Answer:
I think w but I don't know
132 g of C , 22 g of H , 176 g of O
132 + 22 + 176 => 330 g <span>of the substance
</span>Now convert the masses in <span>moles :
</span>
C = 12.0 u H = 1.0 u O = 16.0 u
C = 132 / 12.0 => 11 moles
H = 22 / 1.0 => 22 moles
O = 176 / 16.0 => 11 moles
Using the values obtained the lowest proportion in mols of elements present, simply divide the values found for the least of them<span>:
</span>
C = 11 / 11 => 1
H = 22 / 11 => 2
O = 11 / 11 => 1
formula empirically <span>is : CH</span>₂O
hope this helps!
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:
46.839728 grams of water
Explanation:
1 mole is equal to 1 moles H2O, or 18.01528 grams
18.01528 times 2.6 = 46.839728
There are 46.839728 grams h20 in 2.60 mol of the compound
