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2 years ago

How are electromagnetic waves affected as the frequency of the waves increases?

2 answers:

5 0

As the frequency is how many waves pass a fixed point in certain amount of times. This means that the waves get closer together and, therefore, the wavelength decreases. It also means that the wave has more energy.

Answer: D) Energy increases, wavelength decreases

mrs_skeptik [129]2 years ago

5 0

D is the right answer

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3 years ago

The atom fluorine generally will become _____ stable through the formation of an ionic chemical compound by accepting _______ e

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The atomic number of Fluorine is 9

Valence (outer) electron configuration is : 2s²2p⁵

Therefore, it requires 1 electron in the p-orbital to complete its octet of 8 electrons.

Thus, the atom Fluorine generally will become more stable through the formation of an ionic chemical compound by accepting 1 electron from another atom. This process will fill its outer energy level.

Ans: A) more, 1

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3 years ago

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2 years ago

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An isotope undergoes radioactive decay by emitting radiation that has a –1 charge. What other characteristic does the radiation

MrRissso [65]

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3 years ago

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Chlorine has two naturally occurring isotopes, 35C1 Gsotopic mass 34.9689 amu) and 370 (isotopic mass 36.9659 amu). If chlorine

Lilit [14]

Answer: The percentage abundance of $_{17}^{35}\textrm{Cl}$ and $_{17}^{37}\textrm{Cl}$ isotopes are 75.77% and 24.23% respectively.

Explanation:

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)

Let the fractional abundance of $_{17}^{35}\textrm{Cl}$ isotope be 'x'. So, fractional abundance of $_{17}^{37}\textrm{Cl}$ isotope will be '1 - x'

For $_{17}^{35}\textrm{Cl}$ isotope:

Mass of $_{17}^{35}\textrm{Cl}$ isotope = 34.9689 amu

Fractional abundance of $_{17}^{35}\textrm{Cl}$ isotope = x

For $_{17}^{37}\textrm{Cl}$ isotope:

Mass of $_{17}^{37}\textrm{Cl}$ isotope = 36.9659 amu

Fractional abundance of $_{17}^{37}\textrm{Cl}$ isotope = 1 - x

Average atomic mass of chlorine = 35.4527 amu

Putting values in equation 1, we get:

$35.4527=[(34.9689\times x)+(36.9659\times (1-x))]\\\\x=0.7577$

Percentage abundance of $_{17}^{35}\textrm{Cl}$ isotope = $0.7577\times 100=75.77\%$

Percentage abundance of $_{17}^{37}\textrm{Cl}$ isotope = $(1-0.7577)=0.2423\times 100=24.23\%$

Hence, the percentage abundance of $_{17}^{35}\textrm{Cl}$ and $_{17}^{37}\textrm{Cl}$ isotopes are 75.77% and 24.23% respectively.

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2 years ago