{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Homework 1-solutions

# Homework 1-solutions - kramer(jrk2286 Homework 1...

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

kramer (jrk2286) – Homework 1 – sutcli ff e – (90830) 1 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Which of the following describes a chemical change? 1. gasoline burns in an engine correct 2. alcohol evaporates 3. water is heated from 0 C to 50 C 4. wood is carved Explanation: A chemical change occurs when one or more substances are consumed (at least partially), and one or more new substances are formed. 002 10.0 points What is the volume of a sample of helium that has a mass of 2 . 5 × 10 - 3 g, and a density of 0 . 17847 g / L? Correct answer: 14 . 008 mL. Explanation: m = 2 . 5 × 10 - 3 g d = 0 . 17847 g / L d = m V V = m d = 2 . 5 × 10 - 3 g 0 . 17847 g / L × 1000 mL L = 14 . 008 mL 003 10.0 points What is the nuclear composition of an atom of iodine-131? 1. 131 protons 2. 53 protons, 53 electrons 3. 78 protons, 53 neutrons 4. 53 protons, 78 neutrons correct Explanation: p + = 53 For A Z X , A = mass number = # p + + # n 0 and Z = atomic number = # p + . # n 0 = A - # p + = 131 - 53 = 78 004 10.0 points How many hydrogen atoms appear in the for- mula of ammonium sulfate? 1. 6 atoms 2. 2 atoms 3. 4 atoms 4. 10 atoms 5. 8 atoms correct Explanation: The ammonium ion is NH + 4 and the sulfate ion is SO 2 4 - . Two NH + 4 are needed to balance the charge on each SO 2 - 4 . The formula is (NH 4 ) 2 SO 4 . 005 10.0 points Choose the pair of names and formulae that do not match. 1. SCl 4 : sulfur tetrachloride 2. SnCl 4 : tin(V) chloride correct 3. MgSO 4 : magnesium sulfate 4. N 2 O 3 : dinitrogen trioxide 5. KNO 3 : potassium nitrate Explanation: 006 10.0 points Which of the following is the molecular for- mula for heptane?

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

View Full Document
kramer (jrk2286) – Homework 1 – sutcli ff e – (90830) 2 1. C 7 H 16 correct 2. C 6 H 14 3. C 7 H 14 4. C 7 H 12 5. C 6 H 12 Explanation: 007 10.0 points Calculate the energy of a photon of light with a wavelength of 0.57 km. 1. 3 . 5 × 10 - 25 J 2. 3 . 5 × 10 - 23 J 3. 3 . 8 × 10 - 31 J 4. 5 . 3 × 10 8 J 5. 3 . 5 × 10 - 22 J 6. 5 . 3 × 10 11 J 7. 3 . 8 × 10 - 34 J 8. 3 . 5 × 10 - 28 J correct 9. 530000 J Explanation: λ = 0.57 km Here we use the equation E = h c λ where h is Planck’s constant (6.6262 × 10 - 34 J · s) and c is the speed of light (3.00 × 10 8 m/s). Here we’ll also need to convert km into meters (recall that 1 km = 1 × 10 3 m): E = (6 . 6262 × 10 - 34 J · s)(3 . 00 × 10 8 m / s) 0 . 57 km 1 . 00 × 10 3 m 1 km = 3 . 48747 × 10 - 28 J 008 10.0 points We conduct an experiment by shining 500 nm light on potassium metal. This causes electrons to be emitted from the surface via the photoelectric e ff ect. Now we change our source light to 450 nm at the same intensity level. Which of the following is the result from the 450 nm light source compared to the 500 nm source? 1. The same number of electrons would be emitted, but they would have a lower velocity 2. More electrons would be emitted from the surface. 3. No electrons would be emitted from the surface. 4. The same number of electrons would be emitted, but they would have a higher velocity correct 5. Fewer electrons would be emitted from the surface.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• 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.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• 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.

Dana University of Pennsylvania ‘17, Course Hero Intern

• 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.

Jill Tulane University ‘16, Course Hero Intern