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Oliga [24]
3 years ago
12

The molecular weight of ethanol (CHgCHzOH) is 46 and its density is 0.789 g/cm3. A. \A4rat is the molarity of ethanol in beer th

at is 5% ethanol by volume? [Alcohol content of beer varies from about 4Vo (lite beer) to B% (stout beer).1
Chemistry
1 answer:
mash [69]3 years ago
7 0

Answer:

0.86M

Explanation:

Hello,

In this case, one must consider that the ethanol's concentration in beer is defined as:

\% et =\frac{m_{et}}{m_{beer}}

In such a way, since molarity is defined as:

M=\frac{mol_{et}}{Volume_{solution}}

One is asked to compute the moles, considering a basis of 5 milliliters of ethanol and 100 milliliters of beer, thus:

mol_{et}=5mL*\frac{0.789g}{1mL} *\frac{1mol}{46g}=0.086mol\ et

Now, the volume in molarity is required in liters, therefore:

Volume_{beer}=100mL*\frac{1L}{1000mL}=0.1L

Therefore, the molarity turns out:

M=\frac{0.086mol}{0.1L}=0.86M

Best regards.

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Answer:

pH = 1.32

Explanation:

                 H₂M + KOH ------------------------ HM⁻ + H₂O + K⁺

This problem involves a weak diprotic acid which we can solve by realizing they amount  to buffer solutions.  In the first  deprotonation if all the acid is not consumed we will have an equilibrium of a wak acid and its weak conjugate base. Lets see:

So first calculate the moles reacted and produced:

n H₂M = 0.864 g/mol x 1 mol/ 116.072 g  =  0.074 mol H₂M

54 mL x  1L / 1000 mL x 0. 0.276 moles/L = 0.015 mol KOH

it is clear that the maleic acid will not be completely consumed, hence treat it as an equilibrium problem of a buffer solution.

moles H₂M left = 0.074 - 0.015 = 0.059

moles HM⁻ produced = 0.015

Using the Henderson - Hasselbach equation to solve for pH:

ph = pKₐ + log ( HM⁻/ HA) = 1.92 + log ( 0.015 / 0.059) = 1.325

Notes: In the HH equation we used the moles of the species since the volume is the same and they will cancel out in the quotient.

For polyprotic acids the second or third deprotonation contribution to the pH when there is still unreacted acid ( Maleic in this case) unreacted.

           

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What is the delta H fusion used to calculate volume of a liquid frozen that produces 1 kg of energy?
gulaghasi [49]

Answer:

The correct option is C. 1 kJ x 1/ΔHfus x g/mol x ml/g liquid

Explanation:

The change in enthalpy of fusion (ΔHfus) has units of energy divided into moles of substance:

ΔHfus = energy/mol of substance = J/mol or KJ/mol

Thus, we multiply the kJ of energy produced by the inverse of ΔHfus (in kJ/mol) we obtain the moles of substance:

energy (kJ) x 1/ΔHfus (mol/kJ) = moles of substance(mol)

Then, if we multiply the moles of substance by the molecular weight of the substance (in g/mol) we obtain the grams:

moles of substance(mol) x molecular weight (g/mol) = grams of substance (g)

Finally, if we multiply the grams of substance (in g) by the density of the substance (in g/ml) we obtain the volume (in ml):

grams of substance (g) x 1/density (ml/g) = volume (ml)

We combine all these parts to obtain the equation for the volume required to produce 1 kJ of energy:

energy (kJ) x 1/ΔHfus (mol/kJ) x molecular weight (g/mol) x 1/density (ml/g)

Therefore, the correct option is:

C → 1 kJ x 1/ΔHfus x g/mol x ml/g liquid

5 0
3 years ago
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