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# solution_pdf - beck(kab3335 – HW#1 – Antoniewicz...

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Unformatted text preview: beck (kab3335) – HW #1 – Antoniewicz – (57420) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Assuming that 67 . 9% of the Earth’s surface is covered with water at average depth of 1 . 13 mi, estimate the mass of the water on Earth. One mile is approximately 1.609 km and the radius of the earth is 6 . 37 × 10 6 m. Correct answer: 6 . 29496 × 10 20 kg. Explanation: Let : R = 0 . 679 , h = 1 . 13 mi , ρ = 1000 kg / m 3 , and R e = 6 . 37 × 10 6 m . If the area of Earth covered by water is . 679(4 π R 2 e ), a good approximation of the volume of water covering the earth is V = 0 . 679(4 π R 2 e ) h, and the mass is m = ρ V = 0 . 679 ρ 4 π R 2 e h = 0 . 679 (1000 kg / m 3 ) 4 π (6 . 37 × 10 6 m) 2 × (1 . 13 mi) 1000 m 1 km 1 . 609 km 1 mi = 6 . 29496 × 10 20 kg . 002 (part 1 of 2) 5.0 points A structural I beam is made of iron. A view of its cross-section and dimensions is shown. 22 cm 25 cm 3 cm 3 cm What is the mass of a section 1 . 3 m long? The density of iron is 7560 kg / m 3 , the atomic weight of iron is 55 . 85 g / mol and Avogadro’s number is 6 . 02214 × 10 23 atoms / mol. Correct answer: 185 . 749 kg. Explanation: Let : d = 3 cm = 0 . 03 m , w = 22 cm = 0 . 22 m , h = 25 cm = 0 . 25 m , ℓ = 1 . 3 m , and ρ = 7560 kg / m 3 . w h d d The cross-sectional area of the beam is A = 2 w · d + ( h- 2 d ) d = 2(0 . 22 m) (0 . 03 m) + [0 . 25 m- 2 (0 . 03 m)] (0 . 03 m) = 0 . 0189 m 2 , so the volume of the beam is V = A ℓ and the mass is m = ρ V = ρ A ℓ = (7560 kg / m 3 )(0 . 0189 m 2 )(1 . 3 m) = 185 . 749 kg . 003 (part 2 of 2) 5.0 points How many atoms are there in this section? Correct answer: 2 . 00288 × 10 27 atoms. Explanation: Let : M = 55 . 85 g / mol and N A = 6 . 02214 × 10 23 atoms / mol . beck (kab3335) – HW #1 – Antoniewicz – (57420) 2 The number of moles is n = m M , so the num- ber of atoms is N = n N A = m M N A = 185 . 749 kg 55 . 85 g / mol 1000 g 1 kg × (6 . 02214 × 10 23 atoms / mol) = 2 . 00288 × 10 27 atoms ....
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solution_pdf - beck(kab3335 – HW#1 – Antoniewicz...

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