General Chemistry Reviewer PDF

**GENERAL CHEMISTRY REVIEWER** - **Matter** - is anything that has mass and volume. - **Three states of matter**: 1. Solid - The strongest force of attraction (examples are tables and chair) 2. Liquid - Follows the shape of the container (examples are water and oil) 3. Gas - the...

**GENERAL CHEMISTRY REVIEWER** - **Matter** - is anything that has mass and volume. - **Three states of matter**: 1. Solid - The strongest force of attraction (examples are tables and chair) 2. Liquid - Follows the shape of the container (examples are water and oil) 3. Gas - the weakest force of attraction (examples are smoke and air) - **Physical properties of matter** - density, melting point, boiling point, freezing point - **Chemical properties of matter** - oxidation state, flammability, corrosiveness - **Extensive properties -** depends on the amount or size of the substance (examples are weight, volume, and length) - **Intensive properties -** Depends on the nature or type of the substance (examples luster and color). - **Physical changes** 1. Sublimation - solid to gas (example is dry ice to gas) 2. Evaporation - liquid to gas 3. Condensation - gas to liquid 4. Freezing - liquid to solid 5. Melting - solid to liquid 6. Deposition - gas to solid - boiling point of water - 100° C or 212° F or 373.15 kelvin - boiling point of water - physical intensive property - **Separating mixtures** 1. Distillation - separate two or more liquids with different boiling points 2. Evaporation - separate a solid component from a liquid component 3. Filtration - separate larger components 4. Paper chromatography - the components of the dissolved mixture separate out when they travel up the paper at different rates (inks and dyes) 5. Centrifugation - diagnose blood and urine - **Homogeneous mixtures** - also called a **solution**, has a uniform composition and appearance (example air, seawater, and vinegar) - **Heterogeneous mixtures** - substances are not distributed uniformly (examples mix nuts and salad) - **Precision** - consistency of achieving the same results repeatedly - **Accuracy** - closeness to the target - **Positive error** - measured values are **greater than** the expected value - **Negative error** - measured values are **less than** the expected value - **Percent error** - gives the accuracy of an experimental value - **Random errors** - minimize by making repeated measurements - **For significant figures:** 1. All non-zero digits are significant 2. Zeros between non-zero digits are significant 3. Zeros to the **right** of non-zero digits not significant. 4. If the number is greater than one the zeros to the right of a decimal is significant. 5. Trailing zeros after the decimal point are significant - Scientific notation to decimal notation (example: [1.5 *x* 10^ − 6^]{.math.inline} = 0.0000015 - Decimal notation to scientific notation (example: 0.00000215 = [2.15 *x* 10^ − 6^]{.math.inline} - Significant figures in multiplication and division - when multiplying or dividing numbers, the final answer has the same number of significant figures as the given with the **least number** of significant figures. - Law of conservation of mass - states that during a chemical reaction a**toms are not created nor destroyed.** - Law conservation of mass - **the mass of the reactant is equal to the mass of the product.