ap chemistry - an, Tuesday, January 6, 2015 at 12:59am
HELP ME WITH THIS ONE WHEN DONE WITH THE OTHER THX
calculate the concentration of CO2 in a soft drink after the bottle is opened and equilibrates at 25C under a CO2 partial pressure of 0.0003 atm.
c=KP henry's law
k=?
k for CO2 is 3.4E-2 mols/L*atm
c=KP
Substitute p of 3.4E-2 for K and 3E-4 for P, solve for C (in M or mols/L)
P=0.0003 atm. how to solve
not getting answer of 9.3E-6 M
help me
Don't swith screen names.
Your other post says answer is 0.976E-6M
You need to find out what your teacher used for K; in fact, s/he should have given it to you. A K of 3.1E-2 mol/L*atm will give you 9.3E-6M
no she didn't give it sadly
thanks
To solve this problem, we will use Henry's Law, which relates the concentration of a gas in a solution to its partial pressure. Henry's Law is expressed by the equation c = KP, where c is the concentration of the gas, K is the Henry's Law constant, and P is the partial pressure of the gas.
In this case, you are given K for CO2 as 3.4E-2 mol/L*atm and the partial pressure of CO2 (P) as 0.0003 atm. You need to find the concentration of CO2 (c).
To find c, substitute the given values into the equation and solve for c:
c = KP
Substituting the given values:
c = (3.4E-2 mol/L*atm) * (0.0003 atm)
Now, multiply the two values together:
c = 1.02E-5 mol/L
The concentration of CO2 in the soft drink after the bottle is opened and equilibrates at 25°C under a CO2 partial pressure of 0.0003 atm is approximately 1.02E-5 mol/L or 1.02E-5 M.