Ca(OH)₂ ==> Ca²⁺ + 2 OH<span>-
Ca(OH)</span>₂ is <span>strong Bases</span><span>
</span>Therefore, the [OH-] equals 5 x 10⁻⁴ M. For every Ca(OH)₂ you produce 2 OH⁻<span>.
</span>
pOH = - log[ OH⁻]
pOH = - log [ <span>5 x 10⁻⁴ ]
pOH = 3.30
pH + pOH = 14
pH + 3.30 = 14
pH = 14 - 3.30
pH = 10.7
hope this helps!</span>
Answer:
1) Ca: [Ar]4s²
2) Pm: [Xe]6s²4f⁵
Explanation:
1) Ca:
Its atomic number is 20. So it has 20 protons and 20 electrons.
Since it is in the row (period) 4 the noble gas before it is Ar, and the electron configuration is that of Argon whose atomic number is 18.
So, you have two more electrons (20 - 18 = 2) to distribute.
Those two electrons go the the orbital 4s.
Finally, the electron configuration is [Ar] 4s².
2) Pm
The atomic number of Pm is 61, so it has 61 protons and 61 electrons.
Pm is in the row (period) 6. So, the noble gas before Pm is Xe.
The atomic number of Xe is 54.
Therefore, you have to distribute 61 - 54 = 7 electrons on the orbitals 6s and 4f.
The resultant distribution for Pm is: [Xe]6s² 4f⁵.
the density of a black hole is 22.61 g/cm3 i think
Answer is: Both a fluorine atom and a bromine atom gain one electron, and both atoms become stable.
Fluorine and bromine are in group 17 in Periodic table of elements. Group 17 (halogens) elements are in group 17: fluorine (F), chlorine (Cl), bromine (Br) and iodine (I). They are very reactive and easily form many compounds.
Halogens need to gain one electron to have electron cofiguration like next to it noble gas.
Fluorine has atomic number 9, it means it has 9 protons and 9 electrons.
Fluorine tends to have eight electrons in outer shell like neon (noble gas) and gains one electron in chemical reaction.
Electron configuration of fluorine: ₉F 1s² 2s² 2p⁵.
Electron configuration of neon: ₁₀Ne 1s² 2s² 2p⁶.
