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Ad libitum [116K]
4 years ago
15

Rutherford and his colleagues fired a beam of small, positively charged alpha particles at thin gold foil. Some of these alpha p

articles were deflected at angles as they passed through the foil.
What conclusion did Rutherford and his colleagues draw from their results?
A) The atom must contain some neutral charge spread throughout the atom. He called these particles neutrons.
B) The atom must contain some centrally located negative charge. He called this core of negative charge the nucleus.
C) The atom must contain some centrally located positive charge. He called this core of positive charge the nucleus.
D) Thomson's plum pudding model of the atom was correct. The atom contains positive charge spread throughout its structure.
Chemistry
1 answer:
Darina [25.2K]4 years ago
5 0

Answer:

C

Explanation:

The questions states that some particles were deflected meaning both the nucleus and the particles were the same charge because they repelled. Using common knowledge, we know that the nucleus is positively charged. Using this information, you can immediately eliminate choices A and B. Choice D says that the positive charge was spread throughout its structure. This is false because protons and neutrons are only found in the nucleus. Electrons are the subatomic particles found outside the nucleus, thus you can get rid of choice D, only leaving you with choice C.

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When comparing the colors of the following four compounds, which is most likely to appear green? [Co(CN)6]3− [Co(H2O)6]3+ [Co(en
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How many grams are in 2.3 moles of N?
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Liquid hexane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 6.9 g of hexane is mi
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Answer:

There, there are no leftover for C6H14 instead, an additional mass of 4g of C6H14 is needed to completely react with 38.4g of O2.

Explanation:

Step 1:

We'll begin by writing the balanced equation for the reaction. This is shown below:

2C6H14 + 19O2 —> 12CO2 + 14H2O

Step 2:

Let us calculate the masses of C6H14 and O2 that reacted from the balanced equation. This is illustrated below:

2C6H14 + 19O2 —> 12CO2 + 14H2O

Molar Mass of C6H14 = (12x6) + (14x1) = 72 + 14 = 86g/mol

Mass of C6H14 from the balanced equation = 2 x 86 = 172g

Molar Mass of O2 = 16x2 =32g/mol

Mass of O2 from the balanced equation = 19 x 32 = 608g

From the balanced equation above, 172g of C6H14 reacted with 608g of O2.

Step 3.

Now, let us determine the mass of C6H14 that will react with 38.4 g of oxygen. This is illustrated below:

From the balanced equation above, 172g of C6H14 reacted with 608g of O2.

Therefore, Xg of C6H14 will react with 38.4g of O2 i.e

Xg of C6H14 = (172 x 38.4) /608

Xg of C6H14 = 10.9g

From the calculations made above, we can see clearly that the mass of C6H14 is limited as the reaction requires 10.9g of C6H14 and only 6.9g was given. There, there are no leftover for C6H14 instead, an additional mass ( 10.9 - 6.9 = 4g) of 4g of C6H14 is needed to completely react with 38.4g of O2.

5 0
3 years ago
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