The reactants in the neutralization reaction are an acid and a base while the products are a salt and water.
They are all transioning in states of matter
Answer:
The molar solubility of YF₃ is 4.23 × 10⁻⁶ M.
Explanation:
In order to calculate the molar solubility of YF₃ we will use an ICE chart. We identify 3 stages: Initial, Change and Equilibrium and we complete each row with the concentration of change of concentration. Let's consider the solubilization of YF₃.
YF₃(s) ⇄ Y³⁺(aq) + 3 F⁻(aq)
I 0 0
C +S +3S
E S 3S
The solubility product (Ksp) is:
Ksp = [Y³⁺].[F⁻]³= S . (3S)³ = 27 S⁴
![S=\sqrt[4]{Ksp/27} =\sqrt[4]{8.62 \times 10^{-21} /27}=4.23 \times 10^{-6}M](https://tex.z-dn.net/?f=S%3D%5Csqrt%5B4%5D%7BKsp%2F27%7D%20%3D%5Csqrt%5B4%5D%7B8.62%20%5Ctimes%2010%5E%7B-21%7D%20%20%2F27%7D%3D4.23%20%5Ctimes%2010%5E%7B-6%7DM)
Sulfur will float, and iron will sink because each will retain its properties.
“Iron will float, and sulfur will sink because each will lose its properties” is <em>incorrect</em>. A substance will lose its properties only if it reacts to form a new substance.
“Both will float because they are physically mixed” is <em>incorrect</em>. The substances in a mixture retain their properties.
“Both will float because they react chemically” is <em>incorrect</em>. Iron and sulfur do not react at room temperature. Even if they did, the iron sulfide would sink.
Using E=hν where h is Planck's constant and v is the frequency of the photon. In the question above,the wavelenght is given so we can find the frequency of the photon using c=λν. c is a constant =3*10^8 so frequency is equal to
(3*10^8)/0.135*10^-9. Then use ur frequency in the eqn above using h 6.626*10^-34
