Answer :
Example of polar covalent molecules H-O-H(water), ammonia
Explanation:
The presence of intermolecular Hydrogen bonding makes the boiling point of water unexpectedly high, and the polar covalent nature makes it dissolve polar solute/compound
Answer:
See below
Step-by-step explanation:
Matter is either a <em>pure substance</em> or a <em>mixture.
</em>
Pure substances
- Are composed of one type of atom or molecule.
- Have a constant chemical composition
- Have fixed chemical properties
- Have fixed physical properties
• For example, melting point, boiling point, density, solubility
Mixtures:
- Consist of two or more substances not chemically combined
- Have a variable composition
- Can be separated into two or more components by physical means
• For example, filtration, distillation, centrifugation
- Each component retains its own properties
All of them are important to identify a mineral and get information from it, but I'd say the least important is cleavage.
Answer:
Explanation:
<u>1) Balanced chemical equation:</u>
<u>2) Mole ratio:</u>
- 2 mol S : 3 mol O₂ : 2 mol SO₃
<u>3) Limiting reactant:</u>
n = 6.0 g / 32.0 g/mol = 0.1875 mol O₂
n = 7.0 g / 32.065 g/mol = 0.2183 mol S
Actual ratio: 0.1875 mol O₂ / 0.2183 mol S =0.859
Theoretical ratio: 3 mol O₂ / 2 mol S = 1.5
Since there is a smaller proportion of O₂ (0.859) than the theoretical ratio (1.5), O₂ will be used before all S be consumed, and O₂ is the limiting reactant.
<u>4) Calcuate theoretical yield (using the limiting reactant):</u>
- 0.1875 mol O₂ / x = 3 mol O₂ / 2 mol SO₃
- x = 0.1875 × 2 / 3 mol SO₃ = 0.125 mol SO₃
<u>5) Yield in grams:</u>
- mass = number of moles × molar mass = 0.125 mol × 80.06 g/mol = 10.0 g
<u>6) </u><em><u>Percent yield:</u></em>
- Percent yield, % = (actual yield / theoretical yield) × 100
- % = (7.9 g / 10.0 g) × 100 = 79%