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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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Sunny_sXe [5.5K]
The correct answer is D. N2
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3 years ago
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Calculate the density of sulfuric acid from the information that 35.4 mL of the acid masses 65.14 g.
vladimir1956 [14]

Answer:

<h3>The answer is 1.84 g/mL</h3>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume}  \\

From the question we have

density =  \frac{65.14}{35.4}  \\  = 1.840112994...

We have the final answer as

<h3>1.84 g/mL</h3>

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3 0
3 years ago
What mass of Fe(OH)3 is produced when 35 mL of 0.250 M Fe(NO3)3 solution is mixed with 55 mL of a 0.180 M
Zina [86]

Answer:

0.35 g.

Explanation:

We'll begin by calculating the number of mole of Fe(NO3)3 in 35 mL of 0.250 M Fe(NO3)3 solution.

This is illustrated below:

Molarity of Fe(NO3)3 = 0.250 M

Volume = 35 mL = 35/1000 = 0.035 L

Mole of Fe(NO3)3 =?

Molarity = mole /Volume

0.250 = mole of Fe(NO3)3 / 0.035

Cross multiply

Mole of Fe(NO3)3 = 0.25 x 0.035

Mole of Fe(NO3)3 = 8.75×10¯³ mole

Next, we shall determine the number of mole of KOH in 55 mL of 0.180 M

KOH solution. This is illustrated below:

Molarity of KOH = 0.180 M

Volume = 55 mL = 55/1000 = 0.055 L

Mole of KOH =.?

Molarity = mole /Volume

0.180 = mole of KOH /0.055

Cross multiply

Mole of KOH = 0.180 x 0.055

Mole of KOH = 9.9×10¯³ mole.

Next, we shall write the balanced equation for the reaction. This is given below:

3KOH + Fe(NO3)3 —> Fe(OH)3 + 3KNO3

From the balanced equation above,

3 moles of KOH reacted with 1 mole of Fe(NO3)3 to produce 1 mole of Fe(OH)3.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

3 moles of KOH reacted with 1 mole of Fe(NO3)3.

Therefore, 9.9×10¯³ mole of KOH will react with = (9.9×10¯³ x 1)/3 = 3.3×10¯³ mole of Fe(NO3)3.

From the above illustration, we can see that only 3.3×10¯³ mole out of 8.75×10¯³ mole of Fe(NO3)3 given is needed to react completely with 9.9×10¯³ mole of KOH.

Therefore, KOH is the limiting reactant and Fe(NO3)3 is the excess reactant.

Next, we shall determine the number of mole of Fe(OH)3 produced from the reaction.

In this case, we shall use the limiting reactant because it will give the maximum yield of Fe(OH)3 as all of it is consumed in the reaction.

The limiting reactant is KOH and the mole of Fe(OH)3 produce can be obtained as follow:

From the balanced equation above,

3 moles of KOH reacted to produce 1 mole of Fe(OH)3.

Therefore, 9.9×10¯³ mole of KOH will react to produce = (9.9×10¯³ x 1)/3 = 3.3×10¯³ mole of Fe(OH)3.

Finally, we shall convert 3.3×10¯³ mole of Fe(OH)3 to grams. This can be obtained as follow:

Molar mass of Fe(OH)3 = 56 + 3(16 + 1) = 56 + 3(17) = 107 g/mol

Mole of Fe(OH)3 = 3.3×10¯³ mole

Mass of Fe(OH)3 =?

Mole = mass /Molar mass

3.3×10¯³ = Mass of Fe(OH)3 / 107

Cross multiply

Mass of Fe(OH)3 = 3.3×10¯³ x 107

Mass of Fe(OH)3 = 0.3531 ≈ 0.35 g.

Therefore, 0.35 g of Fe(OH)3 was produced from the reaction.

8 0
3 years ago
Convert 6.25 cm to km.
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Answer:

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Explanation:

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Well a compound is 2 or more different molecules together. So if you see a periodic table you will see the elements to see if they are the same or not.
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