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JulijaS [17]
3 years ago
11

Hello, a little help please guys:( Explain how the series of experiments performed by Crookes, Thomson, Rutherford, and Chadwick

represent the process of developing and refining a scientific theory.
Chemistry
1 answer:
White raven [17]3 years ago
7 0
<span>I did some investigation and summarized the process and made a clearer explanation so those who are confused can imagine the process better :) A scientific theory attempts to explain and describe why things happen. Hypotheses are formed and experiments are done to validate or toss the hypothesis based on the data collected. The Atomic Theory has gone through lots of refining as a scientific theory. For instance, William Crookes conduced an experiment with cathode ray tubes powered by electricity that glowed when powered. Crookes placed an object in between the positive and negative electrode and concluded that the shadow made on the positive side was small particles of matter traveling from the negative side. But more evidence was needed so, later on, J.J. Thomson continued Crookes experiment. He tested what would happen if a negative or positive charged rod was placed along the ray tubes and if it would differ if a different element was used as the negative electrode. Thomson found out that the beam had negatively charged particles and that even if the negative electrode is substituted, the glow is still present, meaning that all elements also had the small negative particles. These particles(now known as electrons) were smaller than the atom and were added to the model of the atom dispersed throughout the neutrally charged atom inside its positive sphere. Now came along Rutherford hoping to support Thomsons model by firing positively charged particles at a thin gold foil thinking it would go straight through the foil, but instead it evenly distributed as they went through the foil, concluding that atoms have a small, dense nucleus(containing positive protons and most of the mass of the atom) that deflected the particles passing through. This was a drastic change in the model now knowing that 1 proton has 2000 times the mass of an electron, but its positive charge cancels the negative electron. After WW1, Chadwick and others were seeing that sometimes the mass of the atom was greater than the mass of the protons and the number of protons was less than the mass of the atom. So it was thought that there were extra electrons and protons adding mass in the nucleus but cancelling their charges, but Rutherford proposed a particle with mass but no charge and called it a neutron; made of paired protons and electrons. But scientists kept studying atoms since there was no evidence of the neutron. Chadwick repeated these experiments though, in hopes to find the neutron and succeeded in 1932, finding it in the nucleus with a close mass to the proton. Thanks to these experiments for refining a scientific theory, we now have a clearer model of the atom.</span>
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An experiment was conducted to see which metal pot would be better for cooking food quickly. The table below shows how quickly s
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Answer:

The term conclusion best illustrates the given statement.

Explanation:

Conclusion refers to a decision or judgment that can be acquired by reasoning. From the observation mentioned in the given table:  

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When a 3.25 g sample of solid sodium hydroxide was dissolved in a calorimeter in 100.0 g of water, the temperature rose from 23.
sertanlavr [38]

Answer : The enthalpy change for the solution is 42.8 kJ/mol

Explanation :

Heat released by the reaction = Heat absorbed by the calorimeter + Heat absorbed by the water

q=[q_1+q_2]

q=[c_1\times \Delta T+m\times c_2\times \Delta T]

where,

q = heat released by the reaction

q_1 = heat absorbed by the calorimeter

q_2 = heat absorbed by the water

c_1 = specific heat of calorimeter = 15.8J/^oC

c_2 = specific heat of water = 4.18J/g^oC

m = mass of water = 100.0 g

\Delta T = change in temperature = T_2-T_1=(32.0-23.9)=8.1^oC

Now put all the given values in the above formula, we get:

q=[(15.8J/^oC\times 8.1^oC)+(100.0g\times 4.18J/g^oC\times 8.1^oC)]

q=3513.8J=3.5138kJ        (1 kJ = 1000 J)

Now we have to calculate the enthalpy change for the solution.

\Delta H=\frac{q}{n}

where,

\Delta H = enthalpy change = ?

q = heat released = 3.5138 kJ

m = mass of NaOH = 3.25 g

Molar mass of NaOH = 40 g/mole

\text{Moles of }NaOH=\frac{\text{Mass of }NaOH}{\text{Molar mass of }NaOH}=\frac{3.25g}{40g/mole}=0.0812mole

Now,

\Delta H=\frac{3.5138kJ}{0.0821mole}=42.8kJ/mol

Therefore, the enthalpy change for the solution is 42.8 kJ/mol

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