Question

1. in order to compare this theoretical value to your experimental heat of combustion, the units of energy need to match. show how to convert your experimental energy released per gram from kj/g to kcal/g. your calculation will start with the heat of combustion (slope) found in the section above. hint: 1 calorie = 1 kilocalorie = 1000 calories and 1 calorie = 4.18 joules. 22.7kj/1g × a/b × c/d × e/f =??kcal/g choices: 4.18 j, 4.18 cal, 1 j, 1000 kj, 1000 kcal, 1 cal, 1000 cal, 1 kcal, 1 kj, 1000 j

Explanation:

Step1: Convert kJ to J

We know that 1 kJ = 1000 J. So to convert 22.7 kJ to J, we multiply by 1000.
$22.7\ kJ\times\frac{1000\ J}{1\ kJ}=22700\ J$

Step2: Convert J to cal

Since 1 cal = 4.18 J, to convert J to cal, we divide by 4.18.
$22700\ J\times\frac{1\ cal}{4.18\ J}\approx5430.62\ cal$

Step3: Convert cal to kcal

As 1 kcal = 1000 cal, to convert cal to kcal, we divide by 1000.
$5430.62\ cal\times\frac{1\ kcal}{1000\ cal}= 5.43062\ kcal$

In the given format $\frac{22.7\ kJ}{1\ g}\times\frac{1000\ J}{1\ kJ}\times\frac{1\ cal}{4.18\ J}\times\frac{1\ kcal}{1000\ cal}=\frac{5.43\ kcal}{g}$ (rounded to two - decimal places)

So $A = 1000\ J$, $B = 1\ kJ$, $C = 1\ cal$, $D = 4.18\ J$, $E = 1\ kcal$, $F = 1000\ cal$

Answer:

$A = 1000\ J$, $B = 1\ kJ$, $C = 1\ cal$, $D = 4.18\ J$, $E = 1\ kcal$, $F = 1000\ cal$