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Marianna [84]
4 years ago
13

Rank the four gases (air, exhaled air, gas produced from the decomposition of H2O2, gas from decomposition of NaHCO3) in order o

f decreasing concentration of oxygen).
Chemistry
1 answer:
charle [14.2K]4 years ago
7 0
H202, NaCHO3, Air, Exhaled Air
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Describe, in detail, to a freshman undergraduate how to make 1 liter of LB + Kan (100 μg/ml final concentration) + Amp (50 μg /m
mario62 [17]

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.

5 0
4 years ago
The mineral vanadinite has the formula pb5(vo4)3cl. what mass percent of chlorine does it contain?
lubasha [3.4K]

The molar mass of vanadinite is 1416.27 g/mol. Looking at the molecular formula we can see that there is 1 mole of Cl for every mole of vanadinite. Therefore:

mass percent = [(1 mole Cl * 35.45 g/mol) / (1 mole Pb5(VO4)3Cl * 1416.27 g/mol)] * 100%

<span>mass percent = 2.50%</span>

5 0
3 years ago
What is the engine piston displacement in liters of an engine who’s displacement is listed as 430 in.²
sergeinik [125]

For an engine whose displacement is listed as 430 in.², the engine piston displacement in liters is mathematically given as

PD= 7.37 L  

<h3>What is the engine piston displacement in liters of an engine whose displacement is listed as 430 in.²?</h3>

Generally, the equation for the dimensional analysis method,\, we convert in to L is mathematically given as

l*(v/l)*l/v

Therefore,  piston displacement

PD=(450 in3) . (1 dm3 / 61.024 in3) . (1 L / 1 dm3)

PD= 7.37 L  

In conclusion, piston displacement

PD= 7.37 L  

Read more about volume

brainly.com/question/1578538

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6 0
3 years ago
Potassium hydroxide is used to precipitate each of the cations from their respective solution. determine the minimum concentrati
MAXImum [283]
This is the three cases that help to determine the minimum concentration of KOH required for precipitation 
Part a) 1.5×10^−2 M K CaCl2 
Part b) 2.3×10^−3 M Fe (NO3)2 
Part c) 2.0×10^−3 M MgBr2

a) CaCl2 + 2KOH --> Ca (OH) 2 + 2KCl Ca (OH) 2 <=> Ca^2+ + 2OH^- 
ksp = 1.5*10^-2 + x^2 
4.68*10^-6 = 1.5*10^-2 + x^2 
x= [KOH] = 0.01766 

b) Fe (NO3)2 +2 KOH--> Fe (OH)2 + 2KNO3 
Fe (OH)2 <=> Fe^2+ + 2OH^- 
ksp = 2.3*10^-3 + x^2 
4.87*10^-17 = 2.3*10^-3 + x^2 
x= 1.46*10^-7 

c) MgBr2 + KOH --> Mg (OH) 2 + 2KBr 
Mg (OH) 2 <=> Mg^2+ + 2OH^- 
ksp = 2.0*10^-3 + x^2 
2.06*10^-13 = 2.0*10^-3 + x^2 
x= 1.015*10^-5
8 0
4 years ago
PLEASE HELP!!
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Grienr rjr or rjr rjr r
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3 years ago
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