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NISA [10]
3 years ago
8

the result was of Rutherford’s gold foil experiment demonstrated that the ____ occupies a very small amount of the total space i

nside an atom.
Chemistry
2 answers:
horsena [70]3 years ago
7 0

the result of Rutherford's gold foil experiment demostrated that the Alpha Particles occupies a very small amount of the total space inside an atom.

Amiraneli [1.4K]3 years ago
3 0

Answer:

The result was of Rutherford’s gold foil experiment demonstrated that the Nucleus occupies a very small amount of the total space inside an atom.

Explanation:

To perform this experiment Rutherford bombarded a very fine gold leaf (approximately 10-4 mm thick) by a beam of alpha (α) particles from a polonium sample. The polonium was inside a lead block with a hole through which only alpha particle emissions would be allowed to escape. In addition, hole plates were placed with holes in their centers that would orient the beam in the direction of the gold blade. And finally, a zinc sulfide-coated bulkhead, which is a fluorescent substance, was placed behind the slide, where the pathway of alpha particles could be seen.

At the end of this experiment, Rutherford noted that most alpha particles crossed the blade, did not deviate, or recede. Some alpha particles deviated, and very few receded. Based on these data, Rutherford concluded that, contrary to what Dalton thought, the atom could not be massive. But in fact, much of the atom would be empty and it would contain a very small, dense, positive nucleus.

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In chemical compounds, atoms like to have 8 electrons in their outer shell, this gives them __________.
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It's A
a. the electron configuration of a noble gas. 
8 0
3 years ago
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Three isotopes of silicon have mass numbers of 28,29, and 30 with an average atomic mass of 28.086 what does this say about the
Sedbober [7]

Answer is: silicon isotope with mass number 28 has highest relative abundance, this isotope is the most common of these three isotopes.

Ar₁(Si) = 28; the average atomic mass of isotope ²⁸Si.  

Ar₂(Si) =29; the average atomic mass of isotope ²⁹Si.  

Ar₃(Si) =30; the average atomic mass of isotope ³⁰Si.  

Silicon (Si) is composed of three stable isotopes, ₂₈Si (92.23%), ₂₉Si (4.67%) and ₃₀Si (3.10%).

ω₁(Si) = 92.23%; mass percentage of isotope ²⁸Si.  

ω₂(Si) = 4.67%; mass percentage of isotope ²⁹Si.

ω₃(Si) = 3.10%; mass percentage of isotope ³⁰Si.

Ar(Si) = 28.086 amu; average atomic mass of silicon.  

Ar(Si) = Ar₁(Si) · ω₁(B) + Ar₂(Si) · ω₂(Si)  + Ar₃(Si) · ω₃(Si).  

28,086 = 28 · 0.9223 + 29 · 0.0467 + 30 · 0.031.

8 0
3 years ago
The Lewis structure of N2H2 shows ________. Group of answer choices a nitrogen-nitrogen single bond each hydrogen has one nonbon
LenKa [72]

Answer:

one bond between nitrogen and hydrogen and a double bond between the nitrogen atoms.

Explanation:

H-N=N-H

7 0
3 years ago
93.2 mL of a 2.03 M potassium fluoride (KF) solution is diluted with 3.92 L of water.
pishuonlain [190]

Answer:

0.047 M

Explanation:

Given data:

Volume of KF = 93.2 mL

Molarity of KF = 2.03 M

volume of water added= 3.92 L

Solution:

First of all we will calculate the number of moles of KF.

number of moles = Molarity × volume in litter

number of moles = 2.03 mol/L × 0.0932 L

number of moles = 0.189 mol

Molarity of solution:

Total volume =  3.92 L + 0.0932 L = 4.0132 L

Molarity = number of moles / volume in litter

Molarity = 0.189 mol/ 4.0132 L

Molarity = 0.047 M

5 0
4 years ago
What tools do scientists use to investigate the natural world
Leya [2.2K]
They use many tools depending on what they want to investigate hopefully this helped!
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3 years ago
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