Chemistry Chapter 7 Answers – Flashcards
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indicates the relative # of atoms of each kind in a chemical compound
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Chemical Formula
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The Simplest ratio of cations and anions
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Formula Unit
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Distribute: Al2(SO4)3
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Al2 S3 O12
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Monatomic Ions
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ions formed from a single atom
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Group 1A Na (balance charge)
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+1 charge Na+1
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Group 4A C (balance charge)
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+0 because it is even distance from noble gasses
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Group 5A N (balance charge)
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-3 charge C-3
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Cation always occurs _______ an anion in a chemical equation
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before
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compounds composed of 2 elements
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binary compounds
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In a binary compound the total number of positive and negative charges must be ______
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equal
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Write the formula for Aluminum Oxide
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1. Al O 2. Al+3 O-2 3. cross charge numbers Al2 O3 (and drop charge)
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Rules in naming binary ionic compounds (3)
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1. name the cation first and anion second, cation is always first in formula 2. Monatomic cations use the element name 3. monatomic anions take their name from the root of the element plus -ide as a suffix
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Write the name of K2O
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Potassium Oxide (K O2) switch numbers and drop sign
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Write the name of CuCl2
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Copper (II) Chloride (Cu Cl2) work backwards Cu+2 Cl-1 use chart given in class to find charge
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write the formula and name of Fe+4 O-2
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Iron (IV) Oxide (Fe O2) Oxygen can never exist as a -1 charge and always -2 so double the cation to balance the doubling of oxygen
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ions that are made up of more than one atom
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polyatomic ions
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the charge given to a polyatomic ion refers to _______________
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the whole group of atoms
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Write the formula for Ammonium Chloride
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NH4Cl NH+1 Cl-1 switch numbers and drop charge signs NH4= ammonium
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Write the name of K2CrO4
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Potassium Chromate K+1 (CrO4)-2 charge switch numbers and drop charge signs CrO4= Chromate
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____________ of naming molecular compounds is based on the use of prefixes (type of system)
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old system
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Rules for the prefix system of nomenclature of binary molecular compounds are (3)
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1. The first element is given a prefix only if it contributes more than one atom to a molecule of the compound ex: K2 O4 = Dipotassium 2. the second element is named by combining a prefix based on the number of atoms contributed, the root name of the element, and ending in -ide 3. The (o or a) at the end of a prefix is usually dropped when the word following the prefix begins with a vowel ex: O4= Tetroxide not Tetraoxide
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Numeral Prefixes 1-10 (name them)
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Mono- Di- Tri- Tetra- Penta- Hexa- Hepta- Octa- Nona- Deca-
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Write the name of As2O5
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Diarsenic Pentoxide As2 O5
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Name the two categories most acids are classified under
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Binary acids and Oxyacids
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Acids that consist of 2 elements, usually hydrogen and one of the halogens
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binary acids
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acids that contain hydrogen, oxygen, and a third element (usually a non metal)
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oxyacids
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Name the acid HF
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hydrofluoric acid
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Name the acid HCl
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hydrochloric acid
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Name the acid HBr
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hydrobromic acid
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Name the acid HI
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hydroiodic acid
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Name the acid H3PO4
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phosphoric acid
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Name the acid HNO2
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nitrous acid
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Name the acid hno3
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nitric acid
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Name the acid H2SO3
