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Llana [10]
2 years ago
9

A solution of acetic acid, CH3CO₂H(aq), is at equilibrium. How would the

Chemistry
1 answer:
s344n2d4d5 [400]2 years ago
3 0

If more acetic acid were added to a solution at equilibrium, [H⁺] and [CH₃CO₂⁻] would increase to counteract the perturbation. (Option C)

<h3>How do systems at equilibrium respond to perturbation?</h3>

When a system at equilibrium suffers a perturbation, it shifts its equilibrium position to counteract such perturbation.

Let's consider a solution of acetic acid at equilibrium.

CH₃CO₂H(aq) = CH₃CO₂⁻(aq) + H⁺(aq)

If more acetic acid were added to the solution, the system will shift toward the products to counteract such an increase.

How would the system change if more acetic acid were added to the solution?

A. [H⁺] would decrease and [CH₃CO₂⁻] would increase. NO.

B. [H⁺] and [CH₃CO₂⁻] would decrease. NO.

C. [H⁺] and [CH₃CO₂⁻] would increase. YES. Both products would increase.

D. [H⁺] would increase and [CH₃CO₂⁻] would decrease. NO.

If more acetic acid were added to a solution at equilibrium, [H⁺] and [CH₃CO₂⁻] would increase to counteract the perturbation.

Learn more about equilibrium here: brainly.com/question/2943338

#SPJ1

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Diano4ka-milaya [45]

Answer:

stay the same.

Explanation:  Period 3 consists of the full 1s, 2s, and 2p electron orbitals, plus the 3s and 3p valence orbitals, which are filled with a total of 8 more electrons as we move from left (Na) to the far right (Ar):

Na:  1s2 2s2 2p6 3s1

Ar:    s2 2s2 2p6 3s2 3p6

As we move from left to right, and ignoring the already-filled 1s, 2s, and 2p orbitals, the period three starting and ending elements have the following:

Na: 3s1

Ar: 3s2, 3p6

All the new electrons electrons filled the third energy level (3s and 3p).  So the energy level does not change, just the orbitals.

5 0
3 years ago
28Which is true?
sergij07 [2.7K]
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3 years ago
How many moles of KOH are required to produce 4.79 g K3PO4 according to the following reaction? 3KOH + H3PO4 -----&gt; K3PO4 + 3
8_murik_8 [283]

Answer:

0.677 moles

Explanation:

Take the atomic mass of K = 39.1, O =16.0, P = 31.0

no. of moles = mass / molar mass

no. of moles of K3PO4 used = 4.79 / (39.1x3 + 31 + 16x4)

= 0.02256 mol

From the equation, the mole ratio of KOH : K3PO4 = 3 :1,

meaning every 3 moles of KOH used, produces 1 mole of K3PO4.

So, using this ratio, let the no. of moles of KOH required to be y.

\frac{3}{1} =\frac{y}{0.02256} \\

y = 0.02256 x3

y = 0.0677 mol

If you don't find exactly 0.677 moles as one of the options, go for the closest one. A very slight error may occur because of taking different significant figures of atomic masses when calculating.

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Molten barium chloride is separeted into two species :
BaCl₂(l) → Ba(l) + Cl₂(g),
but first ionic bonds in this salt are separeted because of heat:
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Reaction of reduction at cathode(-): Ba²⁺(l) + 2e⁻ → Ba(l).

Reaction of oxidation at anode(+): 2Cl⁻(l) → Cl₂(g) + 2e⁻.

<span>The anode is positive and the cathode is negative.</span>

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C10H22 + O2 -- CO2 + H2O balance
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Answer:

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