# hw1 - homework 01 LIN, KEVIN Due: Jan 20 2008, 4:00 am...

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homework 01 – LIN, KEVIN – Due: Jan 20 2008, 4:00 am 1 Question 1, chap 22, sect 1. part 1 of 1 10 points You have 1 . 6 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: 8 . 61266 × 10 7 C (tolerance ± 1 %). Explanation: Let : N A = 6 . 02214 × 10 23 molec / mol , q e = 1 . 6 × 10 19 C / electron , M = 17 . 9 g / mol = 0 . 0179 kg / mol , m = 1 . 6 kg , 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 mN A q e M = (10 electrons / molec) 1 . 6 kg 0 . 0179 kg / mol × (6 . 02214 × 10 23 molec / mol) × ( 1 . 6 × 10 19 C / electron) = 8 . 61266 × 10 7 C . Question 2, chap 22, sect 1. part 1 of 3 10 points We want to Fnd how much charge is on the electrons in a nickel coin. ±ollow this method. A nickel coin has a mass of about 5 g. Each mole (6 . 02 × 10 23 atoms) has a mass of about 58 . 2 g. ±ind the number of atoms in a nickel coin. Correct answer: 5 . 17182 × 10 22 atoms (toler- ance ± 1 %). Explanation: Let : N a = 6 . 02 × 10 23 atoms , M = 58 . 2 g , and m = 5 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 = 5 g 58 . 2 g (6 . 02 × 10 23 atoms) = 5 . 17182 × 10 22 atoms . Question 3, chap 22, sect 1. part 2 of 3 10 points ±ind the number of electrons in the coin. Each nickel atom has 28 electrons / atom. Correct answer: 1 . 44811 × 10 24 electrons (tol- erance ± 1 %). 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 = ( 5 . 17182 × 10 22 atoms ) × (28 electrons / atom) = 1 . 44811 × 10 24 electrons .

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homework 01 – LIN, KEVIN – Due: Jan 20 2008, 4:00 am 2 Question 4, chap 22, sect 1. part 3 of 3 10 points Find the magnitude of the charge of all these electrons. Correct answer: 232013 C (tolerance ± 1 %). 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 . 44811 × 10 24 electrons) × ( 1 . 60218 × 10 19 C / electron ) = 232013 C , which has a magnitude of 232013 C . Question 5, chap 22, sect 2. part 1 of 1 10 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 03/03/2011 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas.

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hw1 - homework 01 LIN, KEVIN Due: Jan 20 2008, 4:00 am...

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