CHAPTER 1 KEYS TO THE STUDY OF CHEMISTRY FOLLOW–UP PROBLEMS 1.1A Plan: The real question is “Does the substance change composition or just change form?” A change in composition is a chemical change while a change in form is a physical change. Solution: The figure on the left shows red atoms and molecules composed of one red atom and one blue atom. The figure on the right shows a change to blue atoms and molecules containing two red atoms. The change is chemical since the substances themselves have changed in composition. 1.1B Plan: The real question is “Does the substance change composition or just change form?” A change in composition is a chemical change while a change in form is a physical change. Solution: The figure on the left shows red atoms that are close together, in the solid state. The figure on the right shows red atoms that are far apart from each other, in the gaseous state. The change is physical since the substances themselves have not changed in composition. 1.2A Plan: The real question is “Does the substance change composition or just change form?” A change in composition is a chemical change while a change in form is a physical change. Solution: a) Both the solid and the vapor are iodine, so this must be a physical change. b) The burning of the gasoline fumes produces energy and products that are different gases. This is a chemical change. c) The scab forms due to a chemical change. 1.2B Plan: The real question is “Does the substance change composition or just change form?” A change in composition is a chemical change while a change in form is a physical change. Solution: a) Clouds form when gaseous water (water vapor) changes to droplets of liquid water. This is a physical change. b) When old milk sours, the compounds in milk undergo a reaction to become different compounds (as indicated by a change in the smell, the taste, the texture, and the consistency of the milk). This is a chemical change. c) Both the solid and the liquid are butter, so this must be a physical change. 1.3A Plan: We need to find the amount of time it takes for the professor to walk 10,500 m. We know how many miles she can walk in 15 min (her speed), so we can convert the distance the professor walks to miles and use her speed to calculate the amount of time it will take to walk 10,500 m. Solution: Time (min) = 10,500 m LN NJ `NJO` `N `LN NJ               = 97.8869 = 98 min