Science questions not worth a thread II

[thetrooper]

Cough cough... a thread with plenty of dust.

Any ICP-OES techs here?

If so; possible interference for Co, Zn, Cr, Pb, Cu in Fe matrix?

 

[Reconciliator]

Dust quickly gathering again!

 

[thetrooper]

All we are is dust in the wind.

This place is in the end stage.

 

[thetrooper]

Could someone tell me the mechanics behind suspended solids, e.g. in wastewater, being environmentally challenging?

 

[thetrooper]

Could someone tell me the mechanics behind suspended solids, e.g. in wastewater, being environmentally challenging?

Nope?

How about buffer capacity of sea water for acidic consumption? Formulas, chemical species, salinity dependence?

 

[Kyriakos]

I have a question. It's tied to this (but not this)

While it is simple to find the area, I was wondering why it is almost (but a tiny bit less) 2 times the area of the square (roughly 1.9635 times). Maybe @Grendeldef ?
(a fast way to calculate the area is to drop a vertical from A, to the line OD..., let's say at a point X, and notice that AX=OD, which leads to a line AO^2=5OD^2. And since AO=r, area is 5OD^2 pi/4, roughly 55).

 

[thetrooper]

Gotta admire your optimism @Kyriakos

 

[Kyriakos]

Gotta admire your optimism @Kyriakos

Wait, isn't this Reddit math?
Oh, it's CFC ^^

 

[uppi]

I have a question. It's tied to this (but not this)

View attachment 660520

While it is simple to find the area, I was wondering why it is almost (but a tiny bit less) 2 times the area of the square (roughly 1.9635 times). Maybe @Grendeldef ?
(a fast way to calculate the area is to drop a vertical from A, to the line OD..., let's say at a point X, and notice that AX=OD, which leads to a line AO^2=5OD^2. And since AO=r, area is 5OD^2 pi/4, roughly 55).

The area of the square is 2OD^2, so the ratio is
5/8*pi = 1.96349

 

[Kyriakos]

The area of the square is 2OD^2, so the ratio is
5/8*pi = 1.96349

Yes, but I don't mean "why" as in "do the calculation" (I already did so ). I mean "why" intuitively. As in, why (if you prefer) would this particular arrangement of an inscribed square, be virtually half of the quarter-circle where it is inscribed?
I hope I conveyed what I mean... Since the area of the quarter-circle is, as noted, 5OD^2 pi/4, of course if you divide that by the area of the square (which is 2OD^2) you get 5/8 pi. But it wasn't the spirit of my question!

Another hint as to how I meant it, was that I thought of arriving at it by some specific integral. Can you think of such, using the inscribed square to define the function? (can have more than one free variables).

 

[CKS]

Densities don't add, so it isn't your formula.
Volumes don't add, either, so this seems pretty complicated to me.

How close do you need to get to reality? Are your salts already dissolved in something, or are they solids dissolving into the acid? If the salts are already dissolved in water and all your solutions are dilute, then the volume won't change much when you mix them, so (m1+m2+m3+macid)/(V1+V2+V3+Vacid) will be close to the right density. If the salts are solids when you start, the volume will potentially change a lot when they dissolve.