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Sav [38]
3 years ago
14

A pharmacist has calculated that a patient requires 30 mmol of phosphate and 80 meq of potassium to be added to the pn. how many

milliliters of potassium phosphate and how many milliliters of potassium chloride will be required? [k-phos is 3 mmol phosphate with 4.4 meq potassium per ml and kcl is 2 meq of potassium per ml.]
Chemistry
1 answer:
aalyn [17]3 years ago
8 0

Since potassium and phosphate is what we are to find for and they are both found in the potassium phosphate solution, therefore we solve for this one first on the basis of the phosphate.

The formula for finding the volume given the concentration and number of moles is:

Volume = number of moles / concentration in Molarity

Volume potassium phosphate required = 30 mmol phosphate / (3 mmol / mL)

<u>Volume potassium phosphate required = 10 mL</u>

This would also contain potassium in amounts of:

Amount of potassium in potassium phosphate = 10 mL (4.4 meg / mL)

Amount of potassium in potassium phosphate = 44 meg

 

Therefore the potassium chloride required is:

Volume of potassium chloride = (80 meg – 44 meg) / (2 meg / mL)

<span><u>Volume of potassium chloride = 72 mL</u></span>

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Calculate Ecell at 80 ºC for a voltaic cell based on the following redox reaction: H2(g, 1.25 atm) + 2AgCl(s) → 2Ag(s) + 2H+(aq,
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Answer:

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the half-cell reactions for a voltanic cell

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anode(oxidation): 2AgCl(s) ------- 2Ag⁺(aq) + 2Cl⁻ + 2e⁻

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Q = [H⁺]/[Cl⁻]

sub for [H+] = 0.10M and [Cl-] = 1.5M

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Q = 0.067

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Unit of measurement
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