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olga_2 [115]
3 years ago
8

How many moles of KCl are required to prepare 1.50 L of 5.4 M KCl?

Chemistry
2 answers:
puteri [66]3 years ago
6 0
5.4 M = moles of solute / 1.50 L 

<span>Multiply both sides by 1.50 L to isolate moles of solute on the right. </span>

<span>8.1 mol = moles of solute </span>


Cerrena [4.2K]3 years ago
6 0

8.1 mol KCI

This would be the answer.

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2NO (g) + O2 (g) →2NO2 (g) At equilibrium [NO] = 2.4 × 10 -3 M, [O2] = 1.4 × 10 -4 M, and [NO2] = 0.95 M.
azamat

Answer:

K=1.12x10^9

Explanation:

Hello there!

Unfortunately, the question is not given in the question; however, it is possible for us to compute the equilibrium constant as the problem is providing the concentrations at equilibrium. Thus, we first set up the equilibrium expression as products/reactants:

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