University:
Arizona State University
Course:
CHM 101 | Introductory Chemistry
Academic year:

2019
Views:

354

Pages:

3
Author:

kundgetanufv

Problem Set 1. Classify each substance as an element, compound, homogeneous mixture or heterogeneous mixture. pecan pie heterogeneous mix Scope mouthwash homogeneous mix Ti element K2 O compound 2. Classify each property as a chemical property or physical property. density phys combustibility chem melting point phys reactivity chem 3. Classify each change as a chemical change or physical change. phys burning chem melting tarnishing chem condensing phys 4. Convert 52 oC to oF. o F = oC x 1.8 + 32 (BTW, 1.8 and 32 are exact numbers) o F = 52 x 1.8 +32 = 130 oC, 2 sig figs (due to 52) 5. What is the mass of aluminum whose volume is 7.9 mL? The density of aluminum is 2.8 g/mL. D = m/V 2.8 g/mL = m/7.9 mL m = 2.8 g/mL x 7.9 mL = 22 g, 2 sig figs (due to both 2.8, 7.9) 6. Convert 56 kg into pounds. 56 kg x (2.20 lbs/1 kg) = 120 lbs, 2 sig figs (due to 56) 7. Convert 39 in into meters. 39 in x (2.54 cm/1 in) x (1 m/100 cm) = 0.99 m, 2 sig figs (due to 39) 8. Convert 1,110 in/s to miles per hour. 1,110 in/s x (60 s/1 min) x (60 min/1 hr) x (1 ft/12 in) x (1 mi/5,280 ft) = 63.1 mph, 3 sig figs (due to 1,110) 9. At what temperature are Celsius and Fahrenheit equal? Start with an expression that relates Celsius to Fahrenheit. o F = oC x 1.8 + 32 If Celsium and Fahrenheit are equal, we can substitute oC for oF. o C = oC x 1.8 + 32 o C = 1.8oC + 32 -0.8 oC = 32 o C = -40 You can also start with oC = (oF – 32)/1.8. You get the same answer. 10. A box contains a spool of copper wire and a lead fishing sinker. The total volume of both metals is 427 mL. The total mass is 4,360 g. What is the mass of each metal in the box? The density of copper is 8.96 g/mL. The density of lead is 11.34 g/mL. There are many ways to solve this problem. I’ll show you two ways. mCu = mass of copper, mPb = mass of lead, VCu = volume of copper, VPb = volume of lead mCu + mPb = 4,360 g VCu + VPb = 427 mL #1. I will rewrite volume in terms of mass. Mass is related to volume by density. D = m/V so V = m/D. VCu + VPb = 427 mL mCu/DCu + mPb/DPb = 427 mL mCu = 4,360 g - mPb (4,360 g – mPb)/DCu + mPb/DPb = 427 mL (4,360 g – mPb)/8.96 g/mL + mPb/11.34 g/mL = 427 mL. Multiply both sides by 8.96 g/mL. 4,360 g – mPb + 8.96 mPb/11.34 = 3,830 g. Multiply both sides by 11.34. 49,400 g – 11.34 mPb + 8.96 mPb = 43,400 g -2.38 mPb = - 6,000 g (2 sig figs) mPb = 2,500 g, 2 sig figs mCu = 1,900 g #2. Or, solve by rewriting mass in terms of volume. DCuVCu + DPbVPb = 4,360 g If VCu + VPb = 427 mL, then VCu = 427 mL - VPb DCu(427 mL - VPb) + DPbVPb = 4,360 g 8.96 g/mL(427 mL – VPb) + 11.34 g/mL(VPb) = 4,360 g 3,830 g - 8.96 g/mL(VPb) + 11.34 g/mL(VPb) = 4,360 g 2.38 g/mL(VPb) = 530 g VPb = 530 g/2.38 g/mL = 220 mL, 2 sig figs If VCu + VPb = 427 mL, then VCu = 427 mL – 220 mL = 210 mL, 2 sig figs mCu = DCuVCu = 8.96 g/mL(210 mL) = 1,900 g, 2 sig figs mPb = DPbVPb = 11.34 g/mL(220 mL) = 2,500 g, 2 sig figs