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Liula [17]
2 years ago
12

Electrons are found in energy levels around the nucleus of an atom. Energy levels can also be called ""shells."" The energy leve

ls are designated by a principal quantum number (n). The lowest energy level is designated as n=1 whereas the highest energy level is n=7. Which energy level is the closest to the nucleus?
Chemistry
1 answer:
melamori03 [73]2 years ago
5 0

Answer:

<u>The first electronic shell or energy level, having n= 1, called the K shell, is the closest to the nucleus.</u>

Explanation:

In an atom, the sub atomic particles called electrons, revolve around the nucleus of the given atom at different <em>energy levels</em>, called <em>shells</em>.

The energy levels or the electronic shells are denoted by the <em>principal quantum number</em>  <em>(n)</em> and has the following values n = 1, 2, 3, 4, 5, 6, 7; <em>from the innermost shell to the outermost.</em>

The electron shells are also labelled by the alphabets, K, L, M, N, O, P, and Q, <em>from the innermost shell to the outermost.</em>

<u><em>Thus the first electronic shell having n= 1, called K shell, is the lowest energy level and is present closest to the nucleus.</em></u>

<u><em>Whereas, the last electronic shell having n =7, called Q shell, is the highest energy level and is farthest from the nucleus.</em></u><u> </u>

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Answer:

At -13 ^{0}\textrm{C} , the gas would occupy 1.30L at 210.0 kPa.

Explanation:

Let's assume the gas behaves ideally.

As amount of gas remains constant in both state therefore in accordance with combined gas law for an ideal gas-

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where P_{1} and P_{2} are initial and final pressure respectively.

           V_{1}  and V_{2} are initial and final volume respectively.

           T_{1} and T_{2} are initial and final temperature in kelvin scale respectively.

Here P_{1}=150.0kPa , V_{1}=1.75L , T_{1}=(273-23)K=250K, P_{2}=210.0kPa and V_{2}=1.30L

Hence    T_{2}=\frac{P_{2}V_{2}T_{1}}{P_{1}V_{1}}

            \Rightarrow T_{2}=\frac{(210.0kPa)\times (1.30L)\times (250K)}{(150.0kPa)\times (1.75L)}

            \Rightarrow T_{2}=260K

            \Rightarrow T_{2}=(260-273)^{0}\textrm{C}=-13^{0}\textrm{C}

So at -13 ^{0}\textrm{C} , the gas would occupy 1.30L at 210.0 kPa.

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Hope it helped!
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