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belka [17]
3 years ago
14

octyl methoxycinnamate and oxybenzone are common ingredients in sunscreen applications. These compounds work by absorbing uv b l

ight (wavelength 280-320 nm), the UV light most associated with sunburn symptoms. What frequency range of light do these compounds absorb?
Chemistry
1 answer:
Rus_ich [418]3 years ago
4 0

Answer:

0.94 PHz – 1.1 PHz

Explanation:

Case 1. <em>280 nm </em>

The formula relating the frequency (f) and wavelength (λ) is

fλ = c                                         Divide both sides by f.

f = c/λ

c = 2.998 × 10⁸ m·s⁻¹

λ = 280 nm = 280 × 10⁻⁹ m     Calculate f.

f = 2.998 × 10⁸/280 × 10⁻⁹

f = 1.1 × 10¹⁵ s⁻¹ = 1.1 PHz

===============

Case 2. 320 nm

λ = 320 nm = 320 × 10⁻⁹ m     Calculate f.

f = 2.998 × 10⁸/320 × 10⁻⁹

f = 9.4 × 10¹⁴ s⁻¹ = 0.94 × 10¹⁵ s⁻¹ = 0.94 PHz

The frequency range is 0.94 PHz to 1.1 PHz.

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Second Blank: Endothermic

Explanation:

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How many grams of NaCl should be added to 250 grams of waterto<br> make a 0.40 m solution?
brilliants [131]

Answer:

5.85 grams of NaCl should be added to 250 grams of water to  make a 0.40 m solution.

Explanation:

molality=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{ Mass of solvent (kg)}}

Mass of NaCl =x

Molar mass of NaCl = 58.5 g/mol

Molality of the solution = 0.40 m

Mass of the solvent = 250 g = 0.250 kg (1g = 0.001 kg)

0.40 m=\frac{x}{58.5 g/mol\times 0.250 kg}

x=0.40 m\times 58.5 g/mol\times 0.250 kg=5.85 g

5.85 grams of NaCl should be added to 250 grams of water to  make a 0.40 m solution.

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3 years ago
Calculate the weighted average
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3 years ago
Suppose you start with a solution of red dye #40 that is 2.3 ✕ 10−5 M. If you do three successive volumetric dilutions pipetting
larisa [96]

Answer: Therefore, the concentration of final solution is 4.0\times 10^{-8}M

Explanation:

According to the neutralization law,

M_1V_1=M_2V_2

where,

M_1 = molarity of stock solution = 2.3\times 10^{-5}M

V_1 = volume of stock solution = 3.00 ml

M_2 = molarity of diluted solution = ?

V_2 = volume of diluted solution = 25.00 ml

2.3\times 10^{-5}M\times 3.00=M_2\times 25.00

M_2=0.28\times 10^{-5}M

Therefore, the concentration of diluted solution is 0.28\times 10^{-5}M

2) On further dilution

M_1V_1=M_2V_2

where,

M_1 = molarity of stock solution = 0.28\times 10^{-5}M

V_1 = volume of stock solution = 3.00 ml

M_2 = molarity of diluted solution = ?

V_2 = volume of diluted solution = 25.00 ml

0.28\times 10^{-5}M\times 3.00=M_2\times 25.00

M_2=0.034\times 10^{-5}M

Therefore, the concentration of diluted solution is 0.034\times 10^{-5}M

3) On further dilution

M_1V_1=M_2V_2

where,

M_1 = molarity of stock solution = 0.034\times 10^{-5}M

V_1 = volume of stock solution = 3.00 ml

M_2 = molarity of diluted solution = ?

V_2 = volume of diluted solution = 25.00 ml

0.034\times 10^{-5}M\times 3.00=M_2\times 25.00

M_2=4.0\times 10^{-8}M

Therefore, the concentration of final solution is 4.0\times 10^{-8}M

5 0
3 years ago
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