Are you asking about the value of Tfinal and why you got it wrong? It looks like you tried to eyeball the y-intercept, "extrapolating" by
drawing a line with your pen. The value you're actually looking for is right there in the equation:
y = -0.0088x + 32.92
You've graphed linear equations using slope-intercept form before, right? Remember this:
y = mx + b
m = slope
x = value of x
b = y-intercept
So, in the first equation, 32.92 is the b variable:
y = mx + b
y = -0.0088x + 32.92
As far as how and why you use that value, remember that you're using constant-pressure calorimetry. The two reactants, in this case aqueous solutions
of HClO3 and KOH, are both initially the same temperature. They are then mixed with one another in the calorimeter and allowed to react. If
you look at the graph, you'll see that at about ~17ish seconds the reaction is complete and no longer producing additional heat. The calorimeter isn't
perfect, though, and is slowly leaking heat to the environment. (notice that the slope is negative)
In a perfect world, the HClO3 and KOH would react instantaneously, so Tfinal would actually be at Time = 0 seconds, which is the
instant they were mixed. Unfortunately, this is not the case, as the reactants took nearly 20 seconds to finish reacting, all the while the
calorimeter is slowly losing heat. So all we can really do is extrapolate back to Time = 0 seconds using linear regression instead. Using the equation
given, you then see that at Time = 0, the temperature is 32.92 °C.
Hope that helped. Just let me know if any of it doesn't make sense. It's been a while since I've done any General Chemistry stuff. After I got to
organic chemistry I never looked back.
[Edited on 3-7-2015 by Darkstar]Yttrium2 - 7-3-2015 at 12:36
I misread the answer as being 32through92 instead of 32.92
I see now.
Your post is kind of confusing I'm a little bad at graphing terminilogy.
I was wondering if thatd be the case with organic, of course id look back bug I mean like maybe id excel with greater enthusiasm and understanding
then.
[Edited on 7-3-2015 by Yttrium2]aga - 7-3-2015 at 14:58
THE most common mistake is Not Understanding The Question.
In an Exam situation, take a little time to consider the question, as failure to fully grasp what it is asking will always result in a wrong answer.
Edit:
That should have read "THE most common mistake ... for gifted and capable students that actually have a grasp of the subject"
A more common mistake is not attending lectures and not giving a shit, thereby learning next to nothing.
[Edited on 7-3-2015 by aga]Yttrium2 - 7-3-2015 at 16:48
At time = zero isn't t around 21c by looking at the graph Darkstar - 7-3-2015 at 17:33
Time = 0 sec is both your Tinitial (21.6 °C) and your Tfinal (32.92 °C). I briefly explained this towards the bottom of my
first post. You're using a linear regression (the line through the dots) to extrapolate what the temperature would've been at Time = 0 if
HClO3 and KOH were to react instantaneously. In a perfect, ideal world, the HClO3 and KOH would completely and fully react with
one another the very instant they are placed in the calorimeter. (hence Time = 0 seconds for both initial temp and final temp. final temp is the
temperature at the end of the reaction) The calorimeter would also remain the exact same temperature at all times instead of slowly cooling down. It
would instantly go from 21.6 °C to 32.92 °C at Time = 0 the moment the HClO3 and KOH are added, then hold steady at 32.92 °C
indefinitely without ever changing temperature or losing heat.
This is not physically possible, however, so a linear regression is used to extrapolate the Tfinal value instead. The Tfinal is
basically where your line would intercept the y-axis. My first post explained how you are able to determine that value by simply looking at the
equation given for the line itself.aga - 8-3-2015 at 14:45
Another key to learning is to read the Answers as well as the Questions.
Edit:
Darkstar's answer pretty much lays it out as flat as you'll ever see.
[Edited on 8-3-2015 by aga]
Yttrium's homework question thread
Yttrium2 - 10-3-2015 at 09:43
How would I use graphing to show when it would pay off having the more gas efficient car.
Circumstances:
Your car, has a value of 7k, gets 24mpg
Car B, costs 15k but gets 45mpg
Is this called a price point comparison?
[Edited on 11-3-2015 by Bert]
what kind of an equation is this? how to solve
Yttrium2 - 10-3-2015 at 09:49
You can buy 28 grams each for $560, but if you buy all 28 grams at once it costs $300.
