chapter 4 review: types of chemical reactions and
... would you use? 4. You want to determine the molar mass of an unknown weak monoprotic acid. You mass out a 2.879 g sample of the pure acid and dissolve it in distilled water. After adding 3 drops of phenolphthalein indicator, you titrate the sample with 0.1704 M NaOH. The pink endpoint is reached aft ...
... would you use? 4. You want to determine the molar mass of an unknown weak monoprotic acid. You mass out a 2.879 g sample of the pure acid and dissolve it in distilled water. After adding 3 drops of phenolphthalein indicator, you titrate the sample with 0.1704 M NaOH. The pink endpoint is reached aft ...
Slide 1
... A positive charge +Q and a negative charge –q are separated by a distance d, and the potential VP = 0 at point P. This means that: ...
... A positive charge +Q and a negative charge –q are separated by a distance d, and the potential VP = 0 at point P. This means that: ...
Name: Per: Date: Unit 1. Materials: Formulating Matter B. Periodic
... 38. Fill in the data table for each ionic compound described below. Number one is filled in as an example. Use the two tables of common ions below. a. Potassium chloride is “lite salt”, used by many people with hypertension. b. CaSO4 is a component of plaster. c. A substance composed of Ca2+ and PO ...
... 38. Fill in the data table for each ionic compound described below. Number one is filled in as an example. Use the two tables of common ions below. a. Potassium chloride is “lite salt”, used by many people with hypertension. b. CaSO4 is a component of plaster. c. A substance composed of Ca2+ and PO ...
Chapter 4: Aqueous Reactions and Solution Stoichiometry
... Solutions are defined as homogeneous mixtures of two or more pure substances. The solvent is present in greatest abundance. All other substances are solutes. ...
... Solutions are defined as homogeneous mixtures of two or more pure substances. The solvent is present in greatest abundance. All other substances are solutes. ...
Chapter 8powerp point for chemical reactions
... The equation must represent known facts The equation must contain the correct formulas for the reactants & products The law of conservation of mass must be satisfied- which means the same # and type of atoms are present on both sides of the equation. ...
... The equation must represent known facts The equation must contain the correct formulas for the reactants & products The law of conservation of mass must be satisfied- which means the same # and type of atoms are present on both sides of the equation. ...
Algebra 1: Test #3 --- REVIEW---3 Show needed work, and write
... 21. What is the product of the solutions of the equation below? 22. What is the solution set of the equation below? 23. When written in standard form, what is the quadratic term of this equation? 8x2 + 6 = –7x 24. Solve by factoring ...
... 21. What is the product of the solutions of the equation below? 22. What is the solution set of the equation below? 23. When written in standard form, what is the quadratic term of this equation? 8x2 + 6 = –7x 24. Solve by factoring ...
Chapter 3
... Remember, these contain only non-metals or a metalloid with a non-metal. Name the first non-metal in the formula. Name the second non-metal in the formula and change the suffix to ide. Add prefixes for all subscripts - except if the first one is a “1”. 1 = mono, 2 = di, 3 = tri, 4 = tetra, 5 = penta ...
... Remember, these contain only non-metals or a metalloid with a non-metal. Name the first non-metal in the formula. Name the second non-metal in the formula and change the suffix to ide. Add prefixes for all subscripts - except if the first one is a “1”. 1 = mono, 2 = di, 3 = tri, 4 = tetra, 5 = penta ...
LN_atoms_etc
... charged particles within an atom. Electrons have a charge of e = 1.6021773 10–19 C and a mass of 9.109390 10–31 kg. Later experiments by Rutherford determined that at the center of an atom is a positively charged, compact, heavy nucleus. The charge on the atomic nucleus is +Ze (Z is the atomic n ...
... charged particles within an atom. Electrons have a charge of e = 1.6021773 10–19 C and a mass of 9.109390 10–31 kg. Later experiments by Rutherford determined that at the center of an atom is a positively charged, compact, heavy nucleus. The charge on the atomic nucleus is +Ze (Z is the atomic n ...
Chapter 7 Notes PowerPoint Version
... the empirical formula and experimental molar mass of a compound. Step 1: Determine the molar mass of the given empirical formula. Step 2: Solve for n by dividing the experimental molar mass by the molar mass of the empirical formula. *Remember: n(empirical formula) = molecular formula ...
... the empirical formula and experimental molar mass of a compound. Step 1: Determine the molar mass of the given empirical formula. Step 2: Solve for n by dividing the experimental molar mass by the molar mass of the empirical formula. *Remember: n(empirical formula) = molecular formula ...
Chapter 06 Notes (PowerPoint) File
... • the relative weights of molecules can be calculated from atomic weights Formula Mass = 1 molecule of H2O = 2(1.01 amu H) + 16.00 amu O = 18.02 amu • since 1 mole of H2O contains 2 moles of H and 1 mole of O Molar Mass = 1 mole H2O = 2(1.01 g H) + 16.00 g O = 18.02 g ...
... • the relative weights of molecules can be calculated from atomic weights Formula Mass = 1 molecule of H2O = 2(1.01 amu H) + 16.00 amu O = 18.02 amu • since 1 mole of H2O contains 2 moles of H and 1 mole of O Molar Mass = 1 mole H2O = 2(1.01 g H) + 16.00 g O = 18.02 g ...
Chapter 6 Chemical Composition
... If ratio ?.5, multiply all by 2; if ratio ?.33 or ?.67, multiply all by 3, etc. Skip if already whole numbers after Step 4. ...
... If ratio ?.5, multiply all by 2; if ratio ?.33 or ?.67, multiply all by 3, etc. Skip if already whole numbers after Step 4. ...
Balancing Equations
... Find the number of atoms for each element on the left side. Compare those against the number of the atoms of the same element on the right side. Determine where to place coefficients in front of formulas so that the left side has the same number of atoms as the right side for EACH element in order t ...
... Find the number of atoms for each element on the left side. Compare those against the number of the atoms of the same element on the right side. Determine where to place coefficients in front of formulas so that the left side has the same number of atoms as the right side for EACH element in order t ...
Examples 2.1 - IHMC Public Cmaps (3)
... The break-even point occurs at a monthly usage of 100 minutes. b. The graph of the equations shows that for monthly usage under 100 minutes, Plan 2 is less expensive. So, Jeremy should probably choose Plan 2. ...
... The break-even point occurs at a monthly usage of 100 minutes. b. The graph of the equations shows that for monthly usage under 100 minutes, Plan 2 is less expensive. So, Jeremy should probably choose Plan 2. ...
2-1 Solving Systems of Equations in Two Variables
... The break-even point occurs at a monthly usage of 100 minutes. b. The graph of the equations shows that for monthly usage under 100 minutes, Plan 2 is less expensive. So, Jeremy should probably choose Plan 2. ...
... The break-even point occurs at a monthly usage of 100 minutes. b. The graph of the equations shows that for monthly usage under 100 minutes, Plan 2 is less expensive. So, Jeremy should probably choose Plan 2. ...