For the past couple weeks, I've been getting hammered by spam in the comments so I have turned on comment moderation (for now). If you have any comments that are on-topic, please be sure that they will be allowed, I'm just trying to avoid a flood of spam-bot garbage in the comments section of my posts. Hopefully this will be a short-term problem...
Have a great day.
2013-05-31
2013-05-14
Email question - van't Hoff factor
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Can you please explain how to determine the i factor. Now matter how much I read the book and look at examples I can't figure it out. Thanks,
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There are a couple things that could be "the i factor", but I'll assume this question is about the van't Hoff factor that's used when determining changes due to colligative properties. Remember, colligative properties are properties of a solution that depend only upon the number of solute particle, not the identity of those solute particles. It doesn't matter if those solute particles are molecules, or cations, or anions, or a big random mixture of them all. When an ionic compound dissolves in water, it dissociates (at least somewhat) into its component ions. The number of ions (particles) that each formula unit breaks into when it dissolves is the van't Hoff factor. For example, if NaCl(s) is dissolved in water, it forms 1 Na+1(aq) ion and 1 Cl-1(aq) ion, so each formula unit (NaCl) forms 2 particles (ions) in solution; the van't Hoff factor is "2". For polyatomic ions, the ions don't break down into their individual atom, they stay polyatomic ions, so if K2SO4(s) is dissolved in water, it forms 2 K+1(aq) ions and 1 SO4-2(aq) ion in solution; the van't Hoff factor is "3".
Can you please explain how to determine the i factor. Now matter how much I read the book and look at examples I can't figure it out. Thanks,
----------
There are a couple things that could be "the i factor", but I'll assume this question is about the van't Hoff factor that's used when determining changes due to colligative properties. Remember, colligative properties are properties of a solution that depend only upon the number of solute particle, not the identity of those solute particles. It doesn't matter if those solute particles are molecules, or cations, or anions, or a big random mixture of them all. When an ionic compound dissolves in water, it dissociates (at least somewhat) into its component ions. The number of ions (particles) that each formula unit breaks into when it dissolves is the van't Hoff factor. For example, if NaCl(s) is dissolved in water, it forms 1 Na+1(aq) ion and 1 Cl-1(aq) ion, so each formula unit (NaCl) forms 2 particles (ions) in solution; the van't Hoff factor is "2". For polyatomic ions, the ions don't break down into their individual atom, they stay polyatomic ions, so if K2SO4(s) is dissolved in water, it forms 2 K+1(aq) ions and 1 SO4-2(aq) ion in solution; the van't Hoff factor is "3".
Exam strategies...
How do you take an exam? The most important thing is, of course, knowing the material, but there are also some "tricks" that can work in your favor, especially on a standardized multiple-choice exam. Some of this comes down to game theory, so it can be useful in settings outside of exams.
1. Know the rules - In a standardized exam, there can be different rules and formulas used for calculating your score. Some of the formulas have the effect of penalizing you for guessing if you truly don't know the answer. Always check to determine whether or not there is a penalty for guessing. Personally, I prefer to just take the total number of correct answers and not mess around with an elaborate formula that penalizes you for guessing.
2. Improve your odds - When you look at a series of multiple choice answers, there are often answers included that have to be obviously wrong if you think about them a little bit. If you're looking at a 4-option multiple choice question, your odds of randomly guessing the correct answer without even reading the answers is 25%. Eliminate one wrong answer, and your odds increase to 33%. Eliminate two, you're at 50%. Keep this in mind when you get to a question that has you stumped; rather than hunt for the right answer, try to eliminate wrong answers. Be like Sherlock Holmes; when you have eliminated all other possibilities, the remaining possibility, no matter how unlikely, must be correct.
3. Build momentum - Don't get hung up on a question you don't know when there are others you know the answer to. Answering a few questions correctly builds confidence and helps you tackle the more challenging questions. This is a challenge for "linear" test-takers who naturally want to start with #1 and work through in order to the end of the exam. {I was a very linear test-taker, and to some extent I still am...} It's OK to skip a question and come back to it later. This also helps with time management on the exam.
