Answer:
10.7L of gasoline overflows
Explanation:
See attached file
What are the condition required for reverberation?
1 answer:
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Hence ,From the Guide there are other parameters which with this equation will give the exact time the athlete's walk back
From the question we are told
If the average velocity during the athlete's walk back to the starting line in Guided Example 2.5 is – 1.50 m/s,
Generally the equation Time spent is mathematically given as
T=\frac{d}{v}
Therefore
Hence ,From the Guide there are other parameters which with this equation will give the exact time the athlete's walk back
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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
