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Lyrx [107]
3 years ago
5

You are asked to prepare 500. mL of a 0.200 M acetate buffer at pH 5.10 using only pure acetic acid (MW=60.05 g/mol, pKa=4.76),

3.00 M NaOH, and water. Answer the following questions regarding the preparation of the buffer.
1. How many grams of acetic acid will you need to prepare the 500 mL buffer? Note that the given concentration of acetate refers to the concentration of all acetate species in solution.
2. What volume of 3.00 M NaOH must you add to the acetic acid to achieve a buffer with a pH of 5.10 at a final volume of 500 mL? (Ignore activity coefficients.)
3. Pour the beaker contents into a 500 mL volumetric flask. Rinse the beaker several times and add the rinses to the flask. Swirl the flask to mix the solution. Add water to the mark to make the volume 500 mL. Stopper the flask and invert how many times to ensure complete mixing?
Chemistry
1 answer:
Alenkasestr [34]3 years ago
6 0

Answer:

1. 6.005 g

2. 22.9 mL

3. Until the mixtures becomes homogeneous.

Explanation:

A buffer is a solution where a weak acid is in equilibrium with its conjugate base (its anion) or a weak base is in equilibrium with its conjugate base (its cation). The buffer remains the pH almost unaltered because it shifts the equilibrium if an acid or base is added.

1. The pH of a buffer can be calculated by the Henderson-Hasselbalch equation:

pH = pKa + log[A⁻]/[HA]

Where [A⁻] is the concentration of the conjugate base (the anion) of the acid, and HA is the acid concentration.

5.10 = 4.76 + log[A⁻]/[HA]

log[A⁻]/[HA] = 5.10 - 4.76

log[A⁻]/[HA] = 0.34

[A⁻]/[HA] = 10^{0.34}

[A⁻]/[HA] = 2.1878

Because the volume is the same, we can replace the concentration by the number of moles (n):

nA⁻/nHA = 2.1878

nA⁻ = 2.1878*nHA

The total number of moles of the substances in the buffer is: 0.200 mol/L * 0.5 L = 0.1 mol

nA⁻ + nHA = 0.1

2.1878*nHA + n HA = 0.1

3.1878nHA = 0.1

nHA = 0.0314 mol

nA⁻ = 0.0686 mol

The total number of moles of acetic acid needed is 0.1 mol (both substances may be from it):

m = MW*mol

m = 60.05*0.1 = 6.005 g

2. NaOH must react with acetic acid to form the anion, so for a 1:1 reaction, it will be needed 0.0686 mol of NaOH:

V = mol/concentration

V = 0.0686/3

V = 0.0229 L = 22.9 mL

3. The buffer must be a homogeneous solution, it means that it can't be noticed phases in the buffer, so the flask must be inverted until all the buffer is diluted in water, and it will be noticed when the solution becomes homogenous.

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Answer:

In the 1770s, the English clergyman Joseph Priestley (who is credited with the discovery of O2) established the production of oxygen by vegetables recognizing that the process was, apparently, the inverse of animal respiration, which consumed such chemical element.

Explanation:

In 1772, Joseph Priestley in his Recherches sur diversces especes d'air differentiated the air of animal respiration from that emitted by vegetables in the presence of light. Of the latter, which he called "dephlogistic air", he highlighted his purifying property of the environment indicating that: plants far from affecting the air in the same way as animal respiration, produce the opposite effects, and tend to preserve the sweet and healthy atmosphere , when it becomes harmful as a result of the life and breathing of the animals or their death and their rot.

In 1780, Jean Ingeshousz in his Experiences sur les vegetaux completed and reaffirmed the observations of Joseph Priestley. At the same time, he could deny Charles Bonnet's hypothesis, by demonstrating that the air expelled from the leaves comes from inside, and that the stimulating factor of the gaseous emission was not the heat produced by the sun, but the intensity of the light .

It was, finally, Jean Senebier that between 1782 and 1784, found that the "fixed air" dissolved in the water favors the vegetation. From these observations, he hypothesized that "fixed air" (carbon dioxide) is absorbed by the plants, which take it from the atmosphere with the humidity it has and in which it is mixed. Once this gas has been captured, both from the atmosphere and from the ground, it is decomposed in the presence of light by the leaves, releasing the "vital air" (oxygen) and leaving the carbon in the plant.

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3 0
3 years ago
N2 + 6e → 2N-3
sveta [45]

Answer:

The correct answer is reduction.

Explanation:

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"You have a solution of glucose in water that has a concentration of 2.50 M and a volume of 0.442 liters. You dilute this soluti
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Answer:

0.737M

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C1V1=C2V2

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7 0
3 years ago
Three kilograms of steam is contained in a horizontal, frictionless piston and the cylinder is heated at a constant pressure of
lakkis [162]

Answer:

Final temperature: 659.8ºC

Expansion work: 3*75=225 kJ

Internal energy change: 275 kJ

Explanation:

First, considering both initial and final states, write the energy balance:

U_{2}-U_{1}=Q-W

Q is the only variable known. To determine the work, it is possible to consider the reversible process; the work done on a expansion reversible process may be calculated as:

dw=Pdv

The pressure is constant, so:  w=P(v_{2}-v_{1} )=0.5*100*1.5=75\frac{kJ}{kg} (There is a multiplication by 100 due to the conversion of bar to kPa)

So, the internal energy change may be calculated from the energy balance (don't forget to multiply by the mass):

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On the other hand, due to the low pressure the ideal gas law may be appropriate. The ideal gas law is written for both states:

P_{1}V_{1}=nRT_{1}

P_{2}V_{2}=nRT_{2}\\V_{2}=2.5V_{1}\\P_{2}=P_{1}\\2.5P_{1}V_{1}=nRT_{2}  

Subtracting the first from the second:

1.5P_{1}V_{1}=nR(T_{2}-T_{1})

Isolating T_{2}:

T_{2}=T_{1}+\frac{1.5P_{1}V_{1}}{nR}

Assuming that it is water steam, n=0.1666 kmol

V_{1}=\frac{nRT_{1}}{P_{1}}=\frac{8.314*0.1666*373.15}{500} =1.034m^{3}

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Answer:

I think the answer would be A.

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