 9 Cu hnh electron ca nguyn tTit 10 Luyn tp: Cu to v electron ca nguyn tTit 11 Luyn tp: Cu to v electron ca nguyn t (tip theo)Tit 12 Kim tra vit (45 pht)CHNG II : BNG TUN HON CC NGUYN T HO HC NH LUT

TUN HON

Tit 13 Bng tun hon cc nguyn t ho hcTit 14 Bng tun hon cc nguyn t ho hc (tip theo)Tit 15 S bin i tun hon cu hnh electron nguyn t cc nguyn t ho

hcTit 16 S bin i tun hon tnh cht cc nguyn t ho hc - nh lut

tun honTit 17 S bin i tun hon tnh cht cc nguyn t ho hc - nh lut

tun hon (tip theo)Tit 18 ngha ca bng tun hon cc nguyn t ho hcTit 19 Luyn tp: Bng tun hon, s bin i tun hon cu hnh electron

nguyn t v tnh cht cc nguyn t ho hcTit 20 Luyn tp: Bng tun hon, s bin i tun hon cu hnh electron

nguyn t v tnh cht cc nguyn t ho hc (tip theo)Tit 21 Kim tra vit (45 pht)

CHNG III : LIN KT HO HC

Tit 22 Lin kt ion Tinh th IonTit 23 Lin kt cng ho trTit 24 Lin kt cng ho tr (tip theo)Tit 25 Tinh th nguyn t - Tinh th phn tTit 26 Ho tr v s oxi hoTit 27 Luyn tp: Lin kt ho hcTit 28 Luyn tp: Lin kt ho hc (tip theo)

CHNG IV : PHN NG OXI HO - KH

Tit 30 Phn ng oxi ho - khTit 30 Phn ng oxi ha - kh (tip theo)Tit 31 Phn loi phn ng trong ho hc v c

Copyright Ng Xun QunhCopyright Ng Xun Qunh [email protected]@gmail.com

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Tit 33 Luyn tp: Phn ng oxi ho khTit 33 Luyn tp: Phn ng oxi ho kh (tip theo)Tit 34 Bi thc hnh s 1: Phn ng oxi ho - kh (kim tra TH)Tit 35 n tp hc k ITit 36 Kim tra hc k I (45 pht)

HC K II

CHNG V : NHM HALOGEN

Tit 37 Khi qut v nhm halogenTit 38 CloTit 39 Hiro clorua Axit clohiric v mui - Luyn tpTit 40 Hiro clorua Axit clohiric v mui - Luyn tp (tip theo)Tit 41 Bi thc hnh s 2: Tnh cht ho hc ca kh clo v hp cht ca cloTit 42 S lc v hp cht c oxi ca cloTit 43 Flo Brm Iot

Tit 44 Flo Brm Iot (tip theo)Tit 45 Luyn tp: Nhm halogenTit 46 Luyn tp: Nhm halogen (tip theo)Tit 47 Bi thc hnh s 3: Tnh cht ho hc ca Brom v IotTit 48 Kim tra vit (45 pht)

CHNG VI : OXI LU HUNH

Tit 49 Oxi Ozon - Luyn tpTit 50 Oxi Ozon - Luyn tp (tip theo)Tit 51 Lu hunh

Tit 52 Bi thc hnh s 4: Tnh cht ca oxi , lu hunhTit 53 Hiro sunfua Lu hunh ioxit Lu hunh trioxitTit 54 Hiro sunfua Lu hunh ioxit Lu hunh trioxit (tip theo)Tit 55 Axit sunfuric - Mui sunfatTit 56 Axit sunfuric - Mui sunfat(tip theo)Tit 57 Luyn tp: Oxi v lu hunhTit 58 Luyn tp: Oxi v lu hunh (tip theo)Tit 59 Bi thc hnh s 5: Tnh cht cc hp cht ca lu hunh (kim tra

TH)

Tit 60 Kim tra vit (45 pht)CHNG VII : TC PHN NG V CN BNG PHN NG

HO HC

Tit 61 Tc phn ng ho hcTit 62 Tc phn ng ho hc (tip theo)Tit 63 Bi thc hnh s 6: Tc phn ng ho hcTit 64 Cn bng ho hcTit 65 Cn bng ho hc (tip theo)Tit 66 Luyn tp: Tc phn ng v cn bng ho hc

Tit 67 Luyn tp: Tc phn ng v cn bng ho hc (tip theo)Tit 68 n tp hc k IITit 69 n tp hc k II (tip theo)Copyright Ng Xun QunhCopyright Ng Xun Qunh [email protected]@gmail.com

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Tit 70 Kim tra hc k II (45 pht)

Tit 1: N TP U NM

Son ngy: 31/09/2007

A. Mc tiu bi gingI. Kin thcQua bi ging hc sinh c c c nhng im sau: Cch h thng li cc

kin thc ha hc c bn c hc THCS c lin quan trc tip ti chngtrnh ha hc lp 10.

