PHY303LHW1S

# PHY303LHW1S - rosen (arr956) – Homework 01 – Chiu –...

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Unformatted text preview: rosen (arr956) – Homework 01 – Chiu – (58295) 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points You have 0 . 8 kg of water. One mole of water has a mass of 17 . 9 g / mol and each molecule of water contains 10 electrons since water is H 2 O. What is the total electron charge contained in this volume of water? Correct answer: − 4 . 30633 × 10 7 C. Explanation: Let : N A = 6 . 02214 × 10 23 molec / mol , q e = − 1 . 6 × 10 − 19 C / electron , m = 0 . 8 kg , M = 17 . 9 g / mol = 0 . 0179 kg / mol , and Z = 10 electrons / molec . The mass is proportional to the number of molecules, so for m grams in n molecules and M grams in N A molecules, m M = n N A n = m M N A Since 10 electrons are in each molecule of water, then the total number of electrons n e in the coin is n e = Z n = Z m M N A and the total charge q for the n e electrons is q = n e q e = Z m N A q e M = (10 electrons / molec) . 8 kg . 0179 kg / mol × ( 6 . 02214 × 10 23 molec / mol ) × ( − 1 . 6 × 10 − 19 C / electron ) = − 4 . 30633 × 10 7 C . 002 (part 1 of 3) 10.0 points We want to find how much charge is on the electrons in a nickel coin. Follow this method. A nickel coin has a mass of about 4 . 2 g. Each mole (6 . 02 × 10 23 atoms) has a mass of about 58 . 2 g. Find the number of atoms in a nickel coin. Correct answer: 4 . 34433 × 10 22 atoms. Explanation: Let : N a = 6 . 02 × 10 23 atoms , M = 58 . 2 g , and m = 4 . 2 g . Mass is proportional to the number of atoms in a substance, so for m grams in N atoms in the nickel coin and M grams in N a atoms in one mole, we have m M = N N a N = m M N a = (4 . 2 g) (58 . 2 g) (6 . 02 × 10 23 atoms) = 4 . 34433 × 10 22 atoms . 003 (part 2 of 3) 10.0 points Find the number of electrons in the coin. Each nickel atom has 28 electrons / atom. Correct answer: 1 . 21641 × 10 24 electrons. Explanation: Let : n Ni = 28 electrons / atom . If n Ni electrons are in each Nickel atom, then the total number of electrons n e in the coin is n e = N n Ni = ( 4 . 34433 × 10 22 atoms ) × (28 electrons / atom) = 1 . 21641 × 10 24 electrons . rosen (arr956) – Homework 01 – Chiu – (58295) 2 004 (part 3 of 3) 10.0 points Find the magnitude of the charge of all these electrons. Correct answer: 1 . 94891 × 10 5 C. Explanation: Let : q e = − 1 . 60218 × 10 − 19 C / electron . The total charge q for the n e electrons is q = n e q e = (1 . 21641 × 10 24 electrons) × ( − 1 . 60218 × 10 − 19 C / electron ) = − 1 . 94891 × 10 5 C , which has a magnitude of 1 . 94891 × 10 5 C . 005 10.0 points Two charges q 1 and q 2 are separated by a distance d and exert a force F on each other....
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## This note was uploaded on 04/13/2009 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas.

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PHY303LHW1S - rosen (arr956) – Homework 01 – Chiu –...

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