The biochem major at my university has similar requirements to the other chem majors here. The straight chem majors need to take specific chem classes depending on which kind of chemistry they're focusing in. For biochem, we take the same prereq's, but also have biology class reqs as well...
Like dustbin said, distractions are a huge factor in getting your work done. Although, I had a computer, I lived without internet for a term and my grades shot up instantly. I had to go the library just to check my email and would usually wind up staying for a while to study.
Diet and...
This fall, I'm about to start my 5th year as an undergraduate pursuing a B.S. in Biochemistry. Long story short, I declared biochemistry in the fall of my senior year without having taken many science or math courses. It's been a tough road cramming 21 credits into last fall, 19 last winter...
So when I studied all of this stuff (literally just last term), it helped me to consider thermodynamics and kinetics as two separate entities. The fundamental difference between the two is the dependence on reaction pathway. While thermo is path-independent, kinetics is path-dependent...
Ok, so after reworking this problem a few more times, I think I'm finally close the right solution:
using y12 as the integrating factor, and Q(x)=1/y3:
=\intQ(y)P(y)
=\inty9+C
5=(1/10)(210)+C
C=(-487/5)
x(y)=(1/P(y))Q(y)P(y)+C]
x(y)=(y-12) \int[((1/10)y10)-(487/5)]...
for future reference:
Formal Charge= valence e-'s - nonbonding e-'s - (.5*bonding e-'s)
for HNC:
-Nitrogen F.C.=5-0-4=+1
-Carbon F.C.=4-2-(.5*6)= -1
for NCH:
-Nitrogen F.C.=5-2-(.5*6)=0
-Carbon F.C.=4-0-(.5*8)=0
Borek's answer is most certainly correct. However, as you already...
So I just tried reworking through my algebra, starting with Sammy's suggest of rewriting the function as :
(dy/dx)=y3+12xy2(dy/dx)
However, I had some confusion about multiplying through using y9. Is that step generating an integrating factor such that:
y= [(\intQ(x)*M(x)dx)]/(M(x))...
It's quite possible that orgo here at UMich is taught a little differently than many other places. Furthermore, it's my experience that everyone has a different approach to the material. In our dreaded coursepack (a spiral-bound book of old exam questions with no answers), they mention that...
that's kind of a simplified, cookie-cutter description of what is actually going on. To really understand what's going on you need to use either the LCAO (linear combination of atomic orbitals) or the VB (valence bond) models.
From either model, you would then need to write out the...
Like pretty much everyone else has said, the amount of math, and the type of math required really depends on what sort of chemistry that you study.
Physical/quantum chemistry is very heavily based in calculus (I just finished taking such a course).
In my experience, analytical chemistry uses...
Simon,
that makes sense, we essentially came to the same solution of the general equation. However, when I plug in for the given x & y values to solve for "C", I get a huge number on the order of 2*10^4
to interconvert between dy/dx, can you just simply invert (given that you invert the...
the problem is as follows:
[1-(12*x*(y^2))]*(dy/dx)=y^3
we need to solve this equation such as x=5 & y=2 by regarding y as the independent variable.
Here is my work for the first attempt:
Step 1)separation & integration
Step 2)ln(x)= (-1/(2y^2))-12ln(y)+C...
Probably could have been clearer in my description.
AmritpalS, thanks for breaking it down for me though. My issue arose when I tried to plug in y(0)=C. I wound up creating a new constant (as C was part of the initial conditions) such that:
ln(8y^4 +14)=32sin(x) +D
then I solved for...
Hey everyone, I'm a long-time visitor, it's my first time posting though.
I have a homework problem that is causing me considerable consternation:
(y^3)*(dy/dx)=(8y^4+14)*cos(x); y(0)=C
Oh, and we're supposed to solve the initial-value problem, and then solve for the particular...