All categories

3 years ago

Oxygen is a diatomic element why?

Chemistry

2 answers:

Tatiana [17]3 years ago

7 0

Answer:

Oxygen exists as a diatomic molecule in nature when it is not combined with any other element. It forms O2 molecule because it is stable when it is uncombined. ... By achieving octet configuration, the diatomic elements become more stable compared to the single atom. Therefore, oxygen and hydrogen are diatomic.

Explanation:

Hope it helps you..

Your welcome in advance..

(ㆁωㆁ)

Dahasolnce [82]3 years ago

3 0

Answer:

Oxygen is a diatomic element because it consists of two atoms of same element O and O that makes O2.Diatomic molecules contain two atoms that are chemically bonded. If the two atoms are identical, as in, for example, the oxygen molecule (O2), they compose a homonuclear diatomic molecule, while if the atoms are different, as in the carbon monoxide molecule (CO), they make up.

You might be interested in

What is the percent composition by mass of chlorine in LiCI

Lelechka [254]

Answer: Lithium (Li) 16.373%

Chlorine (Cl) 83.627%

Explanation:

6 0

3 years ago

Give any two reasons that hydrogen is kept in group 1 in Modern Periodic Table

xeze [42]

Answer:

Hydrogen is usually classified as a group 1 element since it has one valence electron, just like group 1 metals.

Explanation:

3 0

3 years ago

Help ASAP I will mark brainliest for the correct answer !!!⚡️⚡️⚡️

kobusy [5.1K]

Which fractions are equivalent to 0.875?

8 0

3 years ago

Read 2 more answers

Before any reaction occurs, the concentration of A in the reaction below is 0.0510 M. What is the equilibrium constant if the co

Lilit [14]

Answer:

0.0119

Explanation:

There was a part missing. I think this is the whole question:

<em>Before any reaction occurs, the concentration of A in the reaction below is 0.0510 M. What is the equilibrium constant if the concentration of A at equilibrium is 0.0153 M?</em>

A (aq) ⇌ 2B (aq) + C(aq)

<em>Remember to use correct significant figures in your answer. Do not include units in your response.</em>

First, we have to make an ICE Chart, which stands for initial, change and equilibrium. We will call "x" unknown concentrations.

A (aq) ⇌ 2B (aq) + C (aq)

I 0.0510 0 0

C -x +2x +x

E 0.0510-x 2x x

Since the concentration at equilibrium of A is 0.0153 M, we get

$0.0510 - x = 0.0153 \\x = 0.0357$

We can use the value of x to calculate the concentrations at equilibrium.

$[A]e = 0.0153 M \\[B]e = 2x = 2(0.0357) = 0.0714 M \\[C]e = x = 0.0357 M \\$

The equilibrium constant, Kc, is the ratio of the equilibrium concentrations of products over the equilibrium concentrations of reactants each raised to the power of their stoichiometric coefficients.

$Kc = \frac{[B]^{2} \times [C]}{[A]} = \frac{0.0714^{2} \times 0.0357}{0.0153} = 0.0119$