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elena55 [62]
3 years ago
5

What is the pressure of a mixture of oxygen, nitrogen and carbon dioxide gases if the pressures of these gases are as follows: P

O2 = 254 mm Hg, PN2 = 0.351 atm, and
PCO2 = 43.9 kPa?
Chemistry
1 answer:
AleksandrR [38]3 years ago
8 0

Answer:

his is an example of a first-year chemistry question where you must first convert two of the pressures to the units of the third and add them up, per Dalton’s law of additive pressures. There are three possible answers, one for each of the three pressure units.

1 atm = 760 torr …… torr and mm Hg are the same

1 atm = 101.3 kPa

Dalton’s law:

P(total) = P(O2) + P(N2) + P(CO2)

Explanation:

Gases will assume whatever pressure depending on the equation of state of the mixture (in this case) and the volume htey are contained in. That could be the ideal gas law and simple mixing law, If you are quoting the partial pressures which you call simply “the pressure” of each gas, and that these refer to their values in the present mixture, then yes, we would add them up. The pressures are low enough for the ideal gas law to apply provided the temperature is not extremely low as well .

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2 years ago
What is the pH at the equivalence point in the titration of a 25.7 mL sample of a 0.370 M aqueous nitrous acid solution with a 0
expeople1 [14]

Answer:

pH = 8.24

Explanation:

Nitrous acid (HNO₂) reacts with KOH, thus:

HNO₂ + KOH → KNO₂ + H₂O

Moles of HNO₂ are:

0.0257mL ₓ (0.370mol / L) = 0.00951moles.

In equivalence point, the complete moles of nitrous acid reacts with KOH producing potassium nitrite. There are needed:

0.00951mol ₓ (1L / 0.491mol) = 0.01937L ≡ 19.4mL of 0.491M KOH to reach equivalence point.

Total volume in equivalence point is: 19.4mL + 25.7mL = <em>45.1mL</em>

Potassium nitrite is in equilibrium with water, thus:

NO₂⁻ + H₂O ⇄ HNO₂ + OH⁻

Where equilibrium constant, Kb, is defined as:

Kb = 1.41x10⁻¹¹ = \frac{[OH^-][HNO_2]}{[NO_2]}

In equilibrium, molarity of each compound are:

[NO₂⁻]: 0.00951mol/0.00451L - X = 0.211M - X

[HNO₂]: X

[OH⁻]: X

<em>Where X is reaction coordinate</em>

Replacing in Kb:

1.41x10⁻¹¹ = \frac{[X][X]}{[0.211 -X]}

0 = X² + 1.41x10⁻¹¹X - 2.97x10⁻¹²

Solving for X:

X = -1.72x10⁻⁶ <em>FALSE ANSWER. There is no negative concentrations.</em>

X = 1.72x10⁻⁶. <em>Right answer.</em>

That means:

[OH⁻]: 1.72x10⁻⁶M

As pOH is -log [OH⁻] and pH = 14-pOH:

pOH = 5.76; <em>pH = 8.24</em>

3 0
3 years ago
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