Answer:
1.76
Explanation:
There is some info missing. I think this is the original question.
<em>A chemist dissolves 660.mg of pure hydroiodic acid in enough water to make up 300.mL of solution. Calculate the pH of the solution. Be sure your answer has the correct number of significant digits.</em>
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Step 1: Calculate the molarity of HI(aq)
M = mass of solute / molar mass of solute × liters of solution
M = 0.660 g / 127.91 g/mol × 0.300 L
M = 0.0172 M
Step 2: Write the acid dissociation reaction
HI(aq) ⇄ H⁺(aq) + I⁻(aq)
HI is a strong acid, so [H⁺] = 0.0172 M
Step 3: Calculate the pH
pH = -log [H⁺]
pH = -log 0.0172
pH = 1.76
Answer:
Explanation:
Groundwater is stored in the open spaces within rocks and within unconsolidated sediments. Rocks and sediments near the surface are under less pressure than those at significant depth and therefore tend to have more open space. For this reason, and because it’s expensive to drill deep wells, most of the groundwater that is accessed by individual users is within the first 100 m of the surface. Some municipal, agricultural, and industrial groundwater users get their water from greater depth, but deeper groundwater tends to be of lower quality than shallow groundwater, so there is a limit as to how deep we can go.
Mass Molar of
Ca = 3*40 = 120 amu
P = 2*31= 62 amu
O = (16*4)*2 = 64*2 = 128 amu
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Mass Molar of
= 120 + 62 + 128 = 310 g/mol
Therefore: <span>What is the gram formula mass of Ca3(PO4)2 ?
</span>Answer:
310 grams
POH will be -log[conc of OH]
-log (3.9E-08) = 7.409
pH = 14- pOH
pH = 14 - 7.409
pH = 6.59
