This year course engages students in becoming skilled readers of prose written in a variety of periods, disciplines, and
rhetorical contexts and in becoming skilled writers who compose for a variety of purposes. More immediately, the course
prepares the students to perform satisfactorily on the A.P. Examination in Language and Composition given in the spring.
Both their writing and their reading should make students aware of the interactions among a writer’s purposes, audience
expectations, and subjects as well as the way generic conventions and the resources of language contribute to effectiveness
in writing. Students will learn and practice the expository, analytical, and argumentative writing that forms the basis of
academic and professional writing; they will learn to read complex texts with understanding and to write prose of
sufficient richness and complexity to communicate effectively with mature readers. Readings will be selected primarily,
but not exclusively, from American writers. Students who enroll in the class will take the AP examination.
Answer:
9.15×10²³molecules
Explanation:
moles=number of particles/Avogadro's number
1.52=x/6.02×10²³
by cross multiplication;
x=1.52×6.02×10²³
=9.15×10²³
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Find the [OH-] in the solution. The pH is 9.5, so the pOH is 14 - 9.5 = 4.5.
[OH-] = 10^-4.5 M
Now use the dilution equation to find the new [OH-] after the volume is reduced from 150 mL to 50 mL:
M1V1 = M2V2
M1 = 10^-4.5 M
V1 = 150 mL
M2 = ?
V2 = 50 mL
(10^-4.5)(150) = M2(50)
M2 = 9.5 x 10^-5 M ≈ 1 • 10^-4 (We can only use one sig fig, because the pH was given to one decimal place.)
Now use this [OH-] to find pOH:
pOH = -log(1 x 10^-4) = 4.0
14 - pOH = pH, so the expected pH for the new solution is 10.