+ Nguyn t l g? Cu to ca nguyn t nh th no?+ Nguyn t ha hc, ha tr v cch xc nh ha tr ca nguyn t.+ nh lut bo ton khi lng, mol, t khi ca cht kh.

II. K nng- Rn luyn k nng lm mt s bi tp c lin quan n cu to nguyn t- Rn luyn k nng lp cng thc, tnh ton theo cng thc v phng trnh phnng, t khi ca cht kh.- Rn luyn k nng chuyn i gia khi lng mol (M), khi lng cht (m), smol (n), th tch kh ktc (V), v s mol phn t cht (A).

III. Thi - tnh cm- Gy hng th, ham thch hc tp mn ha hc.- Rn luyn thc cn thn, trung thc, kin tr, t m, chnh xc trong cng vic.- C thc trch nhim i vi bn thn, gia nh v x hi.

B. dng hc tp phng php

I. dng hc tp1. Gio vin- Gio n- H thng bi tp v cu hi gi n tp2. Hc sinh- n tp li cc kin thc c lin quan trong chng trnh ha hc lp 8.

II. Phng phpPhng php vn p ti hinC. Tin trnh ging dy

Hot ng ca gio vin Hot ng ca hc sinhHot ng 1

n nh t chc lp v kim tra s sHot ng 2

Vo bi

Tit 1 : N TP U NM

Hot ng 3

1. Nguyn t


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(?) Da vo nhng kin thc hc lp 8, cc em hy cho bit:- Nguyn t l g?

- L ht v cng nh b to nn cc cht.- Nguyn t gm c: ht nhn mang intch dng v c lp v c mt hay nhiuelectron mang in tch m.

Hot ng ca gio vin Hot ng ca hc sinh

- Cu to ca nguyn t?

- c im ca cc ht cu to nnnguyn t?

+ Nghe kin ca HS

+ Nhn xt v a ra kt lun licui cng

- Cu to ca nguyn t:+ Ht nhn nguyn t mang in tchdng nm tm nguyn t, gm ht

proton (p) v ht ntron (n)* Ht p, in tch 1+, khi lng ln hnkhi lng e khong 1836 ln. Trongnguyn t, s ht p bng s ht e.* Ht n, khng mang in, c khi lng

bng khi lng ca ht p.+ V nguyn t gm cc ht e mang intch m (k hiu l e). Cc e c khi lngrt nh b so vi khi lng ca nguyn t

Hot ng 42. Nguyn t ha hc

(?) Da vo cc kin thc hc, emhy cho bit nguyn t ha hc l g?

- Tnh cht ha hc ca cc nguynt thuc cng mt nguyn t ha

hc?

+ Lng nghe, suy ngh tm cch gii quyt- L tp hp nhng nguyn t c cng sht proton trong ht nhn.- Nhng nguyn t ca cng mt nguynt ha hc u c tnh cht ha hc ging

nhau.Hot ng 5

3. Ha tr ca mt nguyn t(?) Ha tr ca mt nguyn t hahc l g?

(?) Cch xc nh ha tr ca mtnguyn t ha hc?

- L con s biu th kh nng lin kt canguyn t nguyn t ny vi nguyn tca nguyn t khc.- Cch xc nh ha tr: trong hp chtAxBy, tch ca ch s v ha tr ca nguynt ny bng tch ch s v ha tr ca

nguyn t kia: a bx yA B a x b y = - Bit c gi tr 3 i lng => ilng th 4

(?) Yu cu HS xc nh ha tr cacc nguyn t trong hp cht Na2O,CH4, SO3, NH3 (bit ha tr ca O l2 v H l 1)?

HS: Lm bi tp vo v

* VD:

2

2

1

4

2

2

1

3

2 2 1 1

1 1 4 4

.1 2.2 4

.1 1.3 3

x

x

x

x

Na O x x

C H x x

S O x x

N H x x

= => =

= => =

= => =

= => =

Hot ng 6


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4. nh lut bo ton khi lng(?) Nu ni dung ca nh lut boton khi lng?

- Ly v d minh ha.

- Trong mt phn ng ha hc, tng khilng ca cc cht sn phm bng tngkhi lng ca cc cht phn ng.VD: Trong phn ng ha hc


(?) Yu cu HS lm bi tp p dng:BT: Cho 1,4 gam hn hp gm Mgv Fe tc dng va vi vi 160ml dung dch HCl 2M, sau phn ngta thu c 0,896 lt H2 ktc. Tnhkhi lng mui khan thu c.

+ Hng dn HS lm bi- Nhn xt v phn ng

- Tnh khi lng ca HCl, H2- p dng nh lut bo ton khilng+ Nhn xt v cho im bi lm cahc sinh.

3 26 2 2 3HCl Al AlCl H + +

Ta c:3 2 HCl Al ACl H

m m m m + = +

HS: Lm bi tp p vo vPhng trnh phn ng:

2 22Mg HCl MgCl H + +

2 22Zn HCl ZnCl H + +

Ta c2

0,896. . .2 0,08

22, 4 22,4HV

m n M M g = = = =

. . . 2.0,16.36,5 11,68 HCl M m n M C V M g = = = =Theo nh lut bo ton khi lng:

2HClkk muoi H

m m m m+ = +

=> 213,44muoi kl HCl H m m m m g

= + =

Hot ng 75. Mol

GV: Yu cu hc sinh nhc li khinim:- Mol l g?

- Khi lng mol l g?

- Khi nim th tch mol cht kh?

GV: Yu cu HS a ra mi quan hgia:- Khi lng cht (m) khi lngmol (M).- Khi lng cht (m) s mol (n).- Khi lng mol (M) s mol (n)

- S mol (n) th tch cht kh (V)- S mol (n) S phn t, snguyn t (A)

HS: Nhc li khi nim mol, khi lngmol, th tch mol.* Mol l lng cht cha 6.1023 nguyn thoc phn t cht .Con s 6.1023 gi l s Avogaro v ck hiu l N = 6.1023.* Khi lng mol (k hiu M) ca mtcht l khi lng tnh bng gan ca Nnguyn t hoc phn t.VD: MO = 16g; MH = 1g; 2OM = 32g;* Th tch mol ca cht kh l th tchchim bi N phn t ca cht kh - ktc, th tch mol ca cc cht kh ul 22,4 lt.HS: a ra mi lin h

mn

M=

.m n M

mM

n

=

=( )

22,4khiV l

n = 22,4.V n =

.A

n A N nN

= =


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(?) Yu cu HS lm bi tp: Hy tnhth tch ( ktc) ca hn hp ccha 1,1 gam CO2 v 1,6 gam O2.+ Hng dn HS lm bi tp- Tnh s mol ca cc cht CO2, O2- Tnh s mol hn hp => Vhh

HS: Lm bi tp vo v.

Ta c2

2

1,10,025( )

441,6

0,05( )32

CO

O

mn mol

Mm

n mol M

= = =

= = =

Vy th tch hn hp l: 1,68( )hhV l=

Hot ng ca gio vin Hot ng ca hc sinhHot ng 8

6. T khi ca cht khGV: Yu cu hc sinh nhc li khinim:- T khi ca cht kh l g?

- Cng thc tnh t khi ca kh A sovi kh B. Gii thch cc k hiu c

trong biu thc tnh.(?) Yu cu HS lm bi tp:- Tnh t khi ca kh CH4, SO2 sovi Hiro ?- Tnh t khi ca kh Cl2, SO3 so vikhng kh ?+ Hng dn HS lm bi+ Nhn xt bi lm ca SH

HS: nhc li khi nim* T khi ca kh A i vi kh B cho bitkh A nng hay nh hn kh B bao nhiuln.

- Cng thc tnh:.

