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Furkat [3]
2 years ago
7

What is the range of 14 and 25

Chemistry
2 answers:
Alexxandr [17]2 years ago
6 0
Range means the difference between the highest and lowest number. So subtract 25 - 14 which gives you 11.
Law Incorporation [45]2 years ago
3 0
For the range, you subtract 25 from 14, which is 11.
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In molecular oxygen (O=O) which atom is partially positive?
saveliy_v [14]

the oxygen atom

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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Ronch [10]

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Group 10

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Hope this helps!

3 0
3 years ago
What is the maximum amount of kcl that can dissolve in 200 g of water? (the solubility of kcl is 34 g/100 g h2o at 20°c.) (1 poi
Solnce55 [7]
The answer is 68g
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5 0
3 years ago
The half-reaction occurring at the cathode in the balanced reaction shown below is __________?2Ag(s) + 1/2 O2(g) + 2H+(aq) → 2Ag
Shtirlitz [24]

Answer:

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D. 1/2 O2 (g) + 2H+ (aq) + 2e- → H2O (l)

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Oxygen from ground state reduce the oxidation state from 0 to -2

3 0
3 years ago
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