PCAT Practice Question

Started by 217933
This forum made possible through the generous support of SDN members, donors, and sponsors. Thank you.
Get help with your application

Use all the free resources available to you from SDN: articles, guides, expert advising, forums discussions, and school research.

2

217933

Advertisement - Members don't see this ad
How do you know if something is a strong electrolyte or a weak electrolyte?

For example sulfuric acid is a strong electrolyte and acetic acid is a weak one but how would you be able to figure that out just by being given the two different acids? 😕
 
Okay I haven't taken calc in like three years and I'm trying to remember how to find the inflection point. I know you have to find the second derivative but the question I'm trying to answer is to find the inflection point for

f(x) = x / ((x^2)+1)

I tried to understand the steps that were given but I can not figure out the "finding of the derivatives" for this fraction. Are there special rules when it comes to fractions or something?
 
im pretty sure there ARE special rules

like in a fraction, its taken as a division problem, like
(numberator)(derivative of the denominator)-(denominator)(derivative of numberator)
IMNOT SURE...thats how it works out for trigonometric functions, but there IS a special rule, i just cant think of it

anyways, in finding things like that
all i know is, the first derivative gives you minimums and maximums
second derivative gives you inflection points
always set them equal to zero
 
Advertisement - Members don't see this ad
im pretty sure there ARE special rules

like in a fraction, its taken as a division problem, like
(numberator)(derivative of the denominator)-(denominator)(derivative of numberator)
IMNOT SURE...thats how it works out for trigonometric functions, but there IS a special rule, i just cant think of it

anyways, in finding things like that
all i know is, the first derivative gives you minimums and maximums
second derivative gives you inflection points
always set them equal to zero

You forgot one small part of that.. (numerator)(derivative of the denominator)-(denominator)(derivative of the numerator) all over (denominator) squared

I like to sing.. Low Dee High minus High Dee Low all over what's below squared... 🙂

Then you will be able to find your second derivative by repeating the process again. Then setting to zero to find your concavity. You will need to apply the chain rule also to find the second derivative on this one.
 
Last edited:
How do you know if something is a strong electrolyte or a weak electrolyte?

For example sulfuric acid is a strong electrolyte and acetic acid is a weak one but how would you be able to figure that out just by being given the two different acids? 😕

How do you know if something is a strong electrolyte or a weak electrolyte?

-Well, you have to ask yourself the basic question... what is an electrolyte. It is something when in water can conduct electricity. A simple test will be to take the salts of your said electrolyte and drop it in water. You can then test the conductance of the solution. The more ionized, the better the conductance.

***Also as per definition in pharmaceutics- a strong electrolyte is something that dissociates completely in water while a weak electrolyte only does so partially based on a calculated constant (takes into consideration ionization due to ph **temp is not a factor when considering ionization)
 
You forgot one small part of that.. (numerator)(derivative of the denominator)-(denominator)(derivative of the numerator) all over (denominator) squared

I like to sing.. Low Dee High minus High Dee Low all over what's below squared... 🙂

Then you will be able to find your second derivative by repeating the process again. Then setting to zero to find your concavity. You will need to apply the chain rule also to find the second derivative on this one.

Or I believe you can use the product rule, move the denominator to the top and make it to the -1 power. Then: F(x)*g '(x)+f '(x)*g(x) Usually makes it easier for some problems.