) In the calibration step of a thermochemistry experiment, a current of 117 mA, from a 24.0 V source was allowed to flow through
1 answer:
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Answer:
0.83 g/cm³
Explanation:
The volume of the alcohol is ...
(33.2 g)/(0.79 g/cm³) ≈ 42.0253 cm³
The density of water is about 1 g/cm³, so the volume of 9 g of it is ...
(9 g)/(1 g/cm³) = 9 cm³
The total volume is ...
42.0253 cm³ +9 cm³ = 51.0253 cm³
The total mass is ...
33.2 g + 9 g = 42.4 g
So, the resulting density is ...
(42.4 g)/(51.0253 cm³) ≈ 0.83 g/cm³
The resulting mixture has a density of about 0.83 g/cm³.
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<em>Additional comment</em>
Alcohol dissolves in water, so the total volume will be slightly less than 51.0253 cm³. The attached curve shows the result of mixing ethanol and water.
The weight of a mole of ethanol is about 46 g, of water, about 18.02 g. Then the mole fraction of alcohol is ...
(33.2/46)/(33.2/46 +9/18.02) ≈ 0.59
The volume of the mix is then estimated to be (-1.05 cc/mol)(1.221 mol), or about 1.28 cm³ less than the volume indicated above. That brings the density up to about 0.85 g/cm³.
We're not completely sure of the relevance of this calculation, since many of the applicable parameters are not specified. The point is that <em>the density of the mix will probably be slightly more than the value calculated above</em>. YMMV
