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Dmitry [639]
3 years ago
8

Objects get their colors from reflecting only certain wavelengths when hit with white light. Light reflected from a green leaf i

s found to have a wavelength of 4.90×10^-7. What is the frequency of the light?​
Chemistry
1 answer:
Vesnalui [34]3 years ago
3 0

Answer:

frequency = 6.12× 10¹⁴ s⁻¹

Explanation:

Given data:

Wavelength = 4.90 ×10⁻⁷ m

Frequency = ?

Solution:

Formula:

<em>Speed of wave = frequency × wavelength</em>

Speed of wave = 3 × 10⁸ m/s

<em>Speed of wave = frequency × wavelength</em>

<em>frequency = Speed of wave/wavelength</em>

frequency <em>= </em>3 × 10⁸ m/s / 4.90 ×10⁻⁷ m

frequency = 0.612 × 10¹⁵ s⁻¹

frequency = 6.12× 10¹⁴ s⁻¹

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Calculate the mass of CO2 that can be produced if the reaction of 54.0 g of propane and sufficient oxygen has a 64.0% yield.
Oduvanchick [21]

Answer:

103.9 g

Explanation:

  • C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

First <u>we convert 54.0 g of propane (C₃H₈) into moles</u>, using its <em>molar mass</em>:

  • 54.0 g ÷ 44 g/mol = 1.23 mol C₃H₈

Then we <u>convert 1.23 moles of C₃H₈ into moles of CO₂</u>, using the <em>stoichiometric coefficients</em>:

  • 1.23 mol C₃H₈ * \frac{3molCO_2}{1molC_3H_8} = 3.69 mol CO₂

We <u>convert 3.69 moles of CO₂ into grams</u>, using its <em>molar mass</em>:

  • 3.69 mol CO₂ * 44 g/mol = 162.36 g

And <u>apply the given yield</u>:

  • 162.36 g * 64.0/100 = 103.9 g
7 0
2 years ago
What mass of the protein gelatin is needed to make 0.5 L of a 3 g/L gelatin solution? Show your work.
REY [17]

Answer:

m = 1.5 gram

Explanation:

Given that,

Density of protein gelatin, d = 3 g/L

The volume of protein gelatin, V = 0.5 L

We need to find the mass of the protein gelatin. The density of an object is given by :

d = m/V

Where

m is mass

m=d\times V\\\\m=3\ g/L\times 0.5\ L\\\\m=1.5\ g

So, the required mass is 1.5 gram.

6 0
2 years ago
What is homeostasis?
Alexandra [31]

Answer:

""the tendency toward a relatively stable equilibrium between interdependent elements, especially as maintained by physiological processes.""

Explanation:

5 0
2 years ago
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How many liters of 15.0 molar NaOH stock solution will be needed to make 17.5 liters of a 1.4 molar NaOH solution? Show the work
strojnjashka [21]
2.0 L
The key to any dilution calculation is the dilution factor

The dilution factor essentially tells you how concentrated the stock solution was compared with the diluted solution.

In your case, the dilution must take you from a concentrated hydrochloric acid solution of 18.5 M to a diluted solution of 1.5 M, so the dilution factor must be equal to

DF=18.5M1.5M=12.333

So, in order to decrease the concentration of the stock solution by a factor of 12.333, you must increase its volume by a factor of 12.333by adding water.

The volume of the stock solution needed for this dilution will be

DF=VdilutedVstock⇒Vstock=VdilutedDF

Plug in your values to find

Vstock=25.0 L12.333=2.0 L−−−−−

The answer is rounded to two sig figs, the number of significant figures you have for the concentration od the diluted solution.

So, to make 25.0 L of 1.5 M hydrochloric acid solution, take 2.0 L of 18.5 M hydrochloric acid solution and dilute it to a final volume of 25.0 L.

IMPORTANT NOTE! Do not forget that you must always add concentrated acid to water and not the other way around!

In this case, you're working with very concentrated hydrochloric acid, so it would be best to keep the stock solution and the water needed for the dilution in an ice bath before the dilution.

Also, it would be best to perform the dilution in several steps using smaller doses of stock solution. Don't forget to stir as you're adding the acid!

So, to dilute your solution, take several steps to add the concentrated acid solution to enough water to ensure that the final is as close to 25.0 L as possible. If you're still a couple of milliliters short of the target volume, finish the dilution by adding water.

Always remember

Water to concentrated acid →.NO!

Concentrated acid to water →.YES!
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Order frome largest to smallest <br>Allele,<br>Chromosome,<br>DNA, <br>nucleus of a cell​
svetlana [45]
Allele, dna, chromosome, nucleus of a cell
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