Given:
M = 0.0150 mol/L HF solution
T = 26°C = 299.15 K
π = 0.449 atm
Required:
percent ionization
Solution:
First, we get the van't Hoff factor using this equation:
π = i MRT
0.449 atm = i (0.0150 mol/L) (0.08206 L atm / mol K) (299.15 K)
i = 1.219367
Next, calculate the concentration of the ions and the acid.
We let x = [H+] = [F-]
[HF] = 0.0150 - x
Adding all the concentration and equating to iM
x +x + 0.0150 - x = <span>1.219367 (0.0150)
x = 3.2905 x 10^-3
percent dissociation = (x/M) (100) = (3.2905 x 10-3/0.0150) (100) = 21.94%
Also,
percent dissociation = (i -1) (100) = (</span><span>1.219367 * 1) (100) = 21.94%</span>
Answer:they will out live us himans
Explanation:
The balanced equation is
4Fe+3O₂⇒2Fe₂O₃
We know that the mole of Fe₂O₃ is 6, and since the ratio between oxygen and <span>Fe₂O₃ is 3:2, we can see that
3:2 = x:6 (3 oxygen moles can make 2 </span>Fe₂O₃ moles = x oxygen moles can make 6 <span>Fe₂O₃ moles)
</span><span>
Multiply outside and inside (3*6 , 2*x) and put them on opposing sides of the equation
2*x = 3*6
2x=18
x=9
Therefore 9 moles of oxygen is needed.
</span>
Answer:
CaO + SO 3 → CaSO. 4
This is an acid-base reaction (neutralization): CaO is a base, SO 3 is an acid.
