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Butoxors [25]
3 years ago
6

A chemist adds of a M mercury(II) iodide solution to a reaction flask. Calculate the mass in milligrams of mercury(II) iodide th

e chemist has added to the flask. Round your answer to significant digits.
Chemistry
1 answer:
Oksi-84 [34.3K]3 years ago
7 0

Answer:

0.42 mg

Explanation:

It seems the question is incomplete, however I'll use values that have been found in a web search:

"A chemist adds 55.0 mL of a 1.7x10⁻⁵ M mercury(II) iodide solution to a reaction flask. Calculate the mass in milligrams of mercury(II) iodide the chemist has added to the flask. Round your answer to 2 significant digits."

<u>Keep in mind that while the process of solving the problem remains the same, if the values in your question are different, your answer will differ as well</u>.

First we <u>calculate the moles of mercury (II) iodide</u> (HgI₂):

  • 1.7x10⁻⁵M * 55.0 mL = 9.35x10⁻⁴ mmol HgI₂

Then we<u> convert mmol to mg, using the molar mass</u> (454.4 g/mol):

  • 9.35x10⁻⁴ mmol HgI₂* 454.4 mg/mmol = 0.42 mg
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3 years ago
Sulfur dioxide and oxygen react to form sulfur trioxide during one of the key steps in sulfuric acid synthesis. An industrial ch
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Answer:

Kp = 0.049

Explanation:

The equilibrium in question is;

2 SO₂ (g)  +  O₂ (g)   ⇄ 2 SO₃ (g)  

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The initial pressures are given, so lets set up the ICE table for the equilibrium:

atm        SO₂         O₂          SO₃

I              3.3        0.79           0

C              -2x           -x          2x

E             3.3 - 2x    0.79 - x    2x

We are told 2x = partial pressure of SO₃ is 0.47 atm at equilibrium, so we can determine the partial pressures of  SO₂ and O₂ as follows:

p SO₂  = 3.3 -0.47 atm = 2.83 atm

p O₂ = 0.79 - (0.47/2) atm = .56 atm

Now we can calculate Kp:

Kp = 0.47² /[ ( 2.83 )² x 0.56 ] = 0.049 ( rounded to 2 significant figures )

Note that we have extra data in this problem we did not need since once we setup the ICE table for the equilibrium we realize we have all the information needed to solve the question.

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