This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: toupal (rgt374) – Homework 01 – Chiu – (58295) 1 This printout should have 17 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points You have 2 . 4 kg of water. One mole of water has a mass of 18 . 1 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: − 1 . 27762 × 10 8 C. Explanation: Let : N A = 6 . 02214 × 10 23 molec / mol , q e = − 1 . 6 × 10 − 19 C / electron , m = 2 . 4 kg , M = 18 . 1 g / mol = 0 . 0181 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) 2 . 4 kg . 0181 kg / mol × ( 6 . 02214 × 10 23 molec / mol ) × ( − 1 . 6 × 10 − 19 C / electron ) = − 1 . 27762 × 10 8 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 5 . 4 g. Each mole (6 . 02 × 10 23 atoms) has a mass of about 57 . 9 g. Find the number of atoms in a nickel coin. Correct answer: 5 . 61451 × 10 22 atoms. Explanation: Let : N a = 6 . 02 × 10 23 atoms , M = 57 . 9 g , and m = 5 . 4 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 . 4 g) (57 . 9 g) (6 . 02 × 10 23 atoms) = 5 . 61451 × 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 . 57206 × 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 = ( 5 . 61451 × 10 22 atoms ) × (28 electrons / atom) = 1 . 57206 × 10 24 electrons . toupal (rgt374) – 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: 2 . 51872 × 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 . 57206 × 10 24 electrons) × ( − 1 . 60218 × 10 − 19 C / electron ) = − 2 . 51872 × 10 5 C , which has a magnitude of 2 . 51872 × 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....
View
Full
Document
This note was uploaded on 04/26/2009 for the course PHY 58310 taught by Professor Chiu during the Spring '09 term at University of Texas at Austin.
 Spring '09
 Chiu
 Physics, Work

Click to edit the document details