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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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Solid NaI is slowly added to a solution that is 0.0079 M Cu+ and 0.0087 M Ag+.Which compound will begin to precipitate first?NaI
NeTakaya

Answer :

AgI should precipitate first.

The concentration of Ag^+ when CuI just begins to precipitate is, 6.64\times 10^{-7}M

Percent of Ag^+ remains is, 0.0076 %

Explanation :

K_{sp} for CuI is 1\times 10^{-12}

K_{sp} for AgI is 8.3\times 10^{-17}

As we know that these two salts would both dissociate in the same way. So, we can say that as the Ksp value of AgI has a smaller than CuI then AgI should precipitate first.

Now we have to calculate the concentration of iodide ion.

The solubility equilibrium reaction will be:

CuI\rightleftharpoons Cu^++I^-

The expression for solubility constant for this reaction will be,

K_{sp}=[Cu^+][I^-]

1\times 10^{-12}=0.0079\times [I^-]

[I^-]=1.25\times 10^{-10}M

Now we have to calculate the concentration of silver ion.

The solubility equilibrium reaction will be:

AgI\rightleftharpoons Ag^++I^-

The expression for solubility constant for this reaction will be,

K_{sp}=[Ag^+][I^-]

8.3\times 10^{-17}=[Ag^+]\times 1.25\times 10^{-10}M

[Ag^+]=6.64\times 10^{-7}M

Now we have to calculate the percent of Ag^+ remains in solution at this point.

Percent of Ag^+ remains = \frac{6.64\times 10^{-7}}{0.0087}\times 100

Percent of Ag^+ remains = 0.0076 %

8 0
3 years ago
key A 0.150 M sodium chloride solution is referred to as a physiological saline solution because it has the same concentration o
lorasvet [3.4K]

Answer:

2.41065 grams

Explanation:

Here we have to apply molarity, particularly in reference to the equation molarity = moles of solute / volume. I would like to rewrite this formula, but with respect to the units - grams = moles / Liters,

We can use molarity to determine the number of moles. After doing so, we can determine the mass of the solute with respect to the formula moles = mass / molar mass. The molar mass of NaCl is 58.44 grams.

_______________________________________________________

275 mL = 0.275 L,

Number of Moles of NaCl = 0.150 * 0.275 = 0.04125 moles,

Mass = 0.04125 * 58.44 = 2.41065 grams,

Solution - Mass of NaCl = 2.41065 grams

<u><em>Hope that helps!</em></u>

5 0
2 years ago
Read 2 more answers
Help me with assigment please
vesna_86 [32]

Answer: I belive the answer is A

Explanation:

8 0
3 years ago
( copper is Cu) How many Copper atoms in 3CuSO 4<br><br> Need help ASAP !!!!!
nydimaria [60]
There are the 3 copper atoms in 3CuSO4. I hope this helps!!
5 0
2 years ago
Imagine that Ana has a block made of pure gold. If she cuts this block into two equal pieces, pick the statement that best descr
dimaraw [331]

Answer : The correct statement is:

The density of each piece is the same as that of the original block.

Explanation :

Intensive property : It is defined as a property of substance which does not change as the amount of substance changes.

Examples: Temperature, refractive index, density, hardness, etc.

According to question, if Ana has a block made of pure gold and she cuts this block into two equal pieces then the density of each piece is the same as that of the original block because density is an intensive property that does not changes until and unless material is changed.

That means density remains same as that of the original piece.

Hence, the correct statement is the density of each piece is the same as that of the original block.

6 0
3 years ago
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