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mario62 [17]
3 years ago
5

The results of Rutherford's gold foil experiment gave him the evidence to arrive at two conclusions: (1) an atom was much more t

han just empty space and scattered electrons and (2) ______________________________________________.
Chemistry
1 answer:
Vedmedyk [2.9K]3 years ago
7 0

Answer: The results of Rutherford's gold foil experiment gave him the evidence to arrive at two conclusions: (1) an atom was much more than just empty space and scattered electrons and (2) an atom consists of a positive charge at the center where most of its mass is placed.

Explanation:

Ernest Rutherford performed an experiment in which he passed alpha particles through a thin gold foil sheet. Through this sheet some of the particles passed on to the other side but some of them were reflected back.

This experiment was done by Rutherford to prove than an atom contains a tiny and heavy nucleus. He concluded that an atom has more than empty space and electrons were present in scattered form.

Also, he concluded that an atom consists of a positive charge at the center where most of its mass is placed.

Thus, we can conclude that the results of Rutherford's gold foil experiment gave him the evidence to arrive at two conclusions: (1) an atom was much more than just empty space and scattered electrons and (2) an atom consists of a positive charge at the center where most of its mass is placed.

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3. Which of the following sentences apply to neutrons? a. have a negative charge. b. circle around the atom nucleus. c. are the
DerKrebs [107]

Answer:

d.have a mass of 1 amu..

Explanation:

Does not match a as the answer for a is electrons B.electron circle around the nucleus because neutrons are inside the nucleus.C.proton and neutrons make the nucleus so the only answer left is .d .which is correct

3 0
2 years ago
13. When a train car traveling at 4 m/s collides with a stationary train car, what happens to the momentum of the first train ca
Solnce55 [7]

Answer:

Momentum of first train car will reduce

Explanation:

When the moving care collides with the stationary car, it will increase the momentum of the stationary car. However, its own momentum will reduce.

It is so because the speed of the first train car will reduce after collision due to loss of energy in the collision while the stationary car may gain some momentum due to rise in velocity from zero (velocity at stationary position).

5 0
2 years ago
For the reaction
s2008m [1.1K]

Answer:

0.558mole of SO₃

Explanation:

Given parameters:

Molar mass of SO₃ = 80.0632g/mol

Mass of S = 17.9g

Molar mass of S = 32.065g/mol

Number of moles of O₂ = 0.157mole

Molar mass of O₂ = 31.9988g/mol

Unknown:

Maximum amount of SO₃

Solution

  We need to write the proper reaction equation.

           2S + 3O₂ → 2SO₃

We should bear in mind that the extent of this reaction relies on the reactant that is in short supply i.e limiting reagent. Here the limiting reagent is the Sulfur, S. The oxygen gas would be in excess since it is readily availbale.

So we simply compare the molar relationship between sulfur and product formed to solve the problem:

First, find the number of moles of Sulfur, S:

   Number of moles of S = \frac{mass }{molar mass}

   Number of moles of S =  \frac{17.9 }{32.065} = 0.558mole

Now to find the maximum amount of SO₃ formed, compare the moles of reactant to the product:

       2 mole of Sulfur produced 2 mole of SO₃

   Therefore; 0.558mole of sulfur will produce 0.558mole of SO₃

5 0
3 years ago
Suppose Highlinium-308 can also undergo positron emission
____ [38]

Answer:

The atomic number of burienium will be 307.

Explanation:

During positron emission proton is converted into the neutron and one electron neutrino with positron is released. It means the atomic number will be reduce by one and atomic mass remain same.

For example:

²³Mg₁₂    →     ₁₁Na²³+ e⁺+ Ve

Similarly, when highlinium-308 undergoes positron emission the new element burienium is produced and the atomic number will be 307 while atomic mass remain same.

Properties of beta radiations:

Beta radiations are result from the beta decay in which electron is ejected. The neutron inside of the nucleus converted into the proton an thus emit the electron which is called β particle.

The mass of beta particle is smaller than the alpha particles.

They can travel in air in few meter distance.

These radiations can penetrate into the human skin.

The sheet of aluminium is used to block the beta radiation

4 0
3 years ago
You are requested to reduce the size of 50 ton/hr of a given solid. The size of the feed is such 80% passes a 4-in (76.2 mm) scr
defon

Answer:

1) The power needed to process 50 ton/hr is 135.4 HP.

2) The void fraction of the bed is 0.37.

Explanation:

1) For this type of milling operations, we can estimate the power needed for an operation according to the work index (Ei), the passing size of the circuit feed (F80) and the passing size of the product (P80).

We assume the units of Ei are kWh/t.

The equation that relates this parameters and the power is (size of particles in μm):

W=Ei*(\frac{10}{\sqrt{P80}} -\frac{10}{\sqrt{F80}} )\\\\W=9.45*(\frac{10}{\sqrt{3175\mu m}} -\frac{10}{\sqrt{76200mm}} )\\\\\\W=9.45*(0.1774+0.0362)=2.019 kWh/t

The power needed to process 50 ton/hor is

P=2.0194\frac{kWh}{Ton}*\frac{50Ton}{h}*\frac{1.341HP}{1kW}=   135.4 \, HP

2) The density of the packed bed can be expressed as

\rho=f_v*\rho_v+f_s*\rho_s

being f the fraction and ρ the density of every fraction. We know that the density of the void is 0 (ρv=0) and that fv=1-fs (the sum  of the fractions ois equal to the total space).

Then we can rearrange

\rho=f_v*\rho_v+f_s*\rho_s\\\\\rho=f_v*0+(1-f_v)*\rho_s\\\\\rho/\rho_s=1-f_v\\\\f_v=1-\rho/\rho_s=1-990/1570=1-0.63=0.37

The void fraction of the bed is 0.37.

6 0
3 years ago
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