Cu 4 : Tnh khi lng ring ca NaCl (g/ ) bit

1) L Th Thanh Hoi 72) Dng Th Huy 93) Nguyn Th Ngc Linh 104) Hong Anh Nam 135) Dng Cng Vnh Phc 186) Nguyn Tr Ti 227) Nguyn Nht Tn 238) Nguyn Th Thm 299) Nguyn Th Thu Trc 3210) L T Tun 33

Cu 1: Silic c cu trc tinh th ging kim cng vi thng s mng a = 0,534 nm. Tnh bn knh nguyn t cng ha tr ca Silic v khi lng ring ca n, bit = 28,086 v kim cng c cu trc lp phng tm din, c 4 nguyn t nm 4 hc t din ca mng c s.Gii

S nguyn t Si = 6. + 8. + 4 = 8 (nguyn t)Cu trc tinh th Si dng lp phng tm din

Bn knh ca nguyn t Si : r = = 0,19 nmKhi lng ring ca Si

= = =

EMBED Equation.DSMT4 2,5 (g/)Cu 2:

a) Tnh s hc t din v bt din trong 1 c s ca mng lp phng tm din.Gii

Trong 1 c s mng LP PHNG TM DIN :

S HC T DIN = 8.1 = 8

S HC BT DIN = 1 + = 4

b) Thit lp cng thc tnh bn knh ca hc t din v hc bt din theo bn knh R ca nguyn t.

GiiGi : r bn knh ca hc ( t din bt din )

R bn knh nguyn t

a cnh d = r + R

HC T DIN

HC BT DIN

p dng nh l Pitago, ta c

=>

=>

Cu 3 : Tnh 1\b khi lng ring ca NaCl (g/) bit

= 0,98 v = 1,81.

Gii

Trong tinh th NaCl, c

S ion = 1 + 12. = 4 (ion)

S ion = 6. + 8. = 4 (ion)

s phn t NaCl = 4(phn t)

Cnh a ca tinh th : a = 2(+)= 2(0,98 + 1,81) = 5,58 = 5,58. cm

Khi lng ring ca NaCl :

= = = 2,236 (g/) Cu 4 : Niken c cu trc lp phng tm din, hi

a) C bao nhiu nguyn t Niken lp y trong 1 t bo c s?

b) Nu mi nguyn t Niken c bn knh bng 1,24 th chiu di cnh ca t bo c s bng bao nhiu?

c) Tnh khi lng ring ca Niken ,cho M = 58,7

Gii

a) S nguyn t Ni = 6. + 8.= 4 (nguyn t)

b) Do tinh th Ni c cu trc lp phng tm din cnh a ca t bo c s : a = = 3,5 = 3,5.cm

c) Khi lng ring ca Ni :

= = =

EMBED Equation.DSMT4 9,04 (g/)Cu 5 : Fe kt tinh mng lp phng tm khi. Khong cch ngn nht gia 2 nguyn t trong t bo l 2,482 .

a) Trong mi t bo c s c bao nhiu nguyn t Fe?

b) Tnh khi lng ring ca Fe, bit = 55,85 ?

Gii

a) S nguyn t Fe trong t bo = 1 + 8. = 2 (nguyn t)

b) Khong cch b nht gia 2 nguyn t = 2,482

( 2r = 2,482 => r = 1,241

Cnh t bo : a = =

= . cmKhi lng ring ca Fe

= = =

EMBED Equation.DSMT4 7,88 (g/)Cu 6 : ng kt tinh mng lp phng tm din. Chiu di cnh ca hnh lp phng l 3,61 .a) Tnh bn knh nguyn t ng ?

b) Tnh khi lng ring ca ng ?

Gii

a) C cnh t bo a = 3,61= 3,61. cmCu kt tinh mng lp phng tm din a = 4r r = =

EMBED Equation.DSMT4 1,28

b) S nguyn t Cu = 6. + 8.= 4 (nguyn t)Khi lng ring ca nguyn t Cu:

= = =

EMBED Equation.DSMT4 9,03 (g/)Cu 7: Bng hnh hc, hy xc nh s n v cu trc (s nguyn t trong 1 mng) trong c s v c kht ca

Lc phng cht kht

Lp phng tm din

Lp phng tm khi

Gii Mng tinh th LC PHNG CHT KHT:

Th tch ca 1 c s = = =

S nguyn t trong 1 c s = 1 + 4. + 4. = 2 (nguyn t) c kht = .100% = . 100% 74%

Mng tinh th LP PHNG TM DIN :

S nguyn t trong 1 c s = 6. + 8.= 4 (nguyn t) c kht = .100% = . 100% 74%

Mng tinh th LP PHNG TM KHI:

S nguyn t trong 1 c s = 1 + 8.= 2 (nguyn t) c kht = .100% = . 100% 68%

Cu 8 : 911, Fe chuyn thnh Fe vi cu trc lp phng tm mt cnh a. Cho = 55,85a) T quan h gia a v r, xc nh bn knh r theo nm ca nguyn t Fe, bit = 7,87 (g/) Giia)

S nguyn t Fe = 6. + 8.= 4 (nguyn t)Tinh th Fe c dng lp phng tm din

a = 4r a =

C khi lng ring ca Fe :

= = =

r = =

EMBED Equation.DSMT4 (cm) = 0,128 (nm)

b) Hc t din v hc bt din l vng khng gian m nguyn t nguyn t khc c kch thc nh b c th xm nhp vo trong iu kin thch hp . Hi ti a c bao nhiu nguyn t nguyn t khc c kch thc nh b c th xm nhp vo 1 mng c s ca tinh th Fe.

