Recent content by UWMpanther

Homework Statement Houdini is in a giant flask and he stands on a block where his feet are shackled and you need to calculate some vital information to help him out. Cross section of flask = r(h)=10/(sqrt(h+1)) water is being pumped into the flask at 22pi Takes Houdini 10 mins to escape Ignore...

I kinda understand. I see how the first term becomes n^4 but why is the last 0?

Homework Statement Evaluate the telescoping sum. a) Sigma i=1 to n [i^4 - (i - 1)^4] Homework Equations sigma i=1 to n [f(1) - f(i+1)] The Attempt at a Solution so i plug in for the eqns. and get 1^4 - n^4 the correct answer should be n^4 but don't know how to get that.

Ok wow I can't believe I did that. So I FOIL it out and get this x^2 = r^2 + 2ax - a^2 - y^2 + 2by - b^2 x= r-a-b+sqrt(2ax)+sqrt(2by) Does this look better?

Homework Statement Find the dimensions of the rectangle of largest area that can be inscribed in a circle of radius r. Homework Equations (x-a)^2 + (y-b)^2 = r^2 max area = 2x(2y) = 4xy The Attempt at a Solution (x-a)^2 + (y-b)^2 = r^2 = y=r-(x-a)+b I then plug...

yeah I got 21661 mins which then I converted to hours and that is 361 hours.

ok so \ln{\frac {[A]_{0}}{[A]_{t}}} = kt is what I'm going to use to calculate k and then i use t_{\frac {1}{2}} = \frac {\ln{2}}{k} to calculate for t_{\frac {1}{2}}

[SOLVED] Half Life Help Homework Statement The activity of a radioisotope is 3000 counts per minute at one time and 2736 counts per minute 48 hours later. What is the half-life of th radioisotope? This is where I'm completely lost.

Thank you very much!

is R not a constant? 0.08206 L\ atm\ mol^-^1\ K^-^1 so M=\frac {grams\ast R\ast T} {P\ast V} convert 3mL to .300L and 789mm Hg to 1.038atm

Ok so I am very confused here. How do I know how many moles I have? And do I need the entire PV= nRT formula?

[SOLVED] Solving for molar mass Homework Statement What is the molar mass of an ideal gas if a 0.622g sample of this gas occupies a volume of 300mL at 35*C and 789mm Hg? Homework Equations PV = nRT The Attempt at a Solution So n= no. moles present So I assumed n=.622/x so : x=...

Ok yeah that makes sense so the difference is .33 Thank you so very much!

Ok cool I got that far then when I was just guessing. So I basically went something like this: since in the equation their relationship is 4FeCl2 to 3O2, I just took the 2O2 x 1.5 so it was equal to the 3.0 mol given. So that put 4FeCl2 to 4.5 mol. But I guess that's all I solved for the...

I think the limiting reagent is the O2 but I don't know how to caculate it since it is given in mol not g.