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Full Answers to Questions - CHAPTER 1 MATTERITS PROPERTIES...

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1 CHAPTER 1 MATTER—ITS PROPERTIES AND MEASUREMENT PRACTICE EXAMPLES 1A Convert the Fahrenheit temperature to Celsius and compare.  5C 9F C = F3 2F =3 5 0F3 = 1 7 7C   . 1B We convert the Fahrenheit temperature to Celsius.   C 2 F = 1 5F3 2 F =2 6 C . The antifreeze only protects to 22 C and thus it will not offer protection to temperatures as low as 15 F = 26.1 C . 2A The mass is the difference between the mass of the full and empty flask. 291.4 g 108.6 g density = =1.46 g/mL 125 mL 2B First determine the volume required. V = (1.000 × 10 3 g) (8.96 g cm -3 ) = 111.6 cm 3 . Next determine the radius using the relationship between volume of a sphere and radius. V = 4 3 r 3 = 111.6 cm 3 = 4 3 (3.1416)r 3 r = 3 111.6 3 4(3.1416) = 2.987 cm 3A The volume of the stone is the difference between the level in the graduated cylinder with the stone present and with it absent. mass 28.4 g rock density = = = 2.76 g/mL volume 44.1 mL rock &water 33.8 mL water = 2.76 g/cm 3 3B The water level will remain unchanged. The mass of the ice cube displaces the same mass of liquid water. A 10.0 g ice cube will displace 10.0 g of water. When the ice cube melts, it simply replaces the displaced water, leaving the liquid level unchanged. 4A The mass of ethanol can be found using dimensional analysis. 1000 mL 0.71 g gasohol 10 g ethanol 1 kg ethanol ethanol mass = 25 L gasohol 1 L 1 mL gasohol 100 g gasohol 1000 g ethanol =1.8 kg ethanol  4B We use the mass percent to determine the mass of the 25.0 mL sample. 100.0 g rubbing alcohol rubbing alcohol mass = 15.0 g (2-propanol) = 21.43 g rubbing alcohol 70.0 g (2-propanol) 21.4 g rubbing alcohol density = = 0.857 g/mL 25.0 mL
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Chapter 1: Matter – Its Properties and Measurement 2 5A For this calculation, the value 0.000456 has the least precision (three significant figures), thus the final answer must also be quoted to three significant figures. 62.356 0.000456 6.422 10 = 21.3 3  5B For this calculation, the value 1.3 10 3 has the least precision (two significant figures), thus the final answer must also be quoted to two significant figures. 8.21 10 1.3 10 0.00236 4.071 10 =1.1 10 43 2 6 6A The number in the calculation that has the least precision is 102.1 (+ 0.1), thus the final answer must be quoted to just one decimal place. 0.236 +128.55 102.1= 26.7 6B This is easier to visualize if the numbers are not in scientific notation.  3 2 1.302 10 + 952.7 1302 + 952.7 2255 = = = 15.6 157 12.22 145 1.57 10 12.22  INTEGRATIVE EXAMPLE A Stepwise Approach: First, determine the density of the alloy by the oil displacement. Mass of oil displaced = Mass of alloy in air – Mass of alloy in oil = 211.5 g – 135.3 g = 76.2 g V Oil = m / D = 76.2 g / 0.926 g/mL = 82.3 mL = V Mg-Al D Mg-Al = 211.5 g / 82.3 mL = 2.57 g/cc Now, since the density is a linear function of the composition, D Mg-Al = mx + b, where x is the mass fraction of Mg, and b is the y-intercept. Substituting 0 for x (no Al in the alloy), everything is Mg and the equation becomes: 1.74 = m · 0 + b. Therefore, b = 1.74 Assuming 1 for x (100% by weight Al): 2.70 = (m × 1) + 1.74, therefore, m = 0.96 Therefore, for an alloy: 2.57 = 0.96x + 1.74 x = 0.86 = mass % of Al Mass % of Mg = 1 – 0.86 = 0.14, 14%
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Chapter 1: Matter – Its Properties and Measurement 3 B Stepwise approach: Mass of seawater = D • V = 1.027 g/mL × 1500 mL = 1540.5 g
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This note was uploaded on 01/04/2012 for the course CHEM 1500 taught by Professor Hameedmmirza during the Fall '11 term at York University.

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Full Answers to Questions - CHAPTER 1 MATTERITS PROPERTIES...

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