Hey student made a model of a river in a sandbox a pile of sand represents soil and bits of rocks in the riverbed water pouring
2 answers:
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Answer:
Explanation:
Given parameters;
pH = 8.74
pH = 11.38
pH = 2.81
Unknown:
concentration of hydrogen ion and hydroxyl ion for each solution = ?
Solution
The pH of any solution is a convenient scale for measuring the hydrogen ion concentration of any solution.
It is graduated from 1 to 14
pH = -log[H₃O⁺]
pOH = -log[OH⁻]
pH + pOH = 14
Now let us solve;
pH = 8.74
since pH = -log[H₃O⁺]
8.74 = -log[H₃O⁺]
[H₃O⁺] = 10⁻
[H₃O⁺] = 1.82 x 10⁻⁹mol dm³
pH + pOH = 14
pOH = 14 - 8.74
pOH = 5.26
pOH = -log[OH⁻]
5.26 = -log[OH⁻]
[OH⁻] = 10
[OH⁻] = 5.5 x 10⁻⁶mol dm³
2. pH = 11.38
since pH = -log[H₃O⁺]
11.38 = -log[H₃O⁺]
[H₃O⁺] = 10⁻
[H₃O⁺] = 4.17 x 10⁻¹² mol dm³
pH + pOH = 14
pOH = 14 - 11.38
pOH = 2.62
pOH = -log[OH⁻]
2.62 = -log[OH⁻]
[OH⁻] = 10
[OH⁻] =2.4 x 10⁻³mol dm³
3. pH = 2.81
since pH = -log[H₃O⁺]
2.81 = -log[H₃O⁺]
[H₃O⁺] = 10⁻
[H₃O⁺] = 1.55 x 10⁻³ mol dm³
pH + pOH = 14
pOH = 14 - 2.81
pOH = 11.19
pOH = -log[OH⁻]
11.19 = -log[OH⁻]
[OH⁻] = 10
[OH⁻] =6.46 x 10⁻¹²mol dm³
Answer:
11.94 grams of carbon dioxide were originally present.
19.94 grams of krypton can you recover.
Explanation:
Mass of carbon dioxide gas = x
Mass of krypton gas = y
x + y = 31.7 g
Moles of carbon dioxide gas =
Moles of krypton gas =
Mole fraction of krpton =
Total pressure of the mixture = P = 0.665 atm
Partial pressure of carbon dioxide gas = p
Partial pressure of krypton gas before removal of carbon dioxide gas = p'
Partial pressure of krypton gas after removal of carbon dioxide gas = p'' = 0.309 atm
p' = p'' = 0.309 atm
0.665 atm = p + 0.309 atm
p = 0.665 atm - 0.306 atm = 0.359 atm
Partial pressure of krypton can also be given by :
..[2]
Solving [1] and [2]:
x = 11.94 g
y = 19.76 g
11.94 grams of carbon dioxide were originally present.
19.94 grams of krypton can you recover.