I'm not sure what you mean. Besides, I feel like you're talking math.
But anyways, if you have 120 let's say
The scientific notation is 1.20 × 10^2
if you have 125000
the scientific notation is 1.25 × 10^ 5
The number of times you go left the decimal, I guess exponent increases
So yea
Because it requires more energy to create a neutron from a proton than it does to create a proton from a neutron, protons were formed more frequently than neutrons in the early universe. The correct answer is option b.
To find the answer, we need to know more about the early universe.
<h3>How the formation of proton over neutrons was favored in the early universe?</h3>
- A neutron is produced with greater energy than a proton.
- However, later on, some of the protons were changed into neutrons.
- Contrary to some claims, the proton is a stable particle that never decays, but the neutron is unstable outside of the nucleus and decays with a half life of around 10.5 minutes.
- However, very few would have had time to decay on the timeline you mention in your question.
- Every matter particle should have been accompanied by an antimatter particle, and every proton, neutron, and electron, by an anti-neutron and a positron, respectively.
- Where did all the antimatter go is the great mystery. There have been a few attempts to explain this, but they have failed.
Thus, we can conclude that, the correct answer is option b.
Learn more about the early universe here:
brainly.com/question/28130096
#SPJ1
Answer:
HOFO = (0, 0, +1, -1)
Explanation:
The formal charge (FC) can be calculated using the following equation:

<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!
Here we have to get the spin of the other electron present in a orbital which already have an electron which has clockwise spin.
The electron will have anti-clockwise notation.
We know from the Pauli exclusion principle, no two electrons in an atom can have all the four quantum numbers i.e. principal quantum number (n), azimuthal quantum number (l), magnetic quantum number (m) and spin quantum number (s) same. The importance of the principle also restrict the possible number of electrons may be present in a particular orbital.
Let assume for an 1s orbital the possible values of four quantum numbers are n = 1, l = 0, m = 0 and s = 
.
The exclusion principle at once tells us that there may be only two unique sets of these quantum numbers:
1, 0, 0, +
and 1, 0, 0, -
.
Thus if one electron in an orbital has clockwise spin the other electron will must be have anti-clockwise spin.