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weqwewe [10]
3 years ago
10

4. You have a very concentrated solution (12 M) of potassium chloride (KCI). You need it to be at the least concentration possib

le for the experiment you are
about to conduct. The problem is you forgot to order lab supplies, so you only have 2 L of distilled water left. What would be the final concentration if you
added the two 2 L of distilled water to the 0.5 L of 12 M KCI?
a 3.0 M
b 24 M
C 2.4M
d 48 M
Chemistry
1 answer:
Alenkinab [10]3 years ago
4 0

Answer:

C. 2.4 M

Explanation:

Because you started with 12M solution of KCl, that means that there was a concentration of 12 mols of KCl per Liter of solution. (12mol/1L =12 M). Since there was only 0.5L of solution, there was only 6 mols of KCl because there is only 12 mols per 1 Liter, so half of that volume would have half the amount of solute to keep that true. (6mol/.5L = 12 M). With the new water added, the formula becomes (6mols KCl/ 2.5L of water =2.4 M)

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The equilibrium constant, k of the reaction in which case, the concentrations of the given reactants and products are as indicated is; Choice A; K = 3.1 x 10⁵

<h3>What is the equilibrium constant , k of the reaction as described in the task content?</h3>

It follows from above that the concentrations of the reactants and products are as follows; [H2] = 0.10 M, [N2] = 0.10 M, and [NH3] = 5.6 M at equilibrium.

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This decomposition is first order with respect to phosphine, and has a half‑life of 35.0 s at 953 K. Calculate the partial press
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