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mr Goodwill [35]
3 years ago
12

Why are chemical equilibria called dynamic equilibria?

Chemistry
1 answer:
Serga [27]3 years ago
4 0

Answer:

Because molecules are moving so they change from reactants to products and vice versa.

Explanation:

Hello,

In this case, it is widely known that chemical equilibrium is related with such condition at which the rate of consumption of reactants and formation of products tends to zero so their concentrations remain the same over the time. For instance, if we consider the formation of hydrogen chloride at equilibrium:

H_2+Cl_2\rightleftharpoons 2HCl

We can also see that the reactants, hydrogen and chlorine are consumed to form hydrogen chloride and it also goes back to its reactants in a dynamic equilibrium (because they are constantly changing).

Best regards.

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Find the pH of the equivalence point(s) and the volume (mL) of 0.0372 M NaOH needed to reach the point(s) in titrations of(a) 42
Neko [114]

The pH of the equivalence points is 8.54

Concept of pH

A solution's acidity or alkalinity can be determined based on the concentration of hydrogen ions in the solution, or pH. Acidic aqueous solutions at 25 °C have a pH under 7, while basic or alkaline aqueous solutions have a pH above 7. Since the concentration of H3O+ is equal to the concentration of OH in pure water, a pH level of 7.0 at 25°C is referred to as "neutral". Strong bases may have a pH above 14, while very strong acids may have a pH below 14.

0.0520 M CH3COOH in 42.2 mL of moles is as follows:

2.194x103 mol CH3COOH = 0.0422L (0.0520mol / L)

that react with NaOH, resulting in:

NaOH + CH3COOH = CH3COO + Na+ + H2O

Thus, 1 mole of acetic acid and 1 mole of NaOH react.

As a result, 2.194x103 mol of NaOH are required to reach the equivalence point in volume:

To attain the equivalency point, 2.194x103 mol (1L / 0.0372mol) = 0.05899L 58.99mL of 0.0372 M NaOH

You will only have CH3COO at the equivalency point because it is in equilibrium with water, so:

H2O(l) + CH3COO(aq) CH3COOH(aq) + OH (aq)

A definition of equilibrium is:

Kb = 5.6x1010 = [OH] / [CH3COO] / [CH3COOH]

2.194x103mol of CH3COO has a molarity of (0.05899L + 0.0422L) = 0.02168M.

Therefore, equilibrium concentrations are:

[CH3COO]=0.02168M-X [CH3COOH]=X [OH]=X

5.6x1010 = [X] [X] / [0.02168M - X] converts to Kb.

1.214x1011 - 5.6x1010X = X2 X2 + 5.6x1010X - 1.214x1011 = 0 Finding the value of X:

False response; there are no negative concentrations. X: -3.48x106

As [OH] = X, [OH] = 3.484x106M, X is 3.484x106.

As 14 = pOH + pH pH = 8.54, pOH = -log [OH], or 5.46.

To know more about pH visit :

brainly.com/question/12546875

#SP4

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2 years ago
By what factor does [h+ ] change for each ph change? (a) 3.20 units
konstantin123 [22]
When a change in PH = 10^-ΔPH
so the change = 10^-3.2
change depends on two factor 0.00063 (10^-3.2)and factor 1585 (10^3.2)depending on the way which the change goes.if PH change from PH=0 to PH= 3.2 so the change is decreasing from concentration from 1 mol to 0.00063 and if PH change from PH = 3.2 to PH=0 so the change is increasing by a factor of 1585.
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3+ because I just know it 8757890
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Answer:

True is the correct answer.

Explanation:

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