Intensive properties and extensive properties are types of physical properties of matter. The terms intensive and extensive were first described by physical chemist and physicist Richard C. Tolman in 1917. Here's a look at what intensive and extensive properties are, examples of them, and how to tell them apart.
Intensive Properties
Intensive properties are bulk properties, which means they do not depend on the amount of matter that is present. Examples of intensive properties include:
Boiling point
Density
State of matter
Color
Melting point
Odor
Temperature
Refractive Index
Luster
Hardness
Ductility
Malleability
Intensive properties can be used to help identify a sample because these characteristics do not depend on the amount of sample, nor do they change according to conditions.
Extensive Properties
Extensive properties do depend on the amount of matter that is present. An extensive property is considered additive for subsystems. Examples of extensive properties include:
Volume
Mass
Size
Weight
Length
The ratio between two extensive properties is an intensive property. For example, mass and volume are extensive properties, but their ratio (density) is an intensive property of matter.
While extensive properties are great for describing a sample, they aren't very helpful identifying it because they can change according to sample size or conditions.
Way to Tell Intensive and Extensive Properties Apart
One easy way to tell whether a physical property is intensive or extensive is to take two identical samples of a substance and put them together. If this doubles the property (e.g., twice the mass, twice as long), it's an extensive property. If the property is unchanged by altering the sample size, it's an intensive property.
Answer:
Step-by-step explanation:
The equation for a line is y=mx + b where m represents the slope
If a line is perpendicular to another line, it's slope is the negative reciprocal of the other line's slope.
The reciprocal of -4 is 
And the negative of that would be
Answer: D
Step-by-step explanation:
in the first equation you need to shade below and for the second equation you need to shade to the right and the only place where the shading matches is in section d
Answer:
- 27 and 7
Step-by-step explanation:
To obtain f(- 6) substitute x = - 6 into f(x), that is
f(- 6) = 4(- 6) - 3 = - 24 - 3 = - 27
To obtain g(- 3) substitute x = - 3 into g(x), that is
g(- 3) = - 3(- 3) - 2 = 9 - 2 = 7
Answer:
(-6,4)
Step-by-step explanation:
If this is the system of equations here is how:
solve for 1 vairuble in the first equation, and plug that it to the other.
