Wednesday

July 30, 2014

July 30, 2014

Posted by **Chris** on Sunday, January 28, 2007 at 9:01pm.

A)What is q for the calorimeter?

In my book it says -q cal= q, so I wrote -q cal = -C(cal) * ΔT

-q cal = -(2.115 E4 J/ºC) * (27.71ºC-23.49ºC) = 8.93 E4 J

Does that seem right?

B) What is q when 4.50 g of fructose is burned?

I started off with the formula q= m*Cp*ΔT, but I have two unknown variables here (ΔT and q) so that doesn't seem right. How do I approach this? I'm assuming they're asking for qC6H12O6, not qCal

C) What is q for the combustion of one mole of fructose?

Not sure how to go about doing this one. I thought that if I could find the value for B, I could find out what percentage 4.50 g is of 1 mole of fructose and go from there.

Fructose is a sugar commonly found in fruit. A sample of fructose, C6H12O6, weighing 4.50 g is burned in a bomb calorimeter. The heat capacity of the colorimeter is 2.115 E4 J/¨¬C. The temperature in the calorimeter rises from 23.49¨¬C to 27.71¨¬C.

A)What is q for the calorimeter?

In my book it says -q cal= q, so I wrote -q cal = -C(cal) * ¥ÄT

-q cal = -(2.115 E4 J/¨¬C) * (27.71¨¬C-23.49¨¬C) = 8.93 E4 J

Does that seem right?

B) What is q when 4.50 g of fructose is burned?

I started off with the formula q= m*Cp*¥ÄT, but I have two unknown variables here (¥ÄT and q) so that doesn't seem right. How do I approach this? I'm assuming they're asking for qC6H12O6, not qCal

C) What is q for the combustion of one mole of fructose?

Not sure how to go about doing this one. I thought that if I could find the value for B, I could find out what percentage 4.50 g is of 1 mole of fructose and go from there.

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