How many grams, at this rate of discount, would have to be bought, for the cost to equal $280DutchChemistryBox - 10-3-2015 at 11:19
Depends on how they calculate the discount, there a lot of ways to do this.
M=grams
G=price per gram
F=discount factor
R=real price
D=discount price
This is what we know for sure:
G*M=R
G*28=560 > G=20
So I would say:
F*G*M=D
F*20*28=300 > F=300/560
This would mean the following:
F*G*M=D
(300/560)*20*M=280 > M=26.1 grams
This was the easy way. But it can also be that they calculate it this way:
F^M*G*M=D
And so we can go on and on with possibilities.
Correct me if I'm wrong. But I would say that you need more information to make a reliable formula about the discount. With this information we can't
do much reliable.Loptr - 10-3-2015 at 11:43
I have heard them referred to price break equations. I don't think there is standard way of doing them, but here is a YouTube that covers it. I
haven't viewed it yet, as my browser is acting up and the Flash plugin keeps crashing.
without specifying the cost per gallon you can't graph or calculate it.
then you have to assume that the price per gallon remains constant over the 'payback' period.Sulaiman - 10-3-2015 at 12:15
27 g
you buy 28 g for $300 then sell 1 g for the market rate of $20
Discount rates never follow a straight line because if they did you would get paid for ordering very large quantities.
If this is a homework problem then go with the 26.1 g method,
which assumes a straight line.
[Edited on 10-3-2015 by Sulaiman]Yttrium2 - 10-3-2015 at 12:36
$2.50, guess I forgot to add that. Assuming it remains constant, (and is constant for both vehicles)Sulaiman - 10-3-2015 at 13:00
You need to draw two straight lines, on a graph of x=miles and y=$
Line 1
when x=0 miles, cost = 7000
gradient = 2.5 ($ per gallon) / 24 (miles per gallon) = $0.10416666.. per mile
Line 2
when x=0 miles, cost = 15000
gradient = 2.5 ($ per gallon) / 45 (miles per gallon) = $0.05555.. per mile
To get the scale of the graph correct it helps to know the answer, which is
15000 - 7000 = miles x (0.10416666 - 0.055555)
8000 = miles x 0.04861111
so miles = 8000/0.04861111 = 164,571.4286
cost = 164,571.4286 x 2.5/24 + 7000 = 24,142.86
or = 164,571.4286 x 2.5/45 + 15,000 = 24,142.86
So your graph should have x axis (miles) = 0 to 200,000
and y axis (cost in dollars) = 0 to 30,000
to make your table for plotting use 10,000 mile increments
miles / cost1 / cost2
0 / 7000 / 15000
10,000 / (7,000+1,041.66) / (15,000+555.55)
20,000 / (7,000+ 2,083.33) / (15,000+1,111.11)
etc.
of course the old $7,000 car will probably not make a further 164,571.4286 miles,
and the new $15,000 car will be worth a lot less,
and running costs (maintenance, tyres, tax ...) will be different
but most importantly,
the girlfriend that you can take for a ride in the $15,000 new car
will probably be higher maintenance than the girlfriend that would go for a ride in the old $7,000 car
but don't let reality confuse the math.
[Edited on 10-3-2015 by Sulaiman]Etaoin Shrdlu - 10-3-2015 at 13:23
Please show your work instead of just asking for solutions, especially if these are a bunch of homework problems.Yttrium2 - 10-3-2015 at 13:42
I'm a little lost on the math, let's just say car A gets 24 mpg at $2.5 a gallon, doesn't cost anything
Car B gets 45 mpg and costs 8 grand.
Question is, how long would I have to drive car B until the savings in gas pay for the car?
I might want to get rid of my car and get a motorcycle, since I have a truck. This might save me a ton on gas mileage
Could you explain the steps of the math there?
[Edited on 10-3-2015 by Yttrium2]aga - 10-3-2015 at 14:01
I have heard them referred to price break equations. I don't think there is standard way of doing them,
If it based on price breaks, then it is almost random.
A price break is basically:-
price A if you buy 1 to x items
price B if you buy (x+1) to y items etc.
E.g.
1.99 per unit if you buy between 1 and 9 units
1.79 per unit if you buy between 10 and 50 units etc.
