{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Homework_01-solutions - toupal(rgt374 Homework 01...

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

View Full Document Right Arrow Icon
toupal (rgt374) – 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 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 0 . 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 .
Image of page 1

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

View Full Document Right Arrow Icon
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. What is the new force F , if charge 1 is increased to q 1 = 5 q , charge 2 is decreased to q 2 = q 2 / 2, and the distance is decreased to d = d/ 2? Choose one 1. F = 5 / 2 F 2. F = 100 F 3. F = 20 F 4. F = 50 F 5. F = 5 F 6. F = 25 / 2 F 7. F = 5 / 4 F 8. F = 25 / 4 F 9. F = 25 F 10. F = 10 F correct Explanation: F = k q 1 q 2 r 2 = k (5 q 1 ) ( q 2 2 ) ( d 2 ) 2 = 10 k q 1 q 2 d 2 = 10 F F = 10 F .
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