<p>2a) 6.09 x 10^-3</p>
<p>b) 1.4-.2 = 1.20 atm
c) .2 atm (5/1) = 1.0 atm</p>
<p>For number 2, C3H8 was the alkane. Hydrocarbon was the limiting reactant. Its partial pressure was .2. This was fully consumed. You know that CO2’s partial p was .6. So the ratio is 1:3 and H2O’s is 1:4. Looking at the number of oxygen, we know that there are 10 moles of O. So there should be 10 on the left too. So the coefficient for O2 should be 5. Therefore .2 * 5 atm should be consumed= 1.00 atm and .2 was left over.
Edit: 10 on the left.</p>
<p>Will .006 moles for part a not be accepted (number 2)</p>
<p>d) C3H8 + 5O2 → 3CO2 + 4H2O
We know this because we know since O2 was left over that the hydrocarbon was the limiting reagent, therefore since .2 atm of hydrocarbon was consumed and .6 atm of CO2 was formed, then it must be a 3:1 CO2 to hydrocarbon ratio meaning the hydrocarbon must have 3 C. Repeat for the 8 H</p>
<p>@flower9898: should be 3 sig figs, so… maybe.</p>
<p>e) .00609 * (44g/mol) = .268 g</p>
<p>How to calculate 1f? </p>
<p>I got 4.67, but people say its 4.8…
My henderson was set up to be pH=4.2 + log ((0.0075)/(0.01-0.0075))</p>
<p>The 0.0075 comes from the 30ml of 0.250M NaoH.</p>
<p>At futuredoc: I’m with you, I used HH equation and pretty sure I got what you got. Of course, have to convert that to H+. 2.1 x 10^-5 M H+</p>
<p>■■■ I bombed this…</p>
<p>I got 4.67 also. Well, 4.68.
What did you guys get for a K in number 3? I’ve heard people say like 254 and .128…and other randoms.</p>
<p>At chemcat12: I got like 3.something 1/min. Used 1/[A]t - 1/[A]0 = kT</p>
<p>Edit: 3.043 min-1</p>
<p>Oh ok cool. I wonder why people were getting 4.8…maybe they were doing a flat out ICE table?</p>
<p>And i dont remember, ill try calculating it now.</p>
<p>Bro, it might have been 4.68. I don’t recall very well. I don’t feel like solving. :D. That rate constant is the right one. SO WHO IS UP FOR A 5?</p>
<p>e) Rate = k [A]
Rate = 3.043min-1 * (4.7x10-3/2)
Rate = 7.15 x 10^-3 M/min</p>
<p>… I think.</p>
<p>i got 0.094. hypno, thats the equation for the second order reaction. the first order reaction equation uses the ln(a) etc.</p>
<p>Anyone know if Form B is posted or when it should be posted??</p>
<p>Are the answers not on the website?</p>
<p>No they are not. I wonder if Form B is the make up one. :O</p>