.A A A A

AB

B B B B

m M n M d

m M n M = = =

HS: lm bi tp vo v.4

4

2 2

2

2

22

168

2

4422

2

CH

CH

H H

CO

COH H

Md

M

Md

M

= = =

= =

;

2

2

3

3

712.45

29

802.76

29

Cl

Cl

kk kk

SO

SO

kk kk

Md

M

Md

M

= = =

= = =

Hot ng 9

Cng c kin thc Bi tp v nh(?) HS h thng ha li cc kin thcquan trng trong bi.- Hng dn hc sinh lm cc bitp.- Nhc hc sinh ni dung cc kinthc s n tp tit 2 v yu cu HSv n tp cc ni dung sau:1. Cc cng thc v dung dch nh:

tan, nng C%, nng CM2. S phn loi cc hp cht v c3. Bng tun hon

HS: H thng ha li cc kin thc quantrng hc trong bi.- Lm cc bi tp theo s hng dn cagio vinHS: Lng nghe cc ni dung cn n tp tit 2 v nh chun b.

NHN XT RT KINH NGHIM


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Tit 2 : N TP U NM (tip theo)

Son ngy 31/09/2007

A. Mc tiu bi gingI. Kin thcHS s tip tc n v cng c li cc kin thc c bn m HS c hc THCS.

- V dung dch ( tan, nng C%, CM) v s dng thnh tho cc cng thc tnh tan, nng C%, nng CM, khi lng ring ca dung dch.- Cc loi cht v c, tnh cht ha hc ca cc cht v c (axit, baz, mui, oxit)- Kin thc v bng h thng tun hon ( nguyn t, chu k, nhm)

II. K nng- Rn luyn k nng lm mt s bi tp c lin quan n dung dch- Vit cc phng trnh phn ng ha hc- K nng lm mt s bi tp lin quan n cu to nguyn t- K nng t duy, phn on, trnh by mt bi ton ha hc

III. Thi d - tnh cm- Thy c vai tr v ngha ca mn ha hc- Gip hc sinh thm yu thch mn ha hc hn.

B. dng hc tp phng phpI. dng hc tp1) Gio vin- Gio n, h thng bi tp v cu hi gi .2) Hc sinh- n tp li cc ni dung m GV nhc nh tit trc.- Gii mt s bi tp vn dng theo s ch o v hng dn ca gio vin.

II. Phng phpPhng php vn p ti hinC. Tin trnh ging dyHot ng ca gio vin Hot ng ca hc sinh

Hot ng 1n nh t chc lp v kim tra s s

Hot ng 2Vo bi

Tit 2: N TP U NM (tip theo)


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Hot ng 3I. Dung dch (15 pht)

(?) yu cu HS tho lun h thng licc ni dung sau:- tan ca mt cht trong nc l g?

- Tho lun a ra cu tr li:- tan ca mt cht trong nc (S)c tnh bng s gam ca cht hatan trong 100g nc to thnh dung

dung dich bo ha mt nhit xcnh

Hot ng ca gio vin Hot ng ca hc sinh- Cng thc tnh tan

- Nhng yu t nh hng n tanca mt cht trong nc?- Nu cc cng thc tnh nng cadung dch m cc em bit?

+ Nng phn trm (C%) ?+ Nng mol (CM) ?

(?) Yu cu HS lm bi tp p dngHa tan 16g NaOH vo nc thuc 200ml dung dch:

a. Tnh nng mol ca dd NaOH?

b. Cn dng bao nhiu gam dung dchaxit H2SO4 19,6% trung ha ht 50mldung dch NaOH ni trn?+ Hng dn hc sinh lm bi

+ Nhn xt bi lm ca HS

.100( )t

dm

mS g

m=

- Cht rn: S tng khi 0t tng- Cht kh: S tng khi 0t gim, p tng- Cc cng thc tnh nng dung dch:+ Nng phn trm (C%)

dd

dd

dd

. %100%% .100%

.100%

%

ct

ct

ct

m Cmm

Cmm

mC

==

=

+ Nng mol (CM) ( )Mn

CV l

=

HS: Lm bi theo s hng dn ca GVa. S mol NaOH trong 200ml dd NaOH

160,4

40NaOHm

n mol M

= = =

0,42

0,2NaOH

M

nC M

V=> = = =

b. Phn ng trung ha2 4 2 4 2

2 2NaOH H SO Na SO H O+ +

. 2.0,05 0,1pu NaOH M

n C V mol = = =

Theo phng trnh phn ng ta c

2 4

1 1. .0,05 0,1

2 2 H SO NaOH n n mol = = =

Vy 2 4 . 0, 05.98 4, 9 H SOm n M g = = =

Hot ng 4S phn loi cc hp cht v c(15 pht)

(?) Hy k cc loi hp cht v c c hc bc THCS?- Ly v d minh ha

- Nu nhng tnh cht ha hc ctrng

Hp cht v c c chia lm 4 loi:+ Oxit (gm oxit baz v oxit axit)- Oxit baz: Tc dng vi dd axit tothnh mui v nc.VD: Na2O; Fe2O3, - Oxit axt: Tc dng vi dd baz tothnh mui v ncVD: SO2; P2O5,


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- Vit phng trnh phn ng minh ha

- Nhn xt cc phng trnh phn ngm HS vit.

+ Axit: tc dng vi baz to thnhmui v ncVD: H2SO4, HNO3, + Baz: tc dng vi axit to thnhmui v ncVD: NaOH, Fe(OH)2,

+ Mui: c th tc dng vi axit to snphm l mui mi v axit mi, c thtc dng vi dung dch baz to sn

Hot ng ca gio vin Hot ng ca hc sinh- H thng ha li phm l mui mi v baz mi.

VD: NaCl, MgCl2, CaCO3, Hot ng 5

Bng tun hon cc nguyn t ha hc (10 pht)(?) Hy cho bit cu to ca bng tun

hon cc nguyn t ha hc v ngha?+ nguyn t l g?+ Chu k l g?+ Nhm l g?

(?) Yu cu HS lm bi tpNguyn t A trong bng HTTH c shiu nguyn t l 12, hy cho bit:a. Cu to nguyn t ca nguyn t A

b. V tr ca A trong bng tun hon.c. Tnh cht ha hc c trng canguyn t A.

- nguyn t cho bit s hiu nguyn

t, k hiu ha hc, tn nguyn t,nguyn t khi ca nguyn t .- Chu k l dy cc nguyn t mnguyn t ca chng c cng s lp ev c sp xp theo chiu tng cain tch ht nhn.- Nhm gm cc nguyn t m nguynt ca chng c cng s e lp ngoicng.

- Lm bi tp theo s hng dna. Cu to nguyn t ca nguyn t A+ Ht nhn c in tch 12++ Trong nhn c 12 proton+ Lp v gm c 12 electron

b. V tr ca A trong bng tun hon- S th t 12- Nhm IIA

+ Chu k 3

c. Tnh cht ha hc c trng l tnh

kim loiHot ng 6Cng c kin thc Bi tp v nh (3 pht)

- Yu cu HS nhc li cc kin thctrng tm hc trong bi.- Nhc HS v n li cac kin thc trngtm c bn ca lp 8, 9 chun b chochng trnh ha hc 10- Yu cu hc sinh v nh lm mt s

bi tp cng c kin thc.

- Nhc li cc kin thc trng tm cnnh trong bi hc- V n li cc kin thc theo s ch oca gio vin

- V nh lm bi tp

NHN XT RT KINH NGHIM


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CHNG I:CHNG I:

NGUYN TNGUYN T

Tit 3: THNH PHN NGUYN T

Son ngy: 04/09/2007

A. Mc tiu bi gingI. Kin thc- Hc sinh hiu c nguyn t cha phi l cu to nh nht ca vt cht.- Nguyn t c cu to phc tp, c cu to bi cc ht nh hn l eeleectron,

proton v ntron.- Nguyn t v cc ht u c khi lng, kch thc v u mang in tr htntron khng mang in v nguyn t trung ha v in

II. K nng- C k nng quan st m hnh hay th nghim m phng, phn tch hin tng rtra kt lun v cu to nguyn t.- Hc sinh bit cch s dng cc n v o lng nh: u, ddvddt, nm, A

o

- Bit cch lm mt s bi tp c lin quanIII. Tnh cm thi - Phn bit c th gii v m v th gii vi m.- hiu c th gii vi m phi t duy trn c s cc kt qu th nghim v cckt qu tnh ton rt ra kt lun.B. dng hc tp Phng Php

I. dng hc tp1. Gio vin- Gio n, h thng cu hi nhm phc v cho bi hc- S tm tt th nghim tm ra tia m cc- M hnh th nghim khm ph ra nguyn t ( nu c)2. Hc sinh- Sch gio khoa ha hc lp 10- Xem nhng kin thc lin quan n phn nguyn t

II. Phng php- Phng php m thoi gi m kt hp dng cc dng dy hc trc quan.


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C. Tin trnh ging dyHot ng ca gio vin Hot ng ca hc sinh

Hot ng 1n nh t chc lp - kim tra s s

Hot ng 2Vo bi

GV t vn : T trc Cng nguyn n th k XIX ngi ta cho rng cc chtu c to nn t nhng ht cc k nh b khng th phn chia c na gi lnguyn t. Ngy nay, ngi ta bit rng nguyn t c cu to phc tp: gm c ht


nhn mang in tch dng v lp v electron mang in tch m.Tit 1: THNH PHN NGUYN T

Hot ng 3I. Thnh phn cu to ca nguyn t

1. Electrona) S tm ra electron

GV t vn : mc ch ca th nghim:Nguyn t khng chia nh hn c nahay nguyn t c to nn t nhng

phn t cn nh hn n. Gi thit

nguyn t c to nn t nhng phnt cn nh hn n th t phi c hintng th hin. Hin tng l g ?GV m t, hng dn HS tm hiu cuto thit b phng in ca th nghimtrong hnh 1.3 (SGK).(?) GV thuyt trnh v th nghim tm ratia m cc. T hin tng xy ra, ta rt

ra c iu g ?GV kt lun: Tia m cc l mt trong scc chng c chng t nguyn t c cuto phc tp.(?) Tia m cc c phi l vt cht cthc hay khng, lm sao chng minhc n ?(?) Trn ng i ca tia m cc nu tat mt chong chng nh thy chongchng quay chng t hin tng g ?GV kt lun: Tia m cc l chm ht vtcht c thc chuyn ng rt nhanh.

- HS nghe v c SGK tm hiu v thnghim ca Tm-xn.

- Mn hunh quang trong ng pht sngdo nhng tia pht ra t cc m v gi l

tia m cc

- HS: tia m cc l vt cht c thc, nc th lm quay chong chng nh v nchuyn ng vi vn tc ln.


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- GV t vn v ht vt cht c trongtia m cc c mang in hay khng.Mang in dng hay in m. Lm thno chng minh c iu ny ?- Minh ha th nghim Tia m cclch v pha bn cc dng.Vy tia m cc l chm ht mang indng hay m ?GV kt lun: Ngi ta gi nhng ht tothnh tia m cc l electron. Electron cmt mi cht, n l mt trong nhngthnh phn cu to nn nguyn t ca

- HS c th t ng phng tia m ccgia hai bn cc mang in tri du. Nu tia m cc mang in th n philch v pha bn cc mang in tri du.

HS: Tia m cc l chm ht mang inm.HS tia m cc l chm ht electron Electron to nn lp v nguyn t cami nguyn t ha hc.


mi nguyn t ha hc.

b) Khi lng v in tch ca electron(?) Yu cu HS c v ghi khi lng vin tch electron vo v.GV Electron c in tch m v c gi tr

191,602.10e

q C= , l in tch nhnht nn c dng lm in tch n v,k hiu l oe

HS: khi lng 319,1094.10 ( )em kg=

in tch 191,602.10 1eq C= = (culng)

Hot ng 4

2. S tm ra ht nhn nguyn tGV t vn : trn chng ta bit nguyn t cha cc ht e mang in tch mm nguyn t th trung ha v in. Vy chc chn phi cha nhng phn t mangin tch dng. Phn mang in tch dng ny phn tn trong c nguyn t haytp trung mt vng no ca nguyn t ? Lm th no chng minh ?.

GV m t th nghim hnh 1.4 (SGK)ca R-d-pho. Dng ht bn phmt l vng mng kt qu th nghim:- Hu ht cc ht u xuyn thng qua

tm vng mng.- C mt s t ht i lch hng ban u.- C s t ht b bt tr li pha sau khigp l vng.(?) Kt qu th nghim ni nn iug ?.

GV hng dn HS kt lun vn :GV nhn mnh: Nguyn t phi cha

phn mang in dng tm l ht nhnc khi lng ln, nhng li c kch

HS nghe GV m ta v xem trong SGK hiu v th nghim.

HS: Hu ht cc ht xuyn qua tm

vng mng chng t nguyn t khngphi l nhng ht c kht m c cu torng.- Cc ht tch in dng, chng lchng i hoc b bt tr li chng ngn cc phn t tch in dng nn by.- V ch c mt phn rt nh cc ht blch hng cc ht tch in dng

trong nguyn t gy nn va chm chchim mt th tch rt nh trong nguyn


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thc rt nh so vi kch thc canguyn t. Do vy, nguyn t c cu torng. Xung quanh ht nhn c cc e tonn nguyn t. S n v in tchdng ca ht nhn ng bng s e quayquanh ht nhn. Khi lng ca nguynt hu nh tp trung ht nhn.

t.

HS ghi kt lun.

Hot ng 53. Cu to ca ht nhn nguyn t

GV t vn : Ht nhn nguyn t l phn t khng cn phn chia c na hayht nhn ht nhn c cu to t nhng ht nh hn. Lm th no chng minh?

a) S tm ra proton


GV m t th nghim ca R-d-pho

nm 1918: Bn ht vo ht nhnnguyn t nit th xut hin ht nhnnguyn t oxi v ht proton mang dindng (1+), khi lng l 271,6726.10 kg chnh l ht proton (p)

14 4 1 17

7 2 1 8N He H O+ +

GV kt lun: ht p l thnh phn cu toca ht nhn nguyn t.(?) khi lng v in tch ht nhn ca

proton l bao nhiu ?

HS ghi kt lun v nhn xt.- Ht nhn p l mt thnh phn cu to

ca ht nhn nguyn t

-19

27

1,602.10 11,6726.10 1

e

p

q Cm kg u

= = +=

b) S tm ra ntron

GV m t th nghim ca Chat-uchnm 1932: bn ht vo ht nhnnguyn t beri th xut hin ht nhnnguyn t cacbon v ht khng mangin: ht ntron

9 4 1 12

4 2 0 6 Be He n C + +

HS nghe v ghi thng tin

c) Cu to ca ht nhn nguyn t

(?) T cc th nghim trn, hy kt lunv cu to ca ht nhn nguyn t ?

HS nu kt lun (SGK Tr.7)

Hot ng 6II. Kch thc v khi lng nguyn t

1. Kch thc

GV hng dn HS nghin cu SGK tm hiu v kch thc ca nguyn t. HS cn nh:- Nguyn t ca cc nguyn t kkhascnhau th c kch thc khc nhau.


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GV lu cc em HS: Vi t l v kchthc nh trn ca nguyn t v htnhn th cc e rt nh b chuyn ngxung quanh ht nhn trong khng gianrng ca nguyn t.

- Nu hnh dung nguyn t nh mt qucu, trong c cc e chuyn ng rtnhanh xung quanh ht nhn th n cng knh khong 10-10m. Con s nyrt nh, do vy cc nh khoa hc ara n v nanomet (nm) hay n vangsstrom ( 0A )

0 09 10 81 10 ;1 10 ;1 10 10nm m nm A A m cm = = = =

- ng knh ca nguyn t khong10-10nm, ng knh ca ht nhn cnnh hn n khong 10-5nm v ngknh ca e, p cn nh hn nhiu, khong10-8nm


Hot ng 72. Khi lng

biu th khi lng ca nguyn t,,phn t v cc ht p, n v e ngi tadng n v khi lng nguyn t khiu l u cn gi l vC.(?) Vy u l g ? v n bng bao nhiu ?GV nhn mnh rng cng thc (1) dng

chuyn i gia n v u v g hocngc li.

(?) Tnh khi lng nguyn t tng ica mt nguyn t H bit

271,67.10Hm g=

HS: 1u l1

12khi lng ca mt nguyn

t ng v cacbon.272719,9206.101 1,6605.10

12

kgu kg

= = (1)

HS: Theo 1 ta c24

24

1,67.10( ) 1

1,66.10KLNT H u

=

Hot ng 8Cng c bi

(?) HS nhc li cc kin thc trng tmca bi hc- HS v lm cc bi tp 1, 2, 3, 4, 5(SGK)

- HS nhc li cc kin thc trng tm cabi hc- HS v lm bi tp

RT KINH NGHIM SAU GI DYRT KINH NGHIM SAU GI DY

Copyright Ng Xun QunhCopyright Ng Xun Qunh netthubuon@gmail comnetthubuon@gmail com

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