Usually they're guesses, so unlikely that an exam question is based on price breaks.Sulaiman - 10-3-2015 at 14:09
It is exactly the same graph except everything is moved down by $8,000
re-arranging the problem
difference in initial cost =8000
difference in running cost = total miles traveled, multiplied by the difference in dollars per mile
$8,000 = miles x ($difference in cost per mile)
8,000 = miles x (2.5/24 - 2.5/45)
8000 = miles x ($0.104166 per mile - $0.0555 per mile)
8000 = miles x 0.0486111
miles = 8000/0.0486111 =164571.4286 miles.
The main difference now will be the hospital costs incurred trying to travel 164571 miles on a motorbike
[Edited on 10-3-2015 by Sulaiman]Amos - 10-3-2015 at 15:40
It's starting to feel like Yahoo answers in here... Besides, doesn't this qualify as asking to be spoonfed?aga - 11-3-2015 at 12:40
It's Beginnings.
Spoonfeeding is kind of allowed.
Also allows us to give answers, which others immediately denounce as totally incorrect (when they are).Invity - 11-3-2015 at 22:53
I misread the answer as being 32through92 instead of 32.92
I see now.
Your post is kind of confusing I'm a little bad at graphing terminilogy.
I was wondering if thatd be the case with organic, of course id look back bug I mean like maybe id excel with greater enthusiasm and understanding
then.
[Edited on 7-3-2015 by Yttrium2]
Hi,
This is Great Answer.. Awesome I am clear.. Thanks for sharing..Darkstar - 17-3-2015 at 10:32
Your post is kind of confusing I'm a little bad at graphing terminilogy.
I was wondering if thatd be the case with organic, of course id look back bug I mean like maybe id excel with greater enthusiasm and understanding
then.
[Edited on 7-3-2015 by Yttrium2]
Organic Chemistry is nothing like General Chemistry I & II. It's an entirely different animal. There's actually very little math involved. Other
than the lab portions, which obviously require some calculations to be made, I honestly don't remember having to use my calculator a single time on an
Organic Chemistry I or II exam. Now, whether or not that turns out to be a good thing for you personally will mostly depend on just how determined you
really are at trying to better understand the material this time around. Unlike the previous two courses, the material in O-Chem is almost entirely
conceptual. You either get it or you don't. No more trying to simply memorize everything needed for the exam like you could in Gen Chem.
Keep in mind that, despite the lack of math, the vast majority of students end up having far more trouble with O-Chem than they ever did with General
Chemistry. Most consider it to be significantly more difficult. Since most students only take O-Chem because they're pre-med/pharmacy/dental/vet and
it's required, O-Chem I is frequently referred to as a "weed-out course," due to its high drop rate and tendency to cause students to seriously
rethink the field they want to go into. It's actually one of the most notorious undergrad courses there are.
But don't let that scare you. O-Chem really isn't all that hard if you just take the time to actually understand the material and why things react the
way they do. You'll definitely excel with greater enthusiasm. Significantly so, in fact. I took very few notes in O-Chem I and literally took zero
notes in O-Chem II, yet made an A in both courses. I even tutored a few of my classmates in the latter. Enthusiasm definitely made a huge difference.
I learned just as much chemistry by reading on my own as I did in O-Chem I. Probably more. By the time I got to O-Chem II, I already knew just about
everything we were learning. Any reaction I didn't already know (O-Chem II is mostly just a bunch of name reactions) immediately made sense with one
look at the mechanism.
Anyway, you're in General Chemistry I right now, correct? If so, I strongly suggest that you brush up on your algebra before taking Gen Chem II. It
can be rather math-heavy at times. Just a friendly heads-up.Loptr - 17-3-2015 at 11:13
1. (3x/2) is mx, where 3/2 is the m, or the slope value.
2. (-4) is b, where it can be a negative or positive value, and determines where on the y-axis the the line will cross when (mx) is equal to 0.
y = (3x/2) + (-4); x = 0
y = [3(0)/2] + (-4)
y = 0 + (-4)
y = -4; also known as the y-intercept, or where it crosses the y-axis.
