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Vedmedyk [2.9K]
3 years ago
10

Which of the following lists elements from lowest to highest atomic number

Chemistry
1 answer:
vitfil [10]3 years ago
8 0
Lowest is Hydrogen highest is <span>Beryllium
-HOPE THIS HELPED </span>
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Which of the following elements is a nonmetal? a. selenium (Se) b. yttrium (Y) c. barium (Ba) d. calcium (Ca) e. All of these ar
san4es73 [151]

<u>Answer:</u>

The correct answer option is a. Selenium (Se).

<u>Explanation:</u>

From the given answer options, Selenium (Se) is the only element which is a non metal.

It has the atomic number 34 with atomic weight 78.96. It is a member of the sulfur group of the non metallic elements and falls in period 4 of the Periodic table.

Selenium has non metallic properties which are intermediate between the elements that lie above and below it in the Periodic table.

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2 years ago
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Explain hydrogen dioxide​
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Answer:

Two molecules of hydrogen combine with two molecules of oxygen to form hydrogen peroxide. Hence, its chemical formula is H2O2. It is the simplest peroxide (since it is a compound with an oxygen-oxygen single bond). Hydrogen peroxide has basic uses as an oxidizer, bleaching agent and antiseptic

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3 years ago
When might a theory be changed
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What is the of 0.068 M of OH^ - ion?
Marina CMI [18]

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4 0
2 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
2 years ago
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