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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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Alinara [238K]

Answer:

\%\ Composition\ of\ iron=69.92\ \%

Explanation:

Percent composition is percentage by the mass of element present in the compound.

The formula for chromium(III) nitrate is Cr(NO_3)_3

Molar mass of chromium(III) nitrate = 238.011 g/mol

1 mole of chromium(III) nitrate contains 9 moles of oxygen

Molar mass of oxygen = 16 g/mol

So, Mass= Molar mass*Moles = 16*9 g = 144 g

\%\ Composition\ of\ iron=\frac{Mass_{iron}}{Total\ mass}\times 100

\%\ Composition\ of\ iron=\frac{144}{238.011}\times 100

\%\ Composition\ of\ iron=69.92\ \%

5 0
2 years ago
The standard molar heat of fusion of ice is 6020 j/mol. calculate q, w, and ∆e for melting 1.00 mol of ice at 0◦c and 1.00 atm p
zysi [14]

Answer :    q = 6020 J, w = -6020 J, Δe = 0

Solution : Given,

Molar heat of fusion of ice = 6020 J/mole

Number of moles = 1 mole

Pressure = 1 atm

Molar heat of fusion : It is defined as the amount of energy required to melt 1 mole of a substance at its melting point. There is no temperature change.

The relation between heat and molar heat of fusion is,

q=\Delta H_{fusion}(\frac{Mass}{\text{ Molar mass}})  (in terms of mass)

or, q=\Delta H_{fusion}\times Moles     (in terms of moles)

Now we have to calculate the value of q.

q=6020J/mole\times 1Mole=6020J

When temperature is constant then the system behaves isothermally and Δe is a temperature dependent variable.

So, the value of \Delta e=0

Now we have to calculate the value of w.

Formula used :    \Delta e=q+w

where, q is heat required, w is work done and \Delta e is internal energy.

Now put all the given values in above formula, we get

0=6020J+w

w = -6020 J

Therefore, q = 6020 J, w = -6020 J, Δe = 0

3 0
2 years ago
A goldsmith melts 12.4 grams of gold to make a ring. The temperature of the gold rises from 26°C to 1064°C, and then the gold me
DiKsa [7]

Problem One

You will use both m * c * deltaT and H = m * heat of fusion.

Givens

m = 12.4 grams

c = 0.1291

t1 = 26oC

t2 = 1204

heat of fusion (H_f) = 63.5 J/grams.

Equation

H = m * c * deltaT + m * H_f

Solution

H = 12.4 * 0.1291 * (1063 - 26) + 12.4 * 63.5

H = 1660.1 + 787.4

H = 2447.5 or 2447.47 is the exact answer. I have to leave the rounding to you. I have no idea where to round it although I suspect 2450 would be right for 3 sig digs.

Problem Two

Formula and Givens

t1 = 14.5

t2 = 50.0

E = 5680

c = 4.186

m = ??

E = m c * deltaT

Solution

5680 = m * 4.186 * (50 - 14.5)

5680 = m * 4.186 * (35.5)

5680 = m * 148.603 * m

m = 5680 / 148.603

m = 38.22 grams That isn't very much. Be very sure you are working in joules. You'd leave that many grams in the kettle after drying it thoroughly.

m = 38.2 to 3 sig digs.

8 0
2 years ago
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Answer:

g NaCl = 424.623 g

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<em>C</em> NaCl = 3.140 m = 3.140 mol NaCl / Kg solvent

∴ solvent: H2O

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