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: The molar mass of each gas
Explanation:
Mole fraction is the ratio of moles of that component to the total moles of solution. Moles of solute is the ratio of given mass to the molar mass.

Suppose if there are three gases A, B and C.
a) 
b) 
c) 
moles of solute =
Thus if mass of each gas is known , we must know the molar mass of each gas to know the moles of each gas.
I think it is correct......
Answer:
100 mL
Explanation:
The reaction that takes place is:
- CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂
First we <u>convert 500 mg of CaCO₃ into mmoles</u>, using its <em>molar mass</em>:
- 500 mg ÷ 100 mg/mmol = 5 mmol CaCO₃
Then we <u>convert 5 mmoles of CaCO₃ into HCl mmoles</u>, using the <em>stoichiometric coefficients of the balanced reaction</em>:
- 5 mmol CaCO₃ *
= 10 mmol HCl
Finally we <u>calculate the volume of a 0.10 M HCl solution (such as stomach acid) that would contain 10 mmoles</u>:
- 10 mmol / 0.10 M = 100 mL