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Ch. 1, #33: These equations test order of operations, but add in the aspect of significant
figures. Simply put, do all the calculations as normal, using the exact numbers gotten
from each calculation to get to the result. Once that’s done, truncate the result so it has
the same number of significant figures as the initial term with the least significant figures.
For instance, in the first question, 33a, you’ll round the result down to two significant
figures on behalf of the 3.1 in the first term of the equation.
o Why this way? The book describes it as certainty, which is the degree to which
measurements’ values can be called exact. It gives an example where people read
volume for some liquid, with the mark resting between 20.1 and 20.2. The 20.1 is
undoubtedly true, but what’s after gets confused due to human error; therefore,
anything in the measured result past the 1 in 20.1 is uncertain. In both
measurements and problems like these, the result is reduced to the number of
significant figures of the value with the least sig figs because that defaults it to the
same level of uncertainty. You can’t use anything after because it’s not
guaranteed to be correct, i.e., certain.
Ch. 1, #35: These are just conversions involving scientific notation. Remember what
each of the prefixes (ex. Pico, nano, micro, centi, deci, milli, kilo) mean, and you’re fine.
Ch. 1, #37a-b: These are just conversions from SI/metric to imperial. You can just look
this up, honestly.
Ch. 1, #39a: More conversions, just remember sig figs, and how the result defaults to the
least amount used in any term.
Ch. 1, #51: Since it’s just Celsius to Kelvin, you’re just adding 273 (or 273.15 if you’re
exact) to each Celsius value. If you can’t do this, I’d be very worried.
Ch. 1, #56, 57, 59, 65: These all concern density, mass, and volume. Density is mass
divided by volume, and is a measure of how much matter is in a given space. From that
conversion ratio, you can solve for any missing value given the other two (ex. Mass from
density and volume by multiplying density by volume).
o 56: Volume is found by multiplying the dimensions. Density is mass/volume.
Should be simple.
o 57: 5.0 carats equal 1.000 gram by the conversion rate given; volume is thus that
mass divided by the density.
o 59: Volume of something placed in a liquid is the difference between the final and
initial volume readings of the whole solution. In this case, it’s 21.6 mL – 12.7 mL.
o 65: Solve for the volume that’d get you the given density (using the given mass,
of course), then solve for the missing dimension by dividing by 16 (4 x 4, the
dimensions given).
Ch. 2, #43, 45, 49: The notation of 𝐴𝑍𝑋 is used on atoms only. The element signified is X
(ex. C, Na, Sn), Z is the number of protons (i.e., atomic number), and A is the atomic
mass, the sum of all protons and neutrons in the atom. Isotopes of an element have more
or less neutrons, and thus different atomic mass, than the most common variant of the
atom.
Ch. 2, #51: You just need the periodic table for this, but…
o Column 1A, the leftmost, has alkali metals. Hydrogen’s the exception.
o Column 2A, starting with beryllium Be, has alkaline metals.
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o Column 3A, starting with boron B, running to column 6A, starting with oxygen
O, has nonmetals and metalloids, but the textbook doesn’t differentiate them.
These and hydrogen are all just grayed out.
o Column 7A, starting with fluorine F, has halogens.
o Column 8A, the rightmost, has noble gases.
Ch. 2, #53: In general, the left side of the periodic table has cations, atoms that have a
positive charge (more protons than electrons), while the right side has anions, atoms that
have a negative charge (less protons than electrons). The difference between protons and
electrons is known as charge, and you’ll just look at the periodic table to figure them out.
For a good portion of them, though, column/group 1A has atoms with charge +1, 2A has
atoms with charge +2, then you jump to 7A with -1, 6A with -2…
Ch. 2, #55-60: The book actually explains this better than I can, since nomenclature, the
naming of chemical compounds, was barely a unit when I was in chem. If I’m to
paraphrase the book, though…
o The first term is the cation, the positively charged ion that’s placed first in the
formula. For instance, for NaCl, the first term is “sodium”.
▪ This holds for every Type I bond, which refers to any bond between
elements which are guaranteed to have a certain number of electrons. For
instance, sodium is always +1, calcium is always +2, and sulfur is always 2.
▪ Type II bonds, referring to bonds with elements with variable amounts of
electrons (ex. iron Fe, which can be +2 or +3), change up the formula a bit
by taking the root and adding “-ic” for the higher charged ion, and “-ous”
for the lower charged ion. So, “ferric oxide” has the formula 𝐹𝑒2 𝑂3 on
behalf of 2 +3 charges matching 3 -2 charges, while “ferrous oxide” has
the formula 𝐹𝑒𝑂 on behalf of iron’s +2 charge matching oxygen’s -2
charge.
o The second term is the anion, the negatively charged ion that’s placed last. Back
to NaCl, the second term is “chloride”, derived from taking the root of the anion
(i.e., “chlor-“) and adding “-ide”.
o These are all monatomic ions used here, by the way. They only have one element;
polyatomic ions are where things get trickier.
Ch. 2, #65-68(a-f): These are why nomenclature was never really too big a unit in the
curriculum. They’re polyatomic ions being used, which have molecules serving as the
cations and anions. Usually the formulae for them will be given, but you can just look up
any formula you need; back in chem they gave us a reference sheet.
o All else is the same, except for Type III bonds, a.k.a. binary covalent bonds.
These are only between two nonmetals and use numerical prefixes to denote the
number of atoms of each element. For instance, 𝑁2 𝑂5 is dinitrogen pentoxide.
Don’t add “mono-“ to the cation though; you don’t say “monocarbon monoxide”.
o Did I ever note that all you need to do when talking about bonds is take the
charges of each ion, and find the combination of each such that the charges cancel
out? Like if you have a +2 cation and a -3 anion, you need 3 cations to get +6, and
2 anions to get -6. +6 and -6 cancel out, making the resulting molecule stable.
Some molecules aren’t stable, though, and have a charge, become ions
themselves.
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Ch. 3, #27: This is just a weighted average; multiply each of the isotopes’ masses by their
percentage, then add them all and divide by the total amount to average all those
products.
Ch. 3, #28: You could do another weighted average, but really you could just look at the
most common isotope and guess based on that.
Ch. 3, #29: This is another weighted average, but you’re solving for a value used in it.
Use the same calculation and solve for the blank term. Alternatively, you could chance it
and just subtract the mass of two neutrons from the first isotope’s mass, but that’s
probably wrong.
Ch. 3, #35: Molar mass is the total mass of a mole of something, which itself is 6.022 ∗
1023 units of that something. It usually is given as some number of grams over 1 mole.
Remember that number, Avogadro’s number, which is derived from how 12 grams of
carbon-12 has 6.022 ∗ 1023 atoms in it… or another derivation involving copper that we
didn’t go over in chem. Anyhow, to solve this, you have to convert from carats to grams,
then from grams to moles using molar mass, then from moles to atoms using Avogadro’s
number.
o This process of using conversion factors, rates that change between two units, is
crucial to chemistry, so be sure to study it well. It also helps to write them all
down as fractions, plugging in values and cancelling units as necessary.
Ch. 3, #39, 40b: Take the molar masses of each element, multiply by the number of
atoms in each molecule, and from there it’s a simple sum.
Ch. 3, #53, 57: Both involve molar mass and conversions from mass to atoms to moles,
though 57 perhaps takes a more step-by-step approach to it.
Good luck :D