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ValentinkaMS [17]
3 years ago
7

A 1 liter solution contains 0.443 M acetic acid and 0.332 M sodium acetate. Addition of 0.365 moles of perchloric acid will: (As

sume that the volume does not change upon the addition of perchloric acid.) Raise the pH slightly Lower the pH slightly Raise the pH by several units Lower the pH by several units Not change the pH Exceed the buffer capacity
Chemistry
1 answer:
Yuki888 [10]3 years ago
4 0

Answer:

Lower the pH several units

Explanation:

Typically, a buffer system would minimize changes in pH values with additions of strong acid or base. However, one should realize buffer systems do have limits. The limits are dictated by the amount of common ion in the buffer solution mix. The chemistry of the buffer and common ion functions to remove the excess strong acid or base added. However, if more strong acid or strong base is added that exceeds the capacity of the buffer solution to remove the an excess acid or base solution would result and pH values would change drastically.

In this problem, perchloric acid (HClO₄) is a strong acid delivering hydronium ion in an amount equal to the amount of perchloric acid noted in the problem.

Since the concentration of perchloric acid (HClO₄) and H⁺ ion exceeds the concentration of available acetate ion in the buffer solution, the buffer effect would be exhausted before removing all of the HClO₄. The remaining HClO₄ would then function to drastically reduce the pH of the remaining solution several units before stabilizing.

The 0.443M HOAc/0.332M NaOAc buffer before addition of the HClO₄ has a pH of 4.62.  Adding 0.365M in HClO₄ would cause the HOAc equilibrium to shift left because of the H⁺ overload. This shift would continue until all of the strong acid was removed or all of the conjugate base is exhausted by converting back to HOAc. That is, adding 0.365M HClO₄ removes all 0.332M OAc⁻ leaving 0.365M - 0.332M = 0.033M excess HClO₄. The pH of a 0.033M HClO₄ solution => pH = -logs[H⁺] = -log(0.033) = 1.48.

=> compare pH before adding acid = 4.62 to pH after adding acid = 1.48.

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In the molecule NO2 which element is pulling the electrons
Andrei [34K]

The bond between the N and 0 (double bond) transfers and gives a -ve charge on O and a +ve charge on N atom at the group . Thus the +vely charged nitrogen is electron-deficient pulling electrons towards itself!

The combination of the +vely charged nitrogen and the electronegative oxygen atom leads to delocalization causing the resonance effect.

7 0
3 years ago
Un mol de amoniaco Tiene una masa molar de 17 g y ocupa un volumen de 22.4 l qué volumen ocupa el 50 g amoniaco en condiciones n
mr_godi [17]

Answer:

V  = 65.81 L

Explanation:

En este caso, debemos usar la expresión para los gases ideales, la cual es la siguiente:

PV = nRT  (1)

Donde:

P: Presion (atm)

V: Volumen (L)

n: moles

R: constante de gases (0.082 L atm / mol K)

T: Temperatura (K)

De ahí, despejando el volumen tenemos:

V = nRT / P   (2)

Sin embargo como estamos hablando de condiciones normales de temperatura y presión, significa que estamos trabajando a 0° C (o 273 K) y 1 atm de presión. Lo que debemos hacer primero, es calcular los moles que hay en 50 g de amoníaco, usando su masa molar de 17 g/mol:

n = 50 / 17 = 2.94 moles

Con estos moles, reemplazamos en la expresión (2) y calculamos el volumen:

V = 2.94 * 0.082 * 273 / 1

<h2>V = 65.81 L</h2>
4 0
3 years ago
Sewage and industrial pollutants dumped into a body of water can reduce the dissolved oxygen concentration and adversely affect
MArishka [77]

Answer:

FALSE

Since 0.385 < 0.526, the value for week 3 is accepted.

Explanation:

Qexp = (|Xq - Xₙ₋₁|)/w

where Xq is the suspected outlier; Xₙ₋₁ is the next nearest data point; w is the range of data

First, the data are arranged in decreasing order, from highest to lowest:

3. 5.6

2. 5.1

8. 5.1

1. 4.9

6. 4.9

5. 4.7

7. 4.5

4. 4.3

Xq = 5.6; Xₙ₋₁ = 5.1; w = 5.6 - 4.3 = 1.3

Qexp = (|5.6 - 5.1|)/1.3 = 0.385

From tables, at 95% confidence level, for n = 8, Qcrit = 0.526

Since 0.385 < 0.526, the value for week 3 is accepted.

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I'm pretty sure it's the same 2mm = 2cc = 2c^2
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