B Lukeiou Ximeia Kateuthinsis Theoria Askiseis

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B Lukeiou Ximeia Kateuthinsis Theoria Askiseis

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    Dalton:

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    : P = pA + pB + ...

    (xA), mol (nA)

    mol (n.).: xA = nA/n.

    : pA = x

    AP pA = (n

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    mol : p

    A/pB = n

    A/nB

    taexeiola.gr

  • -

    13. 2

    .

    :

    . 33: : 19, 20, 21

    . 34: 22, 23, 26, 29

    . 35: 30, 31, 33, 34

    . 1 2

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    (1 )

    . 19: 1

    . 20: 2, 3

    . 21: 1, 2

    . 22: 3

    (2 )

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    taexeiola.gr

  • -

    14. 3

    1. :. CaCl2 . Cl r

    . CO2 HCl

    :

    . CaCl2 2 Ca2+ Cl.

    :

    2.

    - : Ca2+ - H2O Cl - H2O.

    . Cl r , -

    : HCl - HCl, HBr - HBr HCl - HBr.

    . CO2 , HCl . -

    :

    CO2.

    - Cl - CO2.

    - Cl.

    2. N :. Cl2 - KCl . HCl - HF . - 2: rNO = 30, MrO2 = 32.

    :

    . Cl , Cl2, -

    . -

    , KCl .

    . :

    Cl: - F:

    - ,

    HF .

    . :

    : - 2:

    - ,

    .

    taexeiola.gr

  • -

    15. 3

    3. 12,5 g 400 .. 5 L, ;

    . ;

    . 7,5 g ;

    : , Po = 2 atm, R = 0,082 L atm/mol K, MrX = 50.

    :

    . mol : X

    X

    m 12,5gMr 50g / mol

    = = =Xn 0,25mol

    :

    XX

    n RT 0, 25mol 0,082L atm / mol K 400KPV n RT PV 5L

    = = = P = 1,64atm

    .

    , 2 atm.

    V P = Po.

    , P = Po n = n ( ):

    o XX o

    n RT 0,25mol 0,082L atm / mol K 400KP V n RT V2atmP

    = = = V = 4,1L

    4,1 L.

    . , () X(g)

    .

    V 7,5 g . g

    : m. = 12,5g 7,5g = 5 g

    mol : .

    X

    m 5gr 50g/mol

    = =.

    n = 0,1mol

    :

    .o. o

    n RT 0,1mol 0,082L atm / mol K 400KP V n RT V2atmP

    = = = V = 1,64L

    4. 16,4 L 500 , - CO, N2 Ne. H 5 atm, mol CO N2 2/3 Ne 20 g. N :

    . mol . . mol .

    . .

    : R = 0,082 L atm/mol K, Are = 20.

    :

    . :

    taexeiola.gr

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    16. 3

    . .

    PV 5atm 16,4LPV n RT nRT 0,082L atm / mol K 500K

    = = =

    .n = 2mol

    . mol e : Ne

    Ne

    m 20gAr 20g / mol

    = =Nen = 1mol

    :

    2

    CO

    N

    n 2n 3

    = 2

    CON

    3nn =

    2 (1)

    nCO + nN2 + nNe = 2 mol nCO + nN2 + 1 mol = 2 mol nCO + nN2 = 1 mol (2)

    (1) (2) CO3n 3 0,4mol

    2 2

    = =2Nn = 0,6mol nCO = 0,4 mol

    . CO

    .

    n 0,4molP 5atmn 2mol

    = =COP = 1atm .

    n 0,6molP 5atmn 2mol

    = =2

    2

    NNP = 1,5atm

    PCO + PN2 + PNe = P PNe = P - PCO - PN2 = 5atm - 1atm - 1,5atm = 2,5 atm

    5. 2, : 2(g) + 32(g) 23(g)

    , 727 C. 24,6 L.

    A 22,2 atm,

    2.

    : R = 0,082 L atm/mol K, 727 C, Po = 2,2 atm

    :

    mol mol 2.

    :

    /3 mol , 2/3 mol 3

    , : pX = 2,2 atm.

    p + p3 + pX = 22,2 atm p + p3 + 2,2 atm = 22,2 atm pA + pAB3 = 20 atm

    A AB3:

    3A AB 2PV (n n )RT 20atm 24,6L mol 0,082L atm/ mol K 1000K3 3

    = + = +

    = 6mol 6 mol A 6 mol B2.

    taexeiola.gr

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    17. 4

    1. :. 3 . CH4 . ROH . HCl . CO2

    ............................................................................................................................

    ............................................................................................................................

    ............................................................................................................................

    ............................................................................................................................

    ............................................................................................................................

    2. :. CH3CH2OH . O2 . NaCl . HCl

    ............................................................................................................................

    ............................................................................................................................

    ............................................................................................................................

    ............................................................................................................................

    ............................................................................................................................

    3. :

    . CH3CH2CH2CH2CH2CH3 .

    .

    ............................................................................................................................

    ............................................................................................................................

    ............................................................................................................................

    taexeiola.gr

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    18. 4

    4. :. (CH3CH2OH) . (CH3OCH3)

    ............................................................................................................................

    ............................................................................................................................

    ............................................................................................................................

    ............................................................................................................................

    ............................................................................................................................

    5. A SO2, C2H2 2. C2H2 4 atm, :

    . .

    .

    : rS = 32, ArO = 16, ArC = 12, ArH = 1.

    ............................................................................................................................

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