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siniylev [52]
3 years ago
12

From laboratory measurements, we know that a particular spectral line formed by hydrogen appears at a wavelength of 486.1 nanome

ters (nm). The spectrum of a particular star shows the same hydrogen line appearing at a wavelength of 486.0 nm. What can we conclude?
Chemistry
1 answer:
andriy [413]3 years ago
4 0

Answer:

The source of the hydrogen is coming towards us.

Explanation:

The wavelength of the hydrogen is decreasing as it goes from 486.1 nm to 486.0 nm.

<u>The wavelength of a wave is directly proportional to the distance between the observer and the object.</u>

For example, when the observer is far from the source, the observer will observe a longer wavelength which corresponds to red color if it is in the visible region.

<u>As the wavelength is decreasing that means that the hydrogen particle ( source may be a star) is coming towards the observer.</u>

Thus, we can conclude that the source of the hydrogen is coming towards us.

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You will need to prepare 12 mL of 25% Sodium Phosphate Buffer (pH 4) solution for Activity 2. What volume of the stock Sodium Ph
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Assuming the concentration of stock solution is 50% sodium phosphate buffer solution, the volume of stock solution required is 6 mL and the volume of water required is 6 mL.

<h3>What volume of a stock Sodium phosphate buffer and water is needed to 12 mL of 25% sodium phosphate buffer of pH 4?</h3>

The process of preparing solutions from stock solutions of higher concentration is known as dilution.

Dilution is done with the aid of the dilution formula given below:

  • C1V1 = C2V2

where

  • C1 is the concentration of stock solution
  • V1 is the volume of stock solution required to prepare a diluted solution
  • C2 is the concentration of the diluted solution prepared
  • V2 is the final volume of the diluted solution

From the data provided:

C1 is not given

V1 is unknown

C2 = 25%

V2 = 12 mL

  • Assuming C1 is 50% solution

Volume of stock, V1, required is calculated as follows:

V1 = C2V2/C1

V1 = 25 × 12 /50

V1 = 6 mL

Therefore, the volume of stock solution required is 6 mL and the volume of water required is 6 mL.

Learn more about dilution formula at: brainly.com/question/7208546

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2 years ago
Which of the following is not an example of a chemical reaction? Rust Heat from Fire Photosynthesis Melting Ice
VARVARA [1.3K]
Melting ice. Hopefully this helps!
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3 years ago
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How many moles are in 64.8 l of f2 gas
Licemer1 [7]
6.02×[10]1 ×64.8=3900.96 moles
7 0
3 years ago
Given the wavelength of the corresponding emission line, calculate the equivalent radiated energy from n = 4 to n = 2 in both jo
lions [1.4K]

Explanation:

It is given that,

Initial orbit of electrons, n_i=4

Final orbit of electrons, n_f=2

We need to find energy, wavelength and frequency of the wave.

When atom make transition from one orbit to another, the energy of wave is given by :

E=-13.6(\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2})

Putting all the values we get :

E=-13.6(\dfrac{1}{(4)^2}-\dfrac{1}{(2)^2})\\\\E=2.55\ eV

We know that : 1\ eV=1.6\times 10^{-19}\ J

So,

E=2.55\times 1.6\times 10^{-19}\\\\E=4.08\times 10^{-19}\ J

Energy of wave in terms of frequency is given by :

E=hf

f=\dfrac{E}{h}\\\\f=\dfrac{4.08\times 10^{-19}}{6.63\times 10^{-34}}\\\\f=6.14\times 10^{14}\ Hz

Also, c=f\lambda

\lambda is wavelength

So,

\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{6.14\times 10^{14}}\\\\\lambda=4.88\times 10^{-7}\ m\\\\\lambda=488\ nm

Hence, this is the required solution.

4 0
3 years ago
chemist adds of a M iron(III) bromide solution to a reaction flask. Calculate the mass in grams of iron(III) bromide the chemist
Sav [38]

The question is incomplete, here is the complete question:

A chemist adds 370.0 mL of a 2.25 M iron(III) bromide solution to a reaction flask. Calculate the mass in grams of iron(III) bromide the chemist has added to the flask. Round your answer to 3 significant digits

<u>Answer:</u> The mass of iron (III) bromide is 246. grams

<u>Explanation:</u>

To calculate the mass of solute, we use the equation used to calculate the molarity of solution:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

We are given:

Molarity of solution = 2.25 M

Molar mass of iron (III) bromide = 295.6 g/mol

Volume of solution = 370.0 mL

Putting values in above equation, we get:

2.25M=\frac{\text{Mass of solute}\times 1000}{295.6\times 370.0}\\\\\text{Mass of solute}=\frac{2.25\times 295.6\times 370.0}{1000}=246.1g

Hence, the mass of iron (III) bromide is 246. grams

4 0
3 years ago
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