Answer:
When the concentration of F- exceeds 0.0109 M, BaF2 will precipitate.
Explanation:
Ba²⁺(aq) + 2 F⁻(aq) <----> BaF₂(s)
When BaF₂ precipitates, the Ksp relation is given by
Ksp = [Ba²⁺] [F⁻]²
[Ba²⁺] = 0.0144 M
[F⁻] = ?
Ksp = (1.7 × 10⁻⁶)
1.7 × 10⁻⁶ = (0.0144) [F⁻]²
[F⁻]² = (1.7 × 10⁻⁶)/0.0144 = 0.0001180555
[F⁻] = √0.0001180555 = 0.01086 M = 0.0109 M
Hope this Helps!!!
Answer:
the volume of CHCI3 = 7.87 ml
the volume of CHBr3 = 12.13 ml
Explanation:
From the given information:
We all know that 1 g/cm^3 = 1 g/ml
The density of boron = 2.34 g/ml
The Volume of the liquid mixture = 20 ml
Recall that:
Density = mass/volume
Mass = Density × Volume
Mass = 2.34 g/ml × 20 ml
Mass = 46.8 g
Suppose the volume of CHCI3 be Y and the Volume of CHBr3 be 20 - Y
Then :
Y (1.492) + 20-Y(2.890) = 46.8
1.492Y + 57.8 - 2.890Y = 46.8
- 1.398 Y = -11
Y = -11/ - 1.398
Y = 7.87 ml
Therefore, the volume of CHCI3 7.87 ml
the volume of CHBr3 = 20 - Y
= 20 - 7.87
= 12.13 ml
According to this formula:
㏑(K2/K1) = Ea/R(1/T1 - 1/T2)
when K is the rate constant
Ea is the activation energy
R is the universal gas constant
and T is the temperature K
when K is doubled so K2: K1 = 2:1 & R = 8.314 J.K^-1.mol^-1
and T1 = 10 +273 = 283 k & T2 = 21 + 273 = 294 k
So by substitution:
㏑2 =( Ea / 8.314) (1/283 - 1/294 )
∴ Ea = 43588.9 J/mol = 43.6 KJ/mol
Answer:
189.71 secs
Explanation:
We know that decomposition is a first order reaction;
So;
ln[A] = ln[A]o - kt
But;
[A]o = 1.00 M
[A] = 0.250 M
t =135 s
Hence;
ln[A] - ln[A]o = kt
k = ln[A] - ln[A]o/t
k = ln(1) - ln(0.250)/135
k =0 - (-1.386)/135
k = 1.386/135
k= 0.01
So time taken now will be;
ln[A] - ln[A]o = kt
t = ln[A] - ln[A]o/k
t = ln (3) - ln(0.450)/0.01
t = 1.0986 - (-0.7985)/0.01
t = 1.0986 + 0.7985/0.01
t = 189.71 secs
There are 27.67834 g/mol in 3.73 moles of lioh
