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

This preview shows pages 1–4. Sign up to view the full content.

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

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

View Full Document
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%
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

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

## 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.

### Page1 / 16

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

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online