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ira [324]
2 years ago
10

What is the percentage composition of cobalt in cobalt(II) fluoride, CoF2?

Chemistry
1 answer:
emmainna [20.7K]2 years ago
5 0

Answer:

60.8%

Explanation:

We'll begin obtaining the molar mass of cobalt(II) fluoride, CoF2. This can be done as shown below:

Molar mass of CoF2 = 59 + (19x2) = 97g/mol.

The percentage composition of cobalt in cobalt(II) fluoride, CoF2 is given by:

Mass of Co/Molar Mass of CoF2 x 100

=> 59/97 x 100 = 60.8%

Therefore, the percentage composition of cobalt in cobalt(II) fluoride, CoF2 is 60.8%

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What are the smallest sub atomic structure
Anton [14]

Answer:

Electrons

Explanation:

In an atom there would be three subatomic particles: Neutrons, electrons, protons. The smallest and lightest in terms of mass is electrons. This is because the nucleus is comprised of the protons and the neutrons, these have a greater mass than electrons as electrons has very little mass that can considered to be 0.

7 0
2 years ago
The equation below represents a double replacement/displacement reaction;
Naddik [55]
AB+XY->AY+XB
We know that the answer would be KCl because of the switching that takes place during a double displacement reaction. Just like Zn and MnO4 switched and combined, the remaining elements, K and Cl, will combine.
We know that the answer is simply KCl because both K and Cl have an ion of only +/-1, meaning when they cross, no suffixes are made, since their ions are only 1.
For example, if you combined Mg with Cl, you would get MgCl2, because Mg has an ion of +2.
I hope this helps!
4 0
3 years ago
In which of these diatomic molecules would you NOT find an octet of electrons
Semmy [17]

Answer:

The hydrogen molecule is the only one in which can not find an octet of electrons around each atom.

Explanation:

Let's evaluate each case.  

1. Nitrogen (N₂):

With Z = 7, nitrogen has the following electronic configuration

1s²

2s² 2p³  → valence electrons

Since its valence electrons are 5, in the molecule one nitrogen atom shares 3 electrons with the other one, and each remains with an electron pair, so <u>each atom has an octet of electrons.</u>

2. Hydrogen (H₂):

With Z = 1, its electronic configuration is:

1s¹  → valence electron

In the molecule, the hydrogen atoms share the only electron they have, so they will have only 2 electrons around. In this diatomic molecule, <em><u>we can not find an octet.</u></em>

3. Oxygen (O₂):

Z = 8. Electronic configuration:

1s²

2s² 2p⁴  → valence electrons

In the diatomic molecule, each oxygen atom shares 2 electrons with the other one and remains with 2 pairs of electrons, therefore, <u>each oxygen atom has an octet</u>.      

4. Fluorine (F₂)

Z = 9. Electronic configuration:

1s²

2s² 2p⁵  → valence electrons

In this molecule, each fluorine atom shares 1 electron with the other and remains with 3 pairs of electrons, hence, <u>each fluorine atom has an octet of electrons around</u>.

Finally, we can say that the hydrogen molecule is the only one in which can not find an octet of electrons around each atom.

I hope it helps you!  

8 0
2 years ago
A compound is found to contain 38.65 % carbon, 16.25 % hydrogen, and 45.09 % nitrogen by mass. what is the empirical formula for
iren [92.7K]

The molar mass of carbon is 12, hydrogen is 1, and nitrogen is 14, hence the ratio are:

 

C = 38.65 / 12 = 3.22

H = 16.25 / 1 = 16.25

N = 45.09 / 14 = 3.22

 

Divide the three by the lowest ratio which is 3.22:

 

C = 3.22 / 3.22 = 1

H = 16.25 / 3.22 = 5

N = 3.22 / 3.22 = 1

 

So the empirical formula is:

CHN

5 0
3 years ago
λmax for the π → π* transition in ethylene is 170 nm. Is the HOMO-LUMO energy difference in ethylene greater than or less than t
-Dominant- [34]

Answer:

the HOMO-LUMO energy difference in ethylene is greater than that of cis,trans−1,3−cyclooctadiene

Explanation:

The λmax is the wavelength of maximum absorption. We could use it to calculate the HOMO-LUMO energy difference as follows:

For ethylene

E= hc/λ= 6.63×10^-34×3×10^8/170×10^-9= 1.17×10^-18J

For cis,trans−1,3−cyclooctadiene

E= hc/λ=6.63×10^-34×3×10^8/230×10^-9=8.6×10^-19J

Therefore, the HOMO-LUMO energy difference in ethylene is greater than that of cis,trans−1,3−cyclooctadiene

8 0
2 years ago
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