** - The atom of one element differs in properties from the atoms of different elements. - **Three Laws of matter** - law of conservation of mass, definite proportion, and multiple proportion. - Isotope notation - shorthand notation in writing atomic mass and mass number. - **Isotopes** - same atomic number but different mass number - **Atomic number** - number of protons **only** - **Mass number** or atomic weight or molecular weight = proton + neutrons - **Neutrons** = mass number - protons - **Cation** - positively charged ion - **Anion** - negatively charged ion - **Chemical formula** 1. **Molecular formula** - **shorthand notation** indicating the types and number of atoms present in a molecule. 2. **Structural formula** - arrangement of atoms in a form of drawing. 3. **Empirical formula** - the type of atoms presents in the smallest whole number ratio of atoms - Molecular to empirical formula - the simplest form (Divisible to all subscripts) [*C*~5~*H*~10~*O*~5~]{.math.inline} = [*C**H*~2~*O*]{.math.inline} - Glucose - six carbon, 12 hydrogen, 6 oxygen ([*C*~6~*H*~12~*O*~6~]{.math.inline} = [*C**H*~2~*O*]{.math.inline} ) - Ethyl alcohol - two carbons, six hydrogens, one oxygen ( [*C*~2~*H*~6~*O*]{.math.inline} ) - Most common compounds and their name: - Phosphoric acid - [*H*~3~PO~4~]{.math.inline} - Ammonia - [*NH*~3~]{.math.inline} - Methane - [*CH*~4~]{.math.inline} - Sulfur hexafluoride - [SF~6~]{.math.inline} - Dinitrogen tetrahydride - [*N*~2~*H*~4~]{.math.inline} - Dinitrogen pentahydride - - Calcium carbonate - [CO~3~]{.math.inline} - Hydroxide - OH - Barium hydroxide -- BaOH - Lead (IV) nitrate - [*Pb*(*NO*~3~)4]{.math.inline} - Avogadro's number / 1 mole = [6.02 *x* 10^23^]{.math.inline} - Molar mass and molecular weight are the same. - Molar mass of [*H*~2~]{.math.inline}O - Molar mass of [*O*~2~]{.math.inline} O = 2 mol x 16 g = 32g/mol - Molar mass of [CH~4~]{.math.inline} C = 1 mol x 12 g = 12 g/mol H = 4 mol x 1 g = 4 g/mol **16 g/mol** - **Calculate percent composition** 1\. Calculate the percent composition of N and H in NH3. H = 3 mol x 1 g = 3 g/mol N = 1 mol x 14 g = 14 g/mol \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ 17 g/mol H = [\$\\frac{3}{17}\\ x\\ 100\\%\$]{.math.inline} = **17.65 %** N = [\$\\frac{14}{17}\\ x\\ 100\\%\$]{.math.inline} = **82.35 %** 2\. Calculate the percent composition of C and H in CO2. C = 1 mol x 12 g = 12 g/mol O = 2 mol x 16 g = 32 g/mol \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ 44 g/mol C = [\$\\frac{12}{44}\\ x\\ 100\\%\$]{.math.inline} = 27.27 % O = [\$\\frac{32}{44}\\ x\\ 100\\%\$]{.math.inline} = 72.72 % 3\. Calculate the percent composition of H in Sulfuric Acid ([*H*~2~SO~4~]{.math.inline}) H = 2 mol x 1 g = 2 g/mol S = 1 mol x 32 g = 32 g/mol O = 4 mol x 16 g = 64 g/mol \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ 98 g/mol H = [\$\\frac{2}{98}\\ x\\ 100\\%\$]{.math.inline} = 2.04 % O = [\$\\frac{64}{98}\\ x\\ 100\\%\$]{.math.inline} = 65.31 % - **Sample balancing chemical equation:**  \ [*N*~2~+  *H*~2~  →  NH~3~]{.math.display}\ **Answer:** [*N*~2~+  **3***H*~2~  →  **2**NH~3~]{.math.inline} \ [*NaCl* +  *F*~2~  →  *NaF*+ *Cl*~2~]{.math.display}\ **Answer:** [**2***NaCl* +  1*F*~2~  →  **2***NaF*+ 1Cl~2~]{.math.inline}  - **Percent Yield =** You want to produce 20 grams of the product but you only obtain 12 grams of the compound after the experiment. What is the percent yield of the experiment? Actual yield -- 12 Theoretical - 20 **=** [\$\\frac{\\mathbf{12\\ g}}{\\mathbf{20\\ g}}\\mathbf{\\ x\\ 100\\% = 60\\%}\$]{.math.inline} **Limiting reagent -- limited** **Excess reagent -- sobra** 2Fe[(*POS*~2~*HC*~2~)~4~]{.math.inline} C- 16 H -- 8 Fe - 2 P -- 8 O -- 8 S- 16