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2 years ago

Strontium and chlorine combine in only one ratio: one strontium atom for every two chlorine atoms. Based on this

Chemistry

1 answer:

Vinvika [58]2 years ago

4 0

The combinations of Strontium and chlorine that are possible are only those in which strontium and chlorine combine in the ratio of 1:2.

1) The possible combinations are

4 strontium atoms and 8 chlorine

$2.7 * 10^19$ strontium atoms and $5.4 * 10^19$ chlorine atoms

2) The combinations that are not possible are

20 strontium atoms and 60 chlorine atoms

130 billion strontium atoms and 195 billion chlorine

We have to work out the ratio of Strontium and chlorine in each of the given combinations in the question. Only the combinations in which the ratio of Strontium and chlorine is 1:2 is possible.

First case:

4 strontium atoms and 8 chlorine atoms gives a Strontium and chlorine ratio of 1:2 so it is possible.

Second case:

20 strontium atoms and 60 chlorine atoms gives a Strontium and chlorine ratio of 1:3 hence it is not possible.

Third case:

$2.7 * 10^19$ strontium atoms and $5.4 * 10^19$ chlorine atoms gives a Strontium and chlorine ratio of 1:2 hence it is possible.

Fourth case:

130 billion strontium atoms and 195 billion chlorine atoms gives a Strontium and chlorine ratio of 1:1.5 hence it is not possible.

Learn more: brainly.com/question/9743981

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2 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 acid dissociation constant $pK_a$ for acetic acid $\text{HAc}$ ; however, calculating the hydrolysis equilibrium taking place in this basic mixture requires the base 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 $pK_b$ :