2020届福建高三化学大题练-物质结构和性质周练

2020届届届届届届届届届届——届届届届届届届届届届4届

______

______

______

______

______

______

______

______

______

______

______

______

K

______

=186pmK=227pm

______

______

______

______

______

）往浓

______

______

______

AlAlBrIAlBr

AlBr

l

l

l

AlBr

l

______

______

______

______

l

______

LLL

______

L

______

AB

EAB

L

L

p

E

A

_______ _________ _________

_________ B

_________

_________

E

_________

_________

X

Mg

Al

Cl

K A,B,C,D,EAB,C

E______________________E

___________

__________B________________X

E

XX

X-X

XX____

______

XXXeq

\o\ac(△,时)

H=

l

A

XY

XL

KY

2+2+

X

，其中

X

X

Y

Y

pm

bg/

g/cm

ABE

EB_________________________B___________bd

___________________________________________ A______________________

）若

B V____________V

______

______

bb

______V

V__________________

nm B

____________=C=O

____________

YF

KJ

__________________apm______g

______

+

______

______

______

______

______

b

n

______

/

/

______

所示（1pm=10

____________3+3+______________________________

pmA

B

______

3+2+

3+____________%2+

pm______g

A

2+____________

____________

______

______

F___________

______________(

A

B

________________

eq

\o\ac(△,（) +Na

eq

\o\ac(△,（) +Na̲ π(18633×3+2273=5+1−2×3=0=3+0=3 =38×

=112×1 1

33×3+2273

π(1863 3+ 186pm+2×227pm3× 2×

A4(186×2+227×2)3

2

A+NaA

+Na4 8K

K

4×3+32×186pm+227pm），所以晶胞的体积为（2×186pm+2×227pm44π(1863×3+2273(186×2+227×2)34(186×2+227×2)3==0=3+0=3

==0=3+0=3

K

K

NaK

2++K；加

2+

2++2++

KK

）加

K

AlBr

AlBrLIl

AlBr

II

III

AlBr

AlCl

AlClA

A

F

SiFB

FE

F

2+8×1/8+6×1/2=4F

2+

XAlCl

AlClAlCl

AlCl+ABEA

A

AA

B

BBFE

E

E

FF

++B X

EX

=6×

EE

E

EE

A

BA

BB

BA B

NAx3

V=4+

NAx3

V=4+=4

b

b

V

V

VBVBp

p

=3+=3

bbV

V

V=1+8×=2=2+4×=4

V=28 4=1===m=(x×10

3g/cm=

g/cm

83NA

NAx3

NAx3V V

V VVBp

p

V=4+=4

V=4+=4

2

b

2V

V

V=1+8×1=2=2+4×1=4

V=28 24=1==mVMNA

A

1.6×1032NA⋅a3B

A

A 个，（

8×1+6×1=48 2

8×3=124×NA

g4×g÷a×=g/cmNA

NA⋅a3NA⋅a3

IIA

VA

mV）（B

B

IIA

VA

bn

mVB

+

B

IIA

VA

6×nnnn

n

n×=-2n

1+8×1+4×1=4

4×1+6×1=48 2 4 264×4+32×8+56×4g6.02×1023

64×4+32×8+56×46.02×1023

5×10−8cm×5×10−8cm×1.05×10−7cm

42×4

NA⋅(a×10−103

752√3a2NA×10−24

8×1+6×1=4B=4×

8×1+6×1=4B=4× g3+

K

8 NA

g÷a×=NA NA⋅(a×10−10

g/cmNA⋅(a×10−10

B

B

=1+12×1=44=28×1+6×1=4

8

3+

2+

3+

3+

2+

3+

3+

2+

2apm1=×10m

2×m=1×6=162√3×10−24

59+16g=2√3×10−24

2√3×10−24

59+16g=2√3×10−24

=g=NA NA 2√3a2NA×10−24

g752√3a2NA×10−24

3+

BmV

B

B

3+2+

3+3+

2+

3+

2+

2apm

1=×10m

×m=1×6