Any other physical quantities can be written as some combination of these base units, and are sometimes are given their own name.

• $\text{velocity}= \dfrac{\text{distance}}{\text{time}}$ $=\dfrac{\text{m}}{\text{s}}$
• $\text{acceleration}= \dfrac{\text{velocity}}{\text{time}}$ $=\dfrac{\text{m}/\text{s}}{\text{s}}$ $=\dfrac{\text{m}}{\text{s}^2}$
• $\text{force}= \text{mass}\times \text{acceleration}$ $=\text{kg}\times \dfrac{\text{m}}{\text{s}^2}$ $=\text{N}$ $\;\; $(Newton)
• $\text{pressure}= \dfrac{\text{force}}{\text{area}}$ $= \dfrac{\text{N}}{\text{m}^2}$ $= \dfrac{\text{kg}\times \dfrac{\text{m}}{\text{s}^2}}{\text{m}^2}$ $= \dfrac{\text{kg}}{\text{m}\times\text{s}^2}$ $= \text{Pa}$ $\;\;$(Pascal)
• $\text{energy}= \text{mass}\times \text{velocity}^2$ $= \text{kg} \times \dfrac{\text{m}^2}{\text{s}^2}$ $= \dfrac{\text{kg}\times \text{m}^2}{\text{s}^2}$ $= \text{J}$ $\;\;$(Joule)

There are a number of non-SI units accepted for use alongside SI units, e.g.:

• Angle in degrees, minutes and seconds (e.g. latitude and longitude).
• Area in hectares.
• Volumes in litres.
• Time in minutes, hours, days, years.
• Temperature in degrees Celsius.

People in the U.S. uses customary units in daily life (related to the British Imperial system), e.g.:

• Lengths in inches, feet, yards, miles and area in acres.
• Mass in pounds.
• Temperatures in degrees Fahrenheit (F)
• Ounces (oz) for mass and fluid ounces (fl oz) for volume.

1. The length of the stick is measured to be 70 cm. But you want to communicate with your American friend. How many inches it is?
    Hint: 1 inch = 2.54 cm.

2. The pasture is a rectangle with dimensions of 500 m by 2 km. How many hectares it is?
    Hint: 1 ha = 100 m × 100 m.

3. An Olympic swimming pool has dimensions 50 m × 25 m × 2 m. How many liters of water it can fit?
    Hint: 1 L = 10 cm × 10 cm × 10 cm.

4. The conversion from Fahrenheit to Celsius degrees is \(^{\circ}\text{C} = \dfrac{5}{9}\left(^{\circ}\text{F}-32\right).\) Find
      (a) \(\,0^\circ\text{F}\) in \(^\circ \text{C},\qquad\) (b) \(\, 100^\circ\text{F} \) in \(^\circ\text{C},\qquad\) (c) \(\, -40^\circ\text{C} \) in \(^\circ\text{F}\)

1. The length of the stick is measured to be 70 cm. But you want to communicate with your American friend. How many inches it is?
    Hint: 1 inch = 2.54 cm.

We need a conversion factor! Since we know that 1 inch = 2.54 cm, then

Thus we have

2. The pasture is a rectangle with dimensions of 500 m by 2 km. How many hectares is it?
  Hint: 1 ha = 100 m × 100 m.

Let's start by finding the area in square metres. First, we knot that \(2\,\text{km} = 2000\,\text{m}\)

Now use the conversion factor from $\text{m}^2$ → $\text{ha}$:

3. An Olympic swimming pool has dimensions 50 m × 25 m × 2 m. How many litres of water can it fit?
  Hint: 1 L = 10 cm × 10 cm × 10 cm = 1 000 cm³ = 0.001 m³

First, find the volume in cubic metres.

Now use the conversion between cubic metres and litres:

4. The conversion from Fahrenheit to Celsius degrees is \(^{\circ}\text{C} = \dfrac{5}{9}\left(^{\circ}\text{F}-32\right).\) Find:
    (a) \(0^\circ\text{F}\) in \(^{\circ}\text{C},\qquad\) (b) \(100^\circ\text{F}\) in \(^{\circ}\text{C},\qquad\) (c) \(-40^\circ\text{C}\) in \(^{\circ}\text{F}\)

💫(a) For \(0^\circ\text{F}\) in \(^{\circ}\text{C}\): \(\; ^{\circ}\text{C} = \dfrac{5}{9}(0 - 32) \) \(= \dfrac{5}{9}(-32)\) \(= -17.78^{\circ}\text{C}\)

💫(b) For \(100^{\circ}\text{F}\): \(\;^{\circ}\text{C} = \dfrac{5}{9}(100 - 32)\) \( = \dfrac{5}{9}(68)\) \(= 37.78^{\circ}\text{C}\)

💫(c) Now convert \(-40^{\circ}\text{C}\) to \(^{\circ}\text{F}\), use the inverse formula: \(^{\circ}\text{F} = \dfrac{9}{5}{}^{\circ}\text{C} + 32\)

- Cheetahs can reach speeds up to 120 km/h.

