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Vesnalui [34]
3 years ago
6

What contribution did these scientists make to atomic models of the atom?

Chemistry
1 answer:
SashulF [63]3 years ago
7 0

Answer:

Explanation:

Dalton's atomic theory proposed that all matter was composed of atoms, indivisible and indestructible building blocks. While all atoms of an element were identical, different elements had atoms of differing size and mass.

In 1897, J.J. Thomson discovered the electron by experimenting with a Crookes, or cathode ray, tube. He demonstrated that cathode rays were negatively charged. In addition, he also studied positively charged particles in neon gas.

Rutherford overturned Thomson's model in 1911 with his well-known gold foil experiment in which he demonstrated that the atom has a tiny and heavy nucleus. Rutherford designed an experiment to use the alpha particles emitted by a radioactive element as probes to the unseen world of atomic structure.

The Bohr model shows the atom as a small, positively charged nucleus surrounded by orbiting electrons. Bohr was the first to discover that electrons travel in separate orbits around the nucleus and that the number of electrons in the outer orbit determines the properties of an element.

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This is a problem involving heat transfer through radiation. The solution to this problem would be to use the formula for heat flux.

ΔQ/Δt = (1000 W/m²)∈Acosθ

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A = 1.072 m²

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ΔQ = mΔHfus
Let's find its mass knowing that the density of ice is 916.7 kg/m³.
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4 0
3 years ago
The rate constant for a certain reaction is measured at two different temperatures:
Talja [164]

Answer: The activation energy Ea for this reaction is 22689.8 J/mol

Explanation:

According to Arrhenius equation with change in temperature, the formula is as follows.

ln \frac{k_{2}}{k_{1}} = \frac{-E_{a}}{R}[\frac{1}{T_{2}} - \frac{1}{T_{1}}]

k_1 = rate constant at temperature T_1 = 2.3\times 10^8

k_2 = rate constant at temperature T_2 = 4.8\times 10^8

E_a= activation energy = ?

R= gas constant = 8.314 J/kmol

T_1 = temperature = 280.0^0C=(273+280)=553K

T_2 = temperature = 376.0^0C=(273+376)=649K

Putting in the values ::

ln \frac{4.8\times 10^8}{2.3\times 10^8} = \frac{-E_{a}}{8.314}[\frac{1}{649} - \frac{1}{553}]

E_a=22689.8J/mol

The activation energy Ea for this reaction is 22689.8 J/mol

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