Gii

Tinh th Fe c dng lp phng tm din

c 4 hc bt din v 8 hc t din

c ti a 12 nguyn t nguyn t khc c kch thc nh b c th xm nhp vo trong 1 mng c s ca tinh th Fetrong iu kin thch hp.

------(-----

c) Khi sn xut gang, cc nguyn t Cacbon xm nhp vo cc hc tinh th Fe. Tnh bn knh ca qu cu nguyn t xm nhp va kht hc t din (cn gi l bn knh hc t din). Nguyn t Cacbon c bn knh bng 0,077nm, nu Cacbon xm nhp vo hc t din c lm bin dng tinh th Fehay khng?

Gii

Tinh th Fe c dng lp phng tm din

Bn knh ca hc t din : r =R. = 0,128. 0,029 (nm)

M = 0,077nm => nu Cacbon xm nhp vo hc t din s lm bin dng tinh th Fe.------(------

d) S to hp kim gia Fe vi cc n cht Bo hay Lu Hunh c thun li hn so vi Cacbon hay khng? Ti sao?

GiiCc nguyn t Bo hay Lu Hunh u c bn knh ln hn nguyn t Cacbon (= 0,080nm, = 0,103nm) nn khi xm nhp vo cc hc ca mng tinh th st s kh hn => S to hp kim gia Fe vi cc n cht Bo hay Lu Hunh khng thun li hn so vi Cacbon

Cu 9: Tnh bn knh nguyn t gn ng ca Fe 20, bit ti nhit , = 7,87 g/. Gi thit trong tinh th, nguyn t Fe hnh cu v c kht bng 68% v = 55,85Gii c kht ca tinh th Fe bng 68% => tinh th Fe c dng lp phng tm khi

S nguyn t Fe trong t bo = 1 + 8. = 2 (nguyn t)

Khi lng ring ca Fe

= = = = 7,87

a =

2,867. cm = 2,867

C tinh th Fe c dng lp phng tm khi

a = 4r r =

1,241

Cu 10: a) Tnh bn knh gn ng ca nguyn t canxi 20, bit ti nhit , = 1,55 g/. Gi thit trong tinh th, nguyn t Canxi c hnh cu, c kht bng 74% v = 40,08.

Gii c kht ca tinh th Ca bng 74% => tinh th Ca c dng lp phng tm din

S nguyn t Ca = 6. + 8.= 4 (nguyn t)Khi lng ring ca Ca:

= = = =1,55

a =

5,56. cm = 5,56

Tinh th Ca c dng lp phng tm din

a = 4r

r =

1,97

b) Fe kt tinh mng lp phng tm khi, nguyn t Fe c bn knh r =1,24 . Tnh

Cnh ca t bo c s

Khi lng ring ca Fe (g/)Gii

S nguyn t Fe = 1 + 8. = 2 (nguyn t)

Cnh t bo: a = = 2,86 = 2,86.cmKhi lng ring ca Fe :

= = =

EMBED Equation.DSMT4 7,9(g/)nguyn t Ni

nguyn t Cu

nguyn t Ca

nguyn t Fe

nguyn t Fe

nguyn t Fe

EMBED ChemDraw.Document.6.0

nguyn t Si

nguyn t Fe EMBED Equation.DSMT4

a = 2R

EMBED ChemDraw.Document.6.0

EMBED Equation.DSMT4

Xt c s, c dng hnh lng tr ng m cha trong n l mt hnh chp u c y dng hnh tam gic u :

chiu cao y : h = EMBED Equation.DSMT4

Chiu cao ca hnh chp u :

H = EMBED Equation.DSMT4 = EMBED Equation.DSMT4 = EMBED Equation.DSMT4 = EMBED Equation.DSMT4

d= r + R

a = 2R

a

a

d

d

a EMBED Equation.DSMT4

d= r + R

a = R EMBED Equation.DSMT4

p dng nh l Pitago, ta c EMBED Equation.DSMT4

=> EMBED Equation.DSMT4 EMBED Equation.DSMT4

=> EMBED Equation.DSMT4

a = R EMBED Equation.DSMT4

a EMBED Equation.DSMT4

T

T

T

T

T

T

T

O

O

O

O

O

O

O

Hc T Din

Hc Bt Din

O

O

O

O

PAGE 19

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