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Chemistry Review Flashcards
Chemistry Review Flashcards
Chapter 1
Observation vs. Hypothesis
:
An observation is a qualitative or quantitative measurement of an event, a statement of fact without explanation.
A hypothesis is an educated guess explaining the observation; it can be true, untrue, or modified.
Hypothesis to Theory or Law
:
A hypothesis, when proven true on multiple levels, can become a theory or a law.
Scientific Theory vs. Scientific Law
:
A scientific theory explains
why
an observation occurs, based on principles and equations.
A scientific law explains
what
and
how
an observation occurs, based on observations and tangible means.
Examples
:
The leaves turned brown. (Observation)
The water boils because it got hot. (Hypothesis)
The boiling point of water is 212 °F. (Observation)
The Earth is flat because the map I have shows the Earth is flat. (Hypothesis)
Chapter 2
Units of Measurement
:
Liter (L) = volume
Meter (m) = length / distance
Gram (g) = mass
Second (s) = time
Significant Figures
:
1.254 has 4 SF
0.14580 has 5 SF
0.008274 has 4 SF
255000 has 3 SF
5.24 \times 10^6 has 3 SF
Calculations with Significant Figures and Units
:
12.0 - 1.587 + 150 = 160
(1087 \times 1.2) / 10.5 = 120
1.254 \times 12.7 \times 0.0089 = 0.14
124000 / (58.4 - 2.5) = 2200
Unit Conversions
:
2500 cg to L (density = 2.00 g/mL) = 2500 \text{ cg} \times \frac{1 \text{ g}}{100 \text{ cg}} \times \frac{1 \text{ mL}}{2.00 \text{ g}} \times \frac{1 \text{ L}}{1000 \text{ mL}} = 0.013 \text{ L}
mL to L = 100. \text{ mL} \times \frac{1 \text{ L}}{1000 \text{ mL}} = 0.100 \text{ L}
0.0254 kg to mg = 0.0254 \text{ kg} \times \frac{1000 \text{ g}}{1 \text{ kg}} \times \frac{1000 \text{ mg}}{1 \text{ g}} = 25400 \text{ mg}
35 cg to g = 35 \text{ cg} \times \frac{1 \text{ g}}{100 \text{ cg}} = 0.35 \text{ g}
1. 05 g to mL (density = 2.15 g/mL) = 1.05 \text{ g} \times \frac{1 \text{ mL}}{2.15 \text{ g}} = 0.488 \text{ mL}
mph to m/s = 80 \frac{\text{miles}}{\text{hr}} \times \frac{5280 \text{ ft}}{1 \text{ mile}} \times \frac{12 \text{ in}}{1 \text{ ft}} \times \frac{2.54 \text{ cm}}{1 \text{ in}} \times \frac{1 \text{ m}}{100 \text{ cm}} \times \frac{1 \text{ hr}}{60 \text{ min}} \times \frac{1 \text{ min}}{60 \text{ sec}} = 36 \text{ m/s}
Sorting Numbers
:
Ascending order: 2.0 \times 10^{-5} < 2.0 \times 10^{0} < 2.0 \times 10^{5}
Chapter 3
Properties of Matter
:
Matter has mass and takes up space.
States of Matter
:
Solid: Definite shape and definite volume.
Liquid: Definite volume and indefinite shape.
Gas: Indefinite volume and indefinite shape.
Physical vs. Chemical Changes
:
Melting butter: Physical
Boiling water: Physical
Burning wood: Chemical
Rusting steel: Chemical
Digesting food: Chemical
Tearing paper: Physical
Element, Compound, or Mixture
:
Water: Compound
Pure Air: Mixture
Iron bar: Element
Sugar: Compound
Hydrogen Gas: Element
Ocean Water: Mixture
Calorie Calculation
:
Calories needed to heat 1600 g of water from 25 °C to 100 °C:
q = m \times C \times \Delta T = (1600 \text{ g}) \times (1.00 \frac{\text{cal}}{\text{g} \cdot {}^{\circ}\text{C}}) \times (100 {}^{\circ}\text{C} - 25 {}^{\circ}\text{C}) = 120,000 \text{ cal} = 1.2 \times 10^5 \text{ cal}
Calculating Mass of Copper Metal
:
Grams of copper metal (C = 0.385 J/g°C) heated from 16 °C to 27 °C with 275 J applied:
m = \frac{q}{C \times \Delta T} = \frac{275 \text{ J}}{(0.385 \frac{\text{J}}{\text{g} \cdot {}^{\circ}\text{C}}) \times (11 {}^{\circ}\text{C})} = 66 \text{ g}
Temperature Change Calculation
:
Temperature change when 1.5 kg of aluminum is heated with 18 J (C = 0.899 J/g°C):
\Delta T = \frac{q}{m \times C} = \frac{1800 \text{ J}}{(1500 \text{ g}) \times (0.899 \frac{\text{J}}{\text{g} \cdot {}^{\circ}\text{C}})} = 1.3 {}^{\circ}\text{C}
Temperature Conversions to Fahrenheit
:
0°C to F: F = (0 {}^{\circ}\text{C} \times 1.8) + 32 = 32 {}^{\circ}\text{F}
0 K to F:
C = 0 \text{ K} - 273 = -273 {}^{\circ}\text{C}
F = (-273 {}^{\circ}\text{C} \times 1.8) + 32 = -460 {}^{\circ}\text{F}
212 K to F:
C = 212 \text{ K} - 273 = -61 {}^{\circ}\text{C}
F = (-61 {}^{\circ}\text{C} \times 1.8) + 32 = -78 {}^{\circ}\text{F}
23 °C to F: F = (23 {}^{\circ}\text{C} \times 1.8) + 32 = 73 {}^{\circ}\text{F}
Phase Changes
:
Melting: Endothermic
Evaporation: Endothermic
Freezing: Exothermic
Deposition: Exothermic
Sublimation: Endothermic
Condensation: Exothermic
Chapter 4
Types of Elements
:
H: Nonmetal
Mg: Metal
As: Metalloid
P: Nonmetal
Ge: Metalloid
Fe: Metal
Subatomic Particles
:
Proton: Has mass and positive charge.
Neutron: Has mass and no charge.
Electron: Has "no" mass and negative charge.
Location of Subatomic Particles
:
Protons and neutrons are in the nucleus.
Electrons orbit in the electron cloud.
Isotopes
:
¹⁸O: p=8, n=10, e=8
¹³C: p=6, n=7, e=6
¹⁰⁰Mo: p=42, n=58, e=42
Selenium-76: p=34, n=42, e=34
Calcium-42: p=20, n=22, e=20
Iron-55: p=26, n=29, e=26
Determining Isotopes
:
16 protons, 28 neutrons, 16 electrons: ^{44}S
28 protons, 36 neutrons, 28 electrons: ^{64}Ni
98 protons, 142 neutrons, 98 electrons: ^{240}Cf
1 proton, 1 neutron, 1 electron: ^2H
22 protons, 45 neutrons, 22 electrons: ^{67}Ti
48 protons, 73 neutrons, 48 electrons: ^{121}Cd
Identifying Elements
:
(The content about identifying the element by its period and group number is not present in the reference text.)
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