The Cancer Cure

Thanks for all of your help. I appreciate it! 'night.

Hm, so: 24/67 = square root of 32/x so: (24/67)^2 = 32/x .1283136556 = 32/x x = 249.3888889 With the sig figs, x = 250. And that fine sir, is the correct answer. Thank you so much! What about the last question? - Last one for the night.

Thank you. :-p

The ratio of the unknown gas to the oxygen gas should be greater. So...I'm slightly confused. 24^2 + 32 (Oxygen Gas) = 67^2 + X (Unknown) ? Maybe ?

1. A gas of unknown molecular mass was allowed to effuse through a small opening under constant-pressure conditions. It required 67 s for 1.0 L of the gas to effuse. Under identical experimental conditions it required 24 s for 1.0 L of O2 gas to effuse. Calculate the molar mass of the unknown gas. (Remember that the faster the rate of effusion, the shorter the time required for effusion of 1.0 L; that is, rate and time are inversely proportional.) Find g/mol. Don't know what the heck to do. 2. A gaseous mixture of O2 and Kr has a density of 1.110 g/L at 425 torr and 300. K. What is the mole percent O2 in the mixture? Find %. Quote answer from number 1. Specific thanks to herman.

Thank you so much! I'd keep clicking reputation but it won't let me.

Yes!!! I got the correct answer. Thank you!

So it should be 245 + 745 + 244.5 = 1234.5 with three significant figures = 1240 torr? I tried 1240 torr (I'm using WebAssign by the way) and it gave me feedback that the answer was incorrect. What do you think is wrong?

Okay, I'll come back and tell you how it went, thank you very much, made it very clear.

I'm sorry for this second question, but I am very lost in this chapter. I have no idea what to do on this problem. Please advise me. gas pressures N2 245 torr (1 L) Ne 745 torr (1 L) H2 489 torr (.5 L) Each of the bulbs contains a gas at the pressure shown. What is the pressure of the system when all the stopcocks are opened, assuming that the temperature remains constant? (We can neglect the volume of the capillary tubing connecting the bulbs.) Thanks guys.

A sample of 2.00 g of SO2(g) originally in a 5.00-L vessel at 18°C is transferred to a 10.0-L vessel at 25°C. A sample of 3.50 g N2(g) originally in a 2.50-L vessel at 20°C is transferred to this same 10.0 L-vessel. What is the partial pressure of SO2(g) in the larger container? I have NO idea where to begin............please help me get started. I'm not sure how to combine the containers...or how the temperatures or the volumes change.. :-(

(a) Place the following gases in order of increasing average molecular speed at 300. K: CO, SF6, H2S, Cl2, HBr. (Use the appropriate <, =, or > symbol to separate substances in the list.) So, I know how to do (a): SF6 < HBr < Cl2 < H2S < CO SF6 being the slowest and CO being the fastest. (b) Calculate and compare the rms speeds of CO and Cl2 at 300. K. rms speed of CO in m/s rms speed of Cl2 in m/s I have no idea how to do (b). Someone give me a start please? Thanks!

Yep, thanks.

Well, I didn't think I was doing something wrong...but I was getting the answer wrong - so then I thought I had done something wrong. But then I found out it was a couple of decimals.

Okay, I wasn't sure if you minded if I wrote the numbers or not, but I will, sorry. So: Moles 4.14 g N2/28 g mol = 0.147857142857142 moles N2 5.56 g H2/2.0158 g mol = 2.75821013989483 moles H2 7.06 g NH3/17 g mol = 0.415294117647 moles NH3 Total Number of Moles = 3.321361400398972 moles Mole Fractions 0.147857142857142 moles N2/3.321361400398972 moles = ~4.45% 2.75821013989483 moles H2/3.321361400398972 moles = ~83.04% 0.415294117647 moles NH3/3.321361400398972 moles = ~12.50% Partial Pressures .0445 * 5.06 atm = 0.22517 atm of N2 .8304 * 5.06 atm = 4.201824 atm of H2 .1250 * 5.06 atm = 0.6325 atm of NH3 Total atm = 5.059494 atm (Roundoff Error) So...what am I doing wrong? Sorry about not posting numbers earlier! Merged post follows: Consecutive posts mergedWell, now I feel quite dumb. My answers were correct, but I'm using WebAssign, and due to a 1% error because of roundoff error...my answer was incorrect. But since I worked it out again on here with multiple digits held, my answer was correct. I had .224 atm N2 instead of the right answer - .225 atm N2. Well, sorry, thanks for making me right it out though. :doh: Ignore this question, it is resolved.

Hey guys, Here is the question: A mixture of gases contains 4.14 g of N2, 5.56 g of H2, and 7.06 g of NH3. If the total pressure of the mixture is 5.06 atm, what is the partial pressure of each component? Find the partial pressures of N2, H2, and NH3. Here is my attempt: I took the grams of each one, and converted them into their moles via their grams divided by their molar masses. I added up their mole weights to get one large mole weight. I think took each individual mole weight and divided by the total mole weight. With this percentage, I multipled it by the total pressure, 5.06, to get their partial pressures. What did I do wrong? - Because this isn't working. Thanks guys.

Just wanted to say really nice site - looks like a lot of members of whom are smart and nice, nice format, great answers. I'll definitely be back. Hopefully I can educate and be educated.