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Heat Transfer Lab
The following represents a lab set up for heat transfer. The cup on the left started with boiling water at 100 degrees C and the cup on the right has water at 20 degrees C. There is an aluminum bar between the two cups allowing heat to transfer from one cup into the other. The set up will be left alone for 20 minutes and temperatures of each cup of water will be recorded every minute for 20 minutes.
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Answer:
formula units of sodium chloride can be formed from 13.0 gram of ferric chloride.
Explanation:
Mass of ferric chloride = 13.0 g
Moles of ferric chloride = 
1 mole of ferric chloride has three moles of chloride ions.Then 0.08 moles of ferric chloride will have :
of chloride

1 mole of sodium ion reacts with 1 mole of chloride ion to form 1 mole of NaCl. Then 0.24 moles of chloride ion will give:
of NaCl
1 mole =
molecules/ atoms
Number of NaCl molecules in 0.24 moles :

formula units of sodium chloride can be formed from 13.0 gram of ferric chloride.
Answer : The volume of sodium benzoate and benzoic acid solution mixed to prepare this buffer should be, 29.0 mL and 71 mL respectively.
Explanation :
Let the volume of sodium benzoate (salt) be, x
So, the volume of benzoic acid (acid) will be, (100 - x)
Using Henderson Hesselbach equation :
![pH=pK_a+\log \frac{[Salt]}{[Acid]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BSalt%5D%7D%7B%5BAcid%5D%7D)
Now put all the given values in this expression, we get:

x = 29.0
The volume of sodium benzoate = x = 29.0 mL
The volume of benzoic acid (acid) = (100 - x) = (100 - 29.0) = 71 mL
Thus, the volume of sodium benzoate and benzoic acid solution mixed to prepare this buffer should be, 29.0 mL and 71 mL respectively.
Answer:C The final product cannot be converted back to the original ingredients.