Answer:
b) add 130 g of NaCH₃CO₂ to 100 mL of H₂O at 80 °C while stirring until all the solid dissolves, then let the solution cool to room temperature.
Explanation:
The solubility of NaCH₃CO₂ in water is ~1.23 g/mL. This means that at room temperature, we can dissolve 1.23 g of solute in 1 mL of water (solvent).
<em>What would be the best method for preparing a supersaturated NaCH₃CO₂ solution?</em>
<em>a) add 130 g of NaCH₃CO₂ to 100 mL of H₂O at room temperature while stirring until all the solid dissolves.</em> NO. At room temperature, in 100 mL of H₂O can only be dissolved 123 g of solute. If we add 130 g of solute, 123 g will dissolve and the rest (7 g) will precipitate. The resulting solution will be saturated.
<em>b) add 130 g of NaCH₃CO₂ to 100 mL of H₂O at 80 °C while stirring until all the solid dissolves, then let the solution cool to room temperature. </em>YES. The solubility of NaCH₃CO₂ at 80 °C is ~1.50g/mL. If we add 130 g of solute at 80 °C and let it slowly cool (and without any perturbation), the resulting solution at room temperature will be supersaturated.
<em>c) add 1.23 g of NaCH₃CO₂ to 200 mL of H₂O at 80 °C while stirring until all the solid dissolves, then let the solution cool to room temperature.</em> NO. If we add 1.23 g of solute to 200 mL of water, the resulting solution will have a concentration of 1.23 g/200 mL = 0.00615 g/mL, which represents an unsaturated solution.
The best option is a. but it is not complete!
Both liquids have different densities, it's true. But they are immiscible liquids and they don't form a solution.
Answer:
0.098 moles H₂S
Explanation:
The reaction that takes place is
- 2H₂(g) + S₂(g) ⇄ 2H₂S(g) keq = 7.5
We can express the equilibrium constant as:
- keq = [H₂S]² / [S₂] [H₂]² = 7.5
With the volume we can <u>calculate the equilibrium concentration of H₂</u>:
- [H₂] = 0.072 mol / 2.0 L = 0.036 M
<em>The stoichiometric ratio</em> tells us that <u>the concentration of S₂ is half of the concentration of H₂</u>:
- [S₂] = [H₂] / 2 = 0.036 M / 2 = 0.018 M
Now we <u>can calculate [H₂S]</u>:
- 7.5 = [H₂S]² / (0.018*0.036²)
So 0.013 M is the concentration of H₂S <em>at equilibrium</em>.
- This would amount to (0.013 M * 2.0 L) 0.026 moles of H₂S
- The moles of H₂ at equilibrium are equal to the moles of H₂S that reacted.
Initial moles of H₂S - Moles of H₂S that reacted into H₂ = Moles of H₂S at equilibrium
Initial moles of H₂S - 0.072 mol = 0.026 mol
Initial moles of H₂S = 0.098 moles H₂S
Answer:
Air is called a homogeneous mixture because all of the elements that make up air (oxygen, nitrogen, argon, carbon dioxide etc.) are not able to be distinguished between when you look at air.
Explanation:
A heterogeneous mixture would be something like trail mix. You can look at the bowl of trail mix and pick out the raisins from the nuts from the chocolate. A homogeneous mixture is something where the individual elements can't be picked out by the plain eye. When you look at air you can't say oh that part is oxygen and that is argon when you are going on a walk. Air is a solution and thus you can't pick out each part.
