<p>i wonder if i could have all of your permission to begin a homework help thread. if so, to begin with, i have acid/base questions...</p>
<p>VOH is slow cuz my prof doesn't answer the questions too often.</p>
<p>if there is approval, i will post the first question.</p>
<p>MIDTERM TOMORROW AND I FEEL UNPREPARED.</p>
<p>thanks.</p>
<p>You have my permission, and that's all you'll need. However, who knows if anyone will answer your questions.</p>
<p>A good forum for chemistry related questions is:</p>
<p>Chemical</a> Forums: Chemistry Forum, Chemistry Question, Chemistry Help - Index</p>
<p>For physics:</p>
<p>Physics</a> Help and Math Help - Physics Forums</p>
<p>Note: Forum rules states that you must show an attempt to solve the problem, or else they probably won't help you.</p>
<p>Permission is nullified. Puppet's decisions are overridden and his authority delegitimized. Consequently, he has been sentenced to the guillotine for... disposal.</p>
<p>ugh, why even bother asking permish? seriously. just ask. others have done so before.</p>
<p>acid base FTW!! I can help you out if it pertains to ochem.</p>
<p>ok im having probs w/ number 5. i got number 4 tho.</p>
<p>*I know i did the 1st ice box right...there are 2 ice boxes in number 5, i believe?</p>
<p>(4) Aspirin is a weak acid with a Ka of 3.0X10-5. Calculate the pH of a solution by dissolving 0.65g of aspirin in water and diluting it to 50.0ml. You may use the symbol RCOOH to represent aspirin when writing chemical reaction. Molecular weight of aspirin = 180.0g/mol</p>
<p>(5) Calculate the pH of a solution prepared by mixing 20.00ml of the aspirin solution in question #4 with 5.00ml of 0.200M NaOH. BEFORE using ICE box to solve for the pH of this mixture, think about whether the pH should be less than 7, equal to 7 or greater than 7 simply by comparing the moles of aspirin and the NaOH. </p>
<p>thanks! it's from the study guide in chem 14bl.</p>
<p>you can find it here:</p>
<p>click on "practice exam 2"</p>
<p>Hey, dawritingmachine, do you live in Rieber? I got the answer if you want it.</p>
<p>(no icebox, just henderson-hasselbach)</p>
<p>okay, i have to ask. what IS henderson-hasselbach... besides a reference in the spring sing chem rap parody?</p>
<p>thanks, please check ur PM jinobi.</p>
<p>i can't believe this final is 50 mins!! are all finals in summer school like that? 3 hr exams are tough..</p>
<p>liyana--it's to calculate pH if u have Ka and concentrations of acid and its conjugate base. :)</p>
<p>i don't like acids and bases. kinetics makes me smile. but not really after the lab. 14blers-has anyone finished the kinetics lab?</p>
<p>i tried h/h and it still doesn't work. hm.</p>
<p>First, calculate the Molarity of the aspirin solution you made in #4, which should be 0.07222 M. </p>
<p>Now, for #5, you want to find out the pH of the solution after you've reacted 20mL of your aspirin solution with 5mL of 0.2M NaOH. So, find out how many moles of aspirin and how many moles of NaOH you have; determine which one is in excess, and that'll tell you if the solution will be more basic or acidic. </p>
<p>Aspirin: (0.07222 mol/L) * (.02 L) = 0.00144 mol aspirin
NaOH: (0.2 mol/L) * (.005 L) = 0.001 mol NaOH</p>
<p>There's leftover aspirin, so the solution should be more acidic. Assuming aspirin = HA, and the aspirin that's been reacted with NaOH = A-,</p>
<p>HA + -OH -> A- + H2O + HA (excess) </p>
<p>Since you have an acid (HA) and a base (A-) in the same solution, doesn't that make it a buffer solution? Therefore, you can use Henderson-Hasselbach:</p>
<p>pH = pKa + log(A-/HA) </p>
<p>Reacting the 0.001 mol of NaOH with 0.00144 mol of aspirin will create 0.001 mol of A- and 0.00044 mol of excess aspirin (HA). Afterwards, just plug in your values. </p>
<p>pH = -log(3.0X10-5) + log(0.001/0.00044) = 4.88</p>
<p>hey, does anyone know the lab in which hypochlorite reacts with bleach? it's on the VOH link. i still don't know how to calculate the concentration of bleach correctly given that it is 6.15% w/v...</p>
<p>we've measured [dye] but we need to get [bleach] to find the rate of reaction and the overall order. how do we find [bleach]? the VOH guidelines are very confusing.</p>
<p>thanks!</p>
<p>*i mean calculating bleach concentration in molarity from %w/v.</p>
<p>Physics 6A..so i got answers but they seem wrong.. i hope u don't mind..:</p>
<p>A particle moves along the x axis. Its position is given by the equation x = 1.70 + 2.60t - 4.00t2 with x in meters and t in seconds.
