HW1 - nguyen (jn9225) H01: Fundamentals mccord (50950) This...

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nguyen (jn9225) – H01: Fundamentals – mccord – (50950) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – fnd all choices be±ore answering. 001 10.0 points Which compound has the wrong chemical ±or- mula? 1. Mg(OH) 2 2. CaOH correct 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 o± each Ca 2+ , so the ±ormula is Ca(OH) 2 . The magnesium ion is Mg 2+ ; the hydroxide ion is OH . Two OH are needed to balance the charge o± each Mg 2+ , so the ±ormula 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 o± every three Ba 2+ . (This gives a total anion charge o± - 6 and a total cation charge o± +6.) The ±ormula is Ba 3 (PO 4 ) 2 . The ammonium ion is NH + 4 ; the sul±ate ion is SO 2 4 . Two NH + 4 are needed to balance the charge o± each SO 2 4 , so the ±ormula is (NH 4 ) 2 SO 4 . 002 10.0 points Which one has the greatest number o± atoms? 1. All have the same number o± atoms 2. 3.05 moles o± CH 4 correct 3. 3.05 moles o± water 4. 3.05 moles o± helium 5. 3.05 moles o± argon Explanation: For 3.05 moles o± 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 o± 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 o± helium: ? atoms = 3 . 05 mol He × 6 . 02 × 10 23 atoms 1 mol = 1 . 84 × 10 24 atoms For 3.5 moles o± argon: ? atoms = 3 . 05 mol Ar × 6 . 02 × 10 23 atoms 1 mol = 1 . 84 × 10 24 atoms 003 10.0 points I± 100.0 grams o± copper (Cu) completely re- acts with 25.0 grams o± oxygen, how much copper(II) oxide (CuO) will ±orm ±rom 140.0 grams o± copper and excess oxygen? ( Note : CuO is the only product o± this reaction.) 1. 160.0 g 2. 150.0 g 3. 35.0 g 4. 175.0 g correct 5. 200.0 g Explanation: m Cu , ini = 100.0 g m O 2 = 25.0 g m Cu , fn = 140.0 g I± 100 g copper and 25 g oxygen react com- pletely with each other, there must be 125 g o± product ±ormed (law o± conservation o± mass). This product is CuO.
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nguyen (jn9225) – H01: Fundamentals – mccord – (50950) 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 ±nd the mass of CuO that could be formed from 140 g of Cu and excess oxygen. We set our known ratio (100 g Cu : 125 g
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HW1 - nguyen (jn9225) H01: Fundamentals mccord (50950) This...

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