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Arlecino [84]
4 years ago
15

generally speaking electrons are found in shells which we also call principal energy levels located at increasing distances from

the central nucleus the first ionization energy or ie with little sub 1 of an atom of an element describes the amount of energy required to remove the outermost electron from an atom of a_____________________.
Chemistry
1 answer:
xeze [42]4 years ago
7 0

Answer:

Gas

Explanation:

Ionization energy is the energy required to remove the most loosely held electron from a gaseous atom or ion. For the first Ionization energy, it is the energy needed to remove the most loosely bound electron from a gaseous atom

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Figure 2 shows what happens to the electrons in the outer shells when a sodium atom reacts with a fluorine atom.
QveST [7]

Sodium reacts with fluorine to make a covalent bond between them.  A covalent bond is formed when there is a mutual sharing of electrons between the two adjacent atoms.

Here, Sodium is metal and fluorine is non-metal. So, the bond between them may be an ionic bond, but as the Sodium has one valence electron and fluorine is highly electronegative. Sodium wants to lose electrons and fluorine wants to gain electrons to form a chemical bond. Therefore, the sodium donates its valence electron to fluorine, forms a cation, and fluorine accepts the electron and forms an anion and they distribute the difference in the charge between them.

So, we can say that when sodium and fluorine react to form sodium fluoride with a covalent bond.

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2 years ago
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Drag each tile to the correct box.
ElenaW [278]

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3 years ago
At 35°C, K = 1.6 × 10^-5 for the reaction
TEA [102]

Answer:

a) [NOCl] = 0.968 M

[NO] = 0.032M

[Cl²] = 0.016M

b) [NOCl] = 1.992M

[NO] = 0.008 M

[Cl2]  = 1.004 M

Explanation:

Step 1: Data given

Temperature = 35°C = 308K

K = 1.6 × 10^-5

Step 2: The reaction

2 NOCl(g) ⇌ 2 NO(g) + Cl2(g)

For 2 moles NOCl we'll have 2 moles NO and 1 mol Cl2

Step 3

a. 2.0 mol pure NOCl in a 2.0 L flask

Concentration at the start:

Concentration = mol / volume

[NOCl] = mol / volume

[NOCl] = 2.0 / 2.0 L

[NOCl] = 1.0 M

[NO] = 0 M

[Cl] = 0M

Concentration at the equillibrium

[NOCl] = 1.0M - 2x

[NO] = 2x

[Cl2]= x

K = [Cl2][NO]² / [NOCl]² = 1.6*10^-5

1.6*10^-5 = ((2x)² * x) / (1.0-2x)²

x = 0.016

[NOCl] = 1.0 -  2*0.016 = 0.968 M

[NO] = 2*0.016 = 0.032M

[Cl²] = 0.016M

b. 2.0 mol NOCl and 1.0 mol Cl2 in a 1.0 L flask

Concentration at the equillibrium

[NOCl] = 2.0 mol / 1.0 L = 2.0 M

[NO] = 0 M

[Cl2]= 1.0 mol / 1.0 L = 1.0 M

Concentration at the equillibrium

[NOCl] = 2.0M - 2x

[NO] = 2x

[Cl2]= 1.0 + x

K = [Cl2][NO]² / [NOCl]² = 1.6*10^-5

1.6 *10^-5 = (2x)²*(1.0+x) / ((2.0-2x)²)

1.6 *10^-5= (2x)² * 1 )/2.0²

1.6 *10^-5= 4x² / 4 = x²

x = \sqrt{1.6 *10^-5} = 4.0*10^-3

[NOCl] = 2.0 - 2*0.004 = 1.992M

[NO] = 2*0.004 = 0.008 M

[Cl2] = 1+ 0.004M = 1.004 M

5 0
3 years ago
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