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Phoenix [80]
2 years ago
9

You need to produce a buffer solution that has a pH of 4.97. You already have a solution that contains 10. mmol (millimoles) of

acetic acid. How many millimoles of acetate (the conjugate base of acetic acid) will you need to add to this solution?
Chemistry
1 answer:
motikmotik2 years ago
5 0

Answer:

16.9 mmoles are needed to add to the solution

Explanation:

To solve this, we can apply the Henderson Hasselbach equation:

Acetic acid → CH₃COOH

CH₃COOH  +  H₂O  ⇄  CH₃COO⁻   +  H₃O⁺             Ka

That is the acid base equilibrium, to determine the amount:

pKa of acetic acid is 4.74

Henderson Hasselbach formula is:

pH = pKa + log (base/acid)

We replace  →  4.97 = 4.74 + log (B / 10mmoles)

Let's verify mmoles of B

4.97 - 4.74 = log (B/10)

0.23 =  log (B/10)

10^(0.23) = 10^(log (B/10)

1.698 = B/10

B = 1.698 . 10 → 16.9 mmoles

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Answer: It is 5450 mL


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6 0
3 years ago
Carbohydrates are made up of carbon, hydrogen, and __________.
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Oxegan ........
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5 0
2 years ago
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Which aqueous solution has the highest boiling point?a. 0.20 M NaClb. 0.10 M CaCl2c. 0.1 M Ga2(SO4)3d. 0.2 M C6H12O6
abruzzese [7]

Answer:

c. 0.1 M Ga₂(SO₄)₃

Explanation:

The boiling point increasing of a solvent due the addition of a solute follows the formula:

ΔT = K*m*i

<em>Where K is boiling point increasing constant (Depends of the solute), m is molality = molarity when solvent is water, and i is Van't Hoff factor.</em>

<em />

That means the option with the higher m*i will be the solution with the highest boiling point:

a. NaCl has i = 2 (NaCl dissociates in Na⁺ and Cl⁻ ions).

m* i = 0.20*2 = 0.4

b. CaCl₂; i = 3. 3 ions.

m*i= 0.10M * 3 = 0.3

c. Ga₂(SO₄)₃ dissolves in 5 ions. i = 5

m*i = 0.10M*55 = 0.5

d. C₆H₁₂O₆ has i = 1:

m*i = 0.2M*1 = 0.2

The solution with highest boiling point is:

<h3>c. 0.1 M Ga₂(SO₄)₃</h3>
3 0
2 years ago
In the absence of sodium methoxide, the same alkyl bromide gives a different product. Draw an arrowpushing mechanism to account
hoa [83]

Answer:

See explanation below

Explanation:

The question is incomplete, cause you are not providing the structure. However, I found the question and it's attached in picture 1.

Now, according to this reaction and the product given, we can see that we have sustitution reaction. In the absence of sodium methoxide, the reaction it's no longer in basic medium, so the sustitution reaction that it's promoted here it's not an Sn2 reaction as part a), but instead a Sn1 reaction, and in this we can have the presence of carbocation. What happen here then?, well, the bromine leaves the molecule leaving a secondary carbocation there, but the neighbour carbon (The one in the cycle) has a more stable carbocation, so one atom of hydrogen from that carbon migrates to the carbon with the carbocation to stabilize that carbon, and the result is a tertiary carbocation. When this happens, the methanol can easily go there and form the product.

For question 6a, as it was stated before, the mechanism in that reaction is a Sn2, however, we can have conditions for an E2 reaction and form an alkene. This can be done, cause the extoxide can substract the atoms of hydrogens from either the carbon of the cycle or the terminal methyl of the molecule and will form two different products of elimination. The product formed in greater quantities will be the one where the negative charge is more stable, in this case, in the primary carbon of the methyl it's more stable there, so product 1 will be formed more (See picture 2)

For question 6b, same principle of 6a, when the hydrogen migrates to the 2nd carbocation to form a tertiary carbocation the methanol will promove an E1 reaction with the vecinal carbons and form two eliminations products. See picture 2 for mechanism of reaction.

3 0
3 years ago
Need help with the 1st one
lutik1710 [3]

If I’m right this should be the answer

7 0
2 years ago
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