1. The document contains worked chemistry and physics problems involving unit conversions between grams, moles, liters, and other units. 2. Calculations are shown for concentrations in parts per million and billion, molar quantities, and densities in seawater and other samples. 3. Equations are given for thermal expansion coefficients, thermal conductivity, and heat flux with units worked out consistently.

1. 1a • Work out this proportion in grams:
80ppm = 80μg per g
• Multiply by the mass of the sample:
80μg x 50 = 4000μg
• Convert micrograms to milligrams:
4000
1000
= 𝟒𝐦𝐠

2. 1b • Convert to μg per g to find the ppm of 1g:
850ng = 0.85μg
0.85μg in 1g = 0.85ppm
• Multiply by the mass of the sample:
0.85μg in 2g = 2 x 0.85 = 𝟏. 𝟕𝐩𝐩𝐦

3. 1c • Convert this to kg of salt per kg of water:
35ppt = 35mg per g = 35g per kg = 0.035kg per kg
• Multiply by the mass of the sample:
0.035 x 1.4 x 1021 = 𝟒. 𝟗 𝐱 𝟏𝟎 𝟏𝟗 𝐤𝐠

4. 2a • Calculate how many grams are in one mole by multiplying
the number of atoms of each element by their mass:
1 mole = 2 x 1.0079 + 1 x 15.999 = 18.015g
• Multiply this by the number of moles in the sample:
3 x 18.015 = 𝟓𝟒. 𝟎𝟒𝟒𝟒𝐠

5. 2b • Convert ppm to g per kg:
441ppm = 441μg per g
441,000μg per kg = 0.441g per kg
• Divide by the mass of one mole of calcium:
1 mole Ca = 40.078g
0.441
40.078
= 𝟎. 𝟎𝟏𝟐𝐦𝐨𝐥 𝐂𝐚

6. 2c • Convert ppt to g per kg:
10.8ppt = 10.8g per kg
• Divide by the mass of sodium:
1 mole Na = 22.99g
10.8
22.99
= 𝟎. 𝟒𝟕𝐦𝐨𝐥𝐤𝐠−𝟏 𝐍𝐚

7. 2d • Convert from ppb to moles per kg:
45ppb = 45ng per g = 45μg per kg
1 mole PO4
3−
= 30.974 + 4 15.999 = 94.97g
45 x 10−6gkg−1
94.97g
= 4.74 x 10−7
molkg−1
• Multiply by the flux of water:
4.74 x 10−7molkg−1 x 3.6 x 1016kgyr−1 = 𝟏. 𝟕𝟏 𝐱 𝟏𝟎 𝟏𝟎 𝐦𝐨𝐥𝐲𝐫−𝟏 𝐏𝐎 𝟒
𝟑−

8. 3a • Substitute the values for water into the equation:
MORT =
1.4 x 1021L
3.6 x 1016Lyr−1
= 𝟑𝟗 𝐱 𝟏𝟎 𝟑
𝐲𝐫

9. 3b • Substitute the values into the equation:
MORT =
1.4 x 1021L x 1.3 gL−1
3.6 x 1016Lyr−1 x 3.3 mgL−1
• Ensure all units are consistent:
MORT =
1.4 x 1021L x 1.3 gL−1
3.6 x 1016Lyr−1 x 0.0033 gL−1
= 𝟏. 𝟓𝟑 𝐱 𝟏𝟎 𝟕 𝐲𝐫

10. 4a • Convert 1ml into litres:
1ml = 0.001L
• Multiply this by the size of the sample:
5cm3
= 5ml = 0.001L x 5 = 𝟓 𝐱 𝟏𝟎−𝟑
𝐋

11. 4b • Work out the conversion between cm3 and m3:
1cm = 0.01m
1cm3 = cm x cm x cm = 0.01 x 0.01 x 0.01 m3 = 1 x 10−6m3
• Work out the conversion between 1L and 1m3:
1ml = 1cm3
1 x 10−3L = 1 x 10−6m3
1L = 1 x 10−3m3
• Multiply by the volume of the ocean:
1.4 x 1021
L = 1.4 x 1021
x 10−3
m3
= 𝟏. 𝟒 𝐱 𝟏𝟎 𝟏𝟖
𝐦 𝟑

12. 4c • Convert 142 mgL-1 into 142 mgm-3:
1𝐿 = 1 𝑥 10−3
𝑚3
1
1
𝐿−1 =
1
10−3
𝑚−3
1𝐿−1 = 103 𝑚−3
142𝑚𝑔𝐿−1 = 142 𝑥 103 𝑚𝑔𝑚−3 = 1.42 𝑥 105 𝑚𝑔𝑚−3
• Multiply by the sample size and convert mg to g:
1.42 𝑥 105 𝑚𝑔𝑚−3 𝑥 2𝑚3 = 2.84 𝑥 105 𝑚𝑔
2.84 𝑥 105 𝑥 10−3 = 𝟐𝟖𝟒𝒈

13. 4d • Calculate the number of moles of Mg in the ocean:
1 mole Mg = 24.305g
1.81 x 1021
24.305
= 7.45 x 1019mol Mg
• Use the volume of the ocean to work out the overall
concentration of Mg in seawater:
7.45 x 1019
mol Mg
1.4 x 1021L
= 𝟎. 𝟎𝟓𝟑𝐦𝐨𝐥𝐋−𝟏

14. 5 Note: Brackets around α, β and γ represent their units.
• Because the components are being summed, we know their units must all
be equal and equivalent to units of density. Remember α is multiplied by
temperature, units K, and γ is multiplied by pressure, units Pa.
kgm−3
= α K kgm−3
= β kgm−3
= γ Pa
• Rearrange the equations for (α), (β) and (γ):
α =
kgm−3
K
= 𝐤𝐠𝐦−𝟑
𝐊−𝟏
β = 𝐤𝐠𝐦−𝟑
γ =
kgm−3
Pa
= 𝐤𝐠𝐦−𝟑
𝐊𝐏𝐚−𝟏

15. 6 Note: Brackets around Qd and k represent their units.
• Because Qd and A.d are being summed, their units must be equal:
Qd = Wm−3. m = Wm−2
• Because you’re adding two components to make T in Kelvin, both components must
also have units of kelvin:
K =
Wm−3
(k)
m2 =
Wm−2
(k)
m
• Cancel the reoccurring units:
K =
Wm−1
(k)
• Rearrange for (k):
k =
Wm−1
K
= 𝐖𝐦−𝟏 𝐤−𝟏