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Ulleksa [173]
3 years ago
7

Study the electron dot diagrams for lithium, carbon, fluorine, and neon in Figure 6-1. Choose the statement that correctly ident

ifies the most stable of the elements.
A. Lithium is the most stable element because it has to lose only one electron to achieve a stable configuration.
B. Carbon is the most stable element because it can form four bonds.
C. Fluorine is the most stable element because it has to gain only one electron to achieve a stable configuration.
D. Neon is the most stable element because its highest occupied energy level is filled.

Chemistry
2 answers:
snow_lady [41]3 years ago
6 0
D.
The completely filled valence shell of Neon makes it most stable.
statuscvo [17]3 years ago
5 0

Answer is: D. Neon is the most stable element because its highest occupied energy level is filled.

Neon (symbol: Ne) is an element (noble gas) with atomic number 10, which means it has 10 protons and 10 electrons.

Electron configuration of neon atom: ₁₀Ne 1s²2s²2p⁶.

Noble gases are in group 18: helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe) and radon (Rn). They have very low chemical reactivity.

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3 years ago
Write a net ionic equation for the overall reaction that occurs when aqueous solutions of sodium hydroxide and hydrosulfuric aci
Helga [31]

Answer:

2OH-(aq) + 2H+(aq) → 2H2O(l)

Explanation:

Step 1: Data given

sodium hydroxide = NaOH

hydrosulfuric acid = H2S

Step 2: The unbalanced equation

NaOH(aq) + H2S(aq) → Na2S(aq) + H2O(l)

Step 3: Balancing the equation

On the left side we have 1x Na (in NaOH), on the right side we have 2x Na (in Na2S). To balance the amount of Na on both sides, we have to multiply NaOH on the left side by 2.

2NaOH(aq) + H2S(aq) → Na2S(aq) + H2O(l)

On the left side we have 4x H (2x in NaOH and 2x in H2S), on the right side we have 2x H (in H2O). To balance the amount of H on both sides, we have to multiply H2O on the right side by 2. Now the equation is balanced.

2NaOH(aq) + H2S(aq) → Na2S(aq) + 2H2O(l)

Step 4: The net ionic equation

2Na+(aq) + 2OH-(aq) + 2H+(aq) + S^2-(aq) → 2Na+(aq) + S^2-(aq) + 2H2O(l)

The net ionic equation, for which spectator ions are omitted - remember that spectator ions are those ions located on both sides of the equation - will , after canceling those spectator ions in both side, look like this:

2OH-(aq) + 2H+(aq) → 2H2O(l)

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