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Hunter-Best [27]
3 years ago
13

The molecular weight of potassium ferricyanide is 329.24 grams/mol. How would you make 1 liter of a 500mM potassium ferricyanide

solution?
Chemistry
1 answer:
True [87]3 years ago
6 0

Answer:

w = 164.62 g

Explanation:

molarity of a solution is given as -

Molarity (M)  =  ( w / m ) / V ( in L)

where ,

m = molecular mass ,

w = given mass ,

V = volume of solution ,

From the question ,

M = 500 mM = 0.5 M

( since , 1 mM =  1 / 100 M)

As we know , the molecular mass of potassium ferricyanide = 329.24 g/ mol

V = vol.of solution = 1 L

w = ?

<u>To find the value of w , using the above formula , and putting the respected values , </u>

Molarity (M)  =  ( w / m ) / V ( in L)

0.5  =  ( w / 329.24 ) / 1 L

w = 164.62 g

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Answer:

152 kPa = Partial pressure O₂

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Data by percent is the molar fraction . 100.

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Sum of molar fractions in a mixture = 1

0.68 + 0.32 = 1

If we apply the molar fraction, we can determine the partial pressure.

Mole fraction = Partial pressure / Total pressure

0.32 = Partial pressure O₂ / 475kPa →  0.32 . 475 kPa = Partial pressure O₂

152 kPa = Partial pressure O₂

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Formic acid, from the Latin formica, is the acid present in ants sting. What is the
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For the reaction of deprotonation of formic acid, the concentration of HCOO⁻ at equilibrium is 0.0151 M if the initial concentration of formic acid is 1.35 M.  

The reaction of deprotonation of formic acid is the following:

CHOOH + H₂O ⇄ HCOO⁻ + H₃O⁺

At the equilibrium we have:

CHOOH + H₂O ⇄ HCOO⁻ + H₃O⁺    (1)

1.35 - x                         x           x

The acid <em>equilibrium </em>constant for this reaction is:

K_{a} = \frac{[HCOO^{-}][H_{3}O^{+}]}{[CHOOH]} = 1.7 \cdot 10^{-4}  (2)

Entering the values of [CHOOH] = 1.35-x, [HCOO⁻] = [H₃O⁺] = x, into equation (2) we have:

1.7 \cdot 10^{-4} = \frac{[HCOO^{-}][H_{3}O^{+}]}{[CHOOH]} = \frac{x^{2}}{(1.35 - x)}  

1.7 \cdot 10^{-4}(1.35 - x) - x^{2} = 0

After solving the above <em>quadratic equation</em> and taking the positive value for x (<u>concentrations cannot be negative</u>), we have:

x = [HCOO^{-}] = [H_{3}O^{+}] = 0.0151 M

Therefore, the concentration of HCOO⁻ at equilibrium is 0.0151 M.

Learn more about the equilibrium constant here:

  • brainly.com/question/7145687?referrer=searchResults
  • brainly.com/question/9173805?referrer=searchResults

I hope it helps you!

3 0
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