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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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Calculate the volume of a 0.753 M sodium hydroxide solution that is
neutralized by the
addition of 100 mL of a 1.203 M sulfuric acid solution.
This problem requires you to perform the following unit cancellation
calculation:
Problem :
If the volume of a 100 L solution of 1.1 moles of hydrogen in 6.0 moles
argon is suddenly doubled,
what happens to the mole fraction of hydrogen in that solution?
The mole fraction of a solute does not depend on the volume of the
solution. Mole fraction only
depends on the molar amount of each component. Because we did not add or
subtract any hydrogen
or argon, the mole fraction of hydrogen remains unchanged.
Problem :
What is the concentration of HCl stock solution that can create 250 mL of a
1.0 M solution
that is prepared by the dilution of 50 mL of the HCl stock solution?
By rearranging the expression c1 v1 = c2
v2
to solve for c1, we find the concentration of the HCl stock
solution: