Recent content by jools111

Hey thanks for the response. I ended up putting everything into the S domain, and doing a voltage divider as you suggested. Once I had this I did an inverse Laplace transformation back into t. This solved my problem, but I still definitely need to work on my 2nd order DEs.

Yeah Vc is the capacitor voltage, Vr is resistor voltage, Vl is inductor voltage, and Vtot is the source voltage. Is finding V(t) the first step of the process? If so, how do I go about that? Thanks again.

Homework Statement For a series circuit with the following components: 1. Vin(t) 2. 1K resistor 3. 100 mH Inductor 4. 1 uF capacitor 1. Write down the differential eqn. for Vc in the circuit. 2. Solve the DE, calculate the natural and forced response with the following initial...

Ok. Thanks for your help!

It does... But with it being an open system, it not the entropy of the liquid the only factor we would be concerned with? (In reference the soda bottle)

"Large negative values for (delta)H and large positive values for (delta)S are associated with high solubilities. This is consistent with the observation tat acetone and chloroform are completely miscible" Also, when decribing the effect of a carbonated soda bottle being opened: "Entropy...

That's what I thought as well. My text is telling me otherwise though.

Thanks for the replies. I think I might be asking my question wrong. Let's look at it from the other end. O2(g) has a higher entropy than H2O(l), so how come when O2(g) is dissolved into water, the entropy decreases. Should it not go up, as O2(g) contributes a higher entropy to the reaction?

But with a portion of the dissolved oxygen escaped into the atmosphere, shouldn't the H20(l) that is was dissolved in now have a lower entropy, as gas has escaped?

H20(l) + Heat ---> H20(g), in an open system. Is that what you're asking? Thanks.

When water is boiled, and the dissolved oxygen that was in the water escapes, why does this raise the entropy of the system? The O2 gas has a higher entropy than the H20, so when it departs the liquid, shouldn't the entropy decrease as the gas is no longer present? Thanks.

I am working on a lab that deals with ionic concentration and rate of reaction. The data is as follows: Mixture-----[I] mol/L-----[S2O8] mol/L------Time s 1-----------0.10---------0.050-------------20 2-----------0.075--------0.050-------------28...

So would the correct answer to the question be negative? I'm a little confused. I understand the logic, but can I just arbitrarily change the symbol like that?

Homework Statement I am working on a lab that has me a little confused: I diluted 5.5g of NaOH into 200ml of water to disassociate the sodium ions. In the equation below you can see my temperature results. Write the net ionic equation for the reaction, and note the value of ΔH. NaOH...

I'm sorry... I'm not following. In my book the fourth rule states: 4. Most sulfates are soluble except BaSO4, SrSO4, and PbSO4. CaSO4 and Hg2SO4 are moderately soluble. Would the precipitate not be one of the products? Neither of the reactants in my equation are listed there...