Answer:
The concentration of monosodium phosphate is 0.1262M
Explanation:
The buffer of H₂PO₄⁻ / HPO₄²⁻ (Monobasic phosphate and dibasic phosphate has a pKa of 7.2
To determine the pH you must use Henderson-Hasselbalch equation:
pH = pKa + log [A⁻] / [HA]
<em>Where [A⁻] is molarity of the conjugate base of the weak acid, [HA].</em>
For H₂PO₄⁻ / HPO₄⁻ buffer:
pH = 7.2 + log [HPO₄⁻² ] / [H₂PO₄⁻]
As molarity of the dibasic phosphate is 0.2M and you want a pH of 7.4:
7.4 = 7.2 + log [0.2] / [H₂PO₄⁻]
0.2 = log [0.2] / [H₂PO₄⁻]
1.58489 = [0.2] / [H₂PO₄⁻]
[H₂PO₄⁻] = 0.1262M
<h3>The concentration of monosodium phosphate is 0.1262M</h3>
<em />
<span>0.310 moles
First, look up the atomic weights of the elements involved.
Atomic weight carbon = 12.0107
Atomic weight hydrogen = 1.00794
Atomic weight sulfur = 32.065
Molar mass (C3H5)2S = 6 * 12.0107 + 10 * 1.00794 + 32.065
= 114.2086 g/mol
Moles (C3H5)2S = 35.4 g / 114.2086 g/mol = 0.309959145 mol
Since there's just one sulfur atom per (C3H5)2S molecule, the number of moles of sulfur will match the number of moles of (C3H5)2S which is 0.310 when rounded to 3 significant digits.</span>
The answer is A how it react with other chemicals
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2.26%, 26.8%, 2 2/5, 2.62, 271%
Hope this helps! :D
I think you forgot to attach a picture
