MW of K=34 equivalent weight of k=34/1 since the valence of k=1 1mEq of k=1/1000 *34=0.034g 1mEq/0.034=80mEq/X X=2.72g MW of K=34 equivalent weight of k=34/1 since the valence of k=1 1mEq of k=1/1000 *34=0.034g 1mEq/0.034=80mEq/X X=2.72g
20.5
4 milliequivalents of sodium chloride solution is a solution having 0,2338 g in 1 L.
No. You'd need about 4 tablets of 99 mg KCl to equal 10 mEq of prescription-strength KCl, 8 tablets to equal 20 mEq.
100 grams
It is not possible to answer the question since there is no recognised measurement called milliequivalents.
708.1mg
I Don't knows Sorry
16.5
You did not describe the amount of potassium bicarbonate amount in grams in your question. But if you are about 1 gram of potassium bicarbonate it will be 0.0099 moles in one gram of potassium bicarbonate. 0.0199 moles in 2 grams of potassium bicarbonate.
34,7 moles of potassium 1 356,7 g.
530,3 g potassium iodide are needed.
3.99 or 4