This document contains a chemistry review with multiple sections on: 1) temperature conversions between Celsius, Fahrenheit, and Kelvin scales, 2) identifying measurements as relating to length, mass, temperature, or time, 3) performing calculations with significant figures, 4) unit conversions, 5) density and volume calculations, 6) identifying atomic structure and properties of elements, 7) naming common ions and compounds. The review covers a wide range of general chemistry concepts and calculations.
Application of Statistical and mathematical equations in Chemistry -Part 3 Awad Albalwi
Application of Statistical and mathematical equations in Chemistry
Part 3
reaction rate
equilibrium constant
The common ion effect
Activity and Activity Coefficients
The Diverse Ion Effect Theory
Balancing chemical equations - NCERT textbook question of exerciseSantosh Upadhyay
We have taken the topic of balancing chemical equation of class 10th science and tried to make it simpler and easier for you.
We have solved the question of balancing equation for you. You can easily get a solution of Chapter 1 science of class 10th NCERT. We tried to solve the NCERT textbook question of exercise of Chapter 1 science.
Application of Statistical and mathematical equations in Chemistry -Part 3 Awad Albalwi
Application of Statistical and mathematical equations in Chemistry
Part 3
reaction rate
equilibrium constant
The common ion effect
Activity and Activity Coefficients
The Diverse Ion Effect Theory
Balancing chemical equations - NCERT textbook question of exerciseSantosh Upadhyay
We have taken the topic of balancing chemical equation of class 10th science and tried to make it simpler and easier for you.
We have solved the question of balancing equation for you. You can easily get a solution of Chapter 1 science of class 10th NCERT. We tried to solve the NCERT textbook question of exercise of Chapter 1 science.
https://www.deped.gov.ph/wp-content/uploads/2019/01/General-Chemistry-1-and-2.pdf
General Chemistry
GenChem
STEM
Science, Technology, Engineering, and Mathematics
K to 12 Senior High School STEM Specialized Subject – General Chemistry 1 and 2
Quarter 1 – General Chemistry 1
Matter and Its Properties
Measurements
Atoms, Molecules and Ions
Stoichiometry
Percent Composition and Chemical Formulas
Chemical reactions and chemical equations
Mass Relationships in Chemical Reactions
Gases
Dalton’s Law of partial pressures
Gas stoichiometry
Kinetic molecular theory of gases
Quarter 2 – General Chemistry 1
Electronic Structure of Atoms
Electronic Structure and Periodicity
Chemical Bonding
Organic compounds
Quarter 3 – General Chemistry 2
Intermolecular Forces and Liquids and Solids
Physical Properties of Solutions
Thermochemistry
Chemical Kinetics
Quarter 4 – General Chemistry 2
Chemical Thermodynamics
Chemical Equilibrium
Acid-Base Equilibria and Salt Equilibria
Electrochemistry
1. Key
GENERAL CHEMISTRY-I (1411)
S.I. # 2
1. Give the formulas for the following
K = °C +273
°C = (5/9) (F-32)
°F = (9/5 )(°C )+32
2. Do the following temperature conversions:
a. 25°C 77 °F
b.117°F 47.2 °C
c. 45°C 318.15 K
d.430 K 157 °C
e. 600 K 620.6 °F step 1: °C = K-273, step 2: °F = (9/5 °C) +32
f. 30°F 271.89 K step 1: °C = (5/9)( °F-32), step 2: K=°C +273
g. Absolute Zero 0K
3. Using s.i. units identify each of the following measurements as either Length-L,
Mass-m, Temperature-T, Time-t:
a. 1.75 m L
b. 13.32 Kg m
c. 124K T
d. 13 pg m
e. 10.0005 mm L
f. 12 s t
g. 35°F T
h. 112 km L
4. Compute the answer in significant figures:
a. 25 + 1.5 + 0.75 + 0.015 = 27
b. 3 x 12.7772 + 32.066 + 5 x 4.1163 = 90.979
5. Complete the following conversions:
a. 10,000 micrograms = 10 milligrams
-10
b. 7 picometers = 7x10 centimeters
c. 7.6 m/s2 = 7.6x10-3____ mm/ms2
d. 12 kg = 12,000,000 mg
e. 0.001 cm = 0.01 mm
6. A wooden object has a mass of 10.782 g and occupies a volume of 13.72 mL.
What is the Density?
D = m/V => 10.782/13.72 = 0.7859 = 7.859 x 10-1 g/mL
2. Key
7. A liquid has a density of 2.67 g/cm3 and a mass of 1340 g. What volume would the
liquid occupy?
1 cm3 = 1 mL d = 2.67g/0.001 L = 2670 g/L V= m/d so
V = 1340 (g) / 2670 (g/L) = 502 mL
8. Cathode rays are: Electrons
9. What does the atomic number of an element indicate?
The # of protons or electrons in neutral atom.
10. For 13254Xe determine the number of
electrons: 54
protons: 54
neutrons: 78
11. What group are the following in?
a. Halogens 7A
b. Alkali Metals 1A
c. Alkaline earth metals 2A
d. Noble Gases 8A
e. Metalloids middle & bottom
12. Give the charges of the following ions:
a. Na +1 b. Ca +2 c. N -3 d. H +1
e. K +1 f. F -1 g. Br -1 h. O -2
i. Cl -1 j. S -2
13. How many moles are in 14.2 g of carbon?
(14.2 g C)(1mol / 12.01 g C) = 1.18 moles of Carbon
14. How many grams are in 6 moles of carbon?
(6 mole C) ( 12.01 g C / 1 mol C) = 72.06 g C
15. Find the # of H atoms in 0.350 mol of C6H12O6
1 mol = 6.02x1023 molecules or atoms
(0.350 mol C6H12O6) (6.02x1023 atoms/ mol) (12 H atoms) = 2.53x1024 H atoms
3. Key
NOMENCLATURE: Give the name or formula for the following
1. OH- Hydroxide anion
2. H+ Hydrogen Ion
3. Sn Tin
4. Ag Silver
5. O Oxygen
6. alpha particle α or H+ or He2+
7. Sodium Hydroxide Na+OH-
8. Ammonia NH3
9. Chlorine Cl
10. Sulfuric Acid H2SO4 (aq)
11. CaCl2 Calcium Chloride
12. K2SO4 Potassium Sulfate
13. BaS Barium Sulfide