{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

This preview shows pages 1–2. Sign up to view the full content.

CHEM 1212 Review of Chem 1211 material: 1. How many moles are in 21.3 g of O 2 ? 21.3 g * (1 mol / 32 g) = 0.666 mol of O 2 2. Balance the chemical reaction: Al + O 2 Al 2 O 3 What is the mass of O 2 needed to react 72 g of Al? What is the mass of Al 2 O 3 formed? 4 Al + 3 O 2 2 Al 2 O 3 72 g Al * (1 mol Al / 26.98 g) * (3 mol O 2 / 4 mol Al) * (32 g / 1 mol O 2 ) = 64.05 g = 64 g of O 2 3. How much water is added to 12 mL of 1.0 M HCl to obtain a solution that is 0.04 M HCl? Initial number of moles = 0.012 L * 1.0 M = 0.012 moles Final conditions: 0.012 moles of HCl and a Molarity of 0.04 M Volume = moles / M Volume = 0.012 moles / 0.04 M = 0.3 L Final volume is 300 mL; so 288 mL of water must be added to the original 12 mL. 4. After dissolving 73 g of CaCl 2 in 55 L of water, what is the molarity of Cl - ions in solution? CaCl 2 Ca 2+ + 2 Cl - 73 g CaCl 2 * (1 mol CaCl 2 / 110.98 g) * (2 mol Cl - / 1 mol CaCl 2 ) = 1.3156 moles of Cl - Molarity of Cl - ions = 1.3156 moles / 55 L = 0.02392 M = 0.024 M 5. What is the expected shape of CCl 4 ? Is this molecule polar? The expected shape is tetrahedral: The molecules is non -polar, the pull of each Cl cancels each other out 6. Why is water polar?

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern