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QveST [7]
3 years ago
15

Potassium hydroxide molar mass

Chemistry
1 answer:
Mnenie [13.5K]3 years ago
6 0

The answer I will give is an approximate number, but it will be close. The problem is that every periodic table is slightly different. Some round as I will, and some carry the mass of the elements out to 3 decimal places. You should go back and put the correct numbers in to the mass that I use.

Formula

KOH

Givens

  • K = 39
  • O = 16
  • H = 1

Solution

Molar Mass = 39 + 16 + 1

Molar Mass = 56 grams / mole

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2 years ago
Which action would shift this reaction away from solid calcium fluoride and toward the dissolved ions? which action would shift
ANTONII [103]
Hello!

The chemical reaction for the dissolving of calcium fluoride is the following:

CaF₂(s) ⇄ Ca⁺²(aq) + 2F⁻(aq)

In this reaction, and according to Le Chatelier's principle, the action that would shift this reaction away from solid calcium fluoride and towards the dissolved ions is the removing of fluoride ions.

Le Chatelier's principle
states that in an equilibrium reaction, the system would shift in the opposite direction of the changes. If we remove fluoride ions from the system, it will shift towards the formation of more fluoride ions by dissolving more Calcium Fluoride to achieve equilibrium again.

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3 years ago
Which statement best describes how Carl Woese changed the system of classification?
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Answer: C.)

Explanation:

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2 years ago
A 0.753 g sample of a monoprotic acid is dissolved in water and titrated with 0.250 M NaOH. What is the molar mass of the acid i
Alexeev081 [22]

Answer:

MM_{acid}=140.1g/mol

Explanation:

Hello,

In this case, since the acid is monoprotic, we can notice a 1:1 molar ratio between, therefore, for the titration at the equivalence point, we have:

n_{acid}=n_{base} \\\\V_{acid}M_{acid}=V_{base}M_{base}\\\\n_{acid}=V_{base}M_{base}

Thus, solving for the moles of the acid, we obtain:

n_{acid}=0.0215L*0.250\frac{mol}{L}=5.375x10^{-3}mol

Then, by using the mass of the acid, we compute its molar mass:

MM_{acid}=\frac{0.753g}{5.375x10^{-5}mol} \\\\MM_{acid}=140.1g/mol

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7 0
2 years ago
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