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olya-2409 [2.1K]
2 years ago
15

In Rutherford experiment some alpha particles fired at a gold foil bounced backward as a result of ... A.reflection from the sur

faces of gold atoms B.electrostatic repulsion by gold nuclei C.electrostatic repulsion by electronsD.all of the aboveE.none of the above
Chemistry
2 answers:
IRINA_888 [86]2 years ago
6 0

Answer: The correct option is B (electrostatic repulsion by gold nuclei).

Explanation:

In the Rutherford's experiment, he used positively charged particles called alpha particles to bombard an atom in order to find out what is inside the atom. Together with two other scientists, Geiger and Marsden, they used a narrow beam of alpha particles emitted from a radioactive source to bombard a thin gold foil. The scattering of the particles from the gold foil was detected by a movable zinc sulphide screen which could be rotated to various positions around the foil.

Each time an alpha particle hit the screen, a visible flash of light or scintillation was produced. This was observed by a microscope attached to the screen. It was then observed that some of the particles followed a straight path through the gold foil while a few where scattered in a backward direction. This was as a result of electrostatic repulsion by gold nuclei which occurs due to the greater part of the mass of the atom was concentrated in a minute nucleus with positive charge.

Amanda [17]2 years ago
3 0

Answer:

B.electrostatic repulsion by gold nuclei

Explanation:

According to Rutherford's experiment, a thin gold foil was bombarded with alpha particles. Some of the particles passed through the foil undeviated, some were scattered through large angles while some bounced backwards.

It follows that the particles that bounced backwards must have encountered a massive particle of like charge.

The atom is composed of a nucleus which contains positively charged particles. Some of the alpha particles which are positively charged particles bounced back when they encountered the positively charged particles in the nucleus.

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What structures are found in plant cells that are not what structures are found in plant cells that are not me found in animal c
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3 0
2 years ago
When 1 mole of CO(g) reacts with H2O(l) to form CO2(g) and H2(g) according to the following equation, 2.80 kJ of energy are abso
Inga [223]
Answer: Endothermic, 2.80 kJ

Explanation

Since this reaction absorbs heat, it is endothermic.

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Hope this helps:)
6 0
3 years ago
What conversion factor does a balanced chemical equation give you?
Dmitriy789 [7]
From google but i can explain further if needed. <span> The </span>balanced<span> equation for the reaction of interest contains the stoichiometric ratios of the reactants and products; these ratios </span>can<span> be used as </span>conversion factors<span> for mole-to-mole </span>conversions<span>.</span>
7 0
3 years ago
When 500.0 g of water is decomposed by electrolysis and the yield of hydrogen is only 75.3%, how much hydrogen chloride can be m
Evgen [1.6K]

The amount of hydrogen chloride that can be made is 1064 g

Why?

The two reactions are:

2H₂O → 2H₂ + O₂ 75.3 % yield

H₂ + Cl₂ → 2HCl 69.8% yield

We have to apply a big conversion factor to go from grams of water (The limiting reactant), to grams of HCl, the final product. We have to be very careful with the coefficients and percentage yields!

500.0gH_2O*\frac{1moleH_2O}{18.01 gH_2O}*\frac{2 moles H_2}{2 moles H_2O}*\frac{2.015g H_2}{1 mole H_2}*\frac{75.3 actual g}{100 theoretical g}=42.12 g H_2

42.12H_2*\frac{1 mole H_2}{2.015gH_2}*\frac{2 moles HCl}{1 mole H_2}*\frac{36.46g}{1 mole HCl}*\frac{69.8 actualg}{100 theoreticalg} =1064gHCl

Have a nice day!

#LearnwithBrainly

7 0
3 years ago
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