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Fynjy0 [20]
3 years ago
11

What is the [ch3co2-]/[ch3co2h] ratio necessary to make a buffer solution with a ph of 4.34? ka = 1.8 × 10-5 for ch3co2h?

Chemistry
1 answer:
xeze [42]3 years ago
5 0
According to Henderson–Hasselbalch Equation,

                                    pH  =  pKa + log [Acetate] / [Acetic Acid]

As,
           pKa = -log Ka
           pKa = -log (1.8 × 10⁻⁵)
           pKa =  4.74
So,
                               pH  =  4.74 + log [Acetate] / [Acetic Acid]

                                  4.34  =  4.74 + log [Acetate] / [Acetic Acid]

                        4.34 - 4.74  = log [Acetate] / [Acetic Acid]

                                 -0.40  =  log [Acetate] / [Acetic Acid]

Taking Antilog on both sides,

               [Acetate] / [Acetic Acid]  =  0.398
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Which element would make a poor, brittle conductor? use the periodic table for help. a. sulfur b. boron c. cobalt d. calcium?
Greeley [361]
I think the answer is C or B
8 0
3 years ago
Determine Δngas for each of the following reaction:(b) 2PbO(s) + O₂(g) ⇄ 2PbO₂(s)
Elis [28]

2PbO(s) + O₂(g) ⇄ 2PbO₂(s)

Then Δngas = -1

<h3>What is Δngas?</h3>

The number of moles of gas that move from the reactant side to the product side is denoted by the symbol ∆n or delta n in this equation.

Once more, n represents the growth in the number of gaseous molecules the equilibrium equation can represent. When there are exactly the same number of gaseous molecules in the system, n = 0, Kp = Kc, and both equilibrium constants are dimensionless.

<h3>Definition of equilibrium</h3>

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Learn more about Equilibrium

brainly.com/question/11336012

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5 0
2 years ago
The following equation is balanced according to the Law of Conservation of Mass:
igor_vitrenko [27]
The answer would be:
True
6 0
3 years ago
Use the following half-reactions to write three spontaneous reactions, calculate E°cell for each reaction, and rank the oxidizin
Ainat [17]

Answer:

See explaination

Explanation:

1)

we know that

half cell with higher reduction potential is cathode

so

cathode :

N20 + 2H+ + 2e- ---> N2 + H20

anode :

Cr(s) ---> Cr+3 + 3e-

so

overall reaction is

3 N20 + 6H+ + 2 Cr ---> 3N2 + 3H20 + 2Cr+3

now

Eo cell = Eo cathode - Eo anode

so

EO cell = 1.77 + 0.74

Eo cell = 2.51 V

now

in this case

oxidizing agents are N20 and Cr+3

reducing agents are Cr and N2

higher the reduction potential , stronger the oxidizing agent

lower the reduction potential , stronger the reducing agent

so

oxidzing agents

N20 > Cr+3

reducing agents

Cr > N2

2)

cathode :

Au+ + e- --> Au

anode :

Cr ---> Cr+3 + 3e-

overall reaction

3Au+ + Cr ---> 3Au + Cr+3

Eo cell = 1.69 + 0.74

Eo cell = 2.43

now

oxidizing agents :

Au+ > Cr+3

reducing agents :

Cr > Au

3)

cathode :

N20 + 2H+ + 2e- ---> N2 + H20

andoe :

Au ---> Au+ + e-

overall

2 Au + N20 + 2H+ --> 2 Au+ + N2 + H20

Eo cell = 1.77 - 1.69

Eo cell = 0.08

oxidizing agents

N20 > Au+

reducing agents

Au > N2

8 0
3 years ago
6. dissolving ice tea mix in water
Cloud [144]

Answer:

Dissolving ice tea mix in water is a chemical change

Explanation:

It is a chemical change because you can not bring back the ice tea mix back if you mix it in water.

Hope it helps

6 0
3 years ago
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