# Test I – Chemistry Flashcard

 Dimensional Analysis
 A method of conversion using the relation between units of measurement
 Dimensional Analysis Problem: Convert 6.2 grams to ounces
 1 g=0.035274oz   6.2g(0.035274oz/1g)=0.2186988oz   0.22 oz in 6.2 g
 Dimension Analysis Problem: meters/second to miles/hour Convert 18 m/s to mi/hr
 .001 km=1m 1 km=0.621371 mi 1 min=60 s 60 min=1 hr   18m(.001km/1m)(0.621371mi/1km)=0.01118468 mi/s 0.01118468 mi/s (60s/1min)(60mins/1hr)=40.264848 mi/hr 4.0 x 10^1 mi/hr
 Density
 A physical property; mass divided by volume (D=M/V)
 Density Problem:  What is the volume of a piece of aluminum with a mass of  0.002 g? (Density of aluminum is 2.7 g/cc)
 D=M/V   2.7 g/cc=0.002g/V   V=0.002g/2.7g/cc V=0.0007407 V=7 x 10^-4
 Dimensional Analysis Problem:  Convert 6 square inches to square centimeters
 1 in=2.54 cm   6 in^2 (2.54cm/1 in)^2=51.8616 cm^2   50 cm^2
 Density Problem:  Find the mass of a piece of steel with a volume of 6.8 cc   (Density of steel is 7.85g/cc)
 D=M/V   7.85g/cc=M/6.8cc M=53.38g M=53g
 Pure Substance
 A substance that is homogenous and cannot be separated by physical means; compounds, elements
 Element
 A substance containing only one type of atom; hydrogen, oxygen
 Compound
 A substance containing two or more elements that can be separated by chemical, but not physical, change; water, carbon dioxide
 Mixture
 A combination of substances that can be separated through physical means; salt water, a salad
 Heterogeneous Mixture
 A mixture in which the different properties of the substances are visible; salad
 Homogeneous Mixture
 A mixture in which the different properties of the substances are not visible; salt water
 Characteristic Properties of Pure Substances
 Density, boiling point, melting point, solubility, color, etc
 Chemical Property vs Physical Property
 A chemical property can only be observed through a chemical change or reaction, whereas a physical property can be observed without changing the matter’s composition
 Chemical Change vs Physical Change
 A chemical change involves the rearrangement of atoms to form one or more new substances with different properties, whereas a physical change affects the form of a substance rather than its composition;
 Law of Conservation
 Matter is always conserved. The matter before a reaction is equal to the matter after a reaction.;
 Application of the Law of Conservation:; ; Two substances react. There are 6.3 grams of substance A, and the reaction of substances A and B results in 9.7 grams of substance C. How many grams of substance B were involved in the reaction?; ;
 6.3 g+ x g=9.7 g x=3.4g ; 3.4 g of substance B
 Dalton’s Postulates
 1. All matter is composed of small, indivisible particles called atoms. 2. All atoms of a given element are identical in mass and properties.  3.  Compounds are formed by a combination of two or more atoms in definite arrangements in the ratio of small whole numbers.  4. Atoms are not created, destroyed or converted into other kinds of atoms during chemical reactions.  They are simply rearranged into new compounds.
 Protons
 Positively charged subatomic particle inside of the nucleus
 Neutron
 Subatomic particle inside of the nucleus with no charge
 Electron
 Negatively charged subatomic particle orbiting nucleus in the electron cloud
 Atoms vs Ions
 An atom has no charge, so number p+=number e- An ion is charged either positively or negatively. An ion of an element has more (neg ion) or less (pos ion) electrons than an atom of that element (the number of protons does not change, because the number of protons identifies the element)
 AMU
 A unit of mass used to express atomic weights. It is equal to the weight of 1/12 of a carbon-12 atom
 Isotopic Abundance
 The abundance of an isotope in nature as compared to other isotopes of that element
 Isotopic Abundance Problem:  In element X, there are two different isotopes. X-13 has an isotopic abundance of 93.456% and X-12 has an isotopic abundance of 6.544%. What is the atomic mass of this element?
 13x.93456+12x.006544=12.227808   Atomic Mass of 12.227808 amu
 How to determine whether isotopes are of the same element
 Number of protons (or, if a neutral atom, electrons)
 Use Area of Aluminum Foil Lab:  Determine the area of a piece of aluminum foil with a mass of 1.78 g
 D=2.7g/cc   2.7g/cc=1.78g/V V=1.78g/2.7/cc V=0.6592593 cc V=0.66cc
 Isotopes
 Variants of an element wherein the number of neutrons (and mass number) are different but the number of protons remains the same
 Forms of Electromagnetic Radiation (from Left to Right)
 Radio, Microwave, Infrared, Visible (ROYGBIV), Ultraviolet,  X-Ray, Gamma Ray
 From High to Low Energy on the Electromagnetic Spectrum
 -Frequency Decreases -Wavelength Increases
 Law of Multiple Proportions
 When two elements can combine to form more than one compound the amounts of one of them that combines with a fixed amount of the other will exhibit a simple multiple relation.
 Rutherford’s Experiment (Gold Foil)
 Rutherford aimed alpha particles at a piece of gold foil, expecting them all to pass through. He was surprised to find that some of them were deflected back. This was because they hit the dense, positive nucleus; this discovery changed the model of the atom.;
 Thomson’s Model of the Atom
 Thomson created the plum pudding model of the atom: a positively charged mass with negatively charged particles dispersed throughout
 Electron Excitation in Relation to Absorption and Emission
 When an atom absorbs energy in the form of photons (light), an electron can gain enough energy to get excited and move to another energy level. When an electron drops to a lower energy level, this corresponds with the emission of light.
 How to Use the Formula C=wf (And an example problem: What is the frequency of a photon of light with the wavelength of 600 nm?
 speed of light=(wavelength)(frequency) speed of light=3.00×10^8 m/s w is in nm (1×10^-9m) and f in hertz (1/seconds)   3.00×10^8m/s=600nm(1×10^-9nm/1nm)(f) 5×10^16s=f f=2×10^-17 hertz
 Mixture Examples
 Hardwater, soft water, supermarket salt, drugstore hydrogen peroxide