Most fish need at least 4 ppm dissolved O2 in water for survival....

Question

Most fish need at least 4 ppm dissolved O2 in water for survival. (a) What is this concentration in mol>L?

Accepted Answer

Hi everyone for this problem. It reads the maximum allowable amount of nitrate measured as nitrogen and drinking water is 10 parts per million by mass, calculate the corresponding polarity of the nitrate ion. So our goal here is to calculate the malaria t of the nitrate ion. So let's just recall what is malaria T, malaria T is moles over leader. And so our goal is to go from our 10 parts per million. Two moles per liter. Okay? And let's translate what tens parts per million means first. So 10 parts per million of nitrogen. Because it's telling us it's measured as nitrogen. This is equal to 10 g of nitrogen over one times 10 to the six g of water. So this is our starting point. So we want to go from this two moles of nitrate ion per liter of water. And we're going to need the density of water to convert from massive water to volume of water. Okay? And so the density of water is one g per middle leader. Okay, so let's go ahead and get started. So we have 10 g of nitrogen per one times 10 to the 6th g of water. Let's go ahead and convert our grams of water. Two liters of water. Okay? So and Using the density we know that in one gram of water We have one millim of water. So here our grams of water cancel. And we're in milliliters of water. But we want to go to leaders because Armel arat E. Is in the unit of leaders. And so and one mil a liter of water. There's 10 to the negative three leaders of water. Okay. So now are milliliters of water cancel. And we're left with leaders. Which is what we want for our denominator of Armel Arat E. Okay. So now we're going to go from grams of nitrogen. Two moles of nitrate ion. So to go from grams of nitrogen, two moles of nitrogen, we need the molar mass of nitrogen. Okay. And so in one mole of nitrogen using our periodic table to find the mass there's 14.1 g of nitrogen. Okay so our grams of nitrogen here cancels with our grams of nitrogen here and now we need to use a multiple ratio between nitrogen and nitrate to get our molds of nitrate and one more. The ratio is 1 to 1. So for every one mole of nitrogen there's one more of nitrate ion. Okay, So let's go ahead and plug that into our table. So one mole of nitrogen for every one mole of nitrogen there's one more of nitrate. Okay. And so here our molds of nitrogen cancel. And we're left with mold of nitrate ion in the numerator. So now we have our desired units and we can do our calculation and solve. And so when we do that we're going to get a final answer of 7. Times 10 to the -4 moller which is most per liter of our nitrate ion. We can write out actual nitrate ion here. Okay. so this is our malaria T and this is our final answer. That's the end of this problem. I hope this was helpful.

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