{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

hw1 solutions - homework 01 – SMITH TAYLOR – Due 4:00...

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

View Full Document Right Arrow Icon

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

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: homework 01 – SMITH, TAYLOR – Due: Jan 20 2008, 4:00 am 1 Question 1, chap 22, sect 1. part 1 of 1 10 points You have 1 . 4 kg of water. One mole of water has a mass of 18 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: − 7 . 49421 × 10 7 C (tolerance ± 1 %). Explanation: Let : N A = 6 . 02214 × 10 23 molec / mol , q e = − 1 . 6 × 10 − 19 C / electron , M = 18 g / mol = 0 . 018 kg / mol , m = 1 . 4 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 . 4 kg . 018 kg / mol × (6 . 02214 × 10 23 molec / mol) × ( − 1 . 6 × 10 − 19 C / electron) = − 7 . 49421 × 10 7 C . Question 2, chap 22, sect 1. part 1 of 3 10 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 . 8 g. Each mole (6 . 02 × 10 23 atoms) has a mass of about 57 . 8 g. Find the number of atoms in a nickel coin. Correct answer: 4 . 99931 × 10 22 atoms (toler- ance ± 1 %). Explanation: Let : N a = 6 . 02 × 10 23 atoms , M = 57 . 8 g , and m = 4 . 8 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 . 8 g 57 . 8 g (6 . 02 × 10 23 atoms) = 4 . 99931 × 10 22 atoms . Question 3, chap 22, sect 1. part 2 of 3 10 points Find the number of electrons in the coin. Each nickel atom has 28 electrons / atom. Correct answer: 1 . 39981 × 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 = ( 4 . 99931 × 10 22 atoms ) × (28 electrons / atom) = 1 . 39981 × 10 24 electrons . homework 01 – SMITH, TAYLOR – 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: 224274 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 . 39981 × 10 24 electrons) × ( − 1 . 60218 × 10 − 19 C / electron ) = − 224274 C , which has a magnitude of 224274 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....
View Full Document

{[ snackBarMessage ]}

Page1 / 13

hw1 solutions - homework 01 – SMITH TAYLOR – Due 4:00...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon bookmark
Ask a homework question - tutors are online