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Apling LearningAccording to the EPA the maximum contaminant lavel (MCL) of cadmium drinking water is 0.0050 mgl(a) Convert the MCL of cadmium from milligrams per li...

Question

Apling LearningAccording to the EPA the maximum contaminant lavel (MCL) of cadmium drinking water is 0.0050 mgl(a) Convert the MCL of cadmium from milligrams per liter to parts per billion (Ppb) [email protected] faner has recontly 'dec= now well for his property: He sends 0 mL sample of the waler [0 the EPA t0 be lesled lor lhe Praseare of cadmium What is the maximum amount of Cd Sel? Lg) can be present in this 16.0 mL sample based on itsNurberU.OUMM18PG

apling Learning According to the EPA the maximum contaminant lavel (MCL) of cadmium drinking water is 0.0050 mgl (a) Convert the MCL of cadmium from milligrams per liter to parts per billion (Ppb) . Number ppb @h8 faner has recontly 'dec= now well for his property: He sends 0 mL sample of the waler [0 the EPA t0 be lesled lor lhe Praseare of cadmium What is the maximum amount of Cd Sel? Lg) can be present in this 16.0 mL sample based on its Nurber U.OUMM18 PG



Answers

An industrial wastewater contains 3.60 ppb cadmium, $\mathrm{Cd}^{2+} .$ How many $\mathrm{mg}$ of $\mathrm{Cd}^{2+}$ could be recovered from a ton $(1016 \mathrm{kg})$ of this wastewater?

So one part per million is equivalent of one g of solid per one million g of solution or one million g of solid per 1000 g of solution. So if I write this down for you, we have one ppm Is equal to one part Over 10 to the six parts to the maximum contamination level for this substances. Not point not not to PPM that's what we have. Is not point, not not too milligrams. Taiwan kg, that's equal to not point not not two mg Divided by 10 to the six mg. And so one part per billion is equivalent to one g of solid per one billion g of solution. So if we move on here, So what we have is five parts per billion is equal to five micrograms per kilogram which is equal to five micrograms. Multiplied by 10 to the minus three mg, Divided by one microgram Divide that all by 10 kg. Multiplied by 10 to the 3g over 10 kg. Multiplied by 10- three mg, divided by one g. What we get is 5.5 times 10 mg, Divided by 10 to the nine mg, Which gives us 5.0 parts per billion. So because five mg where we have five times 10 to the sit minus six g is greater than not point not, not two mg. The sample of water is not within the maximum contamination level. Yeah.

So in this question we want to determine if the mercury content of a stream is above the minimum, that's considered to be safe, which is one part per billion by weight. So analysis concluded the concentration was .68 parts per billion or PPB. So we want to know how many grams of mercury is present in 15 liters of water, The density of which is .998 g for middle leader. And were given the conversion factor one PPb. Mercury is sick with one nanogram mercury over one g of water. So we can use dimensional analysis to convert from PPB two g per liter of water. So we start with point 6 8 p. b. b. Of mercury and then one PPV mercury as you go through one nanogram of of mercury over one g water because that was the converted factor that were given. So then we have the PPB cancelling out on top and bottom. And then we have .68 milligram of mercury per one g of water. So that is the concentration um in terms of the masses of the mercury and the water. So then what we're going to do is we need to get that in grams of mercury per liter of water. So to go from the mass of water to volume of water, we use the density. So we have the grams of water on the bottom, cancelling out with grams water on top. And then that will convert to millions of water on the bottom. We have to get that in leaders. So they multiply by 1000 ml over one leader. So millions cancel. We're left with leaders of water. And then we want to know specifically how much is in 15 leaders of water. So we're going to multiply this by 15, the leaders will cancel because we want to get this in nanograms of mercury. So we're getting rid of the leaders of water. So then we just need to cancel out the nanogram of mercury and convert grams of mercury because I want to know how many grams of mercury are in the 15 years of water. So we cancel out by using a conversion factor one g is equal to 10 to the nine nanograms of mercury that cancels. And then we get grams of mercury. So if we multiply everything out, we get 1.2 times 10 to the negative five g of mercury. And then we can we can compare that to the safe amount which is given in parts per billion. Um So we know that the concentration was less than one part per billion. Um And so therefore it is going to be safe because we already told it's less than one parts per billion. But now we know how many grams of mercury are in 15 liters of water through using dimensional analysis

Yeah. So in discussing, uh, we have been given a concerned person of mercury in a stream that is required to 0.6 parts per billion. Buyback. And we have to find The amount of Mercury in 15 L of ST Who's the M series. Also given that is 0.998 g per family? No, in this we have to find the amount of mercury. So first we know that the density of stream is 0.9 and it grabs forever. So find the weight of the little stream. So from densely it is evident that one ml of the stream, where's 0.998g. Similarly, 50 m. That is 15,000. So 15,000. Middle of to move. We will apply the United method that is 0.9 and eight by one into 15,000. So we'll get the weight of 15 liters of streamers, 9 14,070 g. So this is the weight of the 15 liter of stream. Now, the concentration of Mercury has given us the upon 6, 8 parts per billion. By the way. By what means that in 10 to over nine parts of the stream, it contains 0.68 part of mercury. That is in terms of weight. It means to say that 10 to about nine g of stream contains 0.68 g of mercury. Now we'll apply the United Matter. That is one grammar stream will contain your 10.68 but into the nine g of mercury. Now, we'll find out how much, this much, this much amount of the stream contents. How much maturity that is 50 m of the stream. That uh that is where this method is. 14,009 and 70 g will content. How much mercury? So 14,009 and 70 g streaming content. Do you want 68? Right into the nine into 14,970. So when you will get 1.018-10 to the minus side, with them? So much So you can say that 15 L of stream will contain this much amount of mercury, and hence this is the answer yes.

In the problem. One along. We're talking about the water being contaminated with mercury and having concentration of twice the legal limit. Oh, which is in this case? Oh, Michael. Oh, for oh, to feed the M and TV. M stands for parts her 1,000,000 or one divided by 10 to the car six. So if you know that we can assigned bbm concentration, I value off, um, Millie graphs for kayla graphs because one milligram is equal to one times 10 to the her off 56 killing grass. Um, so now that we've gotten that out of the way, we can talk about the actual question of the problem. Um, that problem is asking us how much off this contaminated water a person needs to consume in order to get. You're a mass of mercury off 50 Millie grass. So now that we know the concentration on we know the mass, I will need to calculate we can just say that, um, the concentration bbm, which I will mark, uh, see is equal to em, which is mass. Uh oh. Mercury divided by mass off solution. Um, if we solve this for am of the solution that needs to be invested in ingested. We get that the, um solution is equal to and rookery tried it by the concentration so 50 another grabs divided by no point. Oh chlo for co Miller grabs Her kilogram was established before. What that gives us is a mass of order which is equal to bottom 0.3 times 10 to the power for campgrounds. So the answer is in order Thio in just toe total amount off mercury 15 milligrams. You need to drink 13,000 killer grass off this contaminated water, but


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