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jeka94
3 years ago
12

How do the energy and the most probable location of an electron in the third shell of an atom compare to the energy and the most

probable location of an electron in the first shell of the same atom?
(1) In the third shell, an electron has more energy and is closer to the nucleus.
(2) In the third shell, an electron has more energy and is farther from the nucleus.
(3) In the third shell, an electron has less energy and is closer to the nucleus.
(4) In the third shell, an electron has less energy and is farther from the nucleus.
Chemistry
1 answer:
solmaris [256]3 years ago
7 0
The answer is (2). You can think about this question in terms of the Bohr's model of the atom or in terms of quantum chemistry. In the Bohr model, electrons exist in discrete "shells," each respresenting a fixed spherical distance from the nucleus in which electrons of certain energy levels orbit the nucleus. The larger the shell (the greater the "orbit" radius), the greater the energy of the "orbiting" electron (I use quotations because electrons don't actually orbit the nucleus in the traditional sense, as you may know). Thus, according to the Bohr model, a third shell electron should be farther from the nucleus and have greater energy than an electron in the first shell. The quantum model is differs drastically from the Bohr model in many ways, but the essence is the same. A larger principal quantum number indicates 1) greater overall energy and 2) a probability distribution spread a bit more outward.
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If 0.60 L of a solution contains 6.6 g of NaBr, what is its molar concentration?
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Answer:

Molarity = moles ÷ liters

to get moles of NaBr divide grams of NaBr by its molar mass (mass of Na + mass of Bromine)

Na = 22.989769

Br = 79.904

molar mass of NaBr = 102.893769

6.6g ÷ 102.893769 = 0.064143826 moles of NaBr

0.064143826 moles ÷ 0.60 liters = 0.1069 molar concentration or 11 %

7 0
3 years ago
Rolls of foil are 300mm wide and 2.020mm thick. (The density of foil is 2.7 g/cm^3). What maximum length of foil can be made fro
horrorfan [7]

Answer:  8556 mm, or 855.6 cm (8560 mm to 3 sig figs)

Explanation:  Convert mm to cm by dividing by 10 (1cm/10mm)

Find the area of the foil face in cm^2 (30cm*0.2020cm) = 0.606 cm^2

Calculate the volume occupied by 1.40 kg of foil in cm^3.  1.40kg = 1400g

1.400g/(2.7 g/cm^3) = 518.5 cm^3 for 1.40 kg Au

   Volume = Area (of the face) * Length  

We want Length:

Length = Volume/Area

L = (518.5 cm^3/0.606 cm^2)

L = 855.6 cm (8556 mm)  Round to 3 sig figs (856 cm and 8560 mm)

5 0
2 years ago
When aqueous solutions of potassium fluoride and hydrochloric acid are mixed, an aqueous solution of potassium chloride and hydr
Semmy [17]

Answer:

K⁺ (aq)  +  F⁻ (aq)  +  H⁺ (aq)  +  Cl⁻ (aq)  → KCl (aq) + H⁺ (aq)  +  F⁻ (aq)

Explanation:

KF (aq) +  HCl (aq) →  KCl (aq)  + HF (aq)

KF (aq) → K⁺ (aq) +  F⁻ (aq)

HCl (aq) →  H⁺ (aq)  +  Cl⁻ (aq)

KCl (aq) → K⁺ (aq) +  Cl⁻ (aq)

HF (aq) →  H⁺ (aq)  +  F⁻ (aq)

7 0
3 years ago
Flammability is a material’s ability to burn in the presence of
frosja888 [35]

Answer:

Flammability is a material’s ability to burn in the presence of <u><em>oxygen.</em></u>

Explanation:

Flammability can be described as the ability of a substance to get ignited. Flammability will lead to fire or combustion. Some substances are highly flammable like Benzene. Other tend to be just flammable. And there are also compounds which will nor be flammable at all as they won't react with oxygen. Examples of these substances include helium, steel or glass.

The flammability of a substance shall be considered a very important aspect when storing or transporting a substance.

5 0
2 years ago
Suppose that coal of density 1.5 g/cm^3 is pure carbon. (It is, in fact, much more complicated, but this is a reasonable first a
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Answer:

q = -6464.9 kJ

Explanation:

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q =  −394 kJ /mol C x 16.41 mol C = -6464.9 kJ

7 0
3 years ago
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