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grin007 [14]
4 years ago
10

What is the molarity of 1.5 liters of an aqueous solution that contains 52 grams of lithium fluoride, LiF, (gram-formula mass =2

6 grams/mole)?
(1) 1.3 M (3) 3.0 M
(2) 2.0 M (4) 0.75 M
Chemistry
1 answer:
denis23 [38]4 years ago
5 0
The answer is (1) 1.3 M. The first thing you need to do is to convert the unit of gram to mole. The mol number of LiF is 52/26=2 mol. Then using the volume to calculate the molarity: molarity=2/1.5=1.3 M.
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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:

K=\frac{[NO_2]^2}{[NO]^2[O_2]}

Then, we plug in the concentrations at equilibrium to obtain the equilibrium constant as follows:

K=\frac{(0.95)^2}{(0.0024)^2(0.00014)}\\\\K=1.12x10^9

In addition, we can infer this is a reaction that predominantly tends to the product (NO2) as K>>>>1.

Best regards!

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Answer:

Explanatio

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