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# Tutor profile: Cheryl B.

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Cheryl B.
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## Questions

### Subject:Library and Information Science

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Question:

Your professor wants you to find "reputable sources" for your research paper. What does this mean?

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Cheryl B.

### Subject:Chemistry

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Question:

How many grams potassium iodide to we need to create 100.0 mL of 1.0 M aqueous solution?

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Cheryl B.

The first thing I usually do is make certain that all of my units will play nicely together. So: 1.0M = 1.0 mol/L this is the desired concentration, c 100.0 mL = 0.1000 L this is the desired volume, v We could look up the formula to convert between these, but I usually look at the units to determine what additional information I can learn from what I have. If we multiply mol/L by L, then we have how many moles of potassium iodide we need since 1/L and L will cancel out when multiplied. 1.0\frac{mol}{L} \times 0.1000L=0.10mol=n However, we want to know the mass (m) of potassium iodide (KI). To do that we need the molar mass of KI. A periodic table gives the atomic mass of K (39.098 g/mol) and I (126.904 g/mol). There is only one of each atom in KI, so we add them to get 166.002 g/mol. I again look at the units. Multiplying the number of moles n by the molar mass will cancel out moles. 0.10mol \times 166.002\frac{g}{mol} = 16.6002g To the correct number of significant figures, this would be 17g

### Subject:Basic Chemistry

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Question:

What is the balanced reaction between potassium iodide and lead(II) nitrate?

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Cheryl B.

The first thing we need to do is figure out what the reaction is. Potassium (K) normally has a +1 charge when it is part of a salt, and iodide (I) has a -1 charge, so each ion brings enough charge to balance the other. The formula for potassium iodide is therefore KI. Lead (Pb) can have a couple different charges, but the (II) in the compound name tells us, in this case, it will be +2, and nitrate (NO$_{3}$) is -1. To balance the charge of one lead ion we need two nitrates. Therefore, the compound will be Pb(NO$_{3})$_{2} So, we know the reactants. KI + Pb(NO$_{3})$_{2}. What are the products? Since we have two ionic compounds, the most likely reaction type is a "double replacement reaction." The potassium and lead positively charged cations will replace the other, so both negatively charged anions (iodide and nitrate) will have new partners. We determine the chemical formula of newly created potassium nitrate and lead (II) iodide as we did above. Therefore, the unbalanced reaction is: KI + Pb(NO{_3}){_2}\rightarrow KNO{_3} + PbI{_2} A balanced reaction means we have the same number atoms/ions on both the product and reactant side of the reaction arrow. As currently written we have the same number of cations, the anions are unbalanced. We start with one iodide and end up with two. So, there must be two potassium iodides to act as the source for the iodides in the products: 2KI + Pb(NO{_3}){_2}\rightarrow KNO{_3} + PbI{_2} Now the potassium is unbalanced. There must be two potassium nitrates on the product side: 2KI + Pb(NO{_3}){_2}\rightarrow 2KNO{_3} + PbI{_2} After double checking the number of nitrates and lead (II) ions, we can confirm the reaction is balanced. Note: you could have started by balancing the nitrates and gotten to the same place.

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