Answer:
The predominant intermolecular force in the liquid state of each of these compounds:
ammonia (NH3)
methane (CH4)
and nitrogen trifluoride (NF3)
Explanation:
The types of intermolecular forces:
1.Hydrogen bonding: It is a weak electrostatic force of attraction that exists between the hydrogen atom and a highly electronegative atom like N,O,F.
2.Dipole-dipole interactions: They exist between the oppositely charged dipoles in a polar covalent molecule.
3. London dispersion forces exist between all the atoms and molecules.
NH3 ammonia consists of intermolecular H-bonding.
Methane has London dispersion forces.
Because both carbon and hydrogen has almost similar electronegativity values.
NF3 has dipole-dipole interactions due to the electronegativity variations between nitrogen and fluorine.
Be - Beryllium
S - sulfur
K - Potassium
C - Carbon
B - Boron
Ar - Argon
O - oxygen
Ne - Neon
Ca - Calcium
H - Hydrogen
<h3>
Answer:</h3>
91.2 g Mn
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Atomic Structure</u>
- Reading a Periodic Table
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<u>Stoichiometry</u>
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[Given] 1.00 × 10²⁴ atoms Mn
<u>Step 2: Identify Conversions</u>
Avogadro's Numer
[PT] Molar Mass of Mn - 54.94 g/mol
<u>Step 3: Convert</u>
- [DA] Set up:

- [DA] Multiply/Divide [Cancel out units]:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
91.2321 g Mn ≈ 91.2 g Mn
212 ml of lead nitrate is required to prepare a dilute solution of 820.7 ml of lead nitrate.
Answer:
Option A.
Explanation:
Similar to Avagadro's law, there is another law termed as dilution law. As the product of volume and normality of the reactant is equal to the product of volume and normality of the product from the Avagadro's law. In dilution law, it will be as product of volume and concentration of the solute of the reactant is equal to the product of volume and concentration of solution.

So, as per the given question C1 = 5.45 M of lead nitrate and V1 has to be found. While C2 is 1.41 M of lead nitrate and V2 is 820.7 ml.
Then, 

So nearly 212 ml of lead nitrate is required to prepare a dilute solution of 820.7 ml of lead nitrate.
Answer:
Energy equals mass times the speed of light squared.
Explanation:
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