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BabaBlast [244]
3 years ago
5

The nucleus of an atom is dense and positively charged. What was observed when positively charged particles were radiated onto a

gold atom during Rutherford's experiment because of this?
Negative charges were concentrated at the center of the atom.
Particles that struck the center of the atom were repelled.
Particles that struck the edges of the atom were repelled.
Positive charges were distributed all over the atom.
Chemistry
2 answers:
ankoles [38]3 years ago
8 0
<span>According to my knowledge, I feel the answer is -
Particles that struck the center of the atom were repelled.

Hope this helps!
</span>
romanna [79]3 years ago
7 0

<u>Answer:</u> The correct answer is particles that struck the center of the atom were repelled.

<u>Explanation:</u>

Rutherford gave an experiment which is known as gold foil experiment.

In his experiment, he took a gold foil and bombarded it with the alpha particles (carrying positive charge). These alpha particles are also known as  helium nucleus. It is represented as _2^4\alpha.

He thought that these particles will pass straight through the foil, but to his surprise, some of them deflected their path and a few of them bounced back.

From this he concluded that in an atom, a small positive charge in the center is present. Due to this positive charge, the alpha particles deflected their path and some of them bounced straight back from their path.

Hence, the correct answer is particles that struck the center of the atom were repelled.

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7. How many liters of NH3, at STP will react with 10.6 g O2 to form NO2 and water? 4NH3(g) + 7O2(g) → 4NO2(g) + 6H2O(g) *
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7) Answer is: c. 4.24 L.

Balanced chemical reaction: 4NH₃(g) + 7O₂(g) → 4NO₂(g) + 6H₂O(g).

m(O₂) = 10.6 g; mass of oxygen.

n(O₂) = m(O₂) ÷ M(O₂).

n(O₂) = 10.6 g ÷ 32 g/mol.

n(O₂) = 0.33 mol; amount of oxygen.

Vm = 22.4 L/mol; molar volume at STP (Standard Temperature and Pressure).

At STP one mole of gas occupies 22.4 liters of volume.

V(O₂) = n(O₂) · Vm.

V(O₂) = 0.33 mol · 22.4 L/mol.

V(O₂) = 7.42 L; volume of oxygen.

From balanced chemical reaction: n(O₂) : n(NH₃) = 7 : 4.

n(NH₃) = 4 · 0.33 mol ÷ 7.

n(NH₃) = 0.188 mol; amount of ammonia.

V(NH₃) = 0.188 mol · 22.4 L/mol.

V(NH₃) = 4.24 L.

8) Answer is: a. 3.7 g.

Balanced chemical reaction: P₄(g) + 6H₂(g) → 4PH₃(g).

m(P₄) = 3.4 g; mass of phosphorous.

n(P₄) = m(P₄) ÷ M(P₄).

n(P₄) = 3.4 g ÷ 123.9 g/mol.

n(P₄) = 0.0274 mol; limiting reactant.

n(H₂) = 4 g ÷ 2 g/mol.

n(H₂) = 2 mol; amount of hydrogen.

From balanced chemical reaction: n(P₄) : n(PH₃) = 1 : 4.

n(PH₃) = 4 · 0.0274 mol.

n(PH₃) = 0.1096 mol.

m(PH₃) = 0.1096 mol · 34 g/mol.

m(PH₃) = 3.726 g.

9) Answer is: d. Percent yield is the ratio of theoretical yield to actual yield expressed as a percent.

Percent yield = actual yield / theoretical yield.

Theoretical yield is the maximum amount of product that can be produced from limiting reactant and actual yield is a product that is obtained by experimentation.

For example:

the percent yield = 250 g ÷ 294.24 g · 100%.

the percent yield = 84.5 %.

3 0
3 years ago
Which has a geographic North Pole and South Pole?
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Answer:

A. Earth only

Explanation:

The Earth rotates on the geographic north and south poles. The geographic north and south poles are where lines of longitude (meridians) converge in the north. On the other side of the Earth, the South Pole lies on a continental land mass known as Antarctica.

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Answer:

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Answer:

34.9 g/mol is the molar mass for this solute

Explanation:

Formula for boiling point elevation: ΔT = Kb . m . i

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Kb = Ebullioscopic constant → 0.511°C/m

m = molality (moles of solute/1kg of solvent)

i = 2 → The solute is a strong electrolyte that ionizes into 2 ions

For example: AB ⇒ A⁺  +  B⁻

Let's replace → 0.899°C = 0.511 °C/m . m . 2

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