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Base units can be used to check the homogeneity of physical equations. In any correct physical equation, each term much have the same unit.
For example, consider the well know relation E = mc2 i.e. energy = mass × (velocity of light)2
Unit of energy = joule = kg m2 s–2
Unit of mass × (velocity of light)2 = kg × (m s–1)2 = kg m2 s–2
The base units for each term are the same; the equation E =mc2 is therefore homogeneous.
Homogeneity is an important condition for a physical equation to be correct. However, it is not the sole condition. The homogeneity test cannot tell us whether there are missing terms (or superfluous terms) in an equation.
The principle of homogeneity of physical equations can be used to derive the units of unknown constants. e.g. for where is a force,  is a lengthand  is a speed, the constant  has units of N m-2 s = kg m-1 s-1.

One mole (mol) of any substance is defined as that amount of the substance that contains the same number of "elementary entities" as there are atoms in 0.012 kg of carbon-12 (12C).
When the mole is used, the type of "elementary entities" must be specified; these entities may be atoms, molecules, ions, electrons, other particles,or specified groups of such particles.
The Avogadro constant, usually denoted by L or NA, is equal to the number of atoms in 0.012 kg of carbon-12, i.e. the number of atoms in 1 mol of 12C. The Avogadro Number is given by NA = 6.022 141 79(30) × 1023 mol–1 i.e.
mol–1

The atomic mass of a chemical element (or the molecular mass of a compound) is the mass of one mole (in grams). Thus, one moleof 12C has a mass of 12.0 g = 0.0120 kg and contains 6.022 141 79(30) × 1023 atoms of 12C. One mole of O2 has a mass of 32.0 g = 0.0320 kg and contains 6.022 141 79(30) × 1023 molecules of O2. One mole of H2O has a mass of 18.0 g = 0.0180 kg and contains 6.022 141 79(30) × 1023 molecules of H2O.
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Consider the following two measurements of lengths:
L1 = (25 ± 2) mm = (23, 27) mm = {23mm < L1 < 27 mm}
L2 = (75 ± 4) mm = (71, 79) mm
If L = L1 + L2, the mean value of L is simply calculated as 25 mm + 75 mm = 100 mm. What about the error in L?
The minimum possible value of L is 23 mm + 71 mm = 94 mm. The maximum possible value of L is 27 mm + 79 mm = 106 mm. The value of L can be quoted as (94, 106) mm or (100 ± 6) mm.
For a calculated quantity q derived from theaddition or subtraction of two measured quantities q1 and q2, the uncertainty in the calculated quantity q is related to the uncertainties in the measured quantities q1 and q2 by
q = q1 + q2
Therefore, L = L1 + L2 = 2 mm + 4 mm = 6 mm
In general, for a calculated quantity q derived from the addition or subtraction of various measured quantities q1, q2, … qn the uncertainty in the calculated quantity qis related to the uncertainties in the measured quantities q1, q2 … qn.
q = q1 + q2 + … + qn
Fractional uncertainty
The fractional uncertainty in a physical quantity measured as (q ± q) units is .
Percentage uncertainty
The percentage uncertainty in a physical quantity measured as (q ± q) units is ×100%.
Consider the following case: a car travels a distance of (d ± d) m in a time of (t ± t)s. The speed v of the car is given by . What about the uncertainty in the speed v?
The uncertainty can be calculated using the same principle as for the length addition problem. If the distance is measured as (10 ± 1) m and time taken as (1.0 ± 0.1) s, the maximum speed would be = 12 m s–1 and the maximum speed would be = 8.2 m s–1. The speed of the car is quoted as (8.2, 12) m s–1 = (10 ± 2) ms–1.
The uncertainty on the speed can also be calculated using
= + = + = 0.1 + 0.1 = 0.2
v = 0.2 × 10 m s–1 = 2 m s–1.
In general, for a calculated quantity q derived from the product or quotient of various measured quantities q1, q2, … qn the fractional uncertainty in the calculated quantity is related to the fractional uncertainties in the measured quantities , … .
= + + ... +
If a...
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