File Name: chemistry molecular nature of matter and change chapter 2 silberberg.zip
Plan: An element has only one kind of atom; a compound is composed of at least two kinds of atoms. A mixture consists of two or more substances mixed together in the same container. Solution: a There is only one type of atom blue present, so this is an element. Solution: The circle on the left contains molecules with either only orange atoms or only blue atoms. This is a mixture of two different elements. In the circle on the right, the molecules are composed of one orange atom and one blue atom so this is a compound.
Plan: Use the mass fraction of uranium in pitchblende from Sample Problem 2. Subtract the amount of uranium from that amount of pitchblende to obtain the mass of oxygen in that amount of pitchblende. Find the mass fraction of oxygen in pitchblende and multiply the amount of pitchblende by the mass fraction of oxygen to determine the mass of oxygen in the sample.
Solution: Plan: Subtract the amount of silver from the amount of silver bromide to find the mass of bromine in Use the mass fraction of silver in silver bromide to find the mass of silver in 3. Use the mass fraction of bromine in silver bromide to find the mass of bromine in 3.
Solution: Mass g of bromine in Plan: The law of multiple proportions states that when two elements react to form two compounds, the different masses of element B that react with a fixed mass of element A is a ratio of small whole numbers. The law of definite composition states that the elements in a compound are present in fixed parts by mass. The law of mass conversation states that the total mass before and after a reaction is the same.
Solution: The law of mass conservation is illustrated because the number of atoms does not change as the reaction proceeds there are 14 red spheres and 12 black spheres before and after the reaction occurs. The law of multiple proportions is illustrated because two compounds are formed as a result of the reaction.
One of the compounds has a ratio of 2 red spheres to 1 black sphere. The other has a ratio of 1 red sphere to 1 black sphere. The law of definite proportions is illustrated because each compound has a fixed ratio of red-to-black atoms. Solution: Only Sample B shows two different bromine-fluorine compounds. In one compound there are three fluorine atoms for every one bromine atom; in the other compound, there is one fluorine atom for every bromine atom.
The atomic number identifies the element. The superscript gives the mass number A which is the total of the protons plus neutrons. The number of neutrons is simply the mass number minus the atomic number A — Z. Plan: First, divide the percent abundance value found in Figure B2. Plan: To find the percent abundance of each B isotope, let x equal the fractional abundance of 10B and 1 — x equal the fractional abundance of 11B. Plan: Use the provided atomic numbers the Z numbers to locate these elements on the periodic table.
The name of the element is on the periodic table or on the list of elements inside the front cover of the textbook. Classify the element from the color coding in the periodic table. Plan: Locate these elements on the periodic table and predict what ions they will form. Elements after a noble gas lose electrons to become positive ions, while those before a noble gas gain electrons to become negative ions.
Plan: When dealing with ionic binary compounds, the first name is that of the metal and the second name is that of the nonmetal.
If there is any doubt, refer to the periodic table. The metal name is unchanged, while the nonmetal has an -ide suffix added to the nonmetal root. Plan: Use the charges of the ions to predict the lowest ratio leading to a neutral compound.
The sum of the total charges must be 0. Plan: Determine the names or symbols of each of the species present. Then combine the species to produce a name or formula. The metal or positive ions are written first. Review the rules for nomenclature covered in the chapter. This is one of the two common charges for copper ions. This is one of the two common charges on iron ions. This gives the name iron II bromide or ferrous bromide. This gives the formula HgCl 2. The systematic name is tin II fluoride.
The common name is plumbous iodide. To balance the charges, the formula is Cr 2 S 3. The name is iron II oxide. The common name is ferrous oxide. The metal or positive ions always go first. Trihydrate means three water molecules. These combine to give Zn OH 2. The cyanide ion, CN—, has the appropriate charge.
These combine to give lithium cyanide. These combine to give NH 4 2 SO 4. There are 6 water molecules hexahydrate. Therefore, the name is nickel II nitrate hexahydrate. The bicarbonate ion, HCO 3 —, has the appropriate charge to balance out one potassium ion.
