[H+] = 10^[-6.55] = 2.8*10^-7 = [OH-] since it is pure water?
so keq= [2.8*10^-7]^2 = 7.94*10^-14
ΔG°=-(8.31450)(273+50)ln(7.94*10^-14)
= 81.00 KJ mol-
is that correct, are we sure that it is in an equilbrium at 50C?
thanks again
Homework Statement
The pH of pure water is temperature dependent, for example it is 6.55 at 50 C. Determine ΔG° for the ionization of water at 50C
Homework Equations
ΔG=ΔG°+RTln(Q)
at eqm ΔG=0
q=Keq
pure water:
pH 7.00 at 25C
The Attempt at a Solution
Not sure if the process...
Homework Statement
What is the capor pressure of a solution containing .15 mol of glucose (C6H12O6) and 1.65 mol of water at a temperature of 60.0C? the vapor pressure of pure water at this temperature is 149.38 torr
Homework Equations
Ralout's law?
P=PoX
The Attempt at a...
Homework Statement
Find the derivative:
( (X^3-1)/(X^3+1) )^1/3
Homework Equations
d/dx f(g(x)) = f'(g(x)) * g'(x)
quotient rule x/a x'a-xa'/a^2
The Attempt at a Solution
first i used the chain rule and quotient rule to get 1/3 ((x^3-1)/(x^3+1))^-2/3 * ((3x^2(x^3+1) -...
Homework Statement
Find the limit or prove it does not exist.
lim as s->0 (sin2s)^4/s^4
Homework Equations
i know i have to use sinx/x = 1 but I am having trouble manipulating the function into that form
The Attempt at a Solution
my first thought was that i could just take...
wow.. i must have done something wrong in that calculation ...just got what i think is the answer first try.. here's what happened...
(x-x+h)/((x+h)x)
h / x(x+h)
then times that by 1/h from the beginning of the problem to get
h / h(x^2+h)
which is
1 / x^2
Is this right...
I've been working on this one for a while now but just can't figure it out
lim h->0 (1/h) (( 1 / (x + h) ) - ( 1 / x ))
my first thought was to figure out (( 1 / (x + h) ) - ( 1 / x )) first by just putting them togeather and then using the congjigate times by one trick but that just...
1. Homework Statement
http://www.nellilevental.com/calc1.jpg
2. Homework Equations
C1 = \left(x - 1\right) ^{2} + y^{2} = 1
C2 = x^{2} + y^{2} = r^{2}
3. The Attempt at a Solution
Initially I thought that the "value" of R was just going to increase without bound since the...