Answer:
I think that the trend that would be seen in the time column of the data table would be that the number of seconds would increase. I know this because for each flask, the concentration of sodium thiosulfate decreases, since less of it is being mixed with more water. Also, when the concentration of a substance decreases, then the reaction rate also decreases, as there will be fewer collisions with sulfuric acid if there are fewer moles of sodium thiosulfate. When there are fewer collisions in a reaction, the reaction itself will take longer, and so when the sodium thiosulfate is diluted, the reaction takes more time.
Explanation:
<em>I verify this is correct. </em>
I am pretty sure u have pictures and this one should be the one that erika should make
To Find :
The volume of 12.1 moles hydrogen at STP.
Solution :
We know at STP, 1 mole of gas any gas occupy a volume of 22.4 L.
Let, volume of 12.1 moles of hydrogen is x.
So, x = 22.4 × 12.1 L
x = 271.04 L
Therefore, the volume of hydrogen gas at STP is 271.04 L.
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.
