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amid [387]
3 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]3 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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.500 mol of br2 and .500 mol of cl2 are placed in a .500L flask and allowed to reach equilibrium. at equilibrium the flask was f
Tasya [4]

Answer : The value of K_c for the given reaction is, 0.36

Explanation :

Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.

The equilibrium expression for the reaction is determined by multiplying the concentrations of products and divided by the concentrations of the reactants and each concentration is raised to the power that is equal to the coefficient in the balanced reaction.

As we know that the concentrations of pure solids and liquids are constant that is they do not change. Thus, they are not included in the equilibrium expression.

The given equilibrium reaction is,

Br_2(aq)+Cl_2(aq)\rightleftharpoons 2BrCl(aq)

The expression of K_c will be,

K_c=\frac{[BrCl]^2}{[Br_2][Cl_2]}

First we have to calculate the concentration of Br_2,Cl_2\text{ and }BrCl.

\text{Concentration of }Br_2=\frac{Moles}{Volume}=\frac{0.500mol}{0.500L}=1M

\text{Concentration of }Cl_2=\frac{Moles}{Volume}=\frac{0.500mol}{0.500L}=1M

\text{Concentration of }BrCl=\frac{Moles}{Volume}=\frac{0.300mol}{0.500L}=0.6M

Now we have to calculate the value of K_c for the given reaction.

K_c=\frac{[BrCl]^2}{[Br_2][Cl_2]}

K_c=\frac{(0.6)^2}{(1)\times (1)}

K_c=0.36

Therefore, the value of K_c for the given reaction is, 0.36

6 0
3 years ago
Read 2 more answers
Balance the following half reaction in basic conditions. Then, indicate the coefficients for H2O and OH– for the balanced half r
Ugo [173]

Answer:

The ballance half reactions are:

Mg²⁺  + 2e⁻ → Mg

6OH⁻ + Si  → SiO₃²⁻ + 4e⁻ + 3 H₂O

Coefficients for H2O and OH– are 3 for H₂O (in products side) and 6 for OH⁻ (in reactants side)

Explanation:

Si (s) + Mg(OH)₂ (s) → Mg (s) + SiO₃²⁻ (aq)

Let's see the oxidations number.

As any element in ground state, we know that oxidation state is 0, so Si in reactants and Mg in products, have 0.

Mg in reactants, acts with +2, so the oxidation number has decreased.

This is the reduction, so it has gained electrons.

Si in reactants acts with 0 so in products we find it with +4. The oxidation number increased it, so this is oxidation. The element has lost electrons.

Let's take a look to half reactions:

Mg²⁺  + 2e⁻ → Mg

Si  → SiO₃²⁻ + 4e⁻

In basic medium, we have to add water, as the same amount of oxygen we have, IN THE SAME SIDE. We have 3 oxygens in products, so we add 3 H₂O and in the opposite site we can add OH⁻, to balance the hydrogen. The half reaciton will be:

6OH⁻ + Si  → SiO₃²⁻ + 4e⁻ + 3 H₂O

If we want to ballance the main reaction we have to multiply (x2) the half reaction of oxidation. So the electrons can be ballanced.

2Mg²⁺  + 4e⁻ → 2Mg

Now, that they are ballanced we can sum the half reactions:

2Mg²⁺  + 4e⁻ → 2Mg

6OH⁻ + Si  → SiO₃²⁻ + 4e⁻ + 3 H₂O

2Mg²⁺  + 4e⁻  + 6OH⁻ + Si  → 2Mg  +  SiO₃²⁻ + 4e⁻ + 3 H₂O

7 0
3 years ago
What determines an element's chemical properties
Alik [6]
The three factors determine the chemical properties of an element:
<span><span>The number and arrangement of electrons in an atom
</span><span>The number of valence electrons
</span><span>The number and arrangement of electrons</span></span>
5 0
3 years ago
Read 2 more answers
There are four branched isomers of hexane. draw bond-line structures of all four of its isomers.
Finger [1]
C-c-c-c-c
   |
   c

c-c-c-c-c
      |
      c


    c
    |
c-c-c-c
    |
    c



c-c-c-c
   |   |
   c   c




3 0
3 years ago
Assume that a point mutation changes the codon AUU to AUC. Why is this a neutral mutation
lukranit [14]
I used the genetic code table. mRNA codon ===> amino acid

1st base       2nd base           3rd base
A                       U                             U               ===> Isoleucine
A                       U                             C               ===> Isoleucine

The point mutation of codon AUU to AUC is a neutral mutation because it neither benefits nor deter the ability of the organism to survive and reproduce.

As you can see, Both codons result to the Isoleucine amino acid.

Another codon that will still result to the Isoleucin amino acid is AUA.

3 0
3 years ago
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