Answer:
They both have 7 electrons in their outer shell
Answer:
0.185M sulfuric acid
Explanation:
Based on the reaction:
H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O
<em>1 mole of sulfuric acid reacts with 2 moles of KOH</em>
Initial moles of H₂SO₄ and KOH are:
H₂SO₄: 0.750L ₓ (0.470mol / L) = <em>0.3525 moles of H₂SO₄</em>
KOH: 0.700L ₓ (0.240mol / L) = <em>0.168 moles of KOH</em>
The moles of sulfuric acis that react with KOH are:
0.168mol KOH ₓ (1 mole H₂SO₄ / 2 moles KOH) = 0.0840 moles of sulfuric acid.
Thus, moles that remain are:
0.3525moles - 0.0840 moles = <em>0.2685 moles of sulfuric acid remains</em>
As total volume is 0.700L + 0.750L = 1.450L, concentration is:
0.2685mol / 1.450L = <em>0.185M sulfuric acid</em>
Answer:
The effective nuclear charge for a valence electron in oxygen atom: 
Explanation:
Effective nuclear charge ![[Z_{eff}]](https://tex.z-dn.net/?f=%5BZ_%7Beff%7D%5D)
is the net nuclear charge experienced by the electron in a given atom. It is always less than the actual charge of the nucleus [Z], due to shielding by electrons in the inner shells.
<em>It is equal to the difference between the actual nuclear charge or the atomic number (Z) and the shielding constant (s). </em>
<u>For an oxygen atom</u>-
Electron configuration: (1s²) (2s² 2p⁴)
<em>The atomic number (actual nuclear charge): </em>Z = 8
The shielding constant (s) for a valence electron can be calculated by using the Slater's rules:
⇒ s = 5 × 0.35 + 2 × 0.85 = 1.75 + 1.7 = 3.45
<u><em>Therefore, the effective nuclear charge for a valence electron in oxygen atom is:</em></u>
<u>Therefore, the effective nuclear charge for a valence electron in oxygen atom:</u>
Answer:
Explanation:
The Lanthanides were first discovered in 1787 when a unusual black mineral was found in Ytterby, Sweden. This mineral, now known as Gadolinite, was later separated into the various Lanthanide elements. In 1794, Professor Gadolin obtained yttria, an impure form of yttrium oxide, from the mineral.
