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Liula [17]
3 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]3 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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A sample of a gas in a balloon has a volume of 2.22 L and temperature of 23.9 °C. Calculate the volume when the temperature is r
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Answer:

Volume = 4.28L

Explanation:

Charles's gas law relates the volume and temperature of a gas at constant pressure. This law says that at constant pressures if we raise the temperature of a gas it will expand and if we reduce the temperature the volume will decrease. The formula is as follows:

{\displaystyle {\frac {V_{1}}{T_{1}}}={\frac {V_{2}}{T_{2}}}}

So, the initial conditions are 2.22L and 23.9 °C ant the final conditions are 46.1 °C we replace them in the equation. And then we solve it.

{\displaystyle {\frac {2.22}{23.9}}={\frac {V_{2}}{46.1}}

{\displaystyle {\frac {2.22}{23.9}} {46.1}={V_{2}

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Explanation:

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3 years ago
Read 2 more answers
A 1.0 g sample of propane, C3H8, was burned in calorimeter. The temperature rose from 28.5 0C to 32.0 0C and heat of combustion
tensa zangetsu [6.8K]

Answer:

A 1.0 g sample of propane, C3H8, was burned in the calorimeter.

The temperature rose from 28.5 0C to 32.0 0C and the heat of combustion 10.5 kJ/g.

Calculate the heat capacity of the calorimeter apparatus in kJ/0C

Explanation:

Heat of combustion = heat capacity of calorimeter * deltaT\\

Given,

The heat of combustion = 10.5kJ/g.

deltaT = (32.0-28.5)^oC\\deltaT = 3.5^oC

Substitute these values in the above formula to get the value of heat capacity of the calorimeter.

deltaT =heat  capacity of calorimeter   * (change in temperature)\\10.5kJ/g = heat  capacity of calorimeter * (3.5^oC)\\\\=>heat capacity of calorimeter = \frac{10.5kJ/g}{3.5^oC} \\=>heat capacity of calorimeter = 3.0 kJ/g.^oC

Answer:

The heat capacity of the calorimeter is 3.0kJ/g.^oC.

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