Answer:
Fluorine
General Formulas and Concepts:
<u>Chemistry</u>
- Reading a Periodic Table
- Periodic Trends
- Electronegativity - the tendency for an element to attract an electron to itself
- Z-effective and Coulomb's Law, Forces of Attraction
Explanation:
The Periodic Trend for Electronegativity is up and to the right of the Periodic Table.
Fluorine is Element 9 and has 9 protons. Radium is Element 88 and has 88 protons. Therefore, Radium has a bigger Zeff than Flourine.
However, since Radium is in Period 7 while Fluorine is in Period 2, Radium has more core e⁻ than Fluorine does. This will create a much larger shielding effect, causing Radium's outermost e⁻ to have less FOA between them. Fluorine, since it has less core e⁻, the FOA between the nucleus and outershell e⁻ will be much stronger.
Therefore, Fluorine would attract an electron more than Radium, thus bringing us to the conclusion that Fluorine has a higher electronegativity.
<span>Saccharides
Hope it helps!
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I don't even know what the answer is
Solubility
product constants are values to describe the saturation of ionic compounds with
low solubility. A saturated solution is when there is a dynamic equilibrium
between the solute dissolved, the dissociated ions, the undissolved and the
compound. It is calculated from the product of the ion concentration in the
solution. For the generic salt, AB2, the dissociation would be as
follows:<span>
AB2 = A2+ + 2B-
So, the expression for the solubility product would be:
Ksp = [A2+] [B-]^2
</span>Ksp = [x] [2x]^2 = 4x^3
<span>
where x = </span><span>3.72×10^−4 M
</span><span>
Ksp = </span>4( 3.72×10^−4 )^3
Ksp = 2.06x10^-10 M^3
The solubility product constant of AB2 would be Ksp = 2.06x10^-10 M^3.
