Bac2011_M1

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  • 7/25/2019 Bac2011_M1

    1/5

    1

    --2011

    ))3

    )-1

    :

    -

    -

    )

    (

    (

    :3 2 5 3 2 5 2CH COOH C H OH CH COO C H H O + = +

    (

    E:

    .

    :=25t h)-2

    30,6 9 10 /66,7

    acdnv mol hdt

    = = =

    1a cn x=

    ( ) 319 10

    d x

    dt

    =

    3

    9 10 /dx

    mol hdt

    =

    25t h=.

    :)

    0,34a cn mol =

    1 0,34qx =0,66qx mol=

    1m mol=

    0,66

    100 100 66%1

    q

    m

    xr

    x= = =

    3-

    .

    4-([ ] [ ]

    [ ] [ ]

    ( )

    ( )

    23 2 5 2

    2

    3 2 5

    0,663,77

    0,34

    qq ester eq

    ac al q q

    CH COO C H H O n nQ

    CH COOH C H OH n n

    = = = =

    = =3,77qK Q

    (

    0t=

    .

    .

    ( )

    ( )( )

    20,66

    2,370, 34 0, 54

    riQ = =

    >riQ K

    ) .(

    3 2 5 2 5 2CH COOH C H OH HCOO C H H O + = +

    0011

    x11 x

    qxq1 q1 qx

    mxm1 mx1 m

    3 2 5 2 5 2CH COOH C H OH HCOO C H H O + = +

    0,660,660,540 , 3 4

    0,66 +0,66 +0,54 0,34

    50 100( )t h

    ( )aciden mol

    0,2

    25

  • 7/25/2019 Bac2011_M1

    2/5

    2

    ))3

    :-1

    P ma=r

    r

    mg ma=r r

    =a gr r

    Oxuur

    :0a=Ozuur

    :a g=

    :)

    :)(=0t)

    0

    0

    0

    cos

    sin

    vv

    v

    r

    00

    0OG

    z

    uuur

    :

    0- cosxv v =0 sinzv g t v = +

    -

    0 cos .v t=2 0 0

    1sin2z g t v t z= + +

    )()(: .

    :=( )z h x=( )z g t=( )f t:-2

    2

    02 2

    0

    .2 cos

    gz x tn x zv

    = + +20,039 0,7 2z x x= + +

    =0z:-3

    2

    0 0,039 0,7 2x= + +

    .)(=20,45x m

    :)-(M0G

    0 0cM ppM cG ppGE E E E+ = +

    0

    2 2

    0

    1 1

    2 2M Gm v m g h m v mgh+ = +

    ( ) ( )0

    22 2

    0 2 13,7 2 9,8 2 15 /M G Mv v g h h m s= + = + =

    .vv:

    ))3

    -1

    -2 .=N Z)())-3

    (

    :

    :+

    14:( )14) 14 06 7 1C N e +

    ar

    0vr

    vr

    OGuuur

    Oxuur

    0 xdv

    dt= 0 cosv 0 cosv 0 cos .x v t=

    Ozuur

    dv

    gdt

    = 0 sinv 0 sing t v +2

    0 0

    1sin .

    2z g t v t z= + +

    16

    7N14

    6C12

    5B14

    5B

    12

    7N

    13

    7N

    11

    6 C

    8

    5B

    +

    0x

    z

    d vdt

    d v

    dt

    =

    =

    P

    r

    z

    x

    r

    0vr

    0

    G

    O

    0G

    2h m=

    z

    x

  • 7/25/2019 Bac2011_M1

    3/5

    3

    ))3,5

    )1S-1 : (1 1, , ,P f R Trr r r

    )2S : (2 2,P Tr r

    :)-2

    )1S : (1 1 1P f R T m a+ + + =rr r r

    r

    :1 1 1sinP f T m a + =)1(

    )2S : (2 2 2P T m a+ =r r

    r

    :2 2 2P T m a =)2(

    1 2T T=)1()2(:

    ( )2 2 12

    1 2 1 2

    sing m md x fa

    m m m mdt

    = =

    + +)3(

    ) .) .0v=0x=)A=0t)

    2

    2d x ad t

    dx= a t Cd t

    .=0C+ =

    21 '2

    x a t C= +' 0C =( )2 1 2

    1 2 1 2

    sin1

    2

    g m m ft

    m m m m

    =

    + +

    :

    : :2

    0 0

    1

    2a t v t x= + +

    )=)0t=:0v=0x( )2 1 2

    1 2 1 2

    sin1

    2

    g m m ft

    m m m m

    =

    + +

    ))1)-32

    x b t=

    b :.

