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ruslelena [56]
3 years ago
15

How did Thomson discovery change Dalton's atomic theory?

Chemistry
1 answer:
Alja [10]3 years ago
6 0

Answer:

The discovery of electron by j.j Thomson rejected the Dalton theory.

He stated that atom can be divided into its subatomic particles.

Explanation:

Dalton stated that atom is an indivisible particle. It can not be further subdivided.

The discovery of electron by j.j Thomson rejected the Dalton theory.

He stated that atom can be divided into its subatomic particles.

The electron is subatomic particle that revolve around outside the nucleus and has negligible mass. It has a negative charge.

Symbol= e-

Mass= 9.10938356×10-31 Kg

It was discovered by j. j. Thomson in 1897 during the study of cathode ray properties.

He constructed the glass tube and create vacuum in it. He applied electric current between electrodes. He noticed that a ray of particles coming from cathode to wards positively charged anode. This ray was cathode ray.

Properties of cathode ray:

The ray is travel in straight line.

The cathode ray is independent of composition of cathode.

When electric field is applied cathode ray is deflected towards the positively charged plate.

Hence it was consist of negatively charged particles.

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18 An important environmental consideration is the appropriate disposal of cleaning solvents. An environmental waste treatment c
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a) Percentage by mass of carbon: 18.3%

   Percentage by mass of hydrogen: 0.77%

b)  Percentage by mass of chlorine: 80.37%

c) Molecular formula: C_{2} H Cl_{3}

Explanation:

Firstly, the mass of carbon must be determined by using a conversion factor:

0.872g CO _{2} *\frac{12g C}{44g CO_{2} } = 0.238g CO_{2}

The same process is used to calculate the amount of hydrogen:

0.089g H_{2}O*\frac{2g H}{18g H_{2}O }  = 0.010g H

The percentage by mass of carbon and hydrogen are calculated as follows:

%C\frac{0.238g}{1.3g} *100%= 18.3%

%H\frac{0.010g}{1.3g} *100%=0.77%

From the precipation data it is possible obtain the amount of chlorine present in the compound:

1.75 AgCl*\frac{35.45g Cl}{143.45g AgCl}= 0.43g AgCl

Let's calculate the percentage by mass of chlorine:

%Cl=\frac{0.43g}{0.535g} * 100%= 80.37%

Assuming that we have 100g of the compound, it is possible to determine the number of moles of each element in the compound:

18.3g C*\frac{1mol C}{12g C} = 1.52mol C

0.77g H*\frac{1mol H}{1g H} = 0.77mol H

80.37gCl*\frac{1molCl}{35.45g Cl} = 2.27mol Cl

Dividing each of the quantities above by the smallest (0.77mol), the  subscripts in a tentative formula would be

C=\frac{1.52}{0.77} = 1.97 ≈ 2

H = \frac{0.77}{0.77} = 1

Cl =\frac{2.27}{0.77}=2.94≈3

The empirical formula for the compound is:

C_{2} H Cl_{3}

The mass of this empirical formula is:

mass of C + mass of H + mass of Cl= 24g +1+ 106.35 =131.35g

This mass matches with the molar mass, which means that the supscript in the molecular formula are the same of the empirical one.

5 0
3 years ago
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