Answer:
The fluoride which precipitates first is CaF₂
Explanation:
When F⁻ is added, CaF₂ and BaF₂ are produced following the ksp equation:
For CaF₂:
Ksp = 3.2x10⁻¹¹ = [Ca²⁺] [F⁻]²
<em>Where [Ca²⁺] = 0.075M * {35mL / (25mL + 35mL)} = 0.04375M</em>
3.2x10⁻¹¹ = [0.04375M] [F⁻]²
[F⁻]² = 7.31x10⁻¹⁰
[F⁻] = 2.7x10⁻⁵M
<h3>CaF₂ begins precipitation when [F⁻] = 2.7x10⁻⁵M.</h3>
For BaF₂:
Ksp = 1.5x10⁻⁶ = [Ba²⁺] [F⁻]²
<em>Where [Ba²⁺] = 0.090M * {25mL / (25mL + 35mL)} = 0.0375M</em>
1.5x10⁻⁶ = [0.0375M] [F⁻]²
[F⁻]² = 4x10⁻⁵
[F⁻] = 6.3x10⁻³M
BaF₂ begins precipitation when [F⁻] = 6.3x10⁻³M
Thus, the fluoride which precipitates first is CaF₂
Answer:
After 5 second 25% C-15 will remain.
Explanation:
Given data:
Half life of C-15 = 2.5 sec
Original amount = 100%
Sample remain after 5 sec = ?
Solution:
Number of half lives = T elapsed / half life
Number of half lives = 5 sec / 2.5 sec
Number of half lives = 2
At time zero = 100%
At first half life = 100%/2 = 50%
At second half life = 50%/2 = 25%
Thus after 5 second 25% C-15 will remain.
Answer:
"Water and ethyl alcohol will both have dipole-dipole interactions.
Technically they will both have Hydrogen bonding, which is a type of dipole-dipole. This is due to the high electronegativity values of oxygen atoms compared to the carbon and hydrogen atoms the oxygens bond to. This causes regions of both of these molecules that have partial negative charges and other regions wind up with partial positive charges.
Hexane will not have any dipole-dipole interactions because it is a non-polar molecule. The intermolecular forces between hexane molecules will be dispersion forces."
Explanation:
Here is the place I found the answer: https://socratic.org/questions/which-of-these-structures-has-dipole-dipole-interactions-water-h2o-ethyl-alcohol
All of this answer belongs to that person. I do not own any of this information.
<h3><u>Answer and explanation</u>;</h3>
The balanced chemical equation will be;
NaHCO3 + CH3COOH = CH3COONa + H2O + CO2
After balancing the chemical equation the number of atoms of each element is the same on the reactants and on the products side.
Sodium atoms - 1
Hydrogen atoms - 5
Carbon atoms - 3
Oxygen atoms - 5