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Burka [1]
3 years ago
6

A student lives in a place that receives lots of snow and ice during the winter. The student has observed that the highway depar

tment spreads salt on the road after a snowfall. Which of the following correctly explains why the highway department uses salt on the roads?
Salt lowers the freezing point of water, which makes the melted snow on the road less likely to form ice.

Salt raises the freezing point of water, which makes the melted snow on the road less likely to form ice.

Salt raises the vapor pressure of water, which makes the melted snow more likely to evaporate before it freezes.

Salt lowers the osmotic pressure of water, which makes the melted snow more likely to be absorbed into the ground before it freezes.
Chemistry
2 answers:
trapecia [35]3 years ago
7 0

Answer:

Salt lowers the freezing point of water, which makes the melted snow on the road less likely to form ice.

Annette [7]3 years ago
7 0

Answer:

A

Explanation:

Edge.

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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>

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