Explanation:
Metals are the species which readily lose electrons in order to attain stability. This electron lost by the atom is actually present in its outermost shell which is also known as valence shell.
Ionization energy is defined as the energy required to remove the most loosely bound electron from a neutral gaseous atom.
When we move across a period from left to right then there occurs a decrease in atomic size of the atoms. Therefore, ionization energy increases along a period.
But when we move down a group then there occurs an increase in atomic size of the atoms due to addition of number of electrons in the atoms. Hence, ionization energy decreases along a group.
Thus, we can conclude that metals have low ionization energies and readily share their valence or outer electrons with each other to form an electron sea. These electrons are delocalized or shared among all the atoms that are bonded together and can therefore move freely throughout the metal structure.
The atomic number (Z) of the 3 elements F, Ne, and Na, are 9, 10, and 11.
Explanation:
Now Z refers to the number of protons in the element's nucleus, and protons are POSITIVELY charged particles. So a fluoride ion, F−, has 10 electrons rather than 9 (why?), a neutral neon atom has 10 electrons, and a sodium ion, Na+, also has 10 electrons (why?).
So the 3 species are ISOELECTRONIC; they possess the same number of electrons.
You should look at the Periodic Table to confirm the electron number. Elements are (usually) electrically neutral (sometimes they can be ionic if they have lost or gained electrons). If there are 10 positively charged protons in the nucleus, there are NECESSARILY 10 electrons associated with the NEUTRAL atom. I don't know WHY I am capitalizing certain WORDS.
You might ask why sodium will form a positive ion, Na+, whereas F forms a negative ion, F−. This again is a Periodic phenomenon, and explicable on the basis of the electronic structure that the Table formalizes.
Neutral metals tend to be electron-rich species, which have 1 or more electrons in a valence shell remote from the nuclear charge. On the other hand, neutral non-metals have valence electrons in incomplete shells, that do not effectively shield the nuclear charge. The demonstrable consequence is that metals lose electrons to form positive ions, whereas non-metals gain electrons to form negative ions.
Explanation:
Fusion vs Fission
In fission, energy is gained by splitting apart heavy atoms, for example uranium, into smaller atoms such as iodine, caesium, strontium, xenon and barium, to name just a few. However, fusion is combining light atoms, for example two hydrogen isotopes, deuterium and tritium, to form the heavier helium. Both reactions release energy which, in a power plant, would be used to boil water to drive a steam generator, thus producing electricity.
Answer:
11.31g NaClO₂
Explanation:
<em> Is given 250mL of a 1.60M chlorous acid HClO2 solution. Ka is 1.110x10⁻². What mass of NaClO₂ should the student dissolve in the HClO2 solution to turn it into a buffer with pH =1.45? </em>
It is possible to answer this question using Henderson-Hasselbalch equation:
pH = pKa + log₁₀ [A⁻] / [HA]
<em>Where pKa is -log Ka = 1.9547; [A⁻] is the concentration of the conjugate base (NaClO₂), [HA] the concentration of the weak acid</em>
You can change the concentration of the substance if you write the moles of the substances:
[Moles HClO₂] = 250mL = 0.25L×(1.60mol /L) = <em>0.40 moles HClO₂</em>
Replacing in H-H expression, as the pH you want is 1.45:
1.45 = 1.9547 + log₁₀ [Moles NaClO₂] / [0.40 moles HClO₂]
-0.5047 = log₁₀ [Moles NaClO₂] / [0.40 moles HClO₂]
<em>0.3128 = </em>[Moles NaClO₂] / [0.40 moles HClO₂]
0.1251 = Moles NaClO₂
As molar mass of NaClO₂ is 90.44g/mol, mass of 0.1251 moles of NaClO₂ is:
0.1251 moles NaClO₂ ₓ (90.44g / mol) =
<h3>11.31g NaClO₂</h3>
So,
Formate has a resonating double bond.
In molecular orbital theory, the resonating electrons are actually delocalized and are shared between the two oxygens. So the carbon-oxygen bonds can be described as 1.5-bonds (option B). I'm not sure if option C is correct, however, because the likelihood of both delocalized electrons being in the area of one oxygen atom is less than 50%.<span />
