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Levart [38]
3 years ago
11

An element has the following natural abundances and isotopic masses: 90.92% abundance with 19.99 amu, 0.26% abundance with 20.99

amu, and 8.82% abudance with 21.99 amu. Calculate the average atomic mass of this elements.
Chemistry
1 answer:
aksik [14]3 years ago
5 0

<u>Answer:</u> The average atomic mass of element Z is 20.169 amu.

<u>Explanation:</u>

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i .....(1)

  • <u>For isotope 1:</u>

Mass of isotope 1 = 19.99 amu

Percentage abundance of isotope 1 = 90.92 %

Fractional abundance of isotope 1 = 0.9092

  • <u>For isotope 2:</u>

Mass of isotope 2 = 20.99 amu

Percentage abundance of isotope 2 = 0.26 %

Fractional abundance of isotope 2 = 0.0026

  • <u>For isotope 3:</u>

Mass of isotope 3 = 21.99 amu

Percentage abundance of isotope 3 = 8.82 %

Fractional abundance of isotope 3 = 0.0882

Putting values in equation 1, we get:

\text{Average atomic mass of Z}=[(19.99\times 0.9092)+(20.99\times 0.0026)+(21.99\times 0.0882)]

\text{Average atomic mass of Z}=20.169amu

Hence, the average atomic mass of element Z is 20.169 amu.

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2 years ago
could you please explain operations with scientific notation using this example. 5 × 10³ + 4.3 × 10⁴ i dont understand how to so
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Explanation:

The scientific notation:

a\times10^k

where

1\leq a and k is integer.

We have the example:

(5\times10^3)+(4.3\times10^4)

You can write the numbers in a "normal" form:

5\times10^3=5\times1000=5000\\\\4.3\times10^4=4.3\times10000=43000

Make the sum:

5000+43000=48000

And next write it in the scientific notation:

48000=4\underbrace{8000}_{\leftarrow4}=4.8000\times10000=4.8\times10^4

<h3>Other method:</h3>

You can add numbers in scientific notation if the power of tens in both number is the same.

Therefore you must convert the first number:

5\times10^3=0.5\times10\times10^3=0.5\times10^4

Now, you can make the sum:

(5\times10^3)+(4.3\times10^4)=0.5\times10^4+4.3\times10^4=(0.5+4.3)\times10^4=4.8\times10^4

4 0
3 years ago
Iron has density 7.87 g/cm^3. if 52.4 g of iron is added to 75.0 mL of water in a graduated cylinder, to what volume reading wil
vlada-n [284]

Answer:

6.66 mL

Explanation:

The increase in the volume is due to the addition of the iron whose volume can be calculated as:

Using,

Density = Mass / Volume

Given that:

Density of Iron = 7.87 g/cm³

Mass of iron = 52.4 g

Thus, volume is:

Volume = Mass / Density = 52.4 / 7.87 cm³ = 6.66 cm³

Also, 1 cm³ = 1 mL

<u>The rise in the volume = 6.66 mL</u>

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What is the application of the emission spectra regarding elements?
malfutka [58]

The emission spectrum of a chemical element or chemical compound is the spectrum of frequencies of electromagnetic radiation emitted due to an atom or molecule making a transition from a high energy state to a lower energy state. The photon energy of the emitted photon is equal to the energy difference between the two states. There are many possible electron transitions for each atom, and each transition has a specific energy difference. This collection of different transitions, leading to different radiated wavelengths, make up an emission spectrum. Each element's emission spectrum is unique. Therefore, spectroscopy can be used to identify elements in matter of unknown composition. Similarly, the emission spectra of molecules can be used in chemical analysis of substances.

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2 years ago
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