<h3>Answer:</h3>
7.57 × 10⁻²² g of F
<h3>Solution:</h3>
Data Given:
Number of Molecules = 8
M.Mass of BF₃ = 67.82 g.mol⁻¹
Mass of Fluorine atoms = ?
Step 1: Calculate Moles of BF₃
Moles = Number of Molecules ÷ 6.022 × 10²³ Molecules.mol⁻¹
Putting value,
Moles = 8 Molecules ÷ 6.022 × 10²³ Molecules.mol⁻¹
Moles = 1.33 × 10⁻²³ mol
Step 2: Calculate Mass of BF₃:
Moles = Mass ÷ M.Mass
Solving for Mass,
Mass = Moles × M.Mass
Putting values,
Mass = 1.33 × 10⁻²³ mol × 67.82 g.mol⁻¹
Mass = 9.0 × 10⁻²² g
Step 3: Calculate Mass of Fluorine Atoms:
As,
67.82 g BF₃ contains = 57 g of F
So,
9.0 × 10⁻²² g will contain = X g of F
Solving for X,
X = (9.0 × 10⁻²² g × 57 g) ÷ 67.82 g
X = 7.57 × 10⁻²² g of F
A is your answer.
On the periodic table the atomic number is the number of protons inside the nucleus.
First, we must know what happens in the precipitation reaction. This type of reaction is a double replacement reactions. It is consists of two reactant compounds which interchange cations and anions to form two products. One of the products is an insoluble solid called a precipitate. For the precipitation of CaCO₃, there are two consecutive reactions involved:
1. Slaking of quicklime, CaO
CaO + H₂O ⇒ Ca(OH)₂
2. Precipitation
Ca(OH)₂ + CO₂ ⇒ CaCO₃ + H₂O
The ions that make up the H₂O molecule are H⁺ and OH⁻. According to solubility rules, the cation (positively charged ion) is likely to be attracted to an anion (negatively charged ion). Together, they form an ionic bond. This type of bond is when there is a complete transfer of electrons between the two. The Ca²⁺ cation lacks 2 electrons, while the anion OH⁻ has an excess 1 electron. In order to be stable, 1 Ca²⁺ ion and 2 OH⁻ ions must combine.
Therefore, the answer is OH⁻ ion.
Explanation:
It is given that vapor pressure of pure water at 296 K is 2778.5 Pa.These vapors will result in the formation of an ideal gas.
Now, as water is covered with oil and contains only 1% molecules of water. Hence, the vapor pressure of this mixture will also be equal to the vapor pressure of pure water.
So, vapor pressure of mixture = 1% vapor pressure of pure water
Therefore,
=
= 27.785 Pa
Thus, we can conclude that the equilibrium vapor pressure of water above the oil layer is 27.785 Pa.