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sulfurous acid
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Name the acid H2SO4
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sulfuric acid
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Name the acid CH3COOH
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acetic acid
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Name the acid HClO
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hypochlorous acid
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Name the acid HClO2
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chlorous acid
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Name the acid HClO3
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chloric acid
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Name the acid HClO4
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perchloric acid
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Name the acid H2CO3
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carbonic acid
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the mass of any unit represented by a chemical formula, whether the unit is a molecule, a formula unit, or an ion is known as __________
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formula mass (AMU) atomic mass unit
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formula unit or ion is the sum of the average atomic mass of all the atoms in its formula
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formula mass of any molecule
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Find the Formula mass of Potassium Chlorate
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Potassium=K Chlorate=ClO3 K- 1 * 39 =39 Cl-1 * 35=35 O-3 * 16=48 39+35+48=122 122 AMU the 1st column is the element, 2nd is the amount of atoms of that element, 3rd is the atomic mass of those given elements, 4th is the result of the number of atoms multiplied by the atomic mass. Then add all the solutions together to get atomic mass
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What is the molar mass of Barium Nitrate (g/mol)
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Barium=Ba Nitrate=NO3 since oxygen is diatomic it must be even therefor the formula is Ba(NO3)2 Ba-1 * 137=137 N- 2 * 14=28 O- 6 * 16=96 137+28+96=261 261 g/mol
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the _____________ of a compound can be used as a conversion factor to relate an amount in moles to a mass in grams of a given substance
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molar mass
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What is the mass, in grams, of 2.50 mol of Oxygen gas? (O2)
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grams > mole > molecules 2.50 mol O2 / 1 32 grams O2 / 1 mol O2 2.50*32=80 grams sigifgs (3) 80.0 grams 32 came from the atomic mass of oxygen, it was doubled because there were 2 atoms of oxygen in the equation, and the number 2.50 was given
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A: Find the molar mass of Ibuprofen (C13H18O2) B: find out how many moles of ibuprofen are in a 33gram bottle. C: How many molecules are in the bottle? D: What is the total mass, in grams, of carbon in 33 grams of ibuprofen? (use 206.31 as molar mass after step A)
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A: Ibuprofen = C13H18O2 C- 13 * 12= 156 H- 18 * 1 = 18 O- 2 * 16 = 32 156+18+32=(206) for the purpose of the example we are using 206.31 as molar mass because it is 100% accurate and not a rounded representation of the atomic mass of the other elements B: 33grams ibuprofen / 1 1mol ibuprofen / 206.31grams ibuprofen 33grams / 206.31 grams = (0.16 mols) ibuprofen C: 0.16 mol ibuprofen / 1 6.02*10^23 molecules ibu. / 1mol ibu. 0.16* 6.02*10^23= (9.6*10^22 molecules of ibuprofen) D: 0.16mol ibuprofen / 1 13 mol C / 1 mol ibuprofen 12 g C / 1 mol C 12*13*0.16=24.96=25(sigifg) (25grams C)
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mass percentage formula
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mass of element in compound / mass of entire compound *100= percent of element in compound
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the percentage by mass of each element in a compound is known as the _______
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percentage composition
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Find the percentage composition of Copper (I) Sulfide, Cu2S.
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Copper= Cu2 Sulfur = S Copper- 2 *64=128 Sulfur- 1 * 32=32 128+32=160 128 / 160= 0.8=80% Copper 32 / 160= 0.2=20% Sulfur
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the ______________ consists of the symbols for the elements combined in a compound, with subscripts showing the smallest whole number ratio of the elements in a compound
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empirical formula
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Quantitative analysis shows that a compound contains 32.38% sodium, 22.65% sulfur, and 44.99% oxygen. Find the empirical formula for this compound
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32.38g Na / 23 = 1.408mol Na / 0.7078 = 2 22.65g S / 32 = 0.7078mol S / 0.7078 = 1 44.99g O / 16 =2.812mol O / 0.7078 = 4 first # in the column was the percent composition, second was the atomic mass of the element, after dividing the 2 next the answer was divided by the smallest number (0.7078) to the nearest whole number Na2SO4- Sodium sulfate
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the actual formula of the molecular compound
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molecular formula
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molecular formula (formula)
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x (empirical formula) = molecular formula
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the empirical formula of a compound of phosphorous and oxygen was found to be P2O5. experimentation shows that the molar mass of this compound is 283.89 g/mol What is the molecular formula of this compound?
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step 1: find mass of empirical formula step 2: divide molar mass of molecular formula by mass of empirical formula step 3: the answer of step 2 is X in the equation x(empirical formula)= molecular formula step 4: utilize formula 1: P- 2 * 31=62 O- 5 * 16=80 62+80=142 2: 283.89 / 142 = 2 (always a whole number) 3: x=2 4: 2(P2O5)= molecular formula (P4O10)
