Ionic bond involves electrostatic attraction between oppositely charged ions.
The ions are atoms that have gained 1 or more electrons and atoms that have lost 1 or more electrons.
<span>Answer: The type of bond that requires the give and take of electrons is Ionic bond</span>
<h3>Answer:</h3>
a) Moles of Caffeine = 1.0 × 10⁻⁴ mol
b) Moles of Ethanol = 4.5 × 10⁻³ mol
<h3>Solution:</h3>
Data Given:
Mass of Caffeine = 20 mg = 0.02 g
M.Mass of Caffeine = 194.19 g.mol⁻¹
Molecules of Ethanol = 2.72 × 10²¹
Calculate Moles of Caffeine as,
Moles = Mass ÷ M.Mass
Putting values,
Moles = 0.02 g ÷ 194.19 g.mol⁻¹
Moles = 1.0 × 10⁻⁴ mol
Calculate Moles of Ethanol as,
As we know one mole of any substance contains 6.022 × 10²³ particles (atoms, ions, molecules or formula units). This number is also called as Avogadro's Number.
The relation between Moles, Number of Particles and Avogadro's Number is given as,
Number of Moles = Number of Molecules ÷ 6.022 × 10²³
Putting values,
Number of Moles = 2.72 × 10²¹ Molecules ÷ 6.022 × 10²³
Number of Moles = 4.5 × 10⁻³ Moles
Answer:
The molarity of urea in this solution is 6.39 M.
Explanation:
Molarity (M) is <em>the number of moles of solute in 1 L of solution</em>; that is
To calculate the molality, we need to know the number of moles of urea and the volume of solution in liters. We assume 100 grams of solution.
Our first step is to calculate the moles of urea in 100 grams of the solution,
using the molar mass a conversion factor. The total moles of 100g of a 37.2 percent by mass solution is
60.06 g/mol ÷ 37.2 g = 0.619 mol
Now we need to calculate the volume of 100 grams of solution, and we use density as a conversion factor.
1.032 g/mL ÷ 100 g = 96.9 mL
This solution contains 0.619 moles of urea in 96.9 mL of solution. To express it in molarity, we need to calculate the moles present in 1000 mL (1 L) of the solution.
0.619 mol/96.9 mL × 1000 mL= 6.39 M
Therefore, the molarity of the solution is 6.39 M.
Answer:
Filtration works best when the solute isn't dissolve in the solvent. For instance, sand and water can be seperated through filtration because both compounds do not dissolve with each other. However, sugar and water would not be seperated through filtration as they dissolve with each other.
-google :)