(a) Determine its position when it changes direction.
m</p>
<p>(b) What is its velocity when it returns to the position it had at t = 0?
m/s</p>
<hr>
<p>A truck covers 45.0 m in 8.00 s while smoothly slowing down to a final speed of +2.20 m/s.
(a) Find its original speed.
m/s
(b) Find its acceleration.
m/s2</p>
<p>Webassign gives you 6 chances, you know.</p>
<p>MY teacher only gives us three chances for our Webassign :(</p>
<p>yay physics. something i can kinda do lol...</p>
<p>(a)
So you have a position function of x(t) = 1.70 + 2.60t - 4.00t^2... Just so you know... this is a basic kinematics equation in the form of
x(t) = x(0) + v(0)t + (1/2)at^2 (where x(0) is the initial position, v(0) is the initial velocity, and a is the CONST acceleration).
So from this... we can see that the initial velocity is positive and the constant acceleration is negative. This means it is SLOWING down, since acceleration is your change in velocity. Think about something moving... how could it change directions? Well... it would have to come to a stop and then have a negative velocity (the negative velocity meaning it is moving in the opposite direction).
Thus... We want to find out when the velocity is 0. To do this... you could take the derivative of the position function to get a velocity function, or maybe you are supposed to know the basic form for the velocity function:
v(t) = v(0) + at.
Anyway... this yields v(t) = 2.60 -8.00t.
Set it equal to 0... 0 = 2.60 - 8.00t ==> t =0.325.
Now we have the time when the velocity is 0 (changes direction).
Plug this back into x(t) to find that x(t=0.325) = 2.1225 m.
If you graphed this position function... you would see that it is a downwards parabola and this is the maximum point.
(b)
So you know that at t=0 the position is 1.70 m. You could know this from the form of x(t) or by plugging in t=0 for x(t). But... the position function is a quadratic, so there's a chance that it has two solutions. So we want to find out when it comes back to 1.70, so we solve for t as we set x(t) = 1.70:
x(t) = 1.70 + 2.60t - 4.00t^2
1.70 = 1.70 + 2.60t - 4.00t^2
0 = 2.60t - 4.00t^2
t = 0, t = 0.65.
So... the other time that it goes back to 1.70 m is 0.65 sec.
Plug this time back into your velocity function and...
v(t=0.65) = -2.60 m/s.</p>
<p>Ok... so for the second problem. You are given a displacement (change in position) of x(f) - x(0) = 45.0 m. You are given that it took 8.00 s, and that the v(f) = 2.20 m/s.
(a)
I usually don't remember this kinematics equation... but here it is...
x(f) - x(0) = (1/2) * (v(0) + v(f)) * t
This is perfect tho... since we know everything except v(0), which is what we are looking for.
45.0 m = (1/2) * (v(0) + 2.20 m/s) * 8.00s
solve for v(0)... and v(0) = 9.05 m/s.
(b)
Use the velocity function i mentioned earlier:
v(t) = v(0) + at.
You know that v(t=8.00) = 2.20 m/s and v(0) = 9.05 m/s.
2.20 = 9.05 + a(8.00)
a = -0.85625 m/s^2.
A negative acceleration makes sense, since it is slowing down.</p>
<p>Sorry if I made any mistakes =x. I haven't done these for a while ;]. Just remember that these kinematics equations ONLY work when the acceleration is CONSTANT. Good luck.</p>
<p>the ENTIRE worked out solutions to all the 6A hw problems is READILY available....and it's electronic. ask around....(but of course try it on your own first)</p>
<p>thanks guys. this hw is done, so no need to work the problem again unless you really want to.</p>
<p>i have another webassign assignment coming up due monday. might post again. :)</p>
<p>yep, there's a PDF solutions manual for 6A students. make some friends, ask around.</p>