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aniked [119]
3 years ago
11

Try to make only one molecule of water in the simulation. Is it possible?

Chemistry
1 answer:
mel-nik [20]3 years ago
3 0

Answer:

A single molecule of water has been isolated for the first time by trapping it in a fullerene cage. Water molecules are never found alone — they are always hydrogen-bonded to other molecules of water or polar compounds.

While making small volumes of pure water in a lab is possible, it's not practical to “make” large volumes of water by mixing hydrogen and oxygen together. The reaction is expensive, releases lots of energy, and can cause really massive explosions.

While making small volumes of pure water in a lab is possible, it's not practical to “make” large volumes of water by mixing hydrogen and oxygen together. The reaction is expensive, releases lots of energy, and can cause really massive explosions.

A water molecule consists of three atoms; an oxygen atom and two hydrogen atoms, which are bond together like little magnets. The atoms consist of matter that has a nucleus in the centre. The difference between atoms is expressed by atomic numbers.

Explanation:

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If the molar absorptivity constant for the red dye solution is 5.56×104 M-1cm-1, calculate the molarity of the red dye solution
Shtirlitz [24]

Explanation:

a) Using Beer-Lambert's law :

Formula used :

A=\epsilon \times c\times l

where,

A = absorbance of solution = 0.945

c = concentration of solution = ?

l = length of the cell = 1.20 cm

\epsilon = molar absorptivity of this solution =5.56\times 10^4 M^{-1} cm^{-1}

0.945=5.56\times 10^4 M^{-1} cm^{-1}\times 1.20 \times c

c=1.4163\times 10^{-5} M=14.16 \mu M

(1\mu M=10^{-6} M)

14.16 μM is the molarity of the red dye solution at the optimal wavelength 519nm and absorbance value 0.945.

b) c=1.4163\times 10^{-5} mol/L

1 L of solution contains 1.4163\times 10^{-5} moles of red dye.

Mass of 1.4163\times 10^{-5} moles of red dye:

1.4163\times 10^{-5}\times 879.86g/mol=0.01246 g

(w/v)\%=\frac{\text{Mass of solute (g)}}{\text{Volume of solvent (mL)}}\times 100

red(w/v)\%=\frac{0.01246 g}{1000 mL}\times 100=0.001246\%

c) In order to dilute red dye solution by 5 times, we will need to add 1 L of water to solution of given concentration.

Concentration of red dye solution = c=1.4163\times 10^{-5} M

Concentration of red solution after dilution = c'

c=c'\times 5

1.4163\times 10^{-5} M=c'\times 5

c'=2.83\times 10^{-6} M

The final concentration of the diluted solution is 2.83\times 10^{-6} M

8 0
3 years ago
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