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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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Nitrogen dioxide decomposes to nitric oxide and oxygen via the reaction: 2NO2 → 2NO + O2 In a particular experiment at 300 °C, [
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Gastric juice is made up of substances secreted from parietal cells, chief cells, and mucous-secreting cells. The cells secrete
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Answer:

The amount of energy required to transport hydrogen ions from a cell into the stomach is 37.26KJ/mol.

Explanation:

The free change for the process can be written in terms of its equilibrium constant as:

ΔG° = -RTInK_(eq)

where:

R= universal gas constant

T= temperature

K_eq= equilibrum constant for the process

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ΔG= -nFE°

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F = faraday's constant

n = number of electrons exchanged in the process; and  

E = standard cell potential

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ΔG° = -RTInK_(eq)

where;

[texK_eq[/tex]=\frac{[H^+]_(cell)}{[H^+(stomach lumen)]}

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H^+_{inside} ⇒ H^+_{outside}

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K_{eq}=\frac{[H^+]_{outside}}{[H^+]_{inside}}

=\frac{[H^+]_{cell}}{[H^+]_{stomach lumen}}

[H^+]_{cell}= 10⁻⁷⁴

=3.98 * 10⁻⁸M

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=7.94 * 10⁻³M

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K_{eq}=\frac{[H^+]_{cell}}{[H^+]_{stomachlumen}}

=\frac{3.98*10^{-8}}{7.94*10{-3}}

= 5.012 × 10⁻⁶

Furthermore, free energy change for this reaction is related to the equilibrium concentration given as:

ΔG° = -RTInK_(eq)

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=37° C + 273.15K

=310.15K; and

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If the potential difference across the cell membrane= 60.0mV.

Energy required to cross the cell membrane will be:

ΔG₂ = -nFE°_{membrane}

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= 5.79KJ

Therefore, for one mole of electron transfer across the membrane; the energy required is 5.79KJmol⁻¹

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Δ G_{total} = G_{1}+G_{2}

= (31.47+5.79) KJmol⁻¹

= 37.26KJmol⁻¹

We can therefore conclude that;

   The amount of energy required to transport ions from cell to stomach lumen is 37.26KJmol⁻¹

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3 years ago
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