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olya-2409 [2.1K]
3 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]3 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]3 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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How many grams of water react to form 7.21 moles of Ca(OH)2
oksano4ka [1.4K]

You have the stoichiometric equation. This tells you unequivocally that an  

18

⋅

g

mass of water, 1 mole, reacts with a  

56.07

⋅

g

mass of quicklime to form a  

74.09

⋅

g

mass of slaked lime.

If you don't from where I am getting these numbers, you should know, and someone will be willing to elaborate.

Here, you have formed  

6.21

⋅

m

o

l

of quicklime which requires stoichiometric lime AND water. And thus you need a mass of  

6.21

⋅

m

o

l

×

18.01

⋅

g

⋅

m

o

l

−

1

water  

≅

88

⋅

g

.

In practice, of course I would not weigh out this mass. I would just pour  

100

−

200

⋅

m

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of water into the lime.

7 0
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il63 [147K]

Answer:

positive

Explanation:

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yKpoI14uk [10]

Explanation:

When a strong acid, say HCl reacts which a weak base, say NH_3, the reaction is shown below as:-

\text{NH}_3 (\text{aq}) + \text{HCl} (\text{aq}) \rightarrow {\text{NH}_4^+}(\text{aq}) + \text{Cl}^-(\text{aq})

The salt further reacts with water as shown below:-

\text{NH}_4^+ + \text{H}_2\text{O} \rightarrow \text{H}_3\text{O}^+ + \text{NH}_3

Formation of \text{H}_3\text{O}^+ lowers the pH value of the solution as more hydrogen ions leads to less pH.

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3 years ago
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