4. Answer all the questions - This is related to #1 and #2 above. If there is no penalty for guessing, then make sure you answer every question, even if you have no idea what the correct answer is. For a 5-option multiple choice question, there's a 1 in 5 (20%) chance that you'll random guess it correctly. If you don't answer the question at all, there's a 100% chance that you'll get it wrong.
Good luck.
1. Know the rules - In a standardized exam, there can be different rules and formulas used for calculating your score. Some of the formulas have the effect of penalizing you for guessing if you truly don't know the answer. Always check to determine whether or not there is a penalty for guessing. Personally, I prefer to just take the total number of correct answers and not mess around with an elaborate formula that penalizes you for guessing.
2. Improve your odds - When you look at a series of multiple choice answers, there are often answers included that have to be obviously wrong if you think about them a little bit. If you're looking at a 4-option multiple choice question, your odds of randomly guessing the correct answer without even reading the answers is 25%. Eliminate one wrong answer, and your odds increase to 33%. Eliminate two, you're at 50%. Keep this in mind when you get to a question that has you stumped; rather than hunt for the right answer, try to eliminate wrong answers. Be like Sherlock Holmes; when you have eliminated all other possibilities, the remaining possibility, no matter how unlikely, must be correct.
3. Build momentum - Don't get hung up on a question you don't know when there are others you know the answer to. Answering a few questions correctly builds confidence and helps you tackle the more challenging questions. This is a challenge for "linear" test-takers who naturally want to start with #1 and work through in order to the end of the exam. {I was a very linear test-taker, and to some extent I still am...} It's OK to skip a question and come back to it later. This also helps with time management on the exam.
4. Answer all the questions - This is related to #1 and #2 above. If there is no penalty for guessing, then make sure you answer every question, even if you have no idea what the correct answer is. For a 5-option multiple choice question, there's a 1 in 5 (20%) chance that you'll random guess it correctly. If you don't answer the question at all, there's a 100% chance that you'll get it wrong.
Good luck.
2013-05-04
Redox Lab question...
A few people have asked about the chemical reactions for the Redox lab hand-in {Redox Hand-in}, and this is often a source of confusion for students, so let me answer it here. First of all, I think a number of people get confused because you're trying to over-think the question and make it more complex than it really is.
So you need to draw a voltaic cell... you can draw this by hand, or you can draw it electronically. DO NOT just find one online and copy-paste it into your hand-in, if that's all I wanted I would have pasted this {voltaic cell} into the hand-in myself before I posted it.
The part that causes some confusion is the "...write a correctly balanced net ionic equation for the spontaneous process..." Don't over-read that! The metal cation solutions you were using in lab were probably nitrate salts, but nitrate (or whatever anion might have been present) was a spectator in all of your reactions. Net ionic equations are actually easier than full-formula equations because they're not cluttered up with with a bunch of extra stuff, and net ionic equations actually just describe the CHEMISTRY that's happening rather than distracting you with a bunch of spectator ions and species.
Now, I'm not going to write out a net ionic equation that's the exact answer to one that you have to write, but here's an example. Let's say I made up the voltaic cell Fe|Fe+3||Cd+2|Cd with the black/negative lead of my meter hooked up to the Fe(s) electrode and the red/positive lead connected to the Cd(s). The potential I measure is -0.32V. Because the measured potential is negative, the cell is running backwards, so the spontaneous cell reaction is Cd|Cd+2||Fe+3|Fe. Translating that into a reaction, we can write the two half-reactions as:
Cd(s) <=> Cd+2(aq) + 2 e-
3e- + Fe+3(aq) <=> Fe(s)
Adding those up gives the overall (or "net") reaction:
3Cd(s) + 2 Fe+3(aq) <=> 2 Fe(s) + 3 Cd+2(aq)
{Remember to multiply each half-reaction by an appropriate integer to make all the electrons cancel...}
That's a "correctly balanced net ionic equation for the spontaneous process" in this case. Now go do that for all the cells you measured in the experiment.
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