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Natalija [7]
3 years ago
13

Which statement regarding the gold foil experiment is FALSE? 1. The α particles were repelled by electrons. 2. It suggested that

atoms are mostly empty space. 3. It suggested the nuclear model of the atom. 4. Most of the α particles passed through the foil undeflected. 5. It was performed by Rutherford and his research group early in the 20th century.
Chemistry
1 answer:
Alenkinab [10]3 years ago
5 0

Answer:

1. The α particles were repelled by electrons.

Explanation:

The gold foil experiment was performed by Rutherford and his research group in 1911 (at the beginning of the 20th century). In this experiment, α particles were bombed to gold foils, and films were placed surround it to collect the particles.

It was observed that most of the particles passed through of the foil undeflected, and for that, Rutherford stated that the atom was a "huge empty". Some particles were deflected, because they're attracted to the electrons at the electrosphere, and a small number of particles were complete deflected to the origin because they chocked with the small positive nuclei.

Thus, the experiment suggested the nuclear model of the atom, called the planetary model, that was improved after by Bohr and other scientists in the quantum model.

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Which option contains an example of a polyatomic ion? (1 point)
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5 0
2 years ago
How many moles of calcium are there in 5.74 x 1022 atoms of calcium?
igor_vitrenko [27]

Answer:

0.095 moles of Calcium is there in 5.74 x 1022 atoms of calcium.

Explanation:

  • As we know, 6.023*10^23 atoms of an element is equal to its atomic weight.

       And, 6.023*10^23 atoms of an element is also equal to 1 mole of the             element.

We have,

  • 6.023*10^23 atoms of element calcium equals to 1 mole of Calcium
  • 5.74*10^22 atoms of element calcium equals to

         (1/(6.023*10^.23)) * 5.74*10^22 moles of calcium

Therefore,

  • 5.74 x 1022 atoms of calcium= 0.095 moles of calcium.

6 0
3 years ago
Nitroglycerin is a dangerous powerful explosive that violently decomposes when it is shaken or dropped. The Swedish chemist Alfr
Ganezh [65]

Answer:

a. 4 C_3H_5N_3O_9 (l)\rightarrow 6N_2 (g) + O_2 (g) + 10 H_2O (g) + 12 CO_2 (g)

b. 146.0 g

Explanation:

Question 1 (a). Just as the problem states, liquid nitroglycerin decomposes into nitrogen gas N_2, oxygen gas O_2, water vapor H_2O and carbon dioxide CO_2. Let's write the decomposition of nitroglycerin into these 4 components:

C_3H_5N_3O_9 (l)\rightarrow N_2 (g) + O_2 (g) + H_2O (g) + CO_2 (g)

Now we need to balance the equation. Firstly, notice we have 3 carbon atoms on the left and 1 on the right, so let's multiply carbon dioxide by 3:

C_3H_5N_3O_9 (l)\rightarrow N_2 (g) + O_2 (g) + H_2O (g) + 3 CO_2 (g)

Now, we have 3 nitrogen atoms on the left and 2 on the right, so let's multiply nitrogen on the right by \frac{3}{2}:

C_3H_5N_3O_9 (l)\rightarrow \frac{3}{2}N_2 (g) + O_2 (g) + H_2O (g) + 3 CO_2 (g)

We have 5 hydrogen atoms on the left, 2 on the right, so let's multiply the right-hand side by \frac{5}{2}:

C_3H_5N_3O_9 (l)\rightarrow \frac{3}{2}N_2 (g) + O_2 (g) + \frac{5}{2} H_2O (g) + 3 CO_2 (g)

Finally, count the oxygen atoms. We have a total of 9 on the left. On the right we have (excluding oxygen molecule):

\frac{5}{2} + 6 = 8.5

This leaves 9 - 8.5 = 0.5 = \frac{1}{2} of oxygen. Since oxygen is diatomic, we need to take one fourth of it to get one half in total:

C_3H_5N_3O_9 (l)\rightarrow \frac{3}{2}N_2 (g) + \frac{1}{4} O_2 (g) + \frac{5}{2} H_2O (g) + 3 CO_2 (g)

To make it look neater without fractional coefficients, multiply both sides by 4:

4 C_3H_5N_3O_9 (l)\rightarrow 6N_2 (g) + O_2 (g) + 10 H_2O (g) + 12 CO_2 (g)

Question 2 (b). Now we can make use of the balanced chemical equation and apply it for the context of this separate problem. We're given the following variables:

V_{CO_2} = 41.0 L

T = -14.0^oC + 273.15 K = 259.15 K

p = 1 atm

Firstly, we may find moles of carbon dioxide produced using the ideal gas law pV = nRT.

Rearranging for moles, that is, dividing both sides by RT (here R is the ideal gas law constant):

n_{CO_2} = \frac{pV_{CO_2}}{RT} = \frac{1 atm\cdot 41.0 L}{0.08206 \frac{L atm}{mol K}\cdot 259.15 K} = 1.928 mol

According to the stoichiometry of the balanced chemical equation:

4 C_3H_5N_3O_9 (l)\rightarrow 6N_2 (g) + O_2 (g) + 10 H_2O (g) + 12 CO_2 (g)

4 moles of nitroglycerin (ng) produce 12 moles of carbon dioxide. From here we can find moles o nitroglycerin knowing that:

\frac{n_{ng}}{4} = \frac{n_{CO_2}}{12} \therefore n_{ng} = \frac{4}{12}n_{CO_2} = \frac{1}{3}\cdot 1.928 mol = 0.6427 mol

Multiplying the number of moles of nitroglycerin by its molar mass will yield the mass of nitroglycerin decomposed:

m_{ng} = n_{ng}\cdot M_{ng} = 0.6427 mol\cdot 227.09 g/mol = 146.0 g

3 0
3 years ago
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