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No Fear Translations of Shakespeare’s plays (along with audio!) and other classic works
Flashcards
Mastery Quizzes
Infographics
Graphic Novels
AP® Test Prep PLUS
AP® Practice & Lessons
My PLUS Activity
Note-taking
Bookmarking
Dashboard
Testimonials from SparkNotes
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No Fear
provides access to Shakespeare for students who normally couldn’t (or wouldn’t) read his plays.
It’s also a very useful tool when trying to explain Shakespeare’s wordplay!
Erika M.
I
tutor high school students in a variety of subjects. Having access to the literature
translations helps me to stay informed about the various assignments. Your summaries and
translations are invaluable.
Kathy B.
Teaching Shakespeare to today's generation can be challenging. No Fear helps a ton with
understanding the crux of the text.
Kay
H.
Testimonials from SparkNotes Customers
No Fear provides access to Shakespeare for students who normally couldn’t (or wouldn’t) read his plays. It’s also a very useful tool when trying to explain Shakespeare’s wordplay!
Erika M.
I tutor high school students in a variety of subjects. Having access to the literature translations helps me to stay informed about the various assignments. Your summaries and translations are invaluable.
Kathy B.
Teaching Shakespeare to today's generation can be challenging. No Fear helps a ton with understanding the crux of the text.
Kay H.
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To find the normality, we realize that HAc is a monoprotic acid, so the
normality equals the molarity.
So the solution is 0.904 M in HAc.
To calculate the molality of the solution, we find the number of moles of
acetic acid per kilogram of
solvent. Note that we divide by the mass of the solvent and not by
the mass of the solution.
To calculate the mass percent of acetic acid in water we divide the mass of
acetic acid, 14.1 g, by the
total mass of solution, 264.1 g, and multiply by 100%. The solution is
5.34% acetic acid by mass.
The final concentration calculation is to find the mole fraction of acetic
acid in the solution. To do so
we find the number of moles of acetic acid, then divide that by the total
number of moles in solution:
Dilution
There are two common ways to prepare an aqueous solution. The first is to
weigh out a known mass
of solute, then add it to a given amount of solvent to achieve the desired
concentration. The other
method involves the dilution of a concentrated stock solution with more
solvent to achieve a
solution with a lower concentration than the original solution. To
calculate the concentration of the
diluted solution we will use the following formula:
Figure %: The Relationship between Concentration and Volume for Dilution
Problems
Suppose we wished to make 350 mL of a 0.15 M solution of sodium
sulfate by diluting
some 1.2 M sodium sulfate stock solution. To calculate the volume
of stock solution
necessary, we can solve the for
v1:
There are several variants on the dilution problem such as asking for the
volume of solution at a given
concentration produced by diluting a known volume and concentration of a
stock solution. All of
these problems are readily solved by rearranging the concentration-volume
equation then plugging in
the known information.