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max2010maxim [7]
2 years ago
5

Describe, in detail, to a freshman undergraduate how to make 1 liter of LB + Kan (100 μg/ml final concentration) + Amp (50 μg /m

l final concentration) liquid media. [Include things like how many grams of each component that you use, how much antibiotic (in ml) to add (stock solutions – 100 mg/ml ampicillin, 25 mg/ml kanamycin), and in what type of container you perform the sterilization step.] Show your calculations.
Chemistry
1 answer:
mario62 [17]2 years ago
5 0

Answer:

Explanation:

The objective here is to prepare  1 liter of LB + Kan (100 μg/ml final concentration) + Amp (50 μg /ml final concentration) liquid media.

Given that :

the Stock concentration of Amp: 100 mg/ml (since 1 mg/ml = 1000 μg/ml)

it is required that we convert stock concentration in  μg/ml since 100 mg/ml = 100000  μg/ml

However, using formula C₁V₁=C₂V₂ (Ampicilin),

where:

C₁ = 100000 μg/ml,

V₁=?,

C₂= 50  μg/ml,

V₂=1000 ml

100000 μg/ml × V₁ = 50  μg/ml × 1000 ml

V₁ =  50  μg/ml × 1000 ml/100000 μg/ml

V₁ =   0.5 ml

Given that:

the Stock concentration of Kan: 25 mg/ml (since 1 mg/ml = 1000 μg/ml)

it is required that we convert stock concentration in  μg/ml , 25 mg/ml = 25000 μg/ml

Now by using formula C₁V₁=C₂V₂ (Kanamycin),

C₁ = 25000 μg/ml,

V₁=?,

C₂= 100  μg/ml,

V₂=1000 ml

25000 μg/ml × V₁ = 100  μg/ml × 1000 ml

V₁ =  100  μg/ml × 1000 ml/25000 μg/ml

V₁ =   4 ml

Thus; in 1 lite of Lb+ Kan+Amp preparation;

0.5 ml of Amp & 4 ml of kanamycin is used for their stock preparation.

Finally;

Sterilization step should be carried out in flask (Clean dry glass wares) for media in an autoclave, the container size should be twice the volume of media which is prepared.

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Klio2033 [76]

Rate equation for first order reaction is as follows:

t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}

Here, k is rate constant of the reaction, t is time of the reaction, A_{0} is initial concentration and A_{t} is concentration at time t.

The rate constant of the reaction is 0.1 day^{-1}.

(a) Let the initial concentration be 100, If 90% of the chemical is destroyed, the chemical present at time t will be 100-90=10, on putting the values,

t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}=\frac{2.303}{0.1 day^{-1}}log\frac{100}{10}=23.03 days

Thus, time required to destroy 90% of the chemical is 23.03 days.

(b) Let the initial concentration be 100, If 99% of the chemical is destroyed, the chemical present at time t will be 100-99=1, on putting the values,

t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}=\frac{2.303}{0.1 day^{-1}}log\frac{100}{1}=46.06 days

Thus, time required to destroy 99% of the chemical is 46.06 days.

(c)  Let the initial concentration be 100, If 99.9% of the chemical is destroyed, the chemical present at time t will be 100-99.9=0.1, on putting the values,

t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}=\frac{2.303}{0.1 day^{-1}}log\frac{100}{0.1}=69.09 days

Thus, time required to destroy 99.9% of the chemical is 69.09 days.

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What scientific knowledge was necessary for Kroto, Smalley, and Curl to discover buckyballs? Check all that apply. All matter is
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All matter is made of atoms.

Atoms bond together in certain ways.

Carbon atoms have a specific mass.

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Challenge question: This question is worth 6 points. As you saw in problem 9 we can have species bound to a central metal ion. T
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Answer:

CN^- is a strong field ligand

Explanation:

The complex, hexacyanoferrate II is an Fe^2+ specie. Fe^2+ is a d^6 specie. It may exist as high spin (paramagnetic) or low spin (diamagnetic) depending on the ligand. The energy of the d-orbitals become nondegenerate upon approach of a ligand. The extent of separation of the two orbitals and the energy between them is defined as the magnitude of crystal field splitting (∆o).

Ligands that cause a large crystal field splitting such as CN^- are called strong field ligands. They lead to the formation of diamagnetic species. Strong field ligands occur towards the end of the spectrochemical series of ligands.

Hence the complex, Fe(CN)6 4− is diamagnetic because the cyanide ion is a strong field ligand that causes the six d-electrons present to pair up in a low spin arrangement.

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2 years ago
The three‑dimensional structure of a generic molecule is given. Identify the axial and equatorial atoms in the three‑dimensional
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Answer:

Explanation:

CHECK THE ATTACHMENT FOR THE COMPLETE QUESTION AND THE DETAILED EXPLANATION

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Equatorial atoms are referred to atoms that are attached to carbons in the cyclohexane ring which is found at the equator of the ring.

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Sulfur and oxygen react to produce sulfur trioxide. In a particular experiment, 7.9 grams of SO3 are produced by the reaction of
shutvik [7]

Answer:

  • <u>79%</u>

Explanation:

<u>1) Balanced chemical equation:</u>

  • 2S + 3O₂ → 2SO₃

<u>2) Mole ratio:</u>

  • 2 mol S : 3 mol O₂ : 2 mol SO₃

<u>3) Limiting reactant:</u>

  • Number of moles of O₂

        n = 6.0 g / 32.0 g/mol = 0.1875 mol O₂

  • Number of moles of S:

         n = 7.0 g / 32.065 g/mol = 0.2183 mol S

  • Ratios:

        Actual ratio: 0.1875 mol O₂ / 0.2183 mol S =0.859

        Theoretical ratio: 3 mol O₂ / 2 mol S = 1.5

Since there is a smaller proportion of O₂ (0.859) than the theoretical ratio (1.5), O₂ will be used before all S be consumed, and O₂ is the limiting reactant.

<u>4) Calcuate theoretical yield (using the limiting reactant):</u>

  • 0.1875 mol O₂ / x = 3 mol O₂ / 2 mol SO₃

  • x = 0.1875 × 2 / 3 mol SO₃ =  0.125 mol SO₃

<u>5) Yield in grams:</u>

  • mass = number of moles × molar mass = 0.125 mol × 80.06 g/mol =  10.0 g

<u>6) </u><em><u>Percent yield:</u></em>

  • Percent yield, % = (actual yield / theoretical yield) × 100
  • % = (7.9 g / 10.0 g) × 100 = 79%
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