All categories

3 years ago

Which of these elements has an atom with the most stable outer electron configuration? * Cl Ca Ne Na

Chemistry

1 answer:

SashulF [63]3 years ago

6 0

Answer:

Neon (Ne) has the most stable outer electron configuration because the outer electron is completely filled and it has octet structure

Explanation:

The configuration of these elements is as follows;

Cl₁₇ = 2, 8,7 (the outer electron is 7)

Ca₂₀ = 2,8,8,2 (the outer electron is 2)

Ne₁₀ = 2,8 (the outer electron is 8)

Na₁₁ = 2,8,1 (the outer electron is 1)

Based on the outer electron value above, Neon (Ne) has the most stable outer electron configuration because the outer electron is completely filled and it has octet structure.

You might be interested in

if three oxygen particles are needed to form ozone, how many units of ozone could be formed from 6 oxygen particles? from 9? fro

tensa zangetsu [6.8K]

<h3>Answers:</h3>

1) 2 Units of Ozone

2) 3 Units of Ozone

3) 9 Units of Ozone

<h3>Solution:</h3>

1) From 6 Oxygen Particles;

As given,

3 Oxygen Particles form = 1 Unit of Ozone

So,

6 Oxygen Particles will form = X Units of Ozone

Solving for X,

X = (6 O Particles × 1 Unit of Ozone) ÷ 3 O Particles

X = 2 Units of Ozone

2) From 9 Oxygen Particles;

As given,

3 Oxygen Particles form = 1 Unit of Ozone

So,

9 Oxygen Particles will form = X Units of Ozone

Solving for X,

X = (9 O Particles × 1 Unit of Ozone) ÷ 3 O Particles

X = 3 Units of Ozone

3) From 27 Oxygen Particles;

As given,

3 Oxygen Particles form = 1 Unit of Ozone

So,

27 Oxygen Particles will form = X Units of Ozone

Solving for X,

X = (27 O Particles × 1 Unit of Ozone) ÷ 3 O Particles

X = 9 Units of Ozone

3 0

3 years ago

Suppose you have just added 100 ml of a solution containing 0.5 mol of acetic acid per liter to 400 ml of 0.5 m naoh. what is th

Tpy6a [65]

$pH = 13.5$

Explanation:

Sodium hydroxide completely ionizes in water to produce sodium ions and hydroxide ions. Hydroxide ions are in excess and neutralize all acetic acid added by the following ionic equation:

$\text{HAc} + \text{OH}^{-} \to \text{Ac}^{-} + \text{H}_2\text{O}$

The mixture would contain

$0.4 \times 0.5 - 0.1 \times 0.5 = 0.15 \; \text{mol}$ of $\text{OH}^{-}$ and
$0.1 \times 0.5 = 0.05 \; \text{mol}$ of $\text{Ac}^{-}$

if $\text{Ac}^{-}$ undergoes no hydrolysis; the solution is of volume $0.1 + 0.4 = 0.5 \; \text{L}$ after the mixing. The two species would thus be of concentration $0.30 \; \text{mol} \cdot \text{L}^{-1}$ and $0.10 \; \text{mol} \cdot \text{L}^{-1}$ , respectively.

Construct a RICE table for the hydrolysis of $\text{Ac}^{-}$ under a basic aqueous environment (with a negligible hydronium concentration.)

$\begin{array}{cccccccc} \text{R} & \text{Ac}^{-}(aq) &+ & \text{H}_2\text{O}(aq) & \leftrightharpoons & \text{HAc}(aq) & + & \text{OH}^{-} (aq)\\ \text{I} & 0.10 \; \text{M} & & & & & &0.30 \; \text{M}\\ \text{C} & -x \; \text{M}& & & & +x \; \text{M}& & +x \; \text{M} \\ \text{E} & (0.10 - x) \; \text{M} & & & & x \; \text{M} & & (0.30 +x) \; \text{M} \end{array}$

The question supplied the <em>acid</em> dissociation constant $pK_a$ for acetic acid $\text{HAc}$ ; however, calculating the hydrolysis equilibrium taking place in this basic mixture requires the <em>base</em> dissociation constant $pK_b$ for its conjugate base, $\text{Ac}^{-}$ . The following relationship relates the two quantities:

$pK_{b} (\text{Ac}^{-}) = pK_{w} - pK_{a}( \text{HAc})$

... where the water self-ionization constant $pK_w \approx 14$ under standard conditions. Thus $pK_{b} (\text{Ac}^{-}) = 14 - 4.7 = 9.3$ . By the definition of $pK_b$ :