Tarea quimica 4

                                                  Tarea química 4.

Pagina 86:

3.39-    Sn- estaño                O- 16

              SnO2                       Sn- 118

                                               Pm- 134

% elemento=   masa del elemento/ Pm  *100

% Sn=   118q/134q   * 100

% Sn= 88.06

% O= 16*2/134  * 100

%= 23.881

Sn= 88.06%

O= 23.882%

3.41-      Ca H10 O                                                              C= 12*9= 108

                                                                                              H= 10

                                                                                               O= 16

a) C= 80.597%           H= 7.463%                                        O= 11.940%

    % C= 108/134  *100

    %C= 80.597

%H = 10/134  *100                                                       %O= 16/134  *100

%H= 7.463                                                                       %O= 11.940

  3.43-       a)        C= 44.4%                                        O= 9.86%

                           H= 6.21%

                           S= 39.5%                   

1) Convertir a moles:

44.4 g 1mol c/ 12g=   3.7 mol C

6.21g    1 mol H/1g= 6.21 mol H

39.5g     1 mol S/ 32g= 1.234 mol S

9.86g    1 mol O/ 16g= 0.616 mol O

Encontrar factor común:

C= 3.7 mol C/0.616 mol= 6.006 C

H= 6.21 mol H/ 0.616 mol= 10.08 H                        =            C6 H10 S2 0

S= 1.234 mol S/ 0.616 mol= 2.003 S

O= 0.616 mol O/ 0.616 mol= 1

b)  162g/ 162g= 1               =                     C6 H S2 O

Libro pagina 125:

4.53-  # gramos= M     V     Pm

            #gramos= (2.80)     (0.500 lt)     (87)

           #gramos= 121.8 gramos es la masa de IK.

4.55-  MaCl2 =  Cloruro de magnesio

                                      60.0 ml

                                      0.100 M

Ma= 24

Cl= 35*2=70

                 94 Pm                 

            Molaridad= # moles/ 1 lt sal

0.100 M= # moles/ 0.6 lt

0.100/ # moles= 0.6 H

Moles= 0.6/ 0.100 M=  # moles= 6

4.56:   Ko lt

            35.0 ml

            5.50 M

K= 39

O= 16              =56 Pm                            # gramos= 5.50 M     0.035 ml  * 56

H= 1                                                          #gramos= 107.8      

                                                                        107.8= 10.78 gr

4.59-  V= ?                                      Na= 23

                                                        Cl= 35                       =58

a) 2.14- NaCl

    0.270 M

                           # gramos= M  *V  *PM

                           #gramos= 0.270 *V * 58

                           2.14 g= 0.270 * V * 58  

                          2.14/58= 0.270 * V

                       2.14/0.270= 136.6 H

b)  4.30 g de estanol                       V=?

          1.50 M                            Estanol= C2 H S O H

C= 12*2= 24

H= 1*6= 6

O= 16= 16

              46 Pm                            430 g/ 1.50* 46= V

 0.0623 ml= V

623 lt= V

c) 0.85 g – acido acético= CH3 COO H

C= 12*2= 24

H= 1*4= 4                     = 60

O= 16*2= 32

0.85g/ 0.30*60= V

0.0472 ml= V

472 lt= V