Regents Chemistry



Unit Test Review

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|Q# |answer # |

Mixtures can be separated according to their physical and chemical properties. Explain how you would separate the following mixtures into their individual components.

1. sand and water filtration

2. sugar and water 1. evaporate/boil off water (bp water much lower than bp of sugar)

2.crystallization

3. mixture of heptane (bp = 98˚C) and heptanol (bp = 176˚C) distillation

4. mixture of sand and iron pellets magnet (iron is magnetic and sand is not)

ENERGY

The SI unit of energy is called the JOULE.

You put a rock in the oven and heat it up. Then you put the rock in a large pail filled with cold water (about 15˚C). You observe the temperature of the water before and after the rock is placed in it. What happens? Why?

essentially placing a HOT rock in cooler water; heat always travels from hotter object to colder object so heat travels from rock to water causing the water temp to rise and the temp of the rock to decrease; eventually the temp of the water stops changing indicating the water and the rock are the same temperature.

According to the KMT, atoms and molecules are in constant motion. How does temperature affect the movement of particles in a substance? [(temperature ( motion of particles]

Increasing temperature causes particles in a substance to move faster, while decreasing the temperature causes the particles to slow down.

Energy changes accompany all chemical and physical changes. If energy leaves the “system” we say the energy change is EXOTHERMIC because the system had more energy before the change than after. If energy enters the “system” we say the energy change is ENDOTHERMIC because the system had more energy after the change than before.

Identify the following according to their being a physical/chemical change; identify each change as being an endothermic/exothermic change.

|change |physical/chem |exo/endo | |change |physical/chem |exo/endo |

|burning coal |chemical |exo | |melting ice |physical |endo |

|boiling water |physical |endo | |activation of a light |chemical |exo |

| | | | |stick | | |

|burning propane in BBQ |chemical |exo | |evaporation of sweat from|physical |endo |

| | | | |skin | | |

| | | | |l ( g | | |

|drying clothes in |physical |endo | |hammering copper into |physical |exo: hammer |

|clothes dryer | | | |flat sheet | |endo: copper |

A cold pack is placed on your injured leg. Indicate the direction of the flow of energy between your leg and the cold pack. heat travels from leg to cold pack

What is the difference between potential and kinetic energy?

• potential energy (PE) is stored energy of position

• kinetic energy (KE) is the energy of motion

Name the six phase changes and state whether they are endothermic or exothermic.

|ENDOTHERMIC phase changes |EXOTHERMIC phase changes |

|s(l melting |g(l condensation |

|l(g vaporization (boiling) |l(s solidification (freezing) |

|s(g sublimation |g(s deposition |

Show in the particle diagrams below what happens to the particles as they change phase:

need to draw the same number of atoms/molecules in each diagram

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solid liquid gas

Energy changes can be calculated using: Q = mcΔT where: Q = energy

m = mass

c = specific heat

ΔT = change in temperature

Solve the following: [ΔT= Tfinal – Tinitial]

1. What quantity of heat would have to be added to 5000g of water to change its temperature from 20˚C to 80˚C?

Q = mcΔT

= (5000g)(4.18J/gºC)(60ºC)

= 1,254,000J

2. How much heat energy is required to raise the temperature of 8.0g of aluminum from 293K to 298K? [specific heat of aluminum is 0.90J/gK]

Q = mcΔT

= (8g)(0.90J/gK)(5K)

= 36J

3. An exothermic reaction releases 800 Joules of heat to a calorimeter containing 40 grams of water. Calculate the temperature change of the water.

800J = mcΔT

= (40g)(4.18J/gºC)(ΔT)

4.78ºC = ΔT

4. What is the final temperature when 168 joules of heat is added to 4 grams of water at 283K?

168J = mcΔT ΔT= Tfinal – Tinitial

= (4g)(4.18J/gK)(ΔT) 10.0K = Tf – 283K

10.0K = ΔT 293K = Tf

5. 500 gram sample of water loses 1.05 x 104 joules of heat. The temperature of the water, after the heat loss, is found to be 25˚C. What was the initial temperature of the water?

Q = mcΔT ΔT= Tfinal – Tinitial

1.05 x 104J = (500g)(4.18J/gºC)(ΔT) 5.02ºC = 25ºC - Ti

5.02ºC = ΔT 30.02ºC = -x

-30.02ºC = x

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