Calculate the theoretical carbonaceous and nitrogenous oxygen demand for: 100 mg/L Ethylene glycol (C2H6O2) and 100 mg NH3-N/L

The correct answer is 129 mg and 232 mg.

Explanation:

Theoretical carbonaceous oxygen demand:

The reaction will be,  

C₂H₆O₂ + 5/2 O₂ ⇒ 2CO₂ + 3H₂O

Thus, for one mole of C₂H₆O₂ (ethylene glycol), 2.5 moles of O₂ is needed.  

The molecular mass of ethylene glycol is 62 grams per mole.  

The given mass of ethylene glycol is 100 mg or 0.1 grams

The moles of ethylene glycol will be,  

Moles = Weight/Molecular mass

= 0.1/62 = 1.613 × 10⁻³ mol

For 1.613 × 10⁻³ mol, the moles of O₂ will be,  

= 2.5×1.613×10⁻³

= 4.0.×10⁻³ × 32mol

= 0.129 grams or 129 mg.  

The theoretical nitrogenous oxygen demand is:  

The reaction will be,  

2NH₃-N + 9/2O₂ ⇒  4HNO2 + H₂O

Thus, for 2 moles of NH₃-N, 4.5 moles of O₂ is needed,  

Therefore, for 1 mol of NH₃-N, the oxygen required will be,  

= 4.5/2 = 2.25 mol

The given mass of NH₃-N is 100 mg, the moles of NH₃-N will be,  

Moles = 100×10⁻³/31 = 3.225 × 10⁻³ mol (The molecular mass of NH₃-N is 31 gram per mole)

The moles of O₂ is 2.25 × 3.225 × 10⁻³ = 7.258 × 10⁻³ mol.  

Now the mass of O2 will be,  

= 7.258 × 10⁻³ × 32

= 0.232 grams

= 232 mg

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a: 45.9 meters

b: 4.29 seconds

step-by-step explanation:

a: h(t)=1/2(-9.8)(3)^2+0(3)+90

h(t)=-4.9(3)^2+90

h(t)=-44.1+90

h(t)=45.9 meters

b: 0=1/2(-9.8)(t^2)+0t+90

0=-4.9t^2+90

-90=-4.9t^2

18.37=t^2

t=4.29 seconds

in the second question, the numbers aren't exact, but they're a close approximation.