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3 years ago

Which of the following correctly identifies the composition of an oxygen atom?

2 answers:

7 0

Oxygen has the atomic number 8, which eliminates options B and D right away. Any neutral atom, including the oxygen atom, has the same number of protons and electrons, meaning the number of electrons is also 8. If you're still not sold, the oxygen atom has a mass number of 16, which is the sum of the protons and neutrons in the atom, meaning there have to be 8 neutrons as well. The answer is option A.

aleksandrvk [35]3 years ago

3 0

The composition of an oxygen atom is: A) 8 protons, 8 electrons, and 8 neutrons.

<h3>Further Explanation</h3>

To determine how many of the subatomic particles are in the atom, the atomic number is used.

Atomic number is the number that gives the atom its identity and can be found in the periodic table. It gives the number of protons in an atom. It can be observed that the atomic number of the elements are increasing by one unit as you go from left to right.

<em>The atomic number of oxygen is 8, therefore, it has 8 protons.</em>

<em>In a neutral (uncharged atom), the number of protons and electrons are equal.</em>

<em>Hence, oxygen also has 8 electrons.</em>

In the most simple periodic table of elements, the atomic number is usually the smaller number written in the tile for each element. The larger number is the mass number.

The difference between the larger mass number and the smaller atomic number is the number of neutrons:

<em>Number of Neutrons = Mass Number - Atomic Number</em>

The mass number of oxygen is 16. We can use the atomic number and the mass number to get the number of neutrons using the equation above:

<em>Number of Neutrons = 16 – 8</em>

<em>Number of Neutrons = 8</em>

Therefore, it has been determined that the an oxygen atom has 8 protons, 8 electrons and 8 neutrons.

<h3>Learn More</h3>

Periodic Table brainly.com/question/2376134

Isotopes brainly.com/question/2757829

<h3>Keywords: protons, neutrons, electrons, mass number, atomic number</h3>

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At 2000°C, the equilibrium constant for the reaction below is Kc = 4.10 ´ 10–4 . If 0.600 moles of NO is placed in a 1.0-L react

erastova [34]

Answer:

At equilibrium, the concentration of $N_{2 (g)}$ is going to be 0.30M

Explanation:

We first need the reaction.

With the information given we can assume that is:

$N_{2 (g)}$ + $O_{2 (g)}$ ⇄ 2 $NO_{(g)}$

If there is placed 0.600 moles of NO in a 1.0-L vessel, we have a initial concentration of 0.60 M NO; and no $N_{2 (g)}$ nor $O_{2 (g)}$ present. Immediately, $N_{2 (g)}$ and $O_{2 (g)}$ are going to be produced until equilibrium is reached.

By the ICE (initial, change, equilibrium) analysis:

I: [ $N_{2 (g)}$ ]=0 ; [ $O_{2 (g)}$ ]= 0 ; [ $NO_{(g)}$ ]=0.60M

C: [ $N_{2 (g)}$ ]=+x ; [ $O_{2 (g)}$ ]= +x ; [ $NO_{(g)}$ ]=-2x

E: [ $N_{2 (g)}$ ]=0+x ; [ $O_{2 (g)}$ ]= 0+x ; [ $NO_{(g)}$ ]=0.60-2x

Now we can use the constant information:

$K_{c}=\frac{[products]^{stoichiometric coefficient} }{[reactants]^{stoichiometric coefficient} }$

$4.10* 10^{-4} =\frac{(0.60-2x)^{2}}{(x)*(x)}$

$4.10* 10^{-4}$ = $\frac{(0.60-2x)^{2}}{x^{2} }$

$4.10* 10^{-4} * x^{2}$ = $(0.60-2x)^{2}}$

$\sqrt{4.10* 10^{-4} * x^{2}}$ = $\sqrt{(0.60-2x)^{2}}}$

$0.0202 x =0.60 - 2x$

$2x+0.0202x=0.60$

$x=\frac{0.60}{2.0202}$ $= 0.30$

At equilibrium, the concentration of $N_{2 (g)}$ is going to be 0.30M

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