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x(2) YouTube Not secure ng·cengage.com/static/nb/ui/index.html?nbld-737561 &nbNodeld;=278405395&eISBN-97813056575714;parentld-278405501 Labor Force Participi e A Mass M1 3.1kg e A Block with Mass M e A Block With Mass M e A Block With Mass l BOYS SOLID JERSEY Kid\'s Short Sleeve Cl ::: Apps MINDTAP Delio Cunha 16 Challenge Due Tomorrow at 8 AM CST 0 Late Submission until Mar 7 at 8 AM CST is allowed, with one time 10% penalty to submitted score. Question 1 Use the Keferences to access important values if needed for this question. Question 2 A typical coal-fired electric generating planwi bur about 3 metric tons of coal per hour. Question 3 Most of the coal burned in the United States contains 1 to 4 % by weight sulfur in the form of pyrite, which is oxidized as the coal burns: Question Ouce in the atmosphere, the SO is oxidized to SO3, which then reacts with water in the atmosphere to form sulfuric acid: Question 5 68 2 metric tons of coal that contains 2.25 % by weight $ 1 burned and all of the sulfure acc that is formed rans down into a pond of dimensions 371 m x 273 m x 5.42 m what is the pH of the poad? (OK to assume 2 mol H3 per mole HyS04) Question Hint metric ton-1000 kg Question i7 Question 011 Subiriil Ariswer 4 queslion alernpls rernaining Current score: 58 pts (62.5 %) Autosaved at 3:55 AM video-1520018133.mp4 ShowX Type here to search 4:07 AM 3/4/2018

Solution

Ans. Step 1: Amount of S in coal = 2.25 % x Mass of coal

                                                = 2.25 % x (68.2 x 10³ kg)

                                                = 1.5345 x 10³ kg

                                                = 1.5345 x 10⁶ g

                                                = 1.5345 x 10⁶ g / (32.066 g/ mol)

                                                = 4.7854 x 10⁴ moles

# Since 1 mol H₂SO₄ consists of 1 mol S-atom; 1 mol S ultimately produces 1 mol H₂SO₄ as mentioned

So, moles of H₂SO₄ produced = 4.7854 x 10⁴ moles = Moles of S in coal.

# Step 2: Volume of pond = 371 m x 273 m x 5.42 m = 548953.86 m³

                                                = 5.4895 x 10⁸ L

Now,

            Molarity of H₂SO₄ = Moles of H₂SO₄ / Volume of solution in liters

                                                = 4.7854 x 10⁴ moles / (5.4895 x 10⁸ L)

                                                = 8.7174 x 10^-5 M

# Step 3: 1 mol H₂SO₄ donates 2 mol H⁺ (= H₃O⁺).

So,

            Total molarity of H₃O⁺ in pond = 2 x [H₂SO₄]

                                                = 2 x 8.7174 x 10^-5 M

                                                = 1.7435 x 10^-4 M

Now,

            pH = -log [H₃O⁺] = -log (1.7435 x 10^-4) = 3.7586

Therefore, pH of the pond = 3.76