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# Mason - CHEM 211 - General Chemistry 1 Week 2 (reading and in-class

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Mason - CHEM 211 - General Chemistry 1 Week 2 (reading and in-class

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CHEM 211  MW 10:30-11:45 AM  Prof. Gregory Foster  Page 1 of 7     Notes for Week of 9/6  I.  Extra credit quiz  a.  If uncertainty of measurement is .02, what is the correct number of sig figs  reported for the measurement?  i.  1.23, bc sig figs are defined by the uncertainty of your measuring  instrument. Sig figs will be equal to the sig figs of the uncertainty  ii.  Uncertainty- every measurement you make will have error  1.  Accuracy- closeness to the true value
2.  Precision- how close together all of your measurements are;
repeatability  b.  Area of a 12 inch pizza is 156.7 in(sq). express this area in square centimeters. (1  in = 2.54 cm)  i.  Use a conversion factor which is a statement of equivalence with different  units.  1.  2.54𝑐𝑚 1(ⅈ𝑛) = 1  2.  156.7ⅈ𝑛 2 ⋅ ( 2.54𝑐𝑚 1ⅈ𝑛 ) 2 =1011 cm(sq)  ii.  I do this this way. Because we know that 1 inch is equal to 2.54 cm we can  square the centimeters to find cm(sq) per in(sq). We then multiply the
cm(sq) by the in(sq) to get 1,011 cm(sq)
c.  Select the answer with correct number of sig figs for the following sum.  i.  12.9 ⋅ (211.45 + 27.39764)  1.  Correct answer is 3080  a.  For addition and subtraction. Focus on the number of  decimal places. The fewest number of decimal places
determines the number of sig figs
b.  For multiplication and division, use the total sig figs in the  equation and use the one with the least amount of sig figs to
c.  Because this problem can only have 3 sig figs in the answer  due to the multiplication, we have to round down our
answer so that it only has three sig figs.
d.  A sample of mil is found to have arsenic at a concentration of 5.2  𝜇𝑔 ∕ 𝐿. What is  the concentration in oz per gallon? 1 qt= 946.4 mL, 1 gal = 4 qt, 16 oz = 1 lb, 1 kg
= 2.2 lb)
i.  Answer is 6.9e-7  ii.  𝜇𝑔 → 𝑔 → 𝑘𝑔 → 𝑙𝑏𝑠 → 𝑜𝑧  1.  5.2 𝜇𝑔 𝐿  𝑥 10 −6 𝑔 1𝑚ⅈ𝑐𝑟𝑜𝑔𝑟𝑎𝑚 𝑥 1𝑘𝑔 1000𝑔  𝑥 2.2𝑙𝑏𝑠 1𝑘𝑔 𝑥 16𝑜𝑧/1𝑙𝑏  2.  10 −3 𝐿 1𝑚𝐿 𝑥 946.4𝑚𝐿 1𝑞𝑡 𝑥4𝑞𝑡/1𝑔𝑎𝑙=6.9e-7 oz/gal  iii.  𝐿 → 𝑚𝐿 → 𝑞𝑡 → 𝑔𝑎𝑙  II.  Elements, compounds, and mixtures: overview
CHEM 211  MW 10:30-11:45 AM  Prof. Gregory Foster  Page 2 of 7     a.  Substances  i.  Matter with a fixed composition  b.  Elements  i.  Simplest type of matter with unique physical and chemical properties  ii.  Each element is unique because each os its properties is unique  c.  Compounds  i.  Consists of two or more different elements that are chemically bonded  ii.  Have three defining features  1.  Elements are present in fixed parts by mass
2.  Compounds properties are different from the properties of its
elements  3.  Can be broken down into simpler substances by a chemical change  d.  Mixtures   i.  Consists of two or more substances that are physically intermingled  ii.  Different from compounds in these ways  1.  Can vary in their parts by mass
2.  Retains many properties of its components
3.  Can be separated into components by physical changes
III.  Observations that led to an atomic view of matter  a.  Law of mass conservation  i.  Total mass of substances does not change during chemical reactions  ii.  Matter cannot be created or destroyed  b.  Law of definite(or constant) composition  i.  No matter what its source, a particular compound is composed of the same  elements in the same parts (fractions) by mass  ii.  Fraction by mass  1.  Part of the compound that each element contributes  2.  𝑚𝑎𝑠𝑠 𝑓𝑟𝑎𝑐𝑡ⅈ𝑜𝑛 =   𝑚𝑎𝑠𝑠 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 𝑥 ⅈ𝑛 𝑐𝑜𝑚𝑝𝑜𝑢𝑛𝑑 𝑎 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑐𝑜𝑚𝑝𝑢𝑛𝑑 𝐴   3.  Is fixed no matter what the size of the sample is  iii.  Mass percent  1.  Fraction by mass expressed as a percent
2.  Mass percent = mass fraction x 100
iv.  To find the mass of an element in a sample  1.  𝑚𝑎𝑠𝑠 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 ⅈ𝑛 𝑠𝑎𝑚𝑝𝑙𝑒 =   𝑚𝑎𝑠𝑠 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 ⅈ𝑛 𝑐𝑜𝑚𝑝𝑜𝑢𝑛𝑑 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑐𝑜𝑚𝑝𝑜𝑢𝑛𝑑       c.  Law of multiple proportions  i.  If elements A and B react to form two compounds, the different masses of  B that combine with a fixed mass of A can be expressed as a ratio of
whole numbers
1.  Basically, if there are two compounds that have the same elements,  the mass fraction of one element related to another one will have
increments based on ratios of small whole numbers
2.  Look on page 49 for a good example f this
CHEM 211  MW 10:30-11:45 AM  Prof. Gregory Foster  Page 3 of 7     IV.  Daltons Atomic theory  a.  All matter consists of atoms, tiny units of an element that can’t be created or  destroyed  b.  Atoms of one element cannot be converted from one element to another
c.  Atoms of an element are identical in mass and other properties and are different
from atoms of any other element  d.  Compounds result from the chemical combination of a specific ratio of atoms of  different elements  V.  Observations that led to the nuclear atom model  a.  Mass and charge of the electron  i.  Charge of electron   1.  -1.602e-19 coulombs  ii.  Mass of electron  1.  Using the mass charge ratio and then multiplying it by the charge  that we know we can find its mass  a.  9.109e-28  b.  Discovery of atomic nucleus  i.  Matter is electrically neutral, so atoms must be as well  ii.  Rutherford discovered nucleus through the gold foil experiment  1.  Contains all the positive charge and basically the entire mass of the  atom  VI.  Atomic Theory Today   a.  Atom is electrically neutral, spherical entity composed of positively charged  central nucleus  i.  Surrounded by one or more negative electrons  b.  Nucleus contributes 99.7% of atomic mass  i.  Is incredibly dense  1.  10^14  ii.  Consists of protons and neutrons  1.  Proton has a positive charge and the neutron has no charge
2.  Proton charge cancels out electron charge and makes atom neutral
c.  Atomic number  i.  Equals number of protons in the nucleus of each atom  ii.  All atoms of same element have same atomic number  d.  Mass number  i.  Total number of protons and neutrons of an atom  e.  Atomic symbol  i.  Based on the English, latin, or greek name  ii.  Often written with the atomic number as a subscript and the mass sumber  as a superscript on the left  f.  Isotopes  i.  Elements of an atom that have different numbers of neutrons and  therefore different mass numbers

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##### Description: These notes cover the reading from chapter 2 as well as the solutions to the in-class quiz that we took and went over.
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