Temperature is the measure of the average kinetic energy of a system. Kelvin is the SI unit for temperature, and must be used because it is the only scale that does not have any negative numbers possible. At zero Kelvin (absolute zero) and there is absolutely no movement in the system, down to the atomic level.

While temperature is a means of measuring energy in a system, actual heat energy has the unit Joules.

1. Any digit with a value 1-9 is always a significant figure

2. Any 0 between two digits 1-9 is always significant

3. If a number is greater than 1 and does not have a decimal point a 0 at the end is not significant because it is a tailing zero

4. If a number is less than 1 any 0 before a digit 1-9 is not significant because it is a leading zero

5. If there is a decimal point in any number, any 0 that follows a non-zero digit is always significant.

Thus, \(\displaystyle 0.1960g\) has 4 significant figures. The first 0 is not significant because of rule 4. The digits 1, 9, and 6 are always significant (rule 1) and the last 0 is also significant because there is a decimal point and that 0 follows a non-zero digit (rule 5).

The following prefixes are ranked in order from largest to smallest with the power of 10 that the unit represents: peta (P) \(\displaystyle 10^{15}\) > tera (T) \(\displaystyle 10^{12}\) > giga (G) \(\displaystyle 10^9\) > mega (M) \(\displaystyle 10^6\) > kilo (k) \(\displaystyle 10^3\) > hecta (H) \(\displaystyle 10^2\) > deca (D) \(\displaystyle 10^1\) > unit > deci (d) \(\displaystyle 10^{-1}\) > centi (c) \(\displaystyle 10^{-2}\) > milli (m) \(\displaystyle 10^{-3}\) > micro \(\displaystyle \left(\mu \right )\) \(\displaystyle 10^{-6}\) > nano (n) \(\displaystyle 10^{-9}\) > pico (p) \(\displaystyle 10^{-12}\) > femto (f) \(\displaystyle 10^{-15}\). Thus a microsecond is one thousand times longer than a nanosecond.

Units in the metric system can be converted using the correct factor of 10.

\(\displaystyle 0.5 mg \cdot\frac{1000\mu g}{1mg}=500\mu g\)

\(\displaystyle 0.5 mg \cdot\frac{1g}{1000mg}\cdot \frac{1kg}{1000g}=.0000005 kg\)

\(\displaystyle 0.5 mg \cdot\frac{1g}{1000mg}\cdot \frac{100cg}{1g}=.05 cg\)

\(\displaystyle 0.5 mg \cdot\frac{1g}{1000mg}\cdot \frac{10^{9}ng}{1g}= 500000ng\)

\(\displaystyle 0.5 mg \cdot\frac{1g}{1000mg}\cdot \frac{10dg}{1g}=.005 dg\)

Density is equal to the mass of the object divided by the volume.

\(\displaystyle D=\frac{m}{V}\)

The mass of the metal is figured out by taking the mass of the metal and the weighboat \(\displaystyle \left(41.60g \right )\) and subtracting the weighboat \(\displaystyle \left(2.15g \right )\)

\(\displaystyle 41.60 g - 2.15 g =39.45 g\)

The volume of the water is calculated by the displacement of the water so take the final volume and subtract the initial volume.

\(\displaystyle 44.6 ml -28.9 ml = 15.7 ml\)

The density of the metal is calculated by taking the mass divided by the volume.

 \(\displaystyle \frac{39.45g}{15.7ml}=2.51\frac{g}{ml}\) after rounding to 3 significant figures.

You must use equalities and dimensional analysis to solve for the correct answer:

\(\displaystyle 1m = 100 cm\)

\(\displaystyle 2.54 cm = 1in\)

Each equality allows you to write a conversion factor 2 ways: 

Example: \(\displaystyle 1m=100cm\) can be written as 

\(\displaystyle \frac{1m}{100cm}\)  and \(\displaystyle \frac{100cm}{1m}\)

To cancel out units if there is a unit on top it must be cancelled out by the same unit on the bottom.

To solve you must cancel out units until you get to the proper unit

\(\displaystyle \frac{1.65m}{1}\cdot \frac{100cm}{1m}\cdot\frac{1in}{2.54cm}=65.0in\)

It is important to note that the mass of a sample does not tell you the amount of atoms in the sample. The number of atoms in a sample is dependent on the number moles in a sample, given by Avogadro's number. Here is the number of moles for each sample:

\(\displaystyle 50 g\ Li*\frac{1 mol\ Li}{6.94 g\ Li} = 7.2 mol\ Li\)

\(\displaystyle 50 g\ Mg*\frac{1 mol\ Mg}{24.31 g\ Mg} = 2.1 mol\ Mg\)

\(\displaystyle 50 g\ K*\frac{1 mol\ K}{39.10 g\ K} = 1.3 mol\ K\)

\(\displaystyle 50 g\ Cl_2*\frac{1 mol\ Cl_2}{70.90 g\ Cl_2} = 0.71 mol\ Cl_2\)

Remember that chlorine is a diatomic mass, so each molecules contains two atoms. This doubles the molar mass for the conversion.

The sample with the greatest number of moles will also contain the most atoms. In this case, the sample of lithium results in the largest number of moles and, thus, the greatest number of atoms.

In the chemical equation, the ratio of potassium bromide to bromine is 2:1, so for every 2 moles of \(\displaystyle KBr\), 1 mole of \(\displaystyle Br_2\) is produced. Therefore, if we start with 4 moles of \(\displaystyle KBr\), we get 2 moles of \(\displaystyle Br_2\). The number of molecules is equal to the number of moles times Avogadro's Number. Since we've determined the number of moles to be 2, the number of molecules is:

\(\displaystyle 2\cdot 6.02\cdot 10^{23}=1.204\cdot10^{24}molecules\)

Use the periodic table to calculate the molecular weight of sodium hydroxide.

\(\displaystyle MW_{NaOH}= 39.99\frac{g}{mol}\)

Next, use dimensional analysis to find the number of moles.

\(\displaystyle 357g\:NaOH\cdot\frac{1mol}{39.99g}=8.93mol\)

The molecular weight of carbon dioxide is \(\displaystyle 44\tfrac{g}{mol}\), meaning in \(\displaystyle 1mol\) of \(\displaystyle CO_{2}\), there are \(\displaystyle 44g\) of the substance.