practiceexam12 - City College Chemistry Department...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
1 City College, Chemistry Department Chemistry 10301, sections T and T2, Prof. T. Lazaridis Final exam, Dec 22, 2005 Name (last name first): _____________________________________________ I.D. Number: ______________________________________________________ Workshop leader: __________________________________________________ Note: There are 17 questions in this exam. Fill in your answer in the blank space provided immediately following each question. 1/2 point will be subtracted every time you report a numerical result with an incorrect number of significant figures. A Data Sheet with useful information is at the end. 1. (4) Write the names of the elements below next to their atomic symbols: Ti Titanium Mg Magnesium Al Aluminum Kr Krypton 2. (4) Write the molecular formula next to the names of the following compounds: Magnesium Sulfate MgSO 4 Sodium Carbonate Na 2 CO 3 Barium Hydroxide Ba(OH) 2 Potassium Nitrate KNO 3 3. (8) Balance the following chemical equations: 2 C 6 H 6 + 15 O 2 12 CO 2 + 6 H 2 O 3 CaCl 2 + 2 Na 3 PO 4 Ca 3 (PO 4 ) 2 + 6 NaCl 2 KClO 3 2 KCl + 3 O 2 4 FeO + O 2 2 Fe 2 O 3
Image of page 1

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

View Full Document Right Arrow Icon
2 4. (8) How many grams of phosphoric acid will form when 30.0 g of P 4 O 10 is mixed with 75.0 g of water? The reaction is P 4 O 10 + 6H 2 O 4H 3 PO 4 Molar masses: P 4 O 10 : 4X30.97 + 10X16.00 = 283.88 g/mol H 2 O : 2X1.008 + 16.00 = 18.02 g/mol H 3 PO 4 : 30.97 + 4X16.00 + 3X1.008 = 97.99 g/mol P 4 O 10 : 30.0 / 283.88 = 0.106 mol H 2 O : 75.0 / 18.02 = 4.16 mol 4.16/0.106 = 39.24 >> 6 therefore P 4 O 10 is limiting 0.106 moles P 4 O 10 will give 4 X 0.106 = 0.424 mol H 3 PO 4 X 97.99 g/mol = 41.5 g H 3 PO 4 5. (5) It was found that 35.45 mL of a HCl solution was needed to dissolve 1.110 g of pure CaCO
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    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.

    Student Picture

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

  • Left Quote Icon

    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.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    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.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern