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Annette [7]
2 years ago
8

Hi pls answer this i don't get it ​

Chemistry
1 answer:
joja [24]2 years ago
6 0
The answer is P
You can count how many electrons in the picture
Or you can search for the one who has 3 electrons in the outer shell
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Due to the small and highly electronegative nature of fluorine, the oxyacids of the this element are much less common and less s
Molodets [167]

Answer:

HOFO = (0, 0, +1, -1)

Explanation:

The formal charge (FC) can be calculated using the following equation:

FC = V - N - \frac{1}{2}B

<u>Where:</u>

V: are the valence electrons

N: are the nonbonding electrons

B: are the bonding electrons

The arrange of the atoms in the oxyacid is:

H - O₁ - F - O₂

Hence, the formal charge (FC) on each of the atoms is:

H: FC = 1 - 0 - 1/2*(2) = 0            

O₁: FC = 6 - 4 - 1/2*(4) = 0        

F: FC = 7 - 4 - 1/2*(4) = +1

O₂: FC =  6 - 6 - 1/2*(2) = -1

We can see that the negative charge is in the oxygen instead of the most electronegative element, which is the F. This oxyacid is atypical.  

I hope it helps you!

3 0
3 years ago
The number 0.000402 expressed in scientific notation is
goldfiish [28.3K]

Answer:

hope its help you

Explanation:

4.02 x 10-4

3 0
3 years ago
In a science experiment, Javi concludes that a chemical reaction has occurred. What evidence would support this conclusion?
Naily [24]

Answer:

B) a substance's color and odor changed        Explanation: A signal that a chemical change has occurred is when its odor (its smell) or its appearance has changed.

8 0
2 years ago
Read 2 more answers
What is the value of the equilibrium constant at 25 oC for the reaction between the pair: Br2(l) and I-(aq) Use the reduction po
Vikki [24]

Answer:

1.58×10E18

Explanation:

Since we have the reduction potentials we could make decisions regarding which one will be the anode or cathode. Evidently, bromine having the more positive reduction potential will be the cathode while the iodine will be the anode.

E°cell= 1.07- 0.53= 0.54 V

E°cell= 0.0592/n logK

0.54 = 0.0592/2 logK

logK= 0.54/0.0296

logK= 18.2

K= Antilog (18.2)

K= 1.58×10^18

3 0
3 years ago
The most active of all the chemical elements is a halogen known as
rodikova [14]
Astatine. Because it has the smaller shell of electrons. I believe
3 0
3 years ago
Read 2 more answers
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