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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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How many atoms are in a solid piece of iron that has a mass of 24.0g?
Julli [10]
Atomic mass iron ( Fe ) = 55.84 a.m.u

55.84 g ------------ 6.02x10²³ atoms
24.0 g ------------- ??

24.0 x ( 6.02x10²³) / 55.84

=> 2.58x10²³ atoms
5 0
3 years ago
2.(04.01 LC)
Mila [183]

Answer:

2KClO3 —> 2KCl + 3O2

The coefficients are 2, 2, 3

Explanation:

From the question given above, we obtained the following equation:

KClO3 —> 2KCl + 3O2

The above equation can be balance as follow:

There are 2 atoms of K on the right side and 1 atom on the left side. It can be balance by putting 2 in front of KClO3 as shown below:

2KClO3 —> 2KCl + 3O2

Now, the equation is balanced.

Thus, the coefficients are 2, 2, 3

8 0
2 years ago
Calculate the theoretical yield of ammonia produced by the reaction of 100g of H2 gas and 200g of N2 gas
ipn [44]
To get the theoretical yield of ammonia NH3:
first, we should have the balanced equation of the reaction:
3H2(g) + N2(g) → 2NH3(g)
Second, we start to convert mass to moles
moles of N2 = N2 mass / N2 molar mass
                     = 200 / 28 = 7.14 moles
third, we start to compare the molar ratio from the balanced equation between N2 & NH3 we will find that N2: NH3 = 1:2 so when we use every mole of N2 we will get 2 times of that mole of NH3 so,
moles of NH3 = 7.14 * 2 = 14.28 moles 
finally, we convert the moles of NH3 to mass again to get the mass of ammonia:
mass of NH3 = no.moles * molar mass of ammonia
                      = 14.28 * 17 = 242.76 g
6 0
3 years ago
How is a wound healing a chemical change? please explain ​
masya89 [10]

Answer:

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5 0
3 years ago
Read 2 more answers
A hot lump of 30.5 g of iron at an initial temperature of 52.7 °C is placed in 50.0 mL H2O initially at 25.0 °C and allowed to r
slava [35]

Answer:

26.7°C

Explanation:

Using the formula; Q = m × c × ΔT

Where; Q = amount of heat

m = mass

c = specific heat

ΔT = change in temperature

In this question involving iron placed into water, the Qwater = Qiron

For water; m= 50g, c = 4.18 J/g°C, Initial temp= 25°C, final temp=?

For iron; m = 30.5g, c = 0.449J/g°C, Initial temp= 52.7°C, final temp=?

Qwater = -(Qiron)

m × c × ΔT (water) =- {m × c × ΔT (iron)}

50 × 4.18 × (T - 25) = - {30.5 × 0.449 × (T - 52.7)}

209 (T - 25) = - {13.6945 (T - 52.7)}

209T - 5225 = -13.6945T + 721.7

209T + 13.6945T = 5225 + 721.7

222.6945T = 5946.7

T = 5946.7/222.6945

T = 26.7

Hence, the final temperature of water and iron is 26.7°C

8 0
3 years ago
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