Describe Rutherford's contribution to the atomic model.

Chemistry

1 answer:

Nady [450]3 years ago

4 0

Answer:

Rutherford's experiment, also known as

$\alpha - scattering \: experiment$

supports the existence of neutrons and the nucleus.

Explanation:

In the above diagram, Rutherford was trying to explain his contributions using thin foils of gold and other metals as targets for alpha particles from a radioactive source.

He observed that the majority of particles penetrated the foil either undeflected or with only a slight deflection. But, every now and then an alpha particle was scattered(or deflected) at a large angle..

According to Rutherford, most of the atoms must be empty space. This explains why the majority of alpha particles passed through through the gold foil with little or no deflection. The atoms positive charges, Rutherford proposed are all concentrated in the Nucleus, which is a dense central core within the atom.

Whenever an alpha particle came close to a nucleus in the scattering experiment, it experienced a large repulsive force and therefore a large deflection. Moreover, an alpha particle coming towards a nucleus would be completely repelled and its direction would be reversed. The positively charged particles in the Nucleus are called Protons.

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Barium reacts with a polyatomic ion to form a compound with the general formula Ba3(X)2. What would be the most likely formula f

AfilCa [17]

Answer:

D. Na₃X

Explanation:

We have the neutral compound Ba₃(X)₂. The total charge (zero) is equal to the sum of the charges of the ions times the number of ions in the molecule.

3 × qBa + 2 × qX = 0

3 × (+2) + 2 × qX = 0

2 × qX = -6

qX = -3

If we have the cation Na⁺ and X³⁻, a neutral molecule would require 3 Na⁺ and 1 X³⁻. The resulting compound is Na₃X.

3 0

3 years ago

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A gas has a pressure of 5.7 atm at 100.0°C. What is its pressure at20.0°C (Assume volume is unchanged)

son4ous [18]

Answer:

$\large \boxed{\text{4.5 atm}}$

Explanation:

The volume and amount of gas are constant, so we can use Gay-Lussac’s Law:

At constant volume, the pressure exerted by a gas is directly proportional to its temperature.

$\dfrac{p_{1}}{T_{1}} = \dfrac{p_{2}}{T_{2}}$

Data:

p₁ =5.7 atm; T₁ = 100.0 °C

p₂ = ?; T₂ = 20.0 °C

Calculations:

1. Convert the temperatures to kelvins

T₁ = (100.0 + 273.15) K = 373.15

T₂ = (20.0 + 273.15) K = 293.15

2. Calculate the new pressure

$\begin{array}{rcl}\dfrac{5.7}{373.15} & = & \dfrac{p_{2}}{293.15}\\\\0.0153 & = & \dfrac{p_{2}}{293.15}\\\\0.0153\times 293.15 &=&p_{2}\\p_{2} & = & \textbf{4.5 atm}\end{array}\\\text{The new pressure will be $\large \boxed{\textbf{4.5 atm}}$}$