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The boiling points of neon and krypton are \(-245.9^{\circ} \mathrm{C}\) and \(-152.9^{\circ} \mathrm{C}\), respectively. Using these data, estimate the boiling point of argon. (Hint: The properties of argon are intermediate between those of neon and krypton.)

Short Answer

Expert verified
The estimated boiling point of argon is \(-199.4^{\circ}C\).

Step by step solution

01

Understand the Problem

We are given the boiling points of two elements, neon ( -245.9°C) and krypton ( -152.9°C). The task is to estimate the boiling point of argon, knowing that its properties are between those of neon and krypton.
02

Analyze Given Data

Identify that neon has the lowest boiling point and krypton the highest. Since argon properties are intermediate, its boiling point should lie between these two points.
03

Find the Midpoint

Use the average (midpoint) of the given temperatures to estimate the intermediate boiling point. Calculate the midpoint using the formula: \[ \text{Midpoint} = \frac{-245.9 + (-152.9)}{2} \]
04

Perform the Calculation

Calculate the midpoint using the numbers: \[ \text{Midpoint} = \frac{-245.9 - 152.9}{2} = \frac{-398.8}{2} = -199.4 \degree C \] This calculation gives the estimate for argon's boiling point.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

noble gases
Noble gases are a unique group of elements that reside in group 18 on the periodic table. They include helium, neon, argon, krypton, xenon, and radon. These elements are known for their low reactivity, which is due to their full valence electron shells. This inertness makes them very stable in their elemental form.
One key feature of noble gases is how they are typically colorless, odorless, and tasteless under standard conditions. Because of their very low boiling points, they exist as gases at room temperature, which makes them quite different from many other elements in the periodic table.
When examining physical properties like boiling points, noble gases display a trend—boiling points increase as you move down the group. This is due to the increase in atomic number, which increases van der Waals forces. Therefore, knowledge about helium, neon, argon, and krypton can help us predict properties for even less common noble gases, as we use the intermediate features to estimate unknown values.
temperature comparison
Comparing temperatures, especially boiling points, can help us understand a lot about the substances involved. In this particular problem, we're comparing the boiling points of different noble gases. The boiling temperatures tell us at which point a substance changes from a liquid to a gas.
For noble gases like neon and krypton, there's a notable difference in their boiling points, which reflects their atomic sizes and the strength of intermolecular forces. Specifically, neon has a boiling point of -245.9°C while krypton boils at -152.9°C. The larger krypton atom has greater attractive forces between its atoms, resulting in a higher boiling point compared to neon.
Understanding the amount of heat required for this phase change not only provides an insight into their physical properties but also assists in practical applications, such as industrial processes involving these gases. By understanding these differences, we can effectively infer the boiling point for argon, which lies between these two extremes.
interpolation
Interpolation is a valuable mathematical and practical tool used to estimate unknown values that fall within the range of known data points. In this situation, interpolation allows us to predict the boiling point of argon, sitting between neon and krypton in the periodic table.
To perform interpolation, we calculate the midpoint for the boiling temperatures of neon and krypton. This midpoint serves as an estimation for argon's boiling point, assuming its properties are indeed intermediate.
We use the formula: \[ \text{Midpoint} = \frac{-245.9 + (-152.9)}{2} \]This results in a boiling point of -199.4°C. The process of calculating this value illustrates a simple yet powerful use of interpolation to quickly get estimates, particularly when experimentation isn't convenient or possible. Through interpolation, we can handle similar cases efficiently, predicting many other intermediate values across both scientific and everyday applications.

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