H01- Fundamentals-solutions

H01- Fundamentals-solutions - moczulski(ksm935 H01...

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

View Full Document Right Arrow Icon
moczulski (ksm935) – H01: Fundamentals – mccord – (50960) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points Which compound has the wrong chemical for- mula? 1. CaOH correct 2. Mg(OH) 2 3. Ba 3 (PO 4 ) 2 4. (NH 4 ) 2 SO 4 Explanation: The calcium ion is Ca 2+ ; the hydroxide ion is OH . Two OH are needed to balance the charge of each Ca 2+ , so the formula is Ca(OH) 2 . The magnesium ion is Mg 2+ ; the hydroxide ion is OH . Two OH are needed to balance the charge of each Mg 2+ , so the formula is Mg(OH) 2 . The barium ion is Ba 2+ ; the phosphate ion is PO 3 4 . Two PO 3 4 are needed to balance the charge of every three Ba 2+ . (This gives a total anion charge of - 6 and a total cation charge of +6.) The formula is Ba 3 (PO 4 ) 2 . The ammonium ion is NH + 4 ; the sulfate ion is SO 2 4 . Two NH + 4 are needed to balance the charge of each SO 2 4 , so the formula is (NH 4 ) 2 SO 4 . 002 10.0points Which one has the greatest number of atoms? 1. All have the same number of atoms 2. 3.05 moles of argon 3. 3.05 moles of CH 4 correct 4. 3.05 moles of water 5. 3.05 moles of helium Explanation: For 3.05 moles of water: ? atoms = 3 . 05 mol H 2 O × 6 . 02 × 10 23 molec 1 mol × 3 atoms 1 molecule = 5 . 51 × 10 24 atoms For 3.05 moles of CH 4 : ? atoms = 3 . 05 mol CH 4 × 6 . 02 × 10 23 molec 1 mol × 5 atoms 1 molecule = 9 . 18 × 10 24 atoms For 3.05 moles of helium: ? atoms = 3 . 05 mol He × 6 . 02 × 10 23 atoms 1 mol = 1 . 84 × 10 24 atoms For 3.5 moles of argon: ? atoms = 3 . 05 mol Ar × 6 . 02 × 10 23 atoms 1 mol = 1 . 84 × 10 24 atoms 003 10.0points If 100.0 grams of copper (Cu) completely re- acts with 25.0 grams of oxygen, how much copper(II) oxide (CuO) will form from 140.0 grams of copper and excess oxygen? ( Note : CuO is the only product of this reaction.) 1. 150.0 g 2. 35.0 g 3. 175.0 g correct 4. 160.0 g 5. 200.0 g Explanation: m Cu , ini = 100.0 g m O 2 = 25.0 g m Cu , fin = 140.0 g If 100 g copper and 25 g oxygen react com- pletely with each other, there must be 125 g of product formed (law of conservation of mass). This product is CuO.
Image of page 1

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

View Full Document Right Arrow Icon
moczulski (ksm935) – H01: Fundamentals – mccord – (50960) 2 Now we have a ratio: for every 100 g of Cu reacted, 125 g of CuO will be produced (assuming there is enough oxygen). We use this ratio to find the mass of CuO that could be formed from 140 g of Cu and excess oxygen.
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