Answer:
616,0 ng is the right answer.
Explanation:
You should know that 1 mole = 1 .10^9 nanomoles
Get the rule of three.
1 .10^9 nanomoles ...................... 56.0 gr
11 nanomoles .....................
(11 x 56) / 1 .10^9 nanomoles = 6.16 x 10^-7 gr
Let's convert
6.16 x 10^-7 gr x 1 .10^9 = 616 ngr
Answer:
38.75 L
Explanation:
From the question,
Applying Boyles Law,
PV = P'V'....................... Equation 1
Where P = Original pressure of the Argon gas, V = Original Volume of Argon gas, P' = Final pressure of Argon gas, V' = Final Volume of Argon gas.
make V the subject of the equation
V = P'V'/P.................... Equation 2
Given: P = 34.6 atm, V' = 456 L, P' = 2.94 atm.
Substitute these values into equation 2
V = (456×2.94)/34.6
V = 38.75 L
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
Mg(NO3)2 ➡️ Mg2+ + 2 NO3-
（32.0g Mg(NO3)2) / (148.3g Mg(NO3)2/mol)* (2 mol NO3- / 1 mol Mg(NO3)2) / (0.425 L) = 1.02 mol/ L NO3-
