Help me with these calculations and please show your work if possible. Thank you
ID: 532560 • Letter: H
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Help me with these calculations and please show your work if possible.
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11. A 25 ml sample of river water was titrated with 0.001 M K2Cr2O (potassium dichromate) and required 8.3 ml to reach the end point. What is the chemical oxygen demand, in milligram of O2 per litre, of sample? 3pts 12. Gypsum (Calcium sulfate, CaSO4) has a solubility product constant of 2.4x10 5. a What is the solubility of CaSO4. B) Calculate the amount of calcium sulfate (CaSO4) dissolved in the lake in ppm (by mass). Assume a water density of 1.00 g cm 3, that other salts (e.g., limestone CaCO3) are negligible, and that there are no acid-base reactions. 2pts 13. Calculate the ThoD (theoretical oxygen demand of 100 mg/L glucose having the formula C6H1206. 2ptsExplanation / Answer
12) a) Solubility product constant(Ksp) of Gypsum (CaSO4) = 2.4×10-5 .
Let us consider , solubility of CaSO4 = S g/L
CaSO4 Ca2+ + SO42-
S S
Ksp = [Ca2+ ]×[ SO42- ]
= S2
ATQ,
S2 = 2.4×10-5
S2 = 24×10-6
S = 4.9×10-3 g/L
b) 1L of water contains 4.9×10-3 g of CaSO4
1000 cm3 of water contains 4.9×10-3 g of CaSO4
Density of water = 1 g/cm3 i.e weight of water = 1000g, weight of CaSO4 = 4.9×10-3 g
Amount of calsium sulphate(CaSO4) in ppm unit = 4.9×10-3×106/1000
= 4.9
13) Reaction : C6H12O6 + 6 O2 = 6CO2 + 6H2O
MW of glucose = 180 g/mole
glucose = 100mg/l = 0.1g/L
Therefore ThOD(Theoritical Oxygen Demand) = (6×32 g/mole O2 × 0.1 g/L glucose) / 180 g/mole glucose
= 0.107 g/L
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