Recent content by brake4country

  1. B

    Compound Interest: 3.11% Annually, 3x/Year

    I just used common denominator =)
  2. B

    Compound Interest: 3.11% Annually, 3x/Year

    Articulating and defending my answers: 1. Instantaneous rate as a function of t: FIND THE DERIVATIVE (d/dx a^x=a^x*ln(a). Derivative A'(t)=1050*(3.0311/3)^(3t)*ln(3.0311/3) 2. Total amount invested after 17 years: A(t)=350*(3.0311/3)^3t. A(17)=$592.24. 3. Rate at which the investment IS...
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    Compound Interest: 3.11% Annually, 3x/Year

    1. 350*(1+(0.0311)/3)*ln(1+(0.0311)/3) 2 and 3. Not sure how to answer these questions.
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    Max Buffering Capacity w/ 0.6 mol HAc & NaOH

    Right, because of 1 to 1 ratio. Thank you.
  5. B

    Max Buffering Capacity w/ 0.6 mol HAc & NaOH

    I was confused about this problem because my professor did not do a problem with us like this, but I used an ICE table to figure it out. HOAc + OH- ------> H2O + OAc- I 0.6 x 0 C -x -x...
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    Max Buffering Capacity w/ 0.6 mol HAc & NaOH

    Conceptually I understand it but I am having difficulty with the calculations.
  7. B

    Max Buffering Capacity w/ 0.6 mol HAc & NaOH

    I am terribly sorry, the answer from the homework was 0.3 moles. That is why I was confused, that 0.2 moles was a typo. So, is the problem that simple? I thought that if we started with 0.6 moles of acid, then the acid would dissociate (not completely) to 0.6 moles of salt and 0.6 moles of...
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    Max Buffering Capacity w/ 0.6 mol HAc & NaOH

    Yes, I responded to Borek: 0 = log + [salt]/[acid] 1 = [salt]/[acid], which would mean that we would have equal concentrations of salt and acid at maximum buffering capacity. I just don't know how to incorporate NaOH, for which the correct answer is...
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    Max Buffering Capacity w/ 0.6 mol HAc & NaOH

    Not sure if I was clear. I am new to the HH equation, however, I do know that 4.76 = 4.76 + log [salt]/[acid], which would mean that 0 = log [salt]/[acid]. To reverse the log, raise each side to base 10, which would give me: 10^0 = [salt]/[acid] 1 = [salt]/[acid] This is the point where I...
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    Max Buffering Capacity w/ 0.6 mol HAc & NaOH

    It reduces the amount of H+ ions, increasing the pH.
  11. B

    Max Buffering Capacity w/ 0.6 mol HAc & NaOH

    Yes. For acetic acid, that would be 4.76. I think my confusion is based on the question. I know that I need to find the ratio of [salt]/[acid] using the HH equation but I don't know what to do with the NaOH.
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    Max Buffering Capacity w/ 0.6 mol HAc & NaOH

    Homework Statement 0.6 moles of pure acetic acid (no salt) is added to water. How much NaOH would have to be added to get the solution to maximum buffering capacity? Homework Equations pH=pka + log [base]/[acid[ The Attempt at a Solution HAc <---> H+ + Ac- 0.6 mol of H+ 0.6 mol of Ac- All...
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    What are the probabilities for events A and B in a set of 10 cards?

    I need some help in checking my work, especially #4. Problem: You have a set of 10 cards numbered 1-10. You choose a card at random. Event A is choosing a number less than 8. Event B is choosing an even number. Draw a Venn Diagram and calculate each of the following probabilities: 1) P(A)...
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