• How fast is it in m/s?
• What about feet/quarter-hour? (1 foot $\approx$ 30 cm)

- I can paint a wall with the speed of 0.2 $\dfrac{\text{m}^2}{\text{min}}$.

• How long will it take me to paint a 5 m $\times$ 3 m wall?

- Cheetahs can reach speeds up to 120 km/h.

• How fast is it in m/s?
• What about feet/quarter-hour? (1 foot ≈ 30 cm)

👉 \(120 \text{ km/h} = 120{,}000 \text{ m}/3{,}600 \text{ s} = 33.3 \text{ m/s}\)

👉 \(120 \text{ km/h} = 120{,}000 \text{ m/h} = 400{,}000 \text{ ft/h}\) ⇒ \( 100{,}000 \text{ ft}/ \text{quarted-hour}\)

- I can paint a wall with the speed of 0.2 \(\dfrac{\text{m}^2}{\text{min}}\).

• How long will it take me to paint a 5 m × 3 m wall?

👉 Wall area = \(5\text{ m} \times 3\text{ m} = 15 \text{ m}^2\). Then Time = \(\dfrac{15 \text{ m}^2 }{0.2\frac{\text{m}^2}{\text{min}}} = 75 \text{ min} = 1\text{ h }15\text{ min}\)

The Gimli Glider was an Air Canada Boeing 767 that ran out of fuel mid-flight in 1983 due to a metric conversion error—the fuel was loaded in pounds instead of kilograms. The pilots glided the plane safely to an emergency landing at a former airbase in Gimli, Manitoba, with no fatalities.

When values vary across a broad scale, it is helpful to think in terms of the exponent $m$, which we call an order of magnitude — $\log_{10}(10^m)$

$\qquad\qquad 322.5 \text{ m}$ $=322.5 \text{ m} \times \dfrac{ 1 \,000\, 000 \,\mu\text{m}}{1 \text{ m}}$

$\qquad\qquad \qquad \quad =3.225 \times 10^{2} \times 10^{6} \,\mu\text{m}$

$\qquad\qquad \qquad \quad =3.225 \times 10^{8} \,\mu\text{m}$

$3 \times 10^{8}\,\dfrac{\text{m}}{\text{s}}$ $= 3 \times 10^{8}\,\dfrac{\text{m}}{\text{s}} \times 3\,600\,\dfrac{\text{s}}{\text{hour}} $ $= 3 \times 10^{8} \times 3\,600\,\dfrac{\text{m}}{\text{hour}} $

$\qquad = 3 \times 10^{8} \times 3\,600\,\dfrac{\text{m}}{\text{hour}} \times \dfrac{1}{1.5\times 10 ^{11}}\, \dfrac{\text{AU}}{\text{m}} $

$\qquad = 3 \times 10^{8} \times 3\,600 \times \dfrac{1}{1.5\times 10 ^{11}}\, \dfrac{\text{AU}}{\text{hour}} $ $= \dfrac{(3\times 3.6) \times 10^{11}}{1.5\times 10 ^{11}}\, \dfrac{\text{AU}}{\text{hour}} $

$\qquad= \dfrac{3\times 3.6 }{1.5}\, \dfrac{\text{AU}}{\text{hour}} $ $= 7.2\, \dfrac{\text{AU}}{\text{hour}} $

Mole (unit)

The mole (mol) is a unit of measurement, the base unit in the SI for amount of substance, an SI base quantity proportional to the number of elementary entities of a substance.

Molar mass

The molar mass (M) of a chemical substance (element or compound) is defined as the ratio between the mass (m) and the amount of substance (n, measured in moles) of any sample of the substance:

\(\%\text{w/w}\) - weight per weight concentration - how many \(\text{g}\) of the solute there are in \(100 \, \text{g}\) of the solution

\(\%\text{w/v}\) - weight per volume concentration - how many \(\text{g}\) of the solute there are in \(100 \, \text{ml}\) of the solution

\(\%\text{v/v}\) - volume per volume concentration - how many ml of the solute are there in \(100 \, \text{ml}\) of the solution

1. What is the concentration and molar concentration of the 0.444 mol of cobalt chloride $\text{CoCl}_2$ in 0.654 L of solution (129.839 g/mol)?

