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mr Goodwill [35]
3 years ago
15

The size of the orbital is determined by the

Chemistry
1 answer:
Hatshy [7]3 years ago
3 0

Answer: The size of the orbital is determined by the principal quantum number, so the size of the orbital increases as this value increases. Therefore, an electron in a (n-1) orbital is closer to the nucleus than is an electron in a 'n' orbital.

Explanation:

In an atom, the position and energy of an electron is described by a set of numbers and these sets are called quantum numbers.  

There are four quantum numbers. These are as follows.

1). Principal quantum number - This is denoted by "n" and it determines the size and energy of shell in which electron is present. The value of "n" can be 1, 2, 3, and so on but it can never be equal to zero.

2). Azimuthal quantum number - This is denoted by "l" and it determines the shape of an orbital. For s, p, d and f-shell the values of "n" will be 0, 1, 2, 3. The value of l can vary from -n to +n.

3). Magnetic quantum number - This is denoted by "m_{l}." and it determines the orientation of an orbital. The value of ml can vary from -l to +l.

4). Spin quantum number -- This is denoted by "m_{s}" and it determines the spin of an electron. It is independent of the values of n, l and m_{l}.

This means that the size of an orbital is determined by principal quantum number. Lower is the value of 'n' (principal quantum number) more closer will be an electron to the nucleus. Hence, more is the value of 'n' more will be the size of nucleus and vice-versa.

For example, an electron present in a 2s-orbital is closer to the nucleus as compared to the electron present in a 3s-orbital.

Thus, we can conclude that the size of the orbital is determined by the principal quantum number, so the size of the orbital increases as this value increases. Therefore, an electron in a (n-1) orbital is closer to the nucleus than is an electron in a 'n' orbital.

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Burning a magnesium ribbon in the air is an addition reaction while heating potassium manganate 7 is a decomposition reaction.

<h3>Addition and decomposition reactions</h3>

Magnesium burns in air to produce magnesium oxide as follows:

2Mg + O_2 --- > 2MgO

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2 KMnO_4 --- > K_2MnO_4 + MnO_2(s) + O_2

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More on reactions can be found here: brainly.com/question/17434463

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Using the equation below, if you have 4.3 mol of nitrogen tribromide and
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Answer:

sodium hydroxide is the limiting reactant

Explanation:

The first step is usually to put down the balanced reaction equation. This is the first thing to do when solving any problem related to stoichiometry. The balanced reaction equation serves as a guide during the solution.

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Let us pick nitrogen gas as our product of interest. Any of the reactants that gives a lower number of moles of nitrogen gas is the limiting reactant.

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Volgvan

Answer:

Yes

Explanation:

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8 0
3 years ago
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2. How many molecules are contained in 25 L of N₂ at S. T.P.?
irina [24]

Explanation:

How many nitrogen molecules are in 1 liter of nitrogen gas at STP?

Answer

2

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Pete Gannett

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Ph.D. Chemistry, University of Wisconsin-Madison, (1982)2y

Seems to be an ideal gas law question. The relevant equation is:

PV = nRT

where P is the pressure in atmospheres, V is the volume in liters, n is the number of moles of gas, R is the gas constant (0.082 atm-L/mole-deg K), and T is temperature in Kelvins. STP means standard temperature and pressure and this is taken as 1 atm and 0º C or 273 K.

To calculate the number of molecules we will use the constant 6.023 * 10^23 molecules/mole and, therefore, we will need to know the number of moles (n). So, first we’ll rearrange the gas law equation, isolating ’n’ and then put the numbers in.

n = PV/RT = 1 * 1 / (0.082)(273) = 0.0447 moles

So, to calculate the number of molecules, multiple this by the number of molecules in a mole and you get:

# molecules of nitrogen in 1 Liter at STP = 6.023 * 10^23 molecules/mole * 0.0447 moles = 2.6905 * 10^22 molecules

Note, it does not matter what the gas is.

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1 year ago
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