The Atomic Theory - Essay Example

One expects them to be the same and hence have similar properties which is not the case, for example, Ethyl alcohol has higher boiling plinthite and has different solubilities in water.

1. ii) Burning wood leaves ash that is only a small fraction of the mass of the original wood but according to Dalton, all matters is composed of extremely small indivisible and indestructible particles called atoms and chemicals changes are merely rearrangements of these atoms to form new compounds, the total number of atoms (and mass) remaining constant throughout.

iii) Atoms can be broken down into smaller particles for example in atomic bombs but according to Dalton atoms are extremely small, indivisible, and indestructible particles that form matter.

89. iv) According to Dalton when elements combine to form compounds, they do so in proportion, which for any particular compound, are simple and fixed. Thus, since the atoms any element have always the same mass, it follows that the proportion by mass of various elements in the compound is also fixed, which is not the case in liH since one sample is 89.4% of li by mass whereas the other is 74.9% li by mass.

WEEK 2

Q1:

     Cu2O   + 2H2(g)           3Cu(s) + 2H2O(l)

     CuO

     The mixture is   Cu2O

                                 CuO              contain 1.500g

           Mass of li produced =1.252g

           % Composition of li in the mixture;

                 1.252g * 100 = 83.47%

                  1.500g

100%- 83.47% = 16.53%   = % of oxygen in the mixture.

                    Cu2O        CuO     % of oxygen in each of = 16.53 / 2

                                                                                         = 8.27 %

% of Cu in Cu2O = 2/3 * 83.47%

                                 =55.65 li

83.47% - 55.65% =27.82%               % of Cu in Cu2O

% Of Cu in the mixture = 27.82 + 8.27

                                = 36.09%

Q2

GP 6A = 0, S, Se, Te   

     Al                      X

     18.56%              81.44%

18.56                      81.44

18.56                      18.56  

• 4

                         Formula = AlX4       

Q3

M2S3 + 5O2                2MO2 + 3SO3

4.00g                           (4- 0.277g)

(2M+ 3*32)=4.00g          (2M + 2O2  )    =3.723g               (2M + (2* 32))g

Mass of S

              S          (3/5 * 4) g                 96g

              M          (2/5 * 4) g                ?

                  2.4g              96g

                   1.6g?     

                                                      96g * 1.6   = 64g    

                                                         2.4 

                                                 2M = 64

                                                 M   = 32g

2M + O2              M+ O2 (2*16) g = 32g

For oxygen            1    :  2

                                     (2/3 * 3.723)g  = 2.482

 For M       (1/3 * 3.723g) = 1.241g

• 32g

• ?

32*1.241

  2.482

   = 16g

                       32g + 16g       = 48  =24g

                                2                    2

                                               M = 24g