In order to find the x-intercept, you set y to 0, and solve for x. Notice the relationship between the x and y? They are algebraically
linked.
y = (3x/2) + (-4); y = 0
0 = (3x/2) + (-4); add 4 to both sides
4 = (3x/2) + 0; the -4 was canceled out
4 = 3x/2; multiply both sides by 2 to cancel out the 2 on the right
8 = 3x; now divide both sides by 3
8/3 = x; also known as the x-intercept, or where it crosses the x-axis.
Note: I added extra parenthesis to show the mx and b as discrete units, where it can be missed that y=mx-2 is actually y=mx+(-2) making b =
-2.
I hope this helps.
[Edited on 17-3-2015 by Loptr]
An important reason to get a line into slope-intercept form, y = mx+b, is that you can then plug in any value for y or x and get the
corresponding value for the other axis. This allows you to accurately determine the path of a line. Also, if you have two points, you can then derive
the slope using the concept of rise over run, also known as point-slope form.
m = (y2 - y1) / (x2 - x1); in other words, change in y over the change in x from point 1 to point 2. This is your slope.
[Edited on 17-3-2015 by Loptr]
[Edited on 17-3-2015 by Loptr]diddi - 17-3-2015 at 15:23
little learning takes place when the student is not interacting with the teacher(s). the student needs to be involved at a step by step level to
really gain much when there is clearly a significant level of misunderstanding of the concepts. the solutions presented are very clear to me,
obviously not so for the student, which means that no learning is taking place here.
the challenge here is to identify the steps that the student is having trouble with, asking them how or why they perform the step incorrectly, and
changing that misunderstanding so that next time the problem presents the student has the correct conceptual understanding to complete the step
properly.
the student is responsible for their learning; the teacher is the facilitator.Darkstar - 24-3-2015 at 04:02
I might want to get rid of my car and get a motorcycle, since I have a truck. This might save me a ton on gas mileage
What kind of motorcycle did you have in mind, if you don't mind me asking? Is 'Car B' supposed to represent the motorcycle? Because an $8,000 bike
that gets around 45 mpg has "sport bike" written all over it. Specifically a 600. Considering you're my age and a male, I'm guessing that's what you
want? If it is, I'll just tell you right now that you're not going to be saving a "ton" on anything. It might get slightly better gas mileage than
your car, which isn't saying much since it's still pretty bad gas mileage for a motorcycle of that engine size, but they can very well end up costing
you a fortune in speeding tickets and insurance rates if you're not careful. If it's saving money you're after, sport bikes generally aren't the way
to go. They're designed for performance, not economy. They're more of a "toy" than anything else.
That isn't to say they can't be cheaper if you don't ride like an idiot; however, you're likely going to be riding the hell out of it because, let's
face it, bikes are just freaking fun to ride. Don't fool yourself into thinking it's just going to be for transportation to and from work/school and
that's it. There's going to be a lot of joy riding in-between, trust me. Expect to be visiting the gas station a lot more often than you think. Oh,
and you'll need to buy 91 octane gas from now on, too. These are high-performance engines, after all.
/slightly tongue-in-cheek rant
Really, though, in all seriousness, what kind of bike are you interested in? Have you taken the MSF course? Any previous riding experience, either on
the street or in the dirt? If this is your first street bike, don't even think about a liter bike. 600s are terrible first bikes, too, for that matter
(aggressive riding position, insanely strong brakes, touchy throttle, unpredictable powerband, sudden bursts of torque etc), and are not at all
forgiving when you fuck up. Plus they tend to be heavy and intimating, causing you to develop your skills as a rider much more slowly (in addition to
developing bad habits) than you would've if you had started out on a smaller bike.
Anyway, sorry for the long post. I apologize if I have assumed wrong. It's just that I haven't met many 26-year-old males who didn't want a
sport bike for their first motorcycle. Good luck with whatever you decide to do and be safe. And wear some damn gear! Don't be another one of those
dumbass squids on a GSX-R1000 doing triple digits down the highway wearing nothing but flip-flops, a t-shirt and cargo shorts...
And in case you haven't figured it out by now, I ride a sport bike.
My 2010 Yamaha R6:
[Edited on 3-24-2015 by Darkstar]morganbw - 24-3-2015 at 15:19
@ Darkstar
nice ride, siraga - 24-3-2015 at 15:28
It has no Roof.
At my age, i prefer vehicles with air-con, heating, and weather proofing built in.morganbw - 25-3-2015 at 14:24
