Scientific Notation & SigFigs
Dimensional Analysis & Density
Energy & Heat
States, Properties & Changes
Atoms, Isotopes & Ions
100

Write 45,600 in scientific notation with 3 sig figs.

4.56 × 10⁴

100

250 mg → g.

0.250g

100

Convert 225 cal → J. (1 cal = 4.184 J).

225 × 4.184 = 941 J

100

Which state of matter is compressible?

Gas

100

Define atomic number and mass number.

Atomic number = protons; Mass number = protons + neutrons

200

Convert 4.2 × 10⁻³ into decimal form.

0.0042

200

Convert 2.5 hours → seconds.

2.5 hr × 3600 s/hr = 9000 s

200

A 55 g Al can warms 20 °C → 30 °C. How much heat absorbed? (C = 0.897 J/g°C).

q = mCΔT = 55 × 0.897 × 10 = 492 J

200

Classify: sugar dissolving in water — physical or chemical change?

Physical change

200

How many protons, neutrons, and electrons in Cl-37?

17 protons, 20 neutrons, 17 electrons

300

How many sig figs are in 0.002050?

4 sig figs

300

A metal block has mass = 45.0 g and volume = 15.0 mL. Find density.

D = m/V = 45.0 ÷ 15.0 = 3.00 g/mL

300

Heat required to raise 250 g water from 20 °C → 50 °C? (C = 4.184 J/g°C).

q = 250 × 4.184 × 30 = 31,380 J = 31.4 kJ

300

List 2 physical properties and 2 chemical properties.

Physical = density, melting point... Chemical = flammability, reactivity with O₂

300

A chlorine sample is 75.77% Cl-35 (34.97 amu), 24.23% Cl-37 (36.97 amu). Find average atomic mass.

35.45 amu

400

Perform: (3.489 × (5.67 – 2.3)) with correct sig figs.

3.489 × 3.37 = 11.8  

400

Convert 60.0 miles/hr → m/s (1 mile = 1.609 km).

60.0 × 1.609 km / hr = 96.5 km/hr → ÷ 3.6 = 26.8 m/s

400

A 45.0 g piece of iron absorbs 327 J of heat, raising its temperature from 22.0 °C to 37.0 °C. Calculate the specific heat capacity (C) of iron.

q = mCΔT
327 J = (45.0 g)(C)(15.0 °C)
C = 327 ÷ (45 × 15) = 0.484 J/g°C

400

Classify: dry ice subliming, wood burning, iron rusting.

Dry ice = physical; wood burning & rusting = chemical

400

Write the ions formed by: Mg, S, Al. Include charges.

Mg²⁺, S²⁻, Al³⁺

500

Round 0.00076294 to 2 sig figs, then write in scientific notation.

7.6 × 10⁻⁴

500

A sample has mass = 56.2 g and density = 2.80 g/mL. Find volume.

V = m/D = 56.2 ÷ 2.80 = 20.1 mL

500

A 32 g ethanol sample absorbs 562 J. If C = 2.42 J/g°C, calculate ΔT.

7.2 °C

500

Explain difference between heat and temperature

Heat is energy transfer due to temp difference; Temp = average kinetic energy of particles

500

Compare properties of alkali metals, halogens, noble gases. Why are noble gases unreactive?

  • Alkali = very reactive metals, form 1⁺ ions

  • Halogens = reactive nonmetals, form 1⁻ ions

  • Noble gases = inert because they have full valence shells

M
e
n
u