    ( )2 1 21 2 1 2

    sin1

    2

    g m m fx t

    m m m m

    =

    + + .()1

    (1

    0,52

    b= =1

    2b a=21 /a m s=

    :))3)

    ( )( )

    ( )( )2 1 1 2

    1 2

    sin 9,8 0,6 0,8 0,51 1,4 0,56

    1,4

    g m mf a m m N

    m m

    = + = =

    +

    )2(( )2 2 2 0,6 9,8 1 5,3T T P m a N = = =

    ))4

    :)-1 )(

    (( ) ( )d q t

    i tdt

    =( ) ( )Cq t C u t ( )= ( )Cdu t

    i t Cdt

    =

    ))-2 ) ( )R Cu t u t E + =

    ( )( )C C

    du tRC u t E

    dt= +

    ( ) ( )1 C Cdu t u t Adt

    + =

    )6

    1 200 250 10 0,05RC s = = =9A E V= =

    1S

    2S

    1P

    r

    2Pr

    fr

    Rr

    1T

    r

    2Tr

    1 2

    EC

    R

    R

    0

    K

    e

    i

  • 7/25/2019 Bac2011_M1

    4/5

    4

    (

    ( )( )1

    C

    C

    du tE u t

    dt =

    :[ ] [ ]

    [ ] [ ]1

    UU

    T =.

    ] ] [ ]1 T 1= .

    1: .63%

    1)-3 0,63 0,63 9 5,7Cu E V= = =

    1 0,05s ..=

    (t ( ) 0,99Cu t E=0,25t s =

    15t

    :

    (

    .) ) ( ) ( ) 0R R Cu t u t u t + + =

    ( )( )2 0C C

    du tRC u t

    dt+ =

    ( )( )2 0

    C

    C

    du tu t

    dt + =

    )6

    2 2 2 200 250 10 0,1RC s = = =

    2 12 =

    .

    (

    :

    ( )0 ; 9V( )2 ; 0,37E( )25 ; 0,01E

    ( )Cu V

    ( )t s

    3,3

    0,1

    9

    0,5

    :

    1,5V

    1cm

    9E V=.

    ...

    0,1

    1,5

    ( )Cu V

    ( )t s

    5,7

    0,05

    9

    t

  • 7/25/2019 Bac2011_M1

    5/5

    5

    ))3,5

    )-1: (-1

    )-2()-2

    :)2 2/ // /Zn Zn Cu Cu+ +

    3-

    2 2Zn Zn e+ = +)(

    2 -+ 2Cu e Cu+ =)

    (

    ( ) ( ) ( ) ( )2 2

    aq

    s aq sZn Cu Zn Cu+ ++ = +)(

    )-3: (-4

    5-(

    2

    2

    1

    11

    i

    ri

    i

    Zn

    Q Cu

    +

    +

    = = =

    )(>riQ K

    )

    2 2Zn Cu Zn Cu+ ++ = +

    0Cu

    nCVCVZn

    n0t=

    2xCun x+CV x+CV xZnn xt

    t 2x Q I t=

    2I t x F =40,76 2 60

    4,7 102 2 96500

    I tx mol

    F

    = = =

    :

    :I t

    xzF

    =

    .=2z

    -6

    )

    -4(

    2 2

    4,Cu SO+

    2 2

    4,Zn SO+

    CuZn

    -1

    -2

    2 2

    4,Cu SO+

    2 2

    4,Zn SO+

    CuZn

    V +-

    +-

    1iE

    2iE

    eW

    -3

    K+

    24SO

    2

    Cu +

    2Zn +

    24SO

    Cu Zn 3

    NO

    2e

    I

    2e

    2Zn +

    Zn 2e

    Cu

    -4