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.
Answer:
60
Explanation:
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Ok so O3 has a greater electronegativity and is taken into account first, -2*3=-6, so As has to equal 6/2=3, so As has a +3 oxidation number here
The balanced equation would be 2NaOH + H2SO4 → Na2SO4 + 2H2O
Answer:
0.55 g of Cu can produced by this decomposition
Explanation:
Let's think the reaction of decomposition:
2CuO → 2Cu + O₂
Ratio is 2:2. 2 moles of CuO can decompose into 2 moles of Cu and 1 mol of oxygen.
Let's convert the mass of oxyde to moles → 0.695 g. 1 mol/ 79.55 g =
8.74×10⁻³ moles
8.74×10⁻³ mol of CuO decompose to 8.74×10⁻³ moles of Cu
Let's find out the mass → 8.74×10⁻³ mol . 63.55 g/1mol = 0.55 g
