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laiz [17]
3 years ago
8

The concentration of a diluted "wine" sample was found to be 0.24 %(v/v) ethanol. Assuming the wine was diluted by 75x before an

alysis, what was the original concentration of ethanol in the "wine"
Chemistry
1 answer:
Softa [21]3 years ago
3 0

Answer:

The original concentration of ethanol was 18 % (v/v)

Explanation:

For the analysis of a sample, mostly dilutions are made of the original concentrated sample. The analysis results obtained from the diluted sample are then calculated for the concentrated samples.

According to the given data, the wine sample was diluted 75 times. This means that the initial concentration of ethanol was 75 times more than the final concentration i.e. 0.24 % (v/v). So mathematically:

original concentration (v/v)= final concentration (v/v) x times diluted

original concentration (v/v)= 0.24 % x 75

original concentration (v/v) = 18 %

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The earth's mantle will not change in the future because this region is safe from the intervention of human beings.

<h3>What happens in the Earth's mantle?</h3>

The transfer of heat and material in the mantle helps us to determine the landscape of Earth. Activity in the mantle moves the plate tectonics, that contributes to volcanoes, seafloor spreading, earthquakes, and building of mountains.

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2 years ago
A 3.82-g sample of magnesium nitride is reacted with 7.73 g of water. mg3n2 (s) + 3 h2o (â) â 2 nh3 (g) + 3 mgo (s) the yield of
horrorfan [7]
The reaction is properly written as

Mg₃N₂ (s) + 3 H₂O (l) --> 2 NH₃<span> (g) + 3 MgO (s) 

Molar mass of Mg</span>₃N₂ = 100.95 g/mol
Molar mass of H₂O = 18 g/mol
Molar mass of MgO = 40.3 g/mol

Moles Mg₃N₂: 3.82/100.95 = 0.0378
Moles H₂O: 7.73/18 = 0.429 

Theo H₂O required for available Mg₃N₂: 0.0378*3/1 = 0.1134 mol
Hence, the limiting reactant is Mg₃N₂.
Thus,

Theoretical Yield = 0.0378 mol Mg₃N₂ * 3 mol MgO/Mg₃N₂ * 40.3 g/mol
Theo Yield = 4.57 g

Percent Yield = Actual Yield/Theo Yield * 100
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3 years ago
A 20.0 g piece of a metal is heated and place into a calorimeter containing 250.0 g of water initially at 25.0 oC. The final tem
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Answer:

Q_{metal} = -6799\,J

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By the First Law of Thermodynamics, the piece of metal and water reaches thermal equilibrium when water receives heat from the piece of metal. Then:

Q_{metal} = - Q_{w}

Q_{metal} = m_{w} \cdot c_{p,w}\cdot (T_{1}-T_{2})

Q_{metal} = (250\,g)\cdot \left(4.184\,\frac{J}{g\cdot ^{\textdegree}C} \right)\cdot (25\,^{\textdegree}C - 31.5\,^{\textdegree}C)

Q_{metal} = -6799\,J

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4 years ago
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4 years ago
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A student determines that she used 0.0665 mol of sodium hydroxide (NaOH) to completely titrate 25.00 mL of sulfuric acid solutio
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Let us determine the reaction for the titration below:

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The answer is the fourth option: 

<span>2.62 M</span>
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