Week 2 Chemistry Worksheet

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Northeastern University *

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131

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Chemistry

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Dec 6, 2023

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College Chem I CHE 131 Week 2 Worksheet Packet Density – show your work and remember to round your answers to the correct number of significant figures – use additional paper if necessary 1. Indicate how many significant figures there are in each of the following measured values: 1.008 ___ 4 ___ 7,000,000 ___ 1 ___ 107.854 ___ 6 ___ 0.00340 ___ 3 ___ 100.30 ___ 5 ___ 320,001 ___ 6 ___ 0.768 ___ 3 ___ 0.00001 ___ 1 ___ 2. Calculate the answers to the appropriate number of significant figures: 32.567 246.24 859.0 135.0 238.278 23.5478 + 1.4567 + 98.3 + 1345.29 169.0 582.8 2227.8 3. Calculate the answers to the appropriate number of significant figures: a) 23.7 x 3.0 = 71.1 = 71 (2 sig figs) b) 45.76 x 0.25 = 11.44 = 11 (2 sig figs) c) 81.04 x 0.010 = .8104 = .81 (2 sig figs) d) 1.678 / 0.42 = 3.995238095 = 4.0 (2 sig figs) e) 4278 / 0.604 = 7082.781457 = 7080 (3 sig figs) f) 6.47 / 60.50 = .1069421488 = .107 (3 sig figs) 1. Calculate the density of bromine is 586 g occupies 188 mL. D = m/v; m = 586 g; v = 188 mL D = 586 / 188 = 3.117021277 = 3.12 g/mL (3 sig figs) 2. Mercury has a density of 13.6 g/mL. What is the mass of 5.50 mL of mercury? v x D = m; D = 13.6 g/mL; v = 5.50 mL (5.50)(13.6) = m = 74.8 g (3 sig figs)

3. Platinum has a density of 21.5 g/cm 3 . What is the volume of a 120 g piece of platinum? v = m/D; D = 21.5 g/cm 3 ; m = 120 g v = 21.5 / 120 = 5.581395349 = 5.6 cm 3 (2 sig figs) 4. Ethanol has a density of 0.798 g/mL. What is the mass of 0.75 L of ethanol? v x D = m; D = 0.798 g/mL; v = 0.75 L 0.75 L x 1000 mL / 1 L v = 750 mL (750)(0.798) = m = 598.5 = 600 g (2 sig figs) 5. Lead has a density of 11.3 g/cm 3 . What is the volume of a 0.082 kg piece of lead? v = m/D; D = 11.3 g/cm 3 ; m = 0.082 kg 0.082 kg x 1000 g / 1 kg m = 82 g v = 82 / 11.3 = 7.256637168 = 7.3 cm 3 (2 sig figs) 6. Gold has a density of 19.3 g/cm 3 . What is the mass in kg of a gold bar that measures 22.5 cm x 8.70 cm x 3.60 cm? v x D = m; D = 19.3 g/cm 3 ; v = 22.5 cm x 8.70 cm x 3.60 cm v = 22.5 cm x 8.70 cm x 3.60 cm = 704.7 cm 3 = 705 cm 3 (3 sig figs) (705)(19.3) = m = 13606.5 g = 13600 g (3 sig figs) 13600 g / 1000 = 13.6 kg Specific Heat – show your work and remember to round your answers to the correct number of significant figures – use additional paper if necessary. 1. A 860 g gold bar cools from 94°C to -5°C. Calculate the amount of heat released. The specific heat of gold is 0.0923 cal/g°C. q = (specific heat)(mass)(temp change) q = (0.0923 cal/g°C )(860 g )(99°C ) q = 7858.422 q = 7900 cal (2 sig figs)

2. A 150 g sample of ethanol absorbs 5650 calories of energy as it heats from 4°C to 69°C. Calculate the specific heat of ethanol. 5650 = (specific heat)(150 g)(65°C) 5650 / ((150 g)(65°C)) = specific heat specific heat = 0.579487195 = 0.58 cal/g°C (2 sig figs) 3. The specific heat of sand is 0.33 cal/g°C. How much energy is required to heat 3.50 kg of sand from 18°C to 39°C? q = (0.33 cal/g°C)(3.50 kg)(21°C) 3.50 kg x 1000 g / 1 kg = 3500 g q = (0.33 cal/g°C )(3500 g )(21°C ) q = 24255 = 24,000 cal (2 sig figs) 4. An unknown metal sample with a mass of 9560 mg absorbs 25.2 calories of energy as it is heated from 21°C to 98°C. What is the specific heat of the metal? 25.2 cal = (specific heat)(9560 mg)(77°C) 9560 mg x 1 g / 1000 mg = 9.56 g 25.2 cal = (specific heat)(9.56 g)(77°C) specific heat = 25.2 / ((9.56)(77°C)) specific heat = 0.0342335489 = 0.03 cal/g°C (2 sig figs) 5. A 3.10 kg piece of wood absorbs 4.6 x 10 4 calories. What is the final temperature of the wood if it starts at 22°C? The specific heat of wood is 0.42 cal/g°C. 4.6 x 10 4 cal = (0.42 cal/g°C)(3.10 kg)(temp change) 3.10 kg x 1000 g / 1 kg = 3100 g 46000 cal = (0.42 cal/g°C)(3100 g)(temp change) 46000 cal / ((0.42 cal/g°C)(3100 g)) = temp change temp change = 35.33026114 = 35°C (2 sig figs) FINAL TEMPERATURE = 35 + 22 = 57°C Atomic Structure 1. Complete the following chart:

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Particle Location Charge Relative Mass Proton nucleus +1 large Neutron nucleus 0 large Electron orbitals -1 small 2. Complete the following chart: Chemical Symbol Atomic Number Mass Number # Protons # Electrons # Neutrons 8 O 18 8 18 8 8 10 12 Mg 26 12 26 12 12 14 27 Co 60 27 60 27 27 33 3. What are isotopes? Atoms of the same element that have different numbers of neutrons. 4. What is the difference between the mass number and the average atomic mass of an element? Which one appears on the periodic table? The mass of an element is the sum of the number of protons and neutrons in the nucleus. The average atomic mass of an element is an average of an element’s atomic masses, weighted by the natural abundance of each isotope of that element. It is a weighted average because different isotopes have different masses. The atomic number appears on the periodic table. 5. Chlorine has two naturally occurring isotopes. Calculate the average atomic mass of chlorine given the following isotopic data: 35.45 amu (see attached page for work) Mass of Isotope % Abundance 34.97 amu 75.53% 36.97 amu 24.47% 6. Boron has two naturally occurring isotopes. Calculate the average atomic of boron given the following isotopic data: 10.8 amu (see attached page for work) Mass of Isotope % Abundance 10.013 amu 19.6% 11.009 amu 80.4%