Most fish need at least 4 ppm dissolved O₂ in water for survival. (b) What partial pressure of O₂ above water is needed to obtain 4 ppm O₂ in water at 10 °C? (The Henry's law constant for O₂ at this temperature is 1.71⨉10^-3 mol/L-atm.)

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Hello everyone today. We have the following problem. A concentration of as low as 0.5 parts per million of dissolved ammonia and water can already cause damage to the gills of the fish, calculate the partial pressure of ammonia in tour when it reaches a concentration of 0.5 parts per million at 25 degrees Celsius, ammonia has Henry's law constant of 59.8 moles Over L per atmospheres. Assume that the density of the solution is one grand per middle leader. So to calculate that we're gonna take the gas and equal it to this, Henry's constant times our partial pressure of the gas. So in malaria T This is going to be equal to 0.05 parts per million. Equal To our 0. milligram of ammonia divided by our kilogram of solution. Next, what we're gonna do to further calculate that Is we're gonna take what we just saw it for our 0. mg of ammonia per kilogram solution. We're gonna multiply by several different factors. First. We're gonna get rid of milligrams in that. We're gonna use a conversion factor that one mg is equal to 10 to the negative three g. We're gonna multiply by the molar mass of ammonium which states That one mole is equal to 17.03 g. And this is from the periodic table. We're then gonna get rid of grams with the conversion factor that one kg is equal to 10 to the third grams. We're then going to convert our grams to leaders. And in doing that, we're gonna say that one g is equal to one male leader. Given the density and then we're gonna multiply that by one, middle year is equal to one leader. And this is gonna give us our 2.9 times 10 to the power of negative six moles per liter. This is gonna be our polarity. So from our earlier formula, if we rearranged it to get the partial pressure of our gas equal that Polarity of the gas over Henry's constant, it's gonna be our 2.9 times 10 to the negative six moles per liter Divided by Henry's Continent 59.8 moles per liter of times atmosphere. And this is going to give us a value of 4.9 times 10 to the - atmospheres. However, the answer recall wants us to have it in tour. So we're gonna use the conversion factor that one atmosphere is equal to 760 tour. When are units of atmospheric cancels out? We're left with a final partial pressure of 3.73 times to the negative fifth tour. And with that, we have solved the problem overall, I hope this helped hand until next time

Most fish need at least 4 ppm dissolved O₂ in water for survival. (b) What partial pressure of O₂ above water is needed to obtain 4 ppm O₂ in water at 10 °C? (The Henry's law constant for O₂ at this temperature is 1.71⨉10^-3 mol/L-atm.)

Verified Solution

Key Concepts

Henry's Law

Dissolved Oxygen Concentration

Temperature Effects on Gas Solubility

Most fish need at least 4 ppm dissolved O2 in water for survival. (b) What partial pressure of O2 above water is needed to obtain 4 ppm O2 in water at 10 °C? (The Henry's law constant for O2 at this temperature is 1.71⨉10-3 mol/L-atm.)

Verified Solution

Key Concepts

Henry's Law

Dissolved Oxygen Concentration

Temperature Effects on Gas Solubility

Most fish need at least 4 ppm dissolved O₂ in water for survival. (b) What partial pressure of O₂ above water is needed to obtain 4 ppm O₂ in water at 10 °C? (The Henry's law constant for O₂ at this temperature is 1.71⨉10^-3 mol/L-atm.)