#### Transcript Chemistry B Quizzes

Chemistry B Quizzes The number of atoms or molecules in exactly 1.000 mole of a substance, and is equal to 6.022 x 1023 The number of atoms or molecules in exactly 1.000 mole of a substance, and is equal to 6.022 x 1023 Avogrado’s Number atoms of the same element that have different numbers of neutrons, which means that atoms of the same element can have different masses Atoms of the same element that have different numbers of neutrons, which means that atoms of the same element can have different masses 1. Isotope The SI unit for amount of a substance. The SI unit for amount of a substance. Mole a chemical formula that shows the simplest ratio of atoms in a compound a chemical formula that shows the simplest ratio of atoms in a compound Emperical Formula Which will have the greater number of ions, 1 mol of nickel (II) or 1 mol of copper (I)? Which will have the greater number of ions, 1 mol of nickel (II) or 1 mol of copper (I)? They will be the same they both are 1 mole, so each would have 6.022 x 1023 ions Without making a calculation,is 1.11 mol Pt more of less than 6.022 x 1023 atoms? Without making a calculation,is 1.11 mol Pt more of less than 6.022 x 1023 atoms? It would be more because 1 mole would = 6.022 x 1023, and there are 1.11 moles Pt, so it would be larger than avogadro’s number Find the mass in grams in 4.30 x 1016 atoms of He, 4.00 g/mol Find the mass in grams in 16 4.30 x10 atoms of He, 4.00 g/mol 4.30 x 1016 atoms He 1 mol 6.022 x 1023 atoms 4.00 g 1 mol = 2.86 x 10 -7 grams He How many atoms are in two moles of mercury? 1. 2. 3. 4. 1.204 x 1022 1.204 x 1023 1.204 x 1024 6.022 x 1023 How many atoms are in two moles of mercury? 1.204 x 1024 How many atoms are in a mole of barium chloride, BaCl2? 1. 2. 3. 4. 6.022 x 1026 1.806 x 1024 1.204 x 1024 6.022 x 1023 How many molecules are in a mole of barium chloride, BaCl2? 1. 6.022 x 1023 How many grams of cobalt (58.9332 g/mol) are in 4.76 mol of the element? An element has one isotope with an atomic mass of 27.94 amu. Another isotope has an atomic mass of 28.96 amu. The average atomic mass of the element is 28.02 amu. The isotope that makes up the larger percent of a sample of the element is the isotope with a mass of 27.94 amu 28.96 amu 28.02 amu An element has one isotope with an atomic mass of 27.94 amu. Another isotope has an atomic mass of 28.96 amu. The average atomic mass of the element is 28.02 amu. The isotope that makes up the larger percent of a sample of the element is the isotope with a mass of 27.94 amu, because it is closes to average atomic mass of 28.02 amu What is the molar mass of C6H12O6? What is the molar mass of C6H12O6? 180.16 g Simply add up the masses from the periodic table Boron has an isotope with a mass of 10.013 amu that makes up 19.8% of all boron. Its other isotope has a mass of 11.009 amu and makes up 80.2%. What is the average atomic mass of boron. Boron has an isotope with a mass of 10.013 amu that makes up 19.8% of all boron. Its other isotope has a mass of 11.009 amu and makes up 80.2%. What is the average atomic mass of boron. Boron -10 = (.198) x 10.013 = 1.983 Boron – 11 = (.802) x 11.009 = 8.83 1.983+8.83 =10.81 amu How many moles of hydrogen are in one mole of C2H6? How many moles of hydrogen are in one mole of C2H6? 6 How many hydrogen atoms are present in one formula unit of ammonium hydrogen phosphate, (NH4)2HPO4? How many hydrogen atoms are present in one formula unit of ammonium hydrogen phosphate, (NH4)2HPO4? 8 A compound is 5.94% H and 94.06% O. What is its empirical formula? A compound is 5.94% H and 94.06% O. What is its empirical formula? 5.94% = 5.94 g H 94.06% = 94.06 g O 5.94g H 94.06g O 1 mol 1.0079g 1 mole 15.999 g = 5.89 mol H 5.88 =5.88 mol O 5.88 HO The first set of calculations in determining an empirical formula from masses of elements produces the subscripts of 1.67 and 1.00. What do you need to do to find the actual subscripts? a. b. c. d. Round of both numbers Round off 1.67, and use 1.00 as it is Multiply both subscipts by 3 Multiply both subsripts by 6 The first set of calculations in determining an empirical formula from masses of elements produces the subscripts of 1.67 and 1.00. What do you need to do to find the actual subscripts? a. Round off 1.67, and use 1.00 as it is A compound’s empirical formula is CH. If the formula mass is 79.12 g/mol, what is the molecular formula? A compound’s empirical formula is CH. If the molar mass is 79.12 g/mol, what is the molecular formula? Formula mass of compound = 79.12 g Molar mass of CH = 13.019 Molar mass of compound = 79.12 =6.07 Molar mass of CH 13.019 Multiply each by 6 to find the molecular formula = C6H6 What is the percentage of chlorine in NaCl? What is the percentage of chlorine in NaCl? Molar mass of Na = 22.990g Molar mass of Cl = 35.453 g Molar mass of NaCl = 58.443g % of sodium = 22.990/58.443 x 100 = 39% % of chlorine = 35.453/583443 x 100 = 61% Match each equation with the correct type of reaction. 1. Ca(ClO3)2 CaCl2 +O2 2. CaO + H2O Ca(OH)2 3. C8H18 + O2 CO2 + H2O a. Decomposition b. Combustion c. Synthesis Match each equation with the correct type of reaction. 1. Ca(ClO3)2 CaCl2 +O2 2. CaO + H2O Ca(OH)2 3. C8H18 + O2 CO2 + H2O a. Decomposition (1) b. Combustion(3) c. Synthesis ( 2) Balance the following equation: ZnS + O2 ZnO + SO2 Balance the following equation: 2ZnS + 3O2 2ZnO + 2SO2 Match each equation with the correct type of reaction. 1. 2. 3. 4. 5. 6. AgNO3 + AlCl3 AgCl + Al(NO3)2 C8H18 + O2 CO2 + H2O Ca(ClO3)2 CaCl2 +O2 NaOH + HCl NaCl + HOH CaO + H2O Ca(OH)2 Ni(ClO3)2 NiCl2 + O2 a. b. c. d. e. f. Decomposition Combustion Synthesis Single Replacement Double Replacement – Precipitation Double Replacement – Acid/Base Match each equation with the correct type of reaction. 1. 2. 3. 4. 5. 6. AgNO3 + AlCl3 AgCl + Al(NO3)2 C8H18 + O2 CO2 + H2O Ca(ClO3)2 CaCl2 +O2 NaOH + HCl NaCl + HOH CaO + H2O Ca(OH)2 Ni(ClO3)2 NiCl2 + O2 a. b. c. d. e. f. Decomposition (6), (3) Combustion ( 2) Synthesis (5) Single Replacement Double Replacement – Precipitation (1) Double Replacement – Acid/Base (4)