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Ne4ueva [31]
3 years ago
7

Which atomic models does Rutherford’s experimental evidence support? Explain why these models (Dalton, Bohr, Thomson models) are

compatible with the experimental results.
Chemistry
2 answers:
ipn [44]3 years ago
5 0

Answer:

Rutherford was the first scientist who proposed the nuclear model of the atom. According to his atomic model, most of the space of an atom is empty, while the nucleus containing protons and neutrons lie at the center of the atom while electrons revolve around nucleus in definite orbits.

If we talk about studies of some other scientists like Dalton, Neil Bohr and JJ Thomson, they all are compatible with Rutherford's results to a large extent.

For example: Dalton's atomic model assumed that atoms of any substance are similar in size and atoms react to form compounds. Rutherford's concept indicated that atoms contain electrons and they are in a specific number which can be shared to form compounds.

If we talk about Bohr's model, it states that electrons revolve around nucleus in specific shells, this again is compatible with Rutherford's results which gave the concept of shells.

If we talk about Thomson's Plum pudding model, that describe atom as negative particles floating within a soup of diffuse positive charge. This is also compatible with the results of Rutherford that state that negative electrons surround positive nucleus.

Rutherford's model was best atomic model but still it took help from many previous studies and therefore was compatible with the results of old models.


Hope it help!

Gnesinka [82]3 years ago
5 0

Answer:

Rutherford was the first scientist who proposed the nuclear model of the atom. According to his atomic model, most of the space of an atom is empty, while the nucleus containing protons and neutrons lie at the center of the atom while electrons revolve around nucleus in definite orbits.

Explanation:

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Predict the products of each of these reactions and write balanced complete ionic and net ionic equations for each. If no reacti
Bumek [7]

Answer:

Explanation:

Part A : LiCl(aq) + AgNO₃(aq)→

Chemical equation:

LiCl(aq) + AgNO₃(aq)  →  AgCl(s) + LiNO₃(aq)

Ionic equation:

Li⁺(aq)  + Cl⁻(aq) + Ag⁺(aq) + NO₃⁻(aq)  →  AgCl(s) + Li⁻(aq)  + NO⁻₃(aq)

Net ionic equation:

Cl⁻(aq) + Ag⁺(aq) →  AgCl(s)

C = H2SO4(aq)+Li2SO3(aq)→

Chemical equation:

H₂SO₄(aq) + Li₂SO₃(aq)  →  Li₂SO₄(aq) + SO₂(g) + H₂O(l)

Ionic equation:

2H⁺(aq)  + SO²⁻₄(aq) + 2Li⁺(aq)  + SO₃²⁻(aq)  →  2Li⁺ (aq) + SO₄²⁻(aq) + SO₂(g) + H₂O(l)

Net ionic equation:

2H⁺ + SO₃²⁻(aq)  →  SO₂(g) + H₂O(l)

Part E: HClO4(aq)+Ca(OH)2(aq)→

Chemical equation:

HClO₄(aq) + Ca(OH)₂(aq)  →  Ca(ClO₄)₂ (aq) + H₂O(l)

Balanced Chemical equation:

2HClO₄(aq) + Ca(OH)₂(aq)  →  Ca(ClO₄)₂ (aq) + 2H₂O(l)

Ionic equation:

2H⁺(aq) + 2ClO⁻₄(aq) + Ca²⁺(aq) + (OH)²⁻₂(aq)  →  Ca²⁺(aq) +(ClO₄)²⁻₂ (aq) + 2H₂O(l)

Net ionic equation:

2H⁺(aq) + (OH)²⁻₂(aq)  →  2H₂O(l)

Part F: Cr(NO3)3(aq)+LiOH(aq)→

Chemical equation:

Cr(NO₃)₃(aq) + LiOH (aq)  →   LiNO₃(aq) + Cr(OH)₃(s)

Balanced chemical equation;

Cr(NO₃)₃(aq) + 3LiOH (aq)  →   3LiNO₃(aq) + Cr(OH)₃(s)

Ionic equation:

Cr³⁺(aq) + 3NO₃⁻(aq) + 3Li⁺(aq) + 3OH⁻ (aq)  →   3Li⁺(aq) + 3NO⁻₃(aq) + Cr(OH)₃(s)

Net ionic equation:

Cr³⁺(aq) +  3OH⁻ (aq)  →    Cr(OH)₃(s)

Part H: HCl(aq)+Hg2(NO3)2(aq)→

Chemical equation:

HCl (aq) + Hg₂(NO₃)₂(aq)  → Hg₂Cl₂ (s) + HNO₃(aq)

Balanced chemical equation:

2HCl (aq) + Hg₂(NO₃)₂(aq)  → Hg₂Cl₂ (s) + 2HNO₃(aq)

Ionic equation;

2H⁺(aq) + 2Cl⁻ (aq) + 2Hg⁺(aq) + 2NO₃⁻(aq)  → Hg₂Cl₂ (s) + 2H⁺(aq) + 2NO⁻₃(aq)

Net ionic equation:

2Cl⁻ (aq) + 2Hg⁺(aq)   → Hg₂Cl₂ (s)

8 0
2 years ago
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6 0
3 years ago
If 12.5 grams of the original sample of cesium-137 remained after 90.6 years, what was the mass of the original sample?
myrzilka [38]

Answer:

Mass of original sample = 100 g

Explanation:

Half life of cesium-137 = 30.17 years

t_{1/2}=\frac {ln\ 2}{k}

Where, k is rate constant

So,  

k=\frac{\ln2}{t_{1/2}}

k=\frac{\ln2}{30.17}\ year^{-1}

The rate constant, k = 0.02297 year⁻¹

Time = 90.6 years

Using integrated rate law for first order kinetics as:

[A_t]=[A_0]e^{-kt}

Where,  

[A_t] is the concentration at time t

[A_0] is the initial concentration

Initial concentration [A_0] = ?

Final concentration [A_t] = 12.5 grams

Applying in the above equation, we get that:-

12.5\ g=[A_0]e^{-0.02297\times 90.6}

[A_0]=\frac{12.5}{e^{-0.02297\times 90.6}}\ g=100\ g

<u>Mass of original sample = 100 g</u>

8 0
2 years ago
Which of the following structures are accessory structures of the digestive system?
professor190 [17]

the pancreas and the liver

7 0
2 years ago
Read 2 more answers
In the symbol 3p4 
kompoz [17]

Answer:

a. the 3 represents the principal energy level

Explanation:

3 is the principal energy level. The p is the sublevel. 4 is the possible occupying electron.

4 0
3 years ago
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