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olga_2 [115]
3 years ago
10

Find the theoretical oxygen demand for the

Chemistry
1 answer:
never [62]3 years ago
7 0

Answer:

a) 213.3 mg/L

b) 62.61 mg/L

c) 0.0225 mg/L

Explanation:

Theoretical oxygen demand (ThOD)is essentially the amount of oxygen required for the complete degradation of a given compound into the final oxidized products

a) Given:

Concentration of acetic acid,[CH3COOH] = 200 mg/L

CH3COOH + 2O2 \rightarrow 2CO2 + 2H2O

ThOD = \frac{mass\ O2}{mass\ CH3COOH} * conc. CH3COOH

Based on the reaction stoichiometry:

mass of CH3COOH = 60 g

mass of O2= 2(32) = 64 g

ThOD = \frac{64 g}{60 g} * 200mg/L = 213.3 mg/L

b) Given:

Concentration of ethanol, [C2H5OH] = 30 mg/L

C2H5OH + 3O2 \rightarrow 2CO2 + 3H2O

ThOD = \frac{mass\ O2}{mass\ C2H5OH} * conc. C2H5OH

Based on the reaction stoichiometry:

mass of C2H5OH = 46 g

mass of O2= 3(32) = 96 g

ThOD = \frac{96 g}{46 g} * 30mg/L = 62.61 mg/L

c) Given:

Concentration of sucrose, [C12H22O11] = 50 mg/L

C12H22O11 + 12O2 \rightarrow 12CO2 + 11H2O

ThOD = \frac{mass\ O2}{mass\ C22H22O11} * conc. C12H22O11

Based on the reaction stoichiometry:

mass of C12H22O11 = 342 g

mass of O2= 12(32) = 384 g

ThOD = \frac{384 g}{342 g} * 50mg/L = 0.0225 mg/L

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6 0
3 years ago
Lawrencium-262 has a half-life of 4 hr. How much of a 40 mg sample remains after 12 hours?
CaHeK987 [17]

Answer:

5 mg

Explanation:

If one half life is 4 hours, then 3 half lives is 12 hours.

This means that the sample will decay to 1/8 of its original amount.

So, the answer is 40(1/8) = 5 mg.

3 0
2 years ago
Please help I don’t know if my answer is right!!
Temka [501]

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neutralization reaction! Aka: option C!

HOPE THIS HELPS! :)

Explanation:

8 0
3 years ago
A chemist must dilute of aqueous aluminum chloride solution until the concentration falls to . He'll do this by adding distilled
marshall27 [118]

Answer:

0.257 L

Explanation:

The values missing in the question has been assumed with common sense so  that the concept could be applied

Initial volume of the AICI3 solution =23.1 \mathrm{mL}

Initial Molarity of the solution =833 \mathrm{mM}

Final molarity of the solution =75.0 \mathrm{mM}

Final volume of the solution =?

From Law of Dilution, M_{f} V_{f}=M_{i} V_{i}

\Rightarrow V_{f}=\frac{M_{i} V_{i}}{M_{f}}=\frac{833 \mathrm{mM} \times 23.1 \mathrm{mL}}{75.0 \mathrm{mM}}=256.564 \mathrm{mL}=0.256564 \mathrm{L}=0.257 L

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7 0
3 years ago
What is the mass of 0. 513 mol Al2O3? Give your answer to the correct number of significant figures. (Molar mass of Al2O3 = 102.
Margarita [4]

The mass of a 0.513 mol of Al2O3 is 52.33g.

HOW TO CALCULATE MASS:

The mass of a substance can be calculated by multiplying the molar mass of the substance by its number of moles. That is;

mass of Al2O3 = no. of moles of Al2O3 × molar mass of Al2O3

According to this question, there are 0.513 moles of Al2O3.

Mass of Al2O3 = 0.513 × 102

Mass of Al2O3 = 52.33g

Therefore, the mass of a 0.513 mol of Al2O3 is 52.33g.

Learn more about mass calculations at: brainly.com/question/8101390?referrer=searchResults

7 0
2 years ago
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