<span>D. 2.0 M CaCl2 </span><span>will have the lowest vapor pressure at 25°C.
I hope this will help you.
Thank you
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2) 1
3) 1
4) 2
5) 3 or 2, I’m not sure
6) 1
7) 2
8) 1
9) 4
10) 2
11) 3
12) 1
I hope this helps, good luck.
Answer:
When [F⁻] exceeds 0.0109M concentration, BaF₂ will precipitate
Explanation:
Ksp of BaF₂ is:
BaF₂(s) ⇄ Ba²⁺(aq) + 2F⁻(aq)
Ksp = 1.7x10⁻⁶ = [Ba²⁺] [F⁻]²
The solution will produce BaF₂(s) -precipitate- just when [Ba²⁺] [F⁻]² > 1.7x10⁻⁶.
As the concentration of [Ba²⁺] is 0.0144M, the product [Ba²⁺] [F⁻]² will be equal to ksp just when:
1.7x10⁻⁶ = [Ba²⁺] [F⁻]²
1.7x10⁻⁶ = [0.0144M] [F⁻]²
1.18x10⁻⁴ = [F⁻]²
0.0109M = [F⁻]
That means, when [F⁻] exceeds 0.0109M concentration, BaF₂ will precipitate
Answer:
a. KCl, c. BaCl2 and e. LiF.
Explanation:
Hello,
In this case, we can identify the ionic compounds by verifying the difference in the electronegativity between the bonding compounds when it is 1.7 or more (otherwise it is covalent) as shown below:
a. KCl: 3.0-0.9=2.1 -> Ionic.
b. C2H4: 2.5-2.1=0.4 -> Covalent.
c. BaCl2: 3.0-0.8=2.2 -> Ionic.
d. SiCl4: 3.0-1.8=1.2 -> Covalent.
e. LiF: 4.0-1.0=3.0 -> Ionic.
Therefore, ionic compounds are a. KCl, c. BaCl2 and e. LiF.
Regards.
Answer:
If it loses an electron, it becomes positively charged and is known as a cation.
Explanation:
