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Download Chemistry if8766 All Pages with Key Answers

2018-10-28T04:08:00.001+02:00

Chemistry if8766 all pages for direct download in image format, clear images with direct answers
for homework help
includes:
page 9 - significant figures
page 12 - temperature and its measurement
page 13 - freezing and boiling point graph
page 15 - heat and its measurement
page 17 - matter, substances vs mixtures
page 18 - physical vs chemical properties
page 19 - physical vs chemical changes
page 45 - naming ionic compounds
page 46 - naming molecular compounds
page 48 writing formula from names
page 49 - gram formula mass
page 50 - moles and mass
page 52 - the moles and mass key answers
page 70 - the mole and volume

More homework essays
chemical reaction examples for chemistry in everyday life
physics examples application everyday life

moles and mass worksheet chemistry if8766 page 52 answers

page 50 - the moles and mass key answers

CONSTANTS, UNITS, AND CONVERSION FACTORS in Analytical Chemistry

2018-10-06T02:21:00.000+02:00

Analytical Chemistry SELECTED CONSTANTS, UNITS, AND CONVERSION FACTORS
[The uncertainty in the last digit(s) is shown italicized in parentheses]

Atomic mass constant: mu = 1.660 538 73 x 10-27 kg
Avogadro’s number: N = 6.022 141 99 (47) x 1023 mol-1

Boltzmann constant: k = 1.380 6503 (24) x 10-23 J/K
Elementary charge: e = 1.602 176 462 (63) x 10-19 C
Faraday constant: F = 96 485.3415 (39) C/mol
Molar gas constant: R = 8.314 472 (15) J/K-mol = 1.9872 cal/K-mol
= 0.082 057 L-atm/K-mol = 0.022 414 m3/mol at STP
Pi: π = 3.141 592 653 6
Planck’s constant: h = 6.626 068 76 (52) x 10-34 J-s
Speed of light (in a vacuum): c = 2.999 792 458 (exact) x 108 m/s
Stefan-Boltzmann constant: σ = 5.670 400 (40) x 10-8 W/m2-K4

Standard acceleration of gravity: gn = 9.806 65 (exact) m/s2

Wein constant: k = 2.897 7686 (51) x 10-3 m-K

Force: 1 N = 1 kg-m/s2

Joule: 1 J = 1 N-m = 1 kg-m2/s2 = 107

ergs = 1 V x 1 C = 1 V-C = (J/C)(C)

Power: 1 W = 1 J/s = 1 V x 1 A = 1 V-A = (J/C)(C/s)

Electron Volt: 1 eV = 1.602 176 462 (63) x 10-19 J = 3.827 x 10-20 cal

Calorie (thermochemical): 1 cal = 4.184 J [Food “calorie” = 1 Cal = 1000 cal]

Length: 1 km = 1000 m = 0.62137 mi 1 in = 2.54 cm (exactly)

Mass: 1 kg = 1000 g 1 pound = 453.59237 g

Pressure: 101 325 (exact) Pa = 1 atm = 760 mm Hg = 17.70 lb/in2

133.322 Pa = 1 torr = 1 mm Hg 105

Pa = 1 bar 1 Pa = 1 N/m2

Volume: 1 L = 10-3 m3 = 1000 mL = 1000 cm3 = 1.056710 quarts

Nernst factor: (RT/nF) ln = (0.05916 V/n) log10 at 25 C

nF/RT = 38.920 n V-1

RT/nF = 0.02569/n V

∆G = -nFE = -RT(ln Keq) ∆G = -nFE = -RT(ln Q)
µ = x + zσ/n1/2
µ = x + ts/n1/2

Some Less Common Multiplicative Prefixes:
P = peta = 1015
T = tera = 1012
G = giga = 109

n = nano = 10-9
p = pico = 10-12
f = femto = 10-15

a = atto = 10-18
z = zepto = 10-21
y = yocto = 10-24

Ion Product for Water
pKw = 14.3463 at 15 C, 13.9965 at 25 C, 13.5348 at 40 C

Amphiprotic Salts (e.g., for NaHA): [H+] = [(Ka2CNaHA + Kw) / (1 + CNaHA /Ka1)]½

Debye-Huckel Theory
Debye-Huckel Limiting Law (DHLL): - log γi = A zi2 µ½ = 0.5 zi2 µ½
Debye-Huckel Equation (DHE): - log γi = (A zi2 µ½)/(1 + B ao µ½) = (0.51 zi2 µ½)/(1 + 0.33 o
µ½)

A = 0.5115, B = 0.3291 on a volume (molar) basis at 298 K
A = 0.5108, B = 0.3286 on a mass (molal) basis at 298 K

Calculate the theoretical pH of trisodium fictate

2018-10-06T02:11:00.000+02:00

Fictitious acid (H3Fi) is a triprotic acid with pKa values of precisely 4.00, 7.00,
and 10.00. Calculate the theoretical pH of the following solutions.

(a) 0.15 M trisodium fictate, Na3Fi.
(b) 0.15 M disodium fictate, Na2HFi
(c) 0.15 M fictitious acid, H3Fi.

calculate the concentration of calcium ion in a saturated solution

2018-10-05T02:09:00.001+02:00

Assuming there is no hydrolysis or complexation of the ions formed, calculate the concentration of the calcium ion in a solution that is saturated with

(a) Calcium fluoride, CaF2, and also 0.050 M in KF. Ksp = 5.3 x 10−9
(b) Calcium phosphate, Ca3(PO4)2. Ksp = 2.0 x 10−29

Calculate the change in pH of HCL after lactic acid and lactate addition

2018-10-05T02:07:00.001+02:00

Calculate the change in pH that occurs when 0.050 mmol of a strong acid
(such as HCl) is added to 100 mL of 0.0200 M lactic acid plus 0.0800 M sodium lactate.
(CH3CHOHCOOH, Ka = 1.38 x 10−4)

Calculate the 95% confidence interval of a gasoline standard sample

2018-10-05T02:05:00.004+02:00

The concentration of an additive in a standard sample of gasoline was measured 5 times with the following results: 0.13, 0.11, 0.12, 0.20, and 0.14 % by mass.

(a) Calculate the arithmetic average and standard deviation of the data
(b) Calculate the 95% confidence interval
(c) If the accepted value for the standard sample is 0.11 % by mass, are the results for this
set of measurements be significantly different at the 95% confidence level by the t test.
(d) Can any of the data be rejected by the Q test?

Write the charge and mass-balance equations for ferrous ammonium sulfate

2018-10-05T02:03:00.000+02:00

Ferrous ammonium sulfate was used as a standard for iron in the absorbance
experiment in lab. The stable solid weighing form has the formula Fe(NH4)2(SO4)2
Q6H2O.
Write all the charge- and mass-balance equations for a 0.050 M aqueous solution of
“ferrous ammonium sulfate” as written above.

Remember that the ammonium ion is the
conjugate acid of a weak base. In addition, while the first proton on sulfuric acid ionizes
completely, the second proton does not. Technically, the bisulfate ion, HSO4−, is a “weak
acid” and there is a measurable Ka2

calculate the relative uncertainty in the atomic mass of Os

2018-10-05T02:00:00.001+02:00

The atomic mass of fluorine is currently listed on periodic tables as
18.998403 g/mol, the most precisely known atomic mass for a naturally occurring element.
The least well known is that of osmium at 190.2. Using the standard convention for the
uncertainty of a measured number when one is not formally stated, calculate the relative
uncertainty in the atomic mass of Os

Calculate The Molar Mass of a Substance with Density 2kg/dm3

2018-10-05T01:41:00.000+02:00

Q: A substance has density of 2 kg dm-3 and it crystallizes to fcc lattice with edge length equal to 700 pm. The molar mass of the substance is _______.
(A) 55.32 g/mol (B) 130 g/mol
(C) 103.3 g/mol (D) 144 g/mol

A: 103.3 g/mol

Balancing The Chemical Equation NaHCO3 Na2CO3

2018-10-04T14:08:00.000+02:00

N a H C O_{3} = N a_{2} C O_{3} + H_{2} O + C O_{2}

The law of conservation of mass states that matter cannot be created nor destroyed. There are no more or less atoms at the end of a chemical reaction than there were at the beginning. To balance a chemical equation, every element must have the same number of atomson each side of the equation.

To balance a chemical equation:

1. Every element must have the same number of atoms on each side of the equation (law of conservation of mass).

2. Balancing the equation can be done only by adjusting the coefficients.

N a

is not a balanced element. To balance

N a

, multiply

N a H C O_{3}

from the left side by

2

2 N a H C O_{3} \to N a_{2} C O_{3} + H_{2} O + C O_{2}

H

is a balanced element because there is the same number of atoms of

H

in each side of the equation.

H is a balanced element

C

is a balanced element because there is the same number of atoms of

C

in each side of the equation.

C is a balanced element

O

is a balanced element because there is the same number of atoms of

O

in each side of the equation.

O is a balanced element

All elements have the same number of atoms in each side of the equation, which means that the law of conservation of mass is satisfied and the chemical equation is balanced.

2 N a H C O_{3} \to N a_{2} C O_{3} + H_{2} O + C O_{2}

What do Vm & Km stand for in Chemistry of enzymes?

2018-10-02T01:12:00.000+02:00

These terms are from basic enzyme kinetics

Vm is the maximum velocity of an enzymatic reaction.
Km is the dissociation constant of the enzyme-substrate complex

What does Vm stands for in Chemistry (Stoichiometry)?

2018-09-30T01:10:00.000+02:00

In the equation: n=V/Vm , What does the Vm stand for?

Short Answers:

Vm (The molar volume) is the volume occupied by one mole of a substance at a given temperature and pressure.

It has a value of 22.4 dm3/mol for strandard conditions.

What Does dm^3 Stands For in Chemistry?

2018-09-24T01:08:00.000+02:00

cm^3 stands for cubed centimeter, i wonder what does dm^3 in chemistry stands for...i know that in order to get dm^3 you'll have to divide the value in cm^3 by 1000, wat about getting dm from cm?

Short Answer:
dm^3 stands for cubic decimeter. So, if there are 10 centimeters in every decimeter, that explains your having to divide cm^3 by 1000. Bet you can get from cm to dm now,

meter, decimeter, centimeter, millimeter (mm), micrometer (um or 10^-6 m), nanometer (nm or 10^-9 m), picometer (10^-12 m), femtometer (10^-15 m)
Now that we have nanotechnology chemistry and biology see a lot of use of the measurement terms below milli- and micro-.

What Does mM Stand For in Chemistry?

2018-09-17T01:05:00.000+02:00

millimolar
so: 1 M = 1000 mM

M in chemistry stands for molarity, measured in moles per liter. m stands for "milli" in most fields. So this is a 0.005 M solution of sodium nitrate. There are 0.005 moles per liter of solution.

How To Name Covalent Molecular Compounds?

2018-09-10T01:03:00.000+02:00

let's look at how to write names for
compounds like these that are made of
two nonmetals the nonmetals are the
elements to the right of the staircase
on the periodic table now
compounds like these that are made of a
nonmetal and another nonmetal two
nonmetals are called covalent or
molecular compounds and that's because
the elements that are in them are held
together by covalent bonds and they're
held together in groups called molecules
now the way we name covalent or
molecular compounds is different from
how we name ionic compounds ionic
compounds contain a metal and a nonmetal
the metals are the elements here on the
periodic table and I've written in some
of the most common some examples of
ionic compounds and of how we name them
are things like calcium fluoride and
iron ii oxide so if you have to name a
compound the first thing you want to do
is figure out what type of elements are
in it if the compound is made of a metal
and a nonmetal it's an ionic compound I
have a whole bunch of videos on how to
name I on a compounds and how to write
formulas for them if you need to name a
compound that's made of a nonmetal and
another nonmetal well that's a covalent
or molecular compound and naming these
is what this video is going to focus on
okay here's our first chemical formula n
2o 3 it's made of two nonmetals we're
going to write a name for it following
these steps so here's the first one says
for the first element start with the
element name what I'm talking about the
first element I'm talking about the
order that these elements are in in this
chemical formula so the first element
here is nitrogen n so it says start with
the element name okay so I'm going to
put nitrogen down that's the element
name here now for the second element so
that's this here oxygen for the second
element start with the IDE name now the
IDE name is the version of the element
name that ends in IDE these are actually
the same names that we use to name
negative ions so we've got oxygen here
we're going to use its IDE name which is
oxide
okay so we got nitrogen and we got oxide
now use prefixes to show how many atoms
of each type there are the prefixes
here's a list of them are like these
little tags that we put on the front of
each name to tell us how many atoms of
each type we have so n 2 we got two
nitrogens which means that we want to
use this prefix for to die so put that
in front of the element name dinitrogen
now o 3 which means that we have three
oxygens so i'm going to use the prefix
tri here and write that in front of
oxide and so the name for this compound
is dinitrogen trioxide we put the
element name here we put the IBE name
here and then we use the prefixes to
indicate how many atoms of each type we
have let's look at a few more examples P
for s 10 okay for the first element
we'll start with the element name P is
phosphors if you didn't know that you
could look it up on the periodic table
for the second element that's s here
start with a IDE name so s is sulfur and
it's IDE name is sulfide okay so those
are the two names now we're going to use
prefixes to show how many atoms of each
type we have we got P four so that means
we have four phosphorus atoms so four is
tetra we use the tetra prefix tetra and
then four sulfide we got ten so we'll
use this prefix deca deca so the name of
this compound is tetra phosphorus deca
sulfide let's move on we're going to add
a couple more steps to deal with more
complex formulas here's the next step
that we're going to talk about do not
use mono on the first element I'm going
to show you what this means by working
through
naming this formula we'll start it off
just normally so for the first element
start with the element name that's
carbon and then the second element which
is fluorine we use the IE de name which
in this case is fluoride okay now we get
to the prefixed part and this is where
things change a little bit so it says do
not use mono on the first element carbon
here we only have one of them so you
might want to use mono and put it in
front of here that would be a perfectly
logical thing to do but for whatever
reason we don't use mono on the first
element if there's just one of them we
just leave it like this you don't put
anything there so we have one carbon
we're just going to call it carbon and
then for fluorine here in front of
fluoride I'll put the prefix just like I
would any other time so I have four of
these fluorines so i'm going to call it
tetra tetrafluoride so carbon
tetrafluoride not mono carbon
tetrafluoride just carbon tetrafluoride
is how we name this because if you got
just one of the first element you don't
use anything pcl5
let's name this so p the first element
is phosphorus CL the second element is
chlorine and chlorines IDE name is
chloride so there it is now we get to
the prefixes phosphorus here we only
have one of them so we're not going to
use mono here we're just going to keep
it as phosphorus and chlorine we have
five of those so we're going to use ten
two here as the prefix so it's going to
be Penta Penta chloride and phosphorus
pentachloride not mono phosphorus
pentachloride just phosphorus
pentachloride is how we name this now
there's one more step that we're going
to add I'm going to talk about that next
okay there's one more step to keep in
mind and I'm going to show it to you as
I work through this example
okay CL 2 o so the first element here CL
is chlorine and oh here is oxygen its
IDE name is oxide okay so now the next
thing I'm going to do is I'm going to
use these prefixes
so I got two chlorines so I'm going to
use dye dye for two up here and then
oxide I got seven of those so I'm going
to use the prefix hepta now try saying
this compound name
we got dichlorine heptoxide hear that
hep dioxide thing is kind of awkward to
pronounce so to prevent us coming up
with these really weird names there's
one more rule and that's that if you
have a o or o o turn it into oh here's
that means we got this heptoxide here so
to make this easier to pronounce we turn
the a o that we see right there into
just an O so I'm going to get rid of
that a going to get rid of that a and
I'm going to turn this compound name
into chlorine heptoxide so not heptoxide
but just heptoxide so you get rid of
that a if it's next to an O there let's
do one more P 406 so the P phosphorus is
our first element and then oh we got
oxide that is a ID name for oxygen here
all right now let's use the prefixes
switch to our prefix table here we have
four phosphorus so we're going to use
the tetra prefix so tetra tetra
phosphorus and then for oxygen we have
six of those so we're going to use hexa
here and just as before we end up with
this awkward name hex oxide we get an AO
so we're going to get rid of this a-here
soup I always like knock that name out
of the way hexa oxide and that turns it
into tetra phosphorus hex oxide you know
I forgot to circle the name and the last
one and this is just so satisfying so I
did it really slowly to make up for the
fact I didn't circle the last one so we
get rid of this we get rid of this
a-here tetra phosphorus hex oxide so I
wanted to
one more example that pulls in examples
of all of this it's like a good review
for all of these things here's our last
formula this is a really good review
it's just C O so the first element is
carbon
the second element is oxygen its ite
name is oxide ok so now for the prefixes
carbon we've only got one of them so
remember don't put anything in front of
it we just call it carbon but we have
one oh and if you have one of the second
element you do use a prefix so we use
the mono prefix here so mono now mono
oxide gives us the same kind of weird
pronunciation thing it gives us an oo
here so we're going to get rid of this
oh I'm going to try to not knock this
name out here oh just a little bit
really good and we call this we change
the name from mono oxide to carbon
monoxide which is probably something
that you've heard of before here's just
some quick information about how we
change these prefixes to avoid an
awkward pronounciation problem we only
really have this problem with oxide
because it begins in O and so we get rid
of the O here in mono and we get
monoxide then we don't have to worry
about dire try because they don't end in
aro and then for all of these others 4
through 10 we get rid of all of these
A's so we get tetroxide pentoxide hex
oxide heptoxide oxide non oxide and Dec
oxide so here's how we name covalent or
molecular compounds with two nonmetals
just follow each of these steps one
thing to keep in mind this mono for the
first element really confuses a lot of
people so remember not to use mono if
there's just one of the first element
keep this in mind try to prevent
yourself from writing names that have
really awkward pronunciations with a o
or o Oh turn it into o keep these things
in mind you should be all set

How to Balance a Chemical Equation For a Beginners?

2018-09-06T14:15:00.000+02:00

in this video we're going to look at how
to balance chemical equations so I have
an example up here on the screen right
now where we have h2 which is hydrogen
plus CL 2 which is chlorine and when
these two react the reaction yields HCl
which is hydrogen chloride now there's a
problem right now with the way this
equation is written and I'll show you
what the problem is by drawing bringing
in some animations so here's h2 now H
subscript 2 means that you have two
hydrogen atoms bonded to each other so
that's what's drawn here we have two
hydrogen atoms touching because they're
bonded together now over here the CL 2
we need two chlorine atoms again bonded
together and notice the colors match
over here we have the arrow which means
yield so we have the reaction happening
and I've drawn this dashed vertical line
here to kind of separate the reactants
which are over here from the products
which are about to appear over here so
the product here is H bonded to CL the
problem is it's not balanced according
to the law of conservation of mass if
you have two atoms on the product side
there should be two atoms on the
reactant side well let's take a look
let's take a look at the reactant side
for hydrogen we have two atoms of
hydrogen here on the reactant side go
over here to the product side there's
only one atom of hydrogen that's a
problem it's as if a hydrogen atom has
disappeared chlorine has a similar
problem we have two chlorine atoms on
the reactant side but only one on the
product side so to fix this there are a
few rules we have to follow you can only
add things for instance you could add
another h2 right here which I'm circling
around here with my cursor I could add
another CL to which I'm circling here or
I could add another HCl I'm not allowed
to take anything away and I'm also not
allowed to add
parts of molecules I can't just add a
single H atom or a single CL atom so
given those constraints the only thing I
can do here to make this balance the
only thing I can add is to add another
HCl which I'm doing right now I've
brought in another molecule of HCl now
notice again the h and the CL are bonded
together so they're touching but this
HCl molecule is not in contact with this
other one there's some space in between
to show that they're separate they're
not chemically bonded together let's
check to see if this is balanced so over
here on the reactant side we have two
hydrogen atoms on the product side we
have two hydrogen atoms on the reactant
side we have two chlorine atoms and on
the product side we have two chlorine
atoms so yes it is balanced now and we
need to make a coefficient appear in
front of HCl so let's bring in a
coefficient a to the two in front of HCl
tells us we have two molecules of HCl
here there's no coefficient in front of
CL so that means there's just one
molecule of CL CL - and there's a no
coefficient in front of h2 so that means
that there's just one molecule of h2 and
we have now balanced this chemical
equation we have the same number of
atoms of hydrogen on the left side the
reactant side as we have on the right
side the product side and the same thing
is true of chlorine the same number on
both sides notice the way you balance
chemical reaction equations is to put a
coefficient or coefficients in the
proper places we never change these
subscripts you cannot change these small
numbers here the two here or the two
here cannot change also you can't put a
two between the h and the CL or two
afterward because if you did you would
be changing the chemicals in two
different chemicals and that is not
allowed let's take a look at another
example here we have calcium plus water
are reacting and yielding calcium
hydroxide and hydrogen so calcium is all
alone a single atom
here's h2o
now for calcium hydroxide it's calcium
atom and notice the O H here is
hydroxide it's in parentheses so we're
going to bring in a no H in contact with
the CL so at C a so it's bonded to it
but there's a 2 after the parenthesis
which means there are actually 2 OHS two
hydroxide so I'll bring in another one
here so here we have an O H a calcium
and another Oh H so because you have two
O H's this makes sense the Ovation
parenthesis has a subscript two after it
now I just want to warn you here a
little bit of a caution is that in
actuality the calcium hydroxide is not
arranged with all of its atoms in a
straight line like this they're not all
connected exactly the way shown here but
that's okay because right now we're not
concerned about exactly how these atoms
are bonded to each other and exactly
what shape it has it's alright for us
just to put them all in a straight line
because today we're just counting the
atoms and counting molecules so let's go
onto h2 and here we have the h2 now
looking at it I can see it's not
balanced for one thing there's only one
oxygen atom on the reactant side but
there are two on the product side
another problem is on the reactant side
there are two hydrogen atoms and on the
product side there are four one two
three and four so remember the only
thing I can do is add more of something
or some things and I can't add anything
except these four options here I could
add more calcium I could add more water
I could add more calcium hydroxide or I
could add more hydrogen and by hydrogen
I mean a hydrogen molecule I can't just
add a single atom of hydrogen I would
have to add another molecule h2 well
because I have too little oxygen over
here on the reactant side I can tell
right away I'm going to need more and
the only way I can get more oxygen is to
add another h2o molecule
there we go another water molecule now
this has helped because now I have two
oxygen atoms here one and two which
matches on the product side where I have
two the hydrogen situation is better now
as well because I have four hydrogen
atoms over here on the reactant side and
I also have four over on the product
side and even my calcium situation is
now fixed I have one calcium atom here
on the product side and one on the
reactant side so it turns out this
chemical equation is balanced as long as
I put a coefficient of 2 in front of h2o
and so now I have a balanced chemical
equation CA plus 2 h2o yields C a
parenthesis H parenthesis 2 plus h2 next
let's see another example n2 is nitrogen
so here we have n2 h2s hydrogen together
they react and yield ammonia nh3 well I
see a problem it's not balanced I have
two nitrogen atoms here but only one
over here so I definitely need more
nitrogen over here on the product side I
cannot add just a single nitrogen atom
but I can add another nh3 molecule now I
fixed the nitrogen problem I have two
nitrogen atoms here and two over here
however I still have a hydrogen problem
because I my react inside I only have
two hydrogen atoms and on the product
side I have 1 2 3 4 5 6 but this is
fixable I just need some more hydrogen
so here's another hydrogen molecule now
I have 4 atoms of hydrogen on my
reactant side let's bring in another
hydrogen molecule and now it's balanced
I have 6
hydrogen atoms on the reactant side and
six hydrogen atoms on the product side
two nitrogen on each side so I do not
write a number in front of n2 because
there's just one molecule of nitrogen
but I do need a number in front of h2
and that number is of course called the
coefficient and I need a three and in
front of nh3 which is ammonia
I need another coefficient I have two
molecules of ammonia here so I need a
coefficient of two and now I have a
balanced chemical equation n2 plus 3h to
yield two nh3 and that concludes this
video

Understanding Conversion of Units and How To Use Conversion Factors

2018-08-31T01:00:00.000+02:00

in this video we're going to look at how
to convert units using conversion
factors like this in canceling units
some people call this dimensional
analysis some people call the factor
label method but you are going to call
it easy by the end of this video because
we are going to go step by step to show
how to solve these kind of problems so
here is our first one we want to know
what is three point four five pounds
expressed in grams so the conversion
that we're going to be doing is we're
going to be doing pounds to grams and
we're going to be starting with three
point four five pounds okay next thing
we got to do is we got to go and we got
to find some kind of relationship
between pounds and grams so how many
pounds are there in how many grams you
can find this information on the
internet you can find it in a textbook
you can probably find it like in the
back of a notebook where they'll have a
conversion table for the unit's it's
going to look something like this you've
got a bunch of a bunch of these
relationships between different units
and you want to find the equation that
talks about pounds and grams okay so
it's going to be this one down here one
pound equals 450 3.6 grams so this is
the statement that you're going to want
you can get it from a variety of places
but important now it doesn't matter
whether pounds or grams is first it can
be flipped just as long as it has both
pounds in grams so now we have this
statement that tells us how balance
relate to grams we're going to use this
statement now to write two conversion
factors a conversion factor is expressed
as a fraction with a top and a bottom so
here's how we can you take this and
write a conversion factor where you take
this side of the equation one pound and
put it on the top of the fraction and
there's a fraction line and then this
part of the equation is going to be on
the bottom so
going to be 450 3.6 grams now that's one
of the two conversion factors the other
conversion factor that we're going to
write is we just take this and we flip
it
so this 450 3.6 grams we put that now on
the top of the fraction and divide it by
one pound so two conversion factors that
you can write from this statement here
either one of them is correct but only
one of them is what we want to be using
here okay so we're just going to use one
of these the one that I'm going to use
is this one because pounds is up here
and pounds is down here let me tell you
what that means
so three point four or five pounds
that's not part of a fraction right
there's no fraction line here and so if
something does not have the bottom of a
fraction you just assume that it is the
top of a fraction it's the same as it
being on the top of a fraction if it
doesn't have a bottom that's what I mean
now on the other hand pounds is down
here on the bottom of the fraction okay
and when we're using conversion factors
we want to get rid of pounds and we want
to be left with grams and it turns out
that if a unit is on the top on one side
of multiplication sign and then it's on
the bottom on the other side of
multiplication sign it cancels out so
pounds is on top here pounds is on
bottom there so they both cancel out and
that leaves me with units of grams so
just to review I wanted to use this
version of the conversion factor because
pounds was on the bottom and this way
pounds will cancel out now I've
cancelled out pounds I'm ready to do the
math what's it going to be I'm going to
do three point four five times four
hundred and fifty three point six
divided by one or if you have a fancy
scientific calculator you can plug this
whole thing in one expression you can do
three point four five times and then
write this conversion factor in
parentheses four hundred and fifty three
point six divided by one now you don't
really have to worry about the one too
much if you don't want to I'm just
putting it there because sometimes this
won't be a one so just want you to get
used to dividing by whatever is on the
bottom of the fraction
even if happens to be one in this case
so however you decide to plug this into
your calculator when you crank through
it you're going to get the same number
and that's going to be one thousand five
hundred and sixty what are the final
units here
well I cancel out pounds so the final
units that I'm going to be left with our
grams I'm not worrying too much about
significant figures when I'm doing these
unit conversions just for this lesson
because I don't want to add another
thing in here to confuse you there'll be
a lesson later about how to do
significant figures with a unit
cancellation but I don't want you to
worry about sig figs now just worry
about figuring out how to do the unit
conversions so anyway 1560 grams is our
final answer let's do a couple more how
many miles is fifteen thousand one
hundred feet so the conversion we're
going to be doing in this problem is
from units of feet into miles and we'll
be starting with fifteen thousand one
hundred feet we got to go to our unit
conversion table or find the information
on the internet to figure out what the
relationship is between miles and feet
we got it right here so this is going to
be the statement that we're going to use
to write our conversion factors let me
pull this off the table here now don't
freak out that miles is on this side and
feed is on this side it doesn't matter
what unit is on what side of the
equation because you could just easily
flip it okay you don't care what unit is
on what side of the equation all you
care about is being able to have this
and then you can write the conversion
factor so let's write the two conversion
factors that we can get from this
statement I'll take one mile and I'll
put it on top here and then I'll take
the other side five thousand two
80 feet now put that on bottom and now
all right what we could call the
reciprocal of this where we take it and
we flip it up so that 5,280 feet is on
the top and one mile is on the bottom
we're going to be multiplying our
measurement in defeat by one of these
two conversion factors which one is it
going to be
we have feet not as part of a fraction
so you assume that feet is on the top of
a fraction the same as if it's on the
top of the fraction which means that we
are going to want a conversion factor
that has feet on the bottom of the
fraction so they cancel out so it is
going to be this one here and now feet
on the top and feet on the bottom cancel
out and they leave me units in miles
which is what I'm looking for here now
how do I do the math I'll do fifteen
thousand one hundred times one divided
by 5,280 because it's on the bottom of
the fraction or if you can plug larger
expressions into your scientific
calculator you can do fifteen thousand
one hundred times and then in brackets I
mean in parentheses you can do one
divided by 5,280 again you might wonder
why you have to keep doing the one you
can leave the one out if you want but
remember this isn't always going to be a
one so it's a good thing to get in the
habit of multiplying by whatever's on
the top of the fraction and then
dividing by whatever is on the bottom of
the fraction even if it's 1/4 right now
you can do either one of these
expressions and you're going to end up
with an answer of two point eight six
units are in miles and again I'm not
really paying attention to significant
figures for these calculations that's
how you do this let's do two more
problems so you'll really get the hang
of it this problem is about units of
money on a certain day the exchange rate
between US dollars and euros is one US
dollar equals zero point seven eight
euros on that day how much is a hundred
and twenty five euros worth in US
dollars so we will be going from euros
two US dollars here starting with a
hundred and twenty-five euros and the
question gives us this relationship
between dollars and euros which I just
write in bigger letters here in the two
previous problems I took this statement
and I wrote two conversion factors top
of the bottom that I flipped the top of
the bottom what I'm going to do here
though is I'm going to look at what I'm
starting with and I'm just going to
write the one conversion factor that I
need okay so I'm taking 125 euros and
what do I want to multiply that by to
cancel out of yours
I'm going to want the version of the
conversion factor version of the
conversion version of the conversion
factor that has euros on the bottom so
they'll cancel out so I'm going to take
this thing that as your 0.78 euros and
that will be on the bottom which means
then this one u.s. dollar will be on the
top so you don't always have to write
out both of the conversion factors you
can figure out which of the two you need
based on what should be on the top and
what should be on the bottom
now euros up here euros down there they
cancel out leaves us with dollars which
is good because that's what we're
looking for and the math is going to be
a hundred and twenty five times one
divided by zero point seven eight or a
hundred twenty-five times one divided by
zero point seven eight this is going to
give us a hundred and sixty US dollars
so the dollars doing pretty well
compared to the euro on this date one
more how many liters is twenty three
thousand five hundred milliliters these
are both metric units and you know a lot
of times people ask me this this unit
canceling method can I use it for metric
units of course you can use it for any
type of units all you got to do is
figure out what the relationship
we near to Eunice's so we are going from
milliliters ml to liters
l and you may already know this but
there is a thousand milliliters in 1
liter that's what I mean this is all you
need you can convert any two units that
you want just as long as you know the
relationship between so we have this
year 4 liters and Miller's so 23500 M L
which conversion factor am I going to
want to use here I since I want to get
rid of milliliters I want to use a
version of this that puts milliliters on
the bottom so I'll put a thousand
milliliters down here so they'll cancel
out and that means that I'll put 1 liter
on the top cancel cancel I'm left with
leaders and so this is going to be I'm
not even going to write it out because I
think you're getting the hang of it
that's going to be twenty three thousand
five hundred times one divided by a
thousand which is going to be twenty
three point five finally units are in
liters so that's how you can convert
from one unit to another by setting up
your conversion factors and cancelling
your units so where do you go from here
there are two more videos that may be of
interest to you the first is to show how
to string multiple conversion factors
together because you don't just always
have to use one here I'm converting from
days all the way to seconds by setting
up a bunch of conversion factors where
all the units cancel so I'll show you
how to do that in the one of the next
videos and then another video that you
might want to watch is about
understanding unit conversion where I
talked about the rationale the reasoning
behind why you set up conversion factors
away you do why the unit's cancel and
how this relates to things you might be
able to more easily understand

20 Physics Theory Examples Application in Real Everyday Life

2016-03-07T11:50:00.000+02:00

Here is how the physics represented in everyday life activities, I list here some examples but not all, because physics theories and applications cannot be counted all.

Science student must know that Chemistry reactions are the hidden mechanisms of physics.

Physics activities through the Day:

Wake up in the morning - do I pop up or roll to the side an push myself up? Understanding stress due to gravity and use of leveraging as a mechanical advantage to reduce stress on the back.

I take a walk and am climbing a steep hill with a sandy surface - do I climb on my toes or use the full surface of my shoes? Understanding viscosity and friction and mechanics of moving a mass up a grade.

I am making a hot cup of coffee which seems to spill every time I walk with it. Why? Understanding how viscosity of water changes with heat and not filling the cup as high as I would for a capachino.

A boy has run off with my purse. I pick up a rock. Do I throw it at his head or above his head? Understanding mechanics and trajectories.

Examples list for physics applications in Daily Life Activities

1. A car going down the street (friction between the rubber and the road)

2. A leaf falling from a tree (gravitational attraction)

3. An airplane in the sky (airflow keeps the plane aloft)

4. An incandescent light (electrical resistance)

5. A florescent light (ionized particles)

6. A straw in a glass of water (light bends the straw's image)

7. Water from the kitchen faucet (fluid flow)

8. The TV set turned on (electromagnetic waves)

9. A child riding a bicycle (conservation of angular momentum)

10. A piece of wood floating in a river (density and buoyancy)

11. The Space Shuttle launching (Newton's law of action and reaction)

12. A child going down a slide (normal and parallel force components)

13. Drops of water (surface tension)

14. Inside of a clothes washer after the spin cycle (centrifugal force)

15. Water level in a bathtub before & after you get in (Archimedes Principle)

16. A car up on a hoist (hydraulics)

17. Removing the cap from a glass bottle with a bottle opener (lever action)

18. A fast car and a slow car leaving from a red light (acceleration)

19. A supermarket scanner (laser technology)

You've got to think of one involving heat (that's on you)

Also written about: Chemistry Examples from the Daily Life

Hard 7th Grade Math Homework HELP

2016-03-01T11:39:00.000+02:00

3y + 6x = 24

5= 4y - 3x

Solve 7y + 7x = 21x + 35 for y

Solve 4m - (2n to the 2nd power) = 2m - 72 for m

Solve 9g + 7h = 9g - 7h + g for g

Solve 3/4 r - 1/2 = 1_1/4 r + 8s for r

Relative Answer:

1. y=2, x=3
(Did that mentally, Guess &Check)

2. y=5, x=5

*** For the above two, you can solve them using system of equations. If they were a pair in a system you solve like this:

PICK ANY EQUATION:
3y+6x=24
y+3x=8
y= (-3)x +8

THEN, you substitute that entire thing for "Y" in the other equation, like this:

5=4y - 3x
5 = 4(-3x +8) - 3x
5 = -12x + 32 - 3x
5= 32- 15x
-27 = -15x
x = -15/-27

What is meaning of 1 molar solution?

2015-01-24T21:30:00.000+02:00

What does it mean when a solution is 1 molar in a substance?

micro-grams of substance per liter solution
milligrams of substance per gram solution
moles of substance per liter solution
milligrams of substance per liter solution

Correct Answer is (C)

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How to prepare a Solution from a stock of aluminum nitrate?

2015-01-24T13:30:00.000+02:00

You need to make 100. ml of a 0.0500 M solution of aluminum from a 2.00 M stock of aluminum nitrate. What do you do?

Add 5.00 ml of stock to a volumetric flask, and dilute to 100. ml with water.
None of the above.
Add .00250 ml of stock to a volumetric flask, and dilute to 100. ml with water.
Add 2.50 ml of stock to a volumetric flask, and dilute to 100. ml with water.
Add 7.50 ml of stock to a volumetric flask, and dilute to 100. ml with water.

Remember that your stock is 2 M
Question Explanation:
Recall that M1V1=M2V2 (e.g. big M is for molarity and 1 and 2 are the solution volumes before and after dilution)

Complete freshman course of analytical chemistry

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What is the molarity of sodium sulfate in solution of sodium sulfate?

2015-01-24T02:00:00.000+02:00

You have 25 mg of sodium sulfate in 3L of water. What is the molarity of sodium sulfate in this solution?

5.87E-05
5.87E-02
None of the above
1.72E-04
6.89E-05

Do not forget to Check the molecular weight of sodium sulfate (note: its 142 gm/mole)

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How many moles of nitrate in 6.00 moles of aluminum nitrate?

2015-01-23T21:30:00.000+02:00

If you have 6.00 moles of aluminum nitrate how many moles of nitrate do you have?

6.00
12.00
18.0
3.00
1.00

Remember that:

Aluminum is +3 and nitrate is -1 so there are three nitrates for every aluminum.

The Complete course of analytical chemistry

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How many moles of sodium phosphate are in 25 ml Solution?

2015-01-23T14:17:00.002+02:00

You have 25. ml of a solution of trisodium phosphate. After running an atomic analysis, you find a concentration of 10. ppm sodium. How many moles of sodium phosphate are in 25 ml of this solution? (note 1e-6 is equat to .000001 and that the molecular weight of trisodum phosphate is 164 gm/mole although what you really need to do this analysis is the atomic weight of sodium, which is 23 g/mole)

3.6e-6

3.6e-2

1.5e-4

6.8e-7

1.9e-5

remember that:
ppm is milligram per liter
Revise this online Analytical chemistry Course.

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Online Chemistry Course for Analytical Chemistry Instrumental Analysis by Prof Vicki Colvin

2015-01-23T11:53:00.002+02:00

Video (MP4) for Lecture 5: Last One! Semiconductor&Fuel Cell Sensors (14:11)Online

Chemistry Course for Analytical Chemistry and Instrumental Analysis by Professor Vicki Colvin
Represented by Rice University: Located on a 300-acre forested campus in Houston, Rice University is consistently ranked among the top 20 universities in the U.S. and the top 100 in the world. Rice has highly respected schools of Architecture, Business, Continuing Studies, Engineering, Humanities, Music, Natural Sciences and Social Sciences and is home to the Baker Institute for Public Policy.

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CONSTANTS, UNITS, AND CONVERSION FACTORS in Analytical Chemistry

Calculate the theoretical pH of trisodium fictate

calculate the concentration of calcium ion in a saturated solution

Calculate the change in pH of HCL after lactic acid and lactate addition

Calculate the 95% confidence interval of a gasoline standard sample

Write the charge and mass-balance equations for ferrous ammonium sulfate

calculate the relative uncertainty in the atomic mass of Os

Calculate The Molar Mass of a Substance with Density 2kg/dm3

Balancing The Chemical Equation NaHCO3 Na2CO3

What do Vm & Km stand for in Chemistry of enzymes?

What does Vm stands for in Chemistry (Stoichiometry)?

In the equation: n=V/Vm , What does the Vm stand for?

What Does dm^3 Stands For in Chemistry?

What Does mM Stand For in Chemistry?

How To Name Covalent Molecular Compounds?

How to Balance a Chemical Equation For a Beginners?

Understanding Conversion of Units and How To Use Conversion Factors

20 Physics Theory Examples Application in Real Everyday Life

Examples list for physics applications in Daily Life Activities

Hard 7th Grade Math Homework HELP

What is meaning of 1 molar solution?

How to prepare a Solution from a stock of aluminum nitrate?

What is the molarity of sodium sulfate in solution of sodium sulfate?

How many moles of nitrate in 6.00 moles of aluminum nitrate?

How many moles of sodium phosphate are in 25 ml Solution?

Online Chemistry Course for Analytical Chemistry Instrumental Analysis by Prof Vicki Colvin

Who needs this course?

For advanced undergraduate chemistry students which have a basic familiarity with freshman chemistry.