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Give the name or chemical formula, as appropriate, for each of the following binary molecular substances: (a) \(\mathrm{SF}_{6}\), \((\mathrm{b}) \mathrm{IF}_{5}\), (c) \(\mathrm{XeO}_{3}\), (d) dinitrogen tetroxide, (e) hydrogen cyanide, (f) tetraphosphorus hexasulfide.

Short Answer

Expert verified
(a) Sulfur hexafluoride, (b) Iodine pentafluoride, (c) Xenon trioxide, (d) \(\mathrm{N}_{2}\mathrm{O}_{4}\), (e) \(\mathrm{HCN}\), (f) \(\mathrm{P}_{4}\mathrm{S}_{6}\)

Step by step solution

01

(a) Name of \(\mathrm{SF}_{6}\)

Here, we have two elements: Sulfur (S) and Fluorine (F). Since there are six Fluorine atoms, we'll use the prefix "hexa-" for Fluorine. The name of the compound becomes Sulfur hexafluoride.
02

(b) Name of \(\mathrm{IF}_{5}\)

In this case, we have Iodine (I) and Fluorine (F). Since there are five Fluorine atoms, we'll use the prefix "penta-" for Fluorine. The name of the compound is Iodine pentafluoride.
03

(c) Name of \(\mathrm{XeO}_{3}\)

Here, we have Xenon (Xe) and Oxygen (O). With three Oxygen atoms, we'll use the prefix "tri-" for Oxygen. The compound's name is Xenon trioxide.
04

(d) Chemical formula for dinitrogen tetroxide

In this case, the name reveals the elements and their respective prefixes: - "Di-" refers to two Nitrogen (N) atoms - "Tetra-" refers to four Oxygen (O) atoms The chemical formula is \(\mathrm{N}_{2}\mathrm{O}_{4}\).
05

(e) Chemical formula for hydrogen cyanide

Here, the name tells us the elements involved: - Hydrogen (H) - Cyanide, which is a polyatomic ion containing one Carbon (C) and one Nitrogen (N) atom So, the chemical formula for hydrogen cyanide is \(\mathrm{HCN}\).
06

(f) Chemical formula for tetraphosphorus hexasulfide

In this example, the name reveals the elements and their corresponding prefixes: - "Tetra-" refers to four Phosphorus (P) atoms - "Hexa-" refers to six Sulfur (S) atoms The chemical formula is \(\mathrm{P}_{4}\mathrm{S}_{6}\).

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Key Concepts

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

Binary Molecular Compounds
Binary molecular compounds are composed of two nonmetal elements joined together by covalent bonds, which involve the sharing of electrons. It's like a friendship between atoms sharing their resources. The elements come together to create a stable molecule with distinct physical and chemical properties. A well-known example is carbon dioxide, where carbon and oxygen share electrons to form the compound ( CO_2 ).
Chemical Formulas
Chemical formulas are essential for communicating what a compound contains. Think of them as a recipe that describes the exact proportions of ingredients in a compound. Each element is represented by its chemical symbol, like C for carbon, and numbers are used to show how many atoms of each type are in the compound. The subscripts, those little numbers at the bottom like in 102222 ( H_2O), tell you there are two hydrogen atoms for every one oxygen atom in water.
Prefixes in Chemistry
In chemistry, prefixes are like clues telling us how many of each atom are in a compound. They come from the Greek language and make it possible to describe the composition without confusion. For example, ‘mono-’ means one, ‘di-’ means two, ‘tri-’ is three, and so on. When naming compounds like dinitrogen tetroxide, we understand from the prefix ‘di-’ that there are two nitrogen atoms and from ‘tetra-’ that there are four oxygen atoms. With these little hints, anyone can crack the code of a chemical compound's makeup.
Molecular Substances
Molecular substances consist of molecules bound together by covalent bonds. These molecules are often seen huddling together in a specific structure, forming a variety of substances from gases, like oxygen ( O_2), to liquids such as water ( H_2O), to solid substances like sugar ( C_12H_22O_11). They're like teams of atoms working together, each team with a unique arrangement that determines the substance's properties and behavior.

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Most popular questions from this chapter

From the following list of elements \(-\mathrm{Ar}, \mathrm{H}, \mathrm{Ga}, \mathrm{Al}, \mathrm{Ca}\), \(\mathrm{Br}, \mathrm{Ge}, \mathrm{K}, \mathrm{O}-\) pick the one that best fits each description. Use each element only once: (a) an alkali metal, (b) an alkaline earth metal, (c) a noble gas, (d) a halogen, (e) a metalloid, (f) a nonmetal listed in group \(1 \mathrm{~A},(\mathrm{~g})\) a metal that forms a \(3+\) ion, \((h)\) a nonmetal that forms a \(2-\) ion, (i) an element that resembles aluminum.

An atom of tin (Sn) has a diameter of about \(2.8 \times 10^{-8} \mathrm{~cm}\). (a) What is the radius of a tin atom in angstroms \((\AA)\) and in meters \((\mathrm{m}) ?\) (b) How many Sn atoms would have to be placed side by side to span a distance of \(6.0 \mu \mathrm{m}\) ? (c) If you assume that the tin atom is a sphere, what is the volume in \(\mathrm{m}^{3}\) of a singleatom?

Provide the name or chemical formula, as appropriate, for each of the following acids: (a) hydrobromic acid, (b) hydrosulfuric acid, (c) nitrous acid, (d) \(\mathrm{H}_{2} \mathrm{CO}_{3}\), (e) \(\mathrm{HClO}_{3}\), (f) \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\).

In a series of experiments, a chemist prepared three different compounds that contain only iodine and fluorine and determined the mass of each element in each compound: \(\begin{array}{ccc} \text { Compound } & \text { Mass of Iodine (g) } & \text { Mass of Fluorine (g) } \\ \hline 1 & 4.75 & 3.56 \\ 2 & 7.64 & 3.43 \\ 3 & 9.41 & 9.86 \end{array}\) (a) Calculate the mass of fluorine per gram of iodine in each compound. (b) How do the numbers in part (a) support the atomic theory?

The diameter of a rubidium atom is \(4.95 \AA\). We will consider two different ways of placing the atoms on a surface. In arrangement \(\mathrm{A}\), all the atoms are lined up with one another. Arrangement \(\mathrm{B}\) is called a close-packed arrangement because the atoms sit in the "depressions" formed by the previous row of atoms: (a) Using arrangement A, how many Rb atoms could be placed on a square surface that is \(1.0 \mathrm{~cm}\) on a side? (b) How many Rb atoms could be placed on a square surface that is \(1.0 \mathrm{~cm}\) on a side, using arrangement \(\mathrm{B}\) ? (c) By what factor has the number of atoms on the surface increased in going to arrangement B from arrangement A? If extended to three dimensions, which arrangement would lead to a greater density for \(\mathrm{Rb}\) metal?

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