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vladimir2022 [97]
3 years ago
5

-What is the wavelength of a 2.28 x 1017 Hz wave?

Chemistry
1 answer:
Bezzdna [24]3 years ago
6 0
I attached the working and the answer to both questions below.

Please note that C = speed of light, ν = frequency, λ= wavelength

<span>The wavelength of a 2.28 x 10</span>¹⁷<span> Hz wave is 1.31 </span>×<span> 10</span>⁷ cm<span>
The frequency of a 7.30 x 10</span>⁻⁵<span> m wave 4.11 </span>×<span> 10</span>¹² s⁻¹

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A student dissolves powdered tea in water. How can the student separate the parts of the mixture?
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Answer:

The student can evaporate the liquid water.

Explanation:

The student only mixed water with the powdered tea. Therefore, if the water is evaporated from the mixture, the student will be left with the tea mixture only.

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A solution is prepared by combining 5.00 mL of 4.8x10-4 M NaSCN solution, 2.00 mL of 0.21 M Fe(NO3)3 solution and 13.00 mL of 0.
Tcecarenko [31]

Answer:

The analytical concentrations of thiocyanate ions:

[SCN^-]=0.00012 mol/L

The analytical concentrations of ferric ions:

[Fe^{3+}]=0.063 mol/L

Explanation:

Moles (n)=Molarity(M)\times Volume (L)

1) Moles of sodium thiocyanate  = n

Volume of sodium thiocyanate solution = 5.00 mL = 0.005 L

(1 mL = 0.001L)

Molarity of the sodium thiocyanate = 4.8\times 10^{-4} M

n=4.8\times 10^{-4} M\times 0.005 L=2.4\times 10^{-6}mol

1 mole of sodium thiocyanate has 1 mol of thiocyante ions.

So, moles of thioscyanate ions in 2.4\times 10^{-6}mol  of NaSCN.

=1\times 2.4\times 10^{-6}mol=2.4\times 10^{-6}mol

2) Moles of ferric nitrate = n'

Volume of ferric nitrate solution = 2.00 mL = 0.002 L

Molarity of the ferric nitrate = 0.21 M

n'=0.002 M\times 0.21 L=0.00042 mol

1 mole of ferric nitrate has 3 moles of ferric ions.

So number of moles of ferric ions in 0.00042 moles of ferric nitrate is :

3\times 0.00042 mol=0.00126 mol

Volume of nitric acid = 13.00 mL

Total volume by adding all three volumes of solutions = V

V = 5.00 mL + 2.00 mL + 13.00 mL = 20.00 mL = 0.020 L

The analytical concentrations of thiocyanate ions:

[SCN^-]=\frac{2.4\times 10^{-6}mol}{0.020 L}=0.00012 mol/L

The analytical concentrations of ferric ions:

[Fe^{3+}]=\frac{0.00126 mol}{0.020 L}=0.063 mol/L

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I would say the answer is B - Differentiating 
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