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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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In addition to the separation techniques used in this lab (magnetism, evaporation, and filtering), there are other commonly used
DENIUS [597]

Answer:

A. Water and Sugar  can be separated by evaporation and then crystallization

B. Mixture of Hexane and Octane can be separated by distillation

C. Solid Iodine, I₂ and NaCl  can be separated by filtration and then evaporation

D. "Sharpie" permanent marking pen  can be separated by  chromatography

E. Nickel shavings and copper pellets can be separated by magnetic separation

Explanation:

A. A mixture of water and sugar can be separated by employing two separation techniques, evaporation and crystallization. First the sugar solution  is heated to evaporate most of the water. When the solution becomes very saturated, it is allowed to cool and then the sugar molecules are obtained through crystallization induced by seeding or scratching the walls of the container.

B. A mixture of hexane (boiling point = 68 °C) and Octane (boiling point = 125 °C) can be separated by distillation due to their significant difference in boiling points.

The mixture is heated in a flask connected to a Liebig condenser. Hexane with the lower boiling point will distill over first and is collected. Afterwards, octane next distills over and is collected as well.

C. A mixture of solid iodine and NaCl can be seperated by first dissolving in water. Iodine being non- polar does not dissolve and is collected as a residue from filtration using a filter paper, while the NaCl solution is collected as the filtrate. The NaCl is recovered from solution by evaporating to dryness in an evaporating dish.

D. "Sharpie" permanent marking pen contains a mixture of dyes which can be separated by paper chromatography.

A drop of the marker ink is placed on a spot above the solvent level on the paper strip used for the separation. The paper strip is  held vertically inside a jar containing a solvent which serves as the mobile phase. The jar is covered and the different dyes move along the paper which serves as the stationary phase, and is thus separated. The paper strip is removed from the jar when the ascending front of the solvent is approaching the top of the paper. The paper is dried and the various dyes can be identified by comparing the distance each has traveled with those of standards.

E. A mixture of nickel shavings and copper pellets can be separated by magnetic separation.

A magnet is brought near the mixture and the nickel shavings being magnetic is attracted to the magnet leaving copper pellets behind since copper is not magnetic.

4 0
3 years ago
The heat of vaporization for ethanol is 0.826 kJ/j. Calculate the heat energy in joules required to boil 45.65g of ethanol
andrey2020 [161]

Answer:

ethanol is already at the ethanol's boiling point: (0.826 kJ/g) x (70.05 g) = 57.8613 kJ = 5.79 x 10^4 J.

Missing: 45.65 ‎| ‎Must include: ‎45.65

Explanation:

6 0
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Why do boys my age not like me ?
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Because boys are overrated

Explanation:

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3 years ago
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What challenges would you encounter with the titration if you had used apple cider vinegar or balsamic vinegar as the analyte in
Vika [28.1K]

Explanation:

Apple cider vinegar or balsamic vinegar might contain various synthetics which might  interact with or nullified the findings, whereas white vinegar comprises of acidic acid only. In fact, apple cider vinegar and balsamic vinegar are deep in appearance this would make it very difficult to determine the color. This is why it is preferable to use white vinegar in the Titration process.

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3 years ago
Consider the following reaction NHAHS(s)NH3(g) + H2S(g) If a flask maintained at 302 K contains 0.196 moles of NH4HS(s) in equil
quester [9]

Answer:

Kc = 3.72 × 10⁶

Explanation:

Let's consider the following reaction:

NH₄HS(g) ⇄ NH₃(g) + H₂S(g)

At equilibrium, we have the following concentrations:

[NH₄HS] = 0.196 M (assuming a 1 L flask)

[NH₃] = 9.56 × 10² M

[H₂S] = 7.62 × 10² M

We can replace this data in the Kc expression.

Kc=\frac{[NH_{3}] \times [H_{2}S] }{[NH_{4}HS]} =\frac{9.56 \times 10^{2}  \times 7.62  \times 10^{2}}{0.196} =3.72 \times 10^{6}

7 0
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