# As of light decreases productivity decreases

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As % of light decreases, productivity decreases resulting in less carbon assimilated. Less light = Less photosynthesis = Less carbon dioxide used = Less oxygen released into the water Using your data table, what seems to be the trend as the % of light increases? WHY? As % of light increases, productivity increases resulting in more carbon assimilated. More light = more photosynthesis = more carbon dioxide used = more oxygen released into the water Where would you say this organism is using as much energy as they are making? WHY? When Net = 0 (about 9%) This is where photosynthesis and respiration of the plant seem to be equal. No net oxygen is being released into the water, what is being released is equally being used by the plant in respiration. Using your table and graph, explain why most of the time there are bigger plants on land than in the sea? Explain this in terms of evolution. Plants evolved to live on land because it is easier to acquire carbon dioxide (more abundant), and there is more light available as it doesn’t have to % light DO (mg O 2 /L) Gross PP = ottle – Dark (mg O 2 /L) Net PP = Bottle – Light (mg O 2 /L) Gross on fixed in mgC/L Gross PP x 98 x 0.536 Initial 8.4 -- -- -- Dark 6.2 -- -- -- 100% 10.2 4 1.8 1.50 65% 9.7 3.5 1.3 1.31 25% 9.0 2.8 0.6 1.05 10% 8.5 2.3 0.1 0.86 2% 7.1 0.9 -1.3 0.34

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Dilution & pH Review Surviving C 1 V 1 = C 2 V 2 C 1 = original concentration of the solution, before it gets watered down or diluted. C 2 = final concentration of the solution, after dilution. V 1 = volume about to be diluted V 2 = final volume after dilution By drawing the " X " through the equal sign and filling in the formula with letters of a size permitted by the borders of the " X ", it reminds you that : for all dilution problems C 1 > C 2, and V 1 < V 2 . It makes sense because to dilute, we add water. This increases the volume but lowers concentration. Examples by Type : 1. Easiest: Joe has a 2 g/L solution. He dilutes it and creates 3 L of a 1 g/L solution. How much of the original solution did he dilute? Solution: We are looking for V 1 : C 1 V 1 = C 2 V 2 2x = 1(3) x = 1.5 L 2. Joe has 20 L of a 2 g/L solution. He diluted it, and created 3 L of a 1 g/L solution. How did he make such a solution? We're only going to use part of the 20 L. Remember we have to end up with 3 L after dilution, so not only do we have to start with less than 20 L but also less than 3 L. How much is unknown = V 1 , and it amounts to the same problem as in example 1, but don't use the 20L. C 1 V 1 = C 2 V 2 2x = 1(3) x = 1.5 L
3. Joe has 20 L of a 2 g/L solution. To this solution he adds 30 L. What is the final concentration of the solution? If 30 L is added to the 20 L, then the volume about to be diluted, V 1 , is 20 L. And adding 30 L of water to a 20 L solution creates a final volume, V 2 , of 50 L. Our unknown is C 2 . C 1 V 1 = C 2 V 2 2(20) = x(20 + 30) x = 40/50 = 0.80 g/L pH Reminder: pH = -log (H + ) Which is more acidic? (H + ) of 1.0 x 10 -8 or 1.0 x 10 -12 Which is more basic? (H + ) of 1.0 x 10 -6 or 1.0 x 10 -3 Stomach acid has a pH of about 1-2. What would the H + concentration be around?

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