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amid [387]
4 years ago
11

A more experienced colleague mentions to you that you should aim for a dissolved oxygen concentration around 1.0 mg/L at day 5 o

f the experiment. They show you data from a similar wastewater spill that happened last year, in which the BOD5 of the stream water was 60 mg/L. From your previous experiment, you found that the dissolved oxygen concentration at day 0 was 10 mg/L. Based on this information, what dilution factor, P, should you try in your reactor?
Chemistry
1 answer:
Sati [7]4 years ago
4 0

Answer:

3

Explanation:

Lt= Loe^(-kt)

Data:

Lo = 10 mg/mL

Assume k = 0.23/da

1. Calculate L5

L5 = 10e^(-5×0.23) = 10e^-1.15 = 10 × 0.317 = 3.17 mg/mL

2. Calculate the dilution factor

You expect to find L5 to be about 3

You want L5 to be about 1.

You should use a dilute your sample by a factor of 3.

P = 3

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Consider the reaction: CaCO3(s) à CaO(s) + CO2(g)
Vsevolod [243]

Answer:

131.5 kJ

Explanation:

Let's consider the following reaction.

CaCO₃(s) → CaO(s) + CO₂(g)

First, we will calculate the standard enthalpy of the reaction (ΔH°).

ΔH° = 1 mol × ΔH°f(CaO(s)) + 1 mol × ΔH°f(CO₂(g) ) - 1 mol × ΔH°f(CaCO₃(s) )

ΔH° = 1 mol × (-634.9 kJ/mol) + 1 mol × (-393.5 kJ/mol) - 1 mol × (-1207.6 kJ/mol)

ΔH° = 179.2 kJ

Then, we calculate the standard entropy of the reaction (ΔS°).

ΔS° = 1 mol × S°(CaO(s)) + 1 mol × S°(CO₂(g) ) - 1 mol × S°(CaCO₃(s) )

ΔS° = 1 mol × (38.1 J/mol.K) + 1 mol × (213.8 J/mol.K) - 1 mol × (91.7 J/mol.K)

ΔS° = 160.2 J/K = 0.1602 kJ/K

Finally, we calculate the standard Gibbs free energy of the reaction at T = 25°C = 298 K.

ΔG° = ΔH° - T × ΔS°

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ΔG° = 131.5 kJ

6 0
3 years ago
A second unknown substance has a density of 80 g/cm and a mass of 570g. What is the volume?
Salsk061 [2.6K]

Answer:

25 g

10 cm

3

=

2.5 g/cm

3

Explanation:

6 0
3 years ago
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