Therefore, the formula of this compound is KHCO 3. Make corrections accordingly. Parentheses are required around the polyatomic ion. The other ion is HCO 3 —, which is called the hydrogen carbonate or bicarbonate ion.
The correct name is magnesium hydrogen carbonate or magnesium bicarbonate. Nitride is N3—, and nitrate is NO 3 —. This gives the correct name: chromium III nitrate the common name is chromic nitrate. Nitrate is NO 3 —, and nitrite is NO 2 —. The correct name is calcium nitrite.
Perchlorate is ClO 4 —, and chlorate is ClO 3 —. Additionally, parentheses are not needed when there is only one of a given polyatomic ion. The correct formula is KClO 3.
Plan: Use the name of the acid to determine the name of the anion of the acid. The name hydro ic acid indicates that the anion is a monatomic nonmetal. The name ic acid indicates that the anion is an oxoanion with an —ate ending. The name ous acid indicates that the anion is an oxoanion with an —ite ending. Solution: a Chloric acid is derived from the chlorate ion, ClO 3 —.
The —1 charge on the ion requires one hydrogen. These combine to give the formula HClO 3. These combine to give the formula HF. The —1 charge means that one H is needed. These combine to give the formula HNO 2. Plan: Remove a hydrogen ion to determine the formula of the anion.
Identify the corresponding name of the anion and use the name of the anion to name the acid. For the oxoanions, the -ate suffix changes to -ic acid and the -ite suffix changes to -ous acid. For the monatomic nonmetal anions, the name of the acid includes a hydro- prefix and the —ide suffix changes to —ic acid. The corresponding name is sulfurous acid. This gives the name: hypobromous acid. This gives the name: chlorous acid.
This gives the name: hydroiodic acid. Since these are binary compounds consisting of two nonmetals, the number of each type of atom is indicated with a Greek prefix.
Solution: a Sulfur trioxide — one sulfur and three tri oxygens, as oxide, are present. Solution: a Sulfur dichloride — one sulfur and two di chlorines, as chloride, are present. For compounds between nonmetals, the number of atoms of each type is indicated by a Greek prefix. If both elements in the compound are in the same group, the one with the higher period number is named first. Solution: a Suffixes are not used in the common names of the nonmetal listed first in the formula.
Sulfur does not qualify for the use of a suffix. This gives the name disulfur dichloride.
Silberberg and Patricia G. The eighth edition of Chemistry: The Molecular Nature of Matter and Change maintains its standard-setting position among general chemistry textbooks by evolving further to meet the needs of professor and student. The text still contains the most accurate molecular illustrations, consistent step-by-step worked problems, and an extensive collection of end-of-chapter problems. And changes throughout this edition make the text more readable and succinct, the artwork more teachable and modern, and the design more focused and inviting. The three hallmarks that have made this text a market leader are now demonstrated in its pages more clearly than ever.
Plan: An element has only one kind of atom; a compound is composed of at least two kinds of atoms.
Стратмор сощурил. - А ты как думаешь. И уже мгновение спустя ее осенило. Ее глаза расширились. Стратмор кивнул: - Танкадо хотел от него избавиться. Он подумал, что это мы его убили.
В АНБ было только одно помещение, еще более засекреченное, чем шифровалка, и Сьюзан поняла, что сейчас она окажется в святая святых агентства. ГЛАВА 109 Командный центр главного банка данных АНБ более всего напоминал Центр управления полетами НАСА в миниатюре. Десяток компьютерных терминалов располагались напротив видеоэкрана, занимавшего всю дальнюю стену площадью девять на двенадцать метров. На экране стремительно сменяли друг друга цифры и диаграммы, как будто кто-то скользил рукой по клавишам управления. Несколько операторов очумело перебегали от одного терминала к другому, волоча за собой распечатки и отдавая какие-то распоряжения.
Но Танкадо… - размышляла .