OpenStudy (anonymous):

sodium phosphate is commonly found in processed meat and tuna. Calculate the concentration of sodium ions in a solution of .564 M

OpenStudy (anonymous):

@Photon336

OpenStudy (anonymous):

I'm doing a timed thing so I don't exactly have the time to go through the steps. But I can after I've finished.

OpenStudy (anonymous):

@Photon336

OpenStudy (photon336):

let's start by writing the equation

OpenStudy (anonymous):

Hang on. I put the wrong one. Sorry.

OpenStudy (photon336):

\[Na_{3}PO4(aq) --> 3Na ^{+} + PO4^{3-}\]

OpenStudy (photon336):

okay

OpenStudy (anonymous):

C14H9Cl5 = 5.80 x 10^-8 M 5000L of water was tested And it's wanting to know grams of C14H9Cl5 was found in the water sample.

OpenStudy (anonymous):

Sorry about posting the wrong one.

OpenStudy (photon336):

no problem so you're given the molar concentration of this right? and the amount of water

OpenStudy (anonymous):

Is that enough information to work with? I would type the full thing out but this thing is timed.

OpenStudy (anonymous):

Yes

OpenStudy (photon336):

remember M = Mol/L

OpenStudy (anonymous):

I got confused what I'm suppose to convert. Do I do all of the C14etc?

OpenStudy (photon336):

Well let's work with what you're being asked, you need to find the # of grams in your sample so what would you think you would need to do this?

OpenStudy (anonymous):

the moles

OpenStudy (photon336):

exactly

OpenStudy (photon336):

they gave you the amount of volume in liters

OpenStudy (photon336):

what do you think we should do first when given the molarity, remember what molarity means.

OpenStudy (anonymous):

use it to find the moles?

OpenStudy (anonymous):

I mean we got morality and liters.

OpenStudy (photon336):

let me show you

OpenStudy (photon336):

\[Molarity = \frac{ moles }{ liter }\] follow?

OpenStudy (anonymous):

Right

OpenStudy (photon336):

so watch when we multiply the molarity by the volume in liters we get the number of moles. can you see why this would be the case from what is shown below? \[\frac{ moles }{ liter }*(\frac{ liters }{ 1 }) = moles \]

OpenStudy (anonymous):

2.9 x 10^-4

OpenStudy (anonymous):

Then using molar mass to convert it to grams it would be .10280 g. Sorry if I seem to be rushing. The thing I'm doing is timed and the time is running short on it.

OpenStudy (photon336):

\[5.80 x 10^-8 \frac{ mol }{ liters}*\frac{ (5000 liters) }{ 1 } = 2.9x10^{-4} moles\]

OpenStudy (photon336):

no problem the key here is about molarity you can always find the # of moles by multiplying by the volume of solvent by the molarity.

OpenStudy (anonymous):

Is the .10280 grams right?

OpenStudy (photon336):

what's the molar mass that you have?

OpenStudy (anonymous):

354.49

OpenStudy (photon336):

\[moles *\frac{ grams }{ mol } = grams\]

OpenStudy (photon336):

\[2.9x10^{-4} moles * (354.9 \frac{ grams }{ mol })\]

OpenStudy (photon336):

@staldk3 do you see a pattern like this? the unit that you don't want is in the denominator and cancels out and the unit you want is in the numerator.

OpenStudy (anonymous):

Yeah. It always makes me happy when they cancel out. lol

OpenStudy (anonymous):

So when I did it I got .1028021 grams.

OpenStudy (photon336):

yep did they ask for significant figures?

OpenStudy (anonymous):

Yes so .103 grams?

OpenStudy (photon336):

yeah I think so because 2.9 has two significant figures so I think our answer should have two as well.

OpenStudy (anonymous):

Wait should it be .10 then?

OpenStudy (photon336):

if I remember when you multiply numbers or divide your answer has the fewest significant figures.

OpenStudy (photon336):

So I think, that in 2.9 there are two significant figures and in 1.028021 there are like 7 so our answer should be .103 well .103 has 3 so it should have 2 I think it's .10

OpenStudy (anonymous):

Alright. Thanks for helping so quickly!

OpenStudy (photon336):

yeah no problem

OpenStudy (photon336):

take care

OpenStudy (anonymous):

You too