2. A patient has a serum (39.0983 g/mol) level of 3 mmol/L.
  (a) How many mmol are present in a 50 ml sample?
  (b) How many mg are present in a 50 ml sample?

3. A glucose solution contains 10% w/v of glucose in water. How much glucose is in 1L of the solution?

4. What weight of hydrochloric acid is needed to produce 50 g of 30% w/w acid?

1. What is the concentration and molar concentration of the 0.444 mol of cobalt chloride $\text{CoCl}_2$ in 0.654 L of solution (129.839 g/mol)?

  💫Ans. Mass concentration = 88.15 g/L;   Molar concentration = 0.6790 mol/L.

2. A patient has a serum (39.0983 g/mol) level of 3 mmol/L.
  (a) How many mmol are present in a 50 ml sample?
  (b) How many mg are present in a 50 ml sample?

  💫Ans. (a) 0.150 mmol. (b) 5.865 mg.

3. A glucose solution contains 10% w/v of glucose in water. How much glucose is in 1L of the solution?

  💫Ans. 100 g glucose per 1 L.

4. What weight of hydrochloric acid is needed to produce 50 g of 30% w/w acid?

  💫Ans. 15 g HCl (to make 50 g of 30% w/w).

1. What is the concentration and molar concentration of the 0.444 mol of cobalt chloride $\text{CoCl}_2$ in 0.654 L of solution (129.839 g/mol)?

1. Molar concentration (M): \(C = \dfrac{\text{moles}}{\text{volume (L)}} = \dfrac{0.444\ \text{mol}}{0.654\ \text{L}}\).
2. Calculate: \(C = \dfrac{0.444\ \text{mol}}{0.654\ \text{L}}= 0.678899\ \text{mol / L}\). Rounded to four decimals: $0.6790 \,\text{mol / L}$
3. Mass of solute: \(m = n \times M = 0.444\ \text{mol} \times 129.839\ \text{g / mol} = 57.6485\ \text{g}.\)
4. Mass concentration $(\text{g/L})$: \(\rho = \dfrac{m}{V} = \dfrac{57.6485\ \text{g}}{0.654\ \text{L}} = 88.1476\ \text{g / L}.\) Rounded: \(88.15 \,\text{g / L}\)

Thus, Mass concentration = \(88.15 \,\text{g / L}\); Molar concentration = $0.6790 \,\text{mol / L}$.

2. A patient has a serum (39.0983 g/mol) level of 3 mmol/L.
  (a) How many mmol are present in a 50 ml sample?
  (b) How many mg are present in a 50 ml sample?

1. Convert volume to litres: \(50\ \text{mL} = 0.050\ \text{L}.\)
2. (a) Amount in mmol: \(n_{\text{mmol}} = C(\text{mmol / L}) \times V(\text{L}) = 3\ \text{mmol / L}\times 0.050\ \text{L} = 0.150\ \text{mmol}.\)
3. (b) Mass in mg: 1 mmol corresponds to \(M\,/ \,1000\) g = \(39.0983\ \text{mg}\).
   So \(m(\text{mg}) = 0.150\ \text{mmol} \times 39.0983\ \text{mg / mmol} = 5.864745\ \text{mg}.\)
   Rounded: 5.865 mg.

Thus, (a) 0.150 mmol; (b) 5.865 mg.

3. A glucose solution contains 10% w/v of glucose in water. How much glucose is in 1L of the solution?

1. Definition: 10% w/v means 10 g of solute per 100 mL of solution.
2. Scale to 1000 mL (1 L): \( \dfrac{10\ \text{g}}{100\ \text{mL}} \times 1000\ \text{mL} = 100\ \text{g}.\)

Thus, the answer is 100 g glucose per 1 L of solution.

4. What weight of hydrochloric acid is needed to produce 50 g of 30% w/w acid?

1. 30% w/w means mass fraction of solute = 0.30 (i.e. 30 g HCl per 100 g solution).
2. Mass of pure HCl required: \(m_{\text{HCl}} = 0.30 \times 50\ \text{g} = 15\ \text{g}.\)
3. To prepare the 50 g solution you would combine 15 g HCl with 35 g solvent (usually water) to give a total mass of 50 g.

Hence, the answer is 15 g HCl (to make 50 g of 30% w/w solution).