/*! 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 79 There is evidence that radon rea... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 20: Problem 79

There is evidence that radon reacts with fluorine to form compounds similar to those formed by xenon and fluorine. Predict the formulas of these \(\mathrm{RnF}_{x}\) compounds. Why is the chemistry of radon difficult to study?

Short Answer

Expert verified
The predicted formulas of the radon-fluorine \(\mathrm{RnF}_{x}\) compounds, based on the xenon-fluorine compounds, are: RnF鈧� (Radon Difluoride), RnF鈧� (Radon Tetrafluoride), and RnF鈧� (Radon Hexafluoride). Studying the chemistry of radon is difficult primarily due to its radioactive nature and short half-life (3.8 days for radon-222), its chemical inertness as a noble gas, and the controlled laboratory conditions required for its production and study.

Step by step solution

01

Determine the possible oxidation states of radon

Radon, xenon, and fluorine are members of the noble gases group. Radon is a member of the 6th period, Fluorine is a member of the 2nd period, and Xenon is a member of the 5th period. Noble gases (excluding helium) have oxidation states depending on their group number and period. Radon being in the 6th period indicates that it can have a maximum oxidation state of +6.
02

Consider the compounds formed by xenon and fluorine

To predict the formulas of radon-fluorine compounds, let's look at the compounds formed by xenon (Xe) and fluorine (F): 1. XeF鈧�: Xenon Difluoride 2. XeF鈧�: Xenon Tetrafluoride 3. XeF鈧�: Xenon Hexafluoride These compounds show that the oxidation states of xenon in these compounds are +2, +4, and +6.
03

Predict the formulas of the radon-fluorine compounds

Since radon has similar properties to xenon and can also have a maximum oxidation state of +6, the formulas of the radon-fluorine compounds can be predicted by replacing xenon (Xe) with radon (Rn) in the xenon-fluorine compounds: 1. RnF鈧�: Radon Difluoride 2. RnF鈧�: Radon Tetrafluoride 3. RnF鈧�: Radon Hexafluoride These are the predicted formulas of the \(\mathrm{RnF}_{x}\) compounds.
04

Explain why the chemistry of radon is difficult to study

The chemistry of radon is difficult to study for several reasons: 1. Radon is a radioactive element with a very short half-life. Its most stable isotope, radon-222, has a half-life of only 3.8 days. This makes it challenging to work with, as it decays quickly and presents radiation risks. 2. Radon is a noble gas, which means that it has a full valence shell and is chemically inert in many situations. This makes reactions between radon and other elements less frequent and more difficult to study. 3. Due to its radioactive nature, radon is often produced and studied in controlled laboratory settings. These conditions limit the opportunities to study radon's chemical properties over an extended period or in various environments.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Phosphoric acid \(\left(\mathrm{H}_{3} \mathrm{PO}_{4}\right)\) is a triprotic acid, phosphorous acid \(\left(\mathrm{H}_{3} \mathrm{PO}_{3}\right)\) is a diprotic acid, and hypophosphorous acid \(\left(\mathrm{H}_{3} \mathrm{PO}_{2}\right)\) is a monoprotic acid. Explain this phenomenon.

Lead forms compounds in the \(+2\) and \(+4\) oxidation states. All lead(II) halides are known (and are known to be ionic). Only \(\mathrm{PbF}_{4}\) and \(\mathrm{PbCl}_{4}\) are known among the possible lead(IV) halides. Presumably lead(IV) oxidizes bromide and iodide ions, producing the lead(II) halide and the free halogen: Suppose \(25.00 \mathrm{~g}\) of a lead(IV) halide reacts to form \(16.12 \mathrm{~g}\) of a lead(II) halide and the free halogen. Identify the halogen.

An unknown element is a nonmetal and has a valence electron configuration of \(n s^{2} n p^{4}\). a. How many valence electrons does this element have? b. What are some possible identities for this element? c. What is the formula of the compound(s) this element would form with lithium? hydrogen? magnesium? aluminum? fluorine?

a. Many biochemical reactions that occur in cells require relatively high concentrations of potassium ion \(\left(\mathrm{K}^{+}\right) .\) The concentration of \(\mathrm{K}^{+}\) in muscle cells is about \(0.15 \mathrm{M}\). The concentration of \(\mathrm{K}^{+}\) in blood plasma is about \(0.0050 M\). The high internal concentration in cells is maintained by pumping \(\mathrm{K}^{+}\) from the plasma. How much work must be done to transport \(1.0 \mathrm{~mol} \mathrm{~K}^{+}\) from the blood to the inside of a muscle cell at \(37^{\circ} \mathrm{C}\) (normal body temperature)? b. When \(1.0 \mathrm{~mol} \mathrm{~K}^{+}\) is transferred from blood to the cells, do any other ions have to be transported? Why or why not? c. Cells use the hydrolysis of adenosine triphosphate, abbreviated ATP, as a source of energy. Symbolically, this reaction can be represented as $$\operatorname{ATP}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{ADP}(a q)+\mathrm{H}_{2} \mathrm{PO}_{4}^{-}(a q)$$ where ADP represents adenosine diphosphate. For this reaction at \(37^{\circ} \mathrm{C}, K=1.7 \times 10^{5}\). How many moles of ATP must be hydrolyzed to provide the energy for the transport of \(1.0 \mathrm{~mol}\) \(\mathrm{K}^{+}\) ? Assume standard conditions for the ATP hydrolysis reaction.

Silicon is produced for the chemical and electronics industries by the following reactions. Give the balanced equation for each reaction. a. \(\mathrm{SiO}_{2}(s)+\mathrm{C}(s) \longrightarrow \mathrm{Si}(s)+\mathrm{CO}(g)\) b. Silicon tetrachloride is reacted with very pure magnesium, producing silicon and magnesium chloride. c. \(\mathrm{Na}_{2} \mathrm{SiF}_{6}(s)+\mathrm{Na}(s) \longrightarrow \mathrm{Si}(s)+\mathrm{NaF}(s)\)
See all solutions

Recommended explanations on Chemistry Textbooks

Chemistry Branches

Read Explanation

Nuclear Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Kinetics

Read Explanation

Chemical Reactions

Read Explanation

Organic Chemistry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.