Sorry it’s a bit blurry, you should be able to make it out
1, 3, 4, and 2 since rusting takes a long time and an example of rusting taking a long time is the Statue of Liberty.
The given question is incomplete. The complete question is as follows.
The solubility product of calcium fluoride () is at 25 degrees C. Will a fluoride concentration of 1.0 mg/L be soluble in a water containing 200 mg/L of calcium?
Explanation:
Reaction equation for the given chemical reaction is as follows.

Therefore, expression for
will be as follows.
=
Also, moles of per liter = \frac{\text{mass of F^{-} per L}}{\text{molar mass of F}}[/tex]
=
= 
Hence, ![[F^{-}] = \frac{\text{moles of F^{-}}{volume}](https://tex.z-dn.net/?f=%5BF%5E%7B-%7D%5D%20%3D%20%5Cfrac%7B%5Ctext%7Bmoles%20of%20F%5E%7B-%7D%7D%7Bvolume%7D)
=
= M
Now, moles of per L = \frac{\text{mass of Ca^{2+} per L}}{\text{molar mass of Ca}}[/tex]
=
= M
Also, ![[Ca^{2+}] = \frac{moles of Ca^{2+}}{volume}](https://tex.z-dn.net/?f=%5BCa%5E%7B2%2B%7D%5D%20%3D%20%5Cfrac%7Bmoles%20of%20Ca%5E%7B2%2B%7D%7D%7Bvolume%7D)
=
= M
Hence, ionic product =
=
= 
As, the ionic product is less than the
, this means that the fluoride will be soluble in water containing the calcium.
Answer:
197.8 J/(mol•K).
Explanation:
- ∆S° = [Σ bS° products] - [Σ aS° reactants], where b and a are the coefficients of the products and the reactants in the balanced reaction equation.
<em>Br₂(l) ⟶ 2 Br(g)</em>
∴ ∆S° = [2 x S° Br(g)] - [S° Br₂(l)]
∴ ∆S° = [(2)(175.0 J/(mol•K)] - [152.2 J/(mol•K)] = 197.8 J/(mol•K).
The answer is MgBr2 (chemical formula of magnesium)