Suppose you dilute 25 microliters of a chromium (III) acetate Cr(CH3COO)3, stock solution to 2.5 liters final volume...

Suppose you dilute 25 microliters of a chromium (III) acetate Cr(CH3COO)3, stock solution to 2.5 liters final volume. your final solution concentration is 250 nanomolar. What is the concentration of the original stock?
I know this is a simple dilution problem but I cannot determine the units. The answers are
A. 2.5 M
B.250 milimolar
C.25 milimolar
D.2.5 milimolar
E.250 micromolar
Thanks

just a sec lemme grab a pencil

Fuck you for making me remember my chem 2 class

shameless self bump. Got more if you guys are up for it. This teacher is a joke.

it should be 625nanomols/25microliters or 25nanomols/microliter, that would make it 25*10^-9/10^-6 or 2,5 milimolar

25 milimolar, sorry, i screwed it up

sorry my bad again its 2.5 mililolar

I had originally chosen 25 milimolar, thank you.

Calculate the mass of ammonium sulfide (NH4)2S necessary to precipitate all of the chromium from a solution that contains 20g of chromium (III) acetate, Cr(CH3COO)3.
A.20
B.15
C.11
D.8.9
E.5.6
I got the answer E

No it's 25 milimolar, here's the calculation, I really did made an honest mistake the first time

wrong pic, from earlier thread

Well according to my calculations its D.

Thank you again, those look similar to my calculations, I just was not able to transfer the units, but I see what you did. I could use help on another problem if you would.

Could you show me how you calculated? I have taken a Chem 2 class before, but this teacher uses handwritten notes and doesn't have practice problems. Nor does he instruct how to solve, he will just show.

No, it's 8, 92 so it's D. You have to account for the fact that acetate and ammonium react in the 1:1 ratio, but you will get 2 ammonium ions for each ammonium sulphide molecule and three acetate ions for each cromium acetate molecule, so when you even that out you get 8,9234 grams of ammonium sulfide

>Sup Forums is full of degenerates with lack of intelligence.

What the actual fuck?!? I don't get any of this shit. No. Let me drive my truck and haul some goods. That I know... But this?! It's all black to me.

C: 25 mM

25 microliters to 2.5 liters, that's 1:100.000

Now multiply 250 nM by 100.000. First by 100, you get 25 microM, now by 1.000, that's 25 mM. Not very difficult.

OP, you here still? I will go to sleep soon, so hurry if you have any more problems

Suppose you dissolve 78 mL of acetonitrile CH3CN, to 1.0 L with deionized water at room temperature. What is the molarity of the acetonitrile solution? Assume density of .786 g/ml
A..77M
B..92M
C.1.5M
D.2.6M
E.3.1M

It's C, you have 78mL times .786 g, that gives you 61,308 g of acetonitrile in total, that in turn amounts to 1,493 mol which diluted to 1L gives 1,493mol/L

Calculate molarity of pure Phenol. Assume a density of 1.07 g/ml
A.13.7M
B.12.5M
C.11.4M
D.10.1M
I got the answer C 11.36
Let me know what you get

Okay, I also got C. It's just that of the 10 questions, I have gotten 5 Cs, and 3 Ds, just worries me having so many similar letter answers.

Yes, the answer is C, so I suppose you know the process

if the math checks out you shouldn't worry

Got another. I only ask this because my instructor likes to show off his knowledge of terms; therefore, we never get the official definition. Thanks

And another one. Do you guys want to see my instructors hand-written lecturing notes we are suppose to learn from?

This one is a, the solute is always present in lesser amount, regardless of the size of the molecule

Okay, I also got A. I will post his lecture notes here in a minute

This gives you 5*10^10 molecules per mililiter, and according to avogadros number that gives you 8,3*10^-14 mol/mililiter or 83*10^-12 mol/L, the answer is A

This is his solubility chart we are suppose to memorize. The instructor is a 5'6 dweeb, I don't know how to better describe him.

Also I really don't care about the lectures, I'm quite tired, do you need anymore help?

Probably hardest question on the homework, I got answer D.

getting more shameful self bump

You need 900L, the answer is A I will explain

So first you calculate the volume of the sphere with the given formula. Then you multiply it by 0,049 to get the volume of the gold. Now you calculate the substance of gold from the volume. Form the substance using the equation you can calculate the substance of NaCN needed and that in turn will give you the volume of 100% v/v NaCN needed.