/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 81 Novelty devices for predicting r... [FREE SOLUTION] | 91Ó°ÊÓ

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Novelty devices for predicting rain contain cobalt(II) chloride and are based on the following equilibrium:$$\mathrm{CoCl}_{2}(s)+6 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CoCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}(s)$$ Purple Pink.What color will such an indicator be if rain is imminent?

Short Answer

Expert verified
The indicator will be Pink if rain is imminent, due to the equilibrium shift favoring the formation of cobalt(II) chloride hexahydrate (CoCl₂·6H₂O) in response to increased water vapor concentration.

Step by step solution

01

Determine the color of the indicator

Due to the equilibrium shift favoring the formation of cobalt(II) chloride hexahydrate (CoCl₂·6H₂O), the color of the indicator will be Pink, which is the color of CoCl₂·6H₂O. So, the indicator will be Pink if rain is imminent.

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Key Concepts

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

Chemical Equilibrium
Understanding the dynamics of chemical reactions is fundamental to the study of chemistry, and chemical equilibrium plays a key role in this. At equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction, meaning the reactants and products are formed at the same rate. In our exercise, we have the equilibrium between anhydrous cobalt(II) chloride and its hydrated form, cobalt(II) chloride hexahydrate.

The reaction can be represented as: \[\text{CoCl}_2(s) + 6\text{H}_2\text{O}(g) \rightleftharpoons \text{CoCl}_2 \cdot 6\text{H}_2\text{O}(s)\]. In the presence of moisture (like high humidity before rain), the reaction shifts to the right, forming more of the pink hydrate. Conversely, in dry conditions, the equilibrium shifts to the left, and the anhydrous form, which has a purple color, predominates. Through Le Chatelier's Principle, we understand that a change in conditions leads to an equilibrium shift to counteract that change. This shift is what causes the color change in our hygroscopic weather predictor.
Hygroscopic Substances
Hygroscopic substances have a knack for attracting and holding water molecules from the surrounding environment through absorption or adsorption. Cobalt(II) chloride is a particularly well-known hygroscopic material that undergoes a noticeable physical change upon water absorption.

The reaction, \[\text{CoCl}_2(s) + 6\text{H}_2\text{O}(g) \rightleftharpoons \text{CoCl}_2 \cdot 6\text{H}_2\text{O}(s)\], showcases cobalt(II) chloride's hygroscopic nature as it bonds with water to form its hydrate. This property is leveraged in applications such as moisture indicators in novelty devices, desiccants for keeping goods dry, and even in laboratories to control humidity levels. Their ability to react with water vapor and change color based on the degree of hydration or dehydration makes them invaluable tools.
Predicting Weather Changes
While modern meteorology uses complex instruments and computer models to predict the weather, simple chemical principles can still provide clues about upcoming changes. Substances like cobalt(II) chloride serve as rudimentary weather prediction tools. The color change from purple to pink in the presence of moisture is a visual cue of increased humidity, which often precedes rain.

Novelty weather-indicating devices capitalize on this chemistry to provide a handy, albeit basic, forecast tool. As moisture in the air increases, these devices, containing cobalt(II) chloride, absorb water and transition to a pink state. Such rudimentary methods can be surprisingly accurate in the short term and provide a fun way to engage with the principles of science and nature.

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Most popular questions from this chapter

At a particular temperature, \(K=4.0 \times 10^{-7}\) for the reaction $$\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)$$,In an experiment, 1.0 mole of \(\mathrm{N}_{2} \mathrm{O}_{4}\) is placed in a 10.0 -L vessel. Calculate the concentrations of \(\mathrm{N}_{2} \mathrm{O}_{4}\) and \(\mathrm{NO}_{2}\) when this reaction reaches equilibrium.

The synthesis of ammonia gas from nitrogen gas and hydrogen gas represents a classic case in which a knowledge of kinetics and equilibrium was used to make a desired chemical reaction economically feasible. Explain how each of the following conditions helps to maximize the yield of ammonia. a. running the reaction at an elevated temperature b. removing the ammonia from the reaction mixture as it forms c. using a catalyst d. running the reaction at high pressure

Le Châtelier's principle is stated (Section \(12-7\) ) as follows: "If a change is imposed on a system at equilibrium, the position of the equilibrium will shift in a direction that tends to reduce that change." The system \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)\) is used as an example in which the addition of nitrogen gas at equilibrium results in a decrease in \(\mathrm{H}_{2}\) concentration and an increase in \(\mathrm{NH}_{3}\) concentration. In the experiment the volume is assumed to be constant. On the other hand, if \(\mathrm{N}_{2}\) is added to the reaction system in a container with a piston so that the pressure can be held constant, the amount of \(\mathrm{NH}_{3}\) actually could decrease, and the concentration of \(\mathrm{H}_{2}\) would increase as equilibrium is reestablished. Explain how this can happen. Also, if you consider this same system at equilibrium, the addition of an inert gas, holding the pressure constant, does affect the equilibrium position. Explain why the addition of an inert gas to this system in a rigid container does not affect the equilibrium position.

At a particular temperature, \(K_{\mathrm{p}}=0.25\) for the reaction $$\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)$$. a. A flask containing only \(\mathrm{N}_{2} \mathrm{O}_{4}\) at an initial pressure of 4.5 atm is allowed to reach equilibrium. Calculate the equilibrium partial pressures of the gases. b. A flask containing only \(\mathrm{NO}_{2}\) at an initial pressure of 9.0 atm is allowed to reach equilibrium. Calculate the equilibrium partial pressures of the gases. c. From your answers to parts a and b, does it matter from which direction an equilibrium position is reached?

The reaction $$2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g)$$ has \(K_{\mathrm{p}}=109\) at \(25^{\circ} \mathrm{C}\). If the equilibrium partial pressure of \(\mathrm{Br}_{2}\) is 0.0159 atm and the equilibrium partial pressure of NOBr is 0.0768 atm, calculate the partial pressure of NO at equilibrium.

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