assuming an experimental value of enthalpy of -655 kj/mol and a true value of -603, what is the percent…

assuming an experimental value of enthalpy of -655 kj/mol and a true value of -603, what is the percent error of the experiment? use a calculator and round to the nearest hundredth as well as the absolute value of the percentage.
Answer
Explanation:
Step1: Recall percent - error formula
Percent error = $\left|\frac{\text{Experimental value}-\text{True value}}{\text{True value}}\right|\times100%$
Step2: Substitute given values
The experimental value is $- 655$ kJ/mol and the true value is $-603$ kJ/mol. Percent error = $\left|\frac{-655 - (-603)}{-603}\right|\times100%=\left|\frac{-655 + 603}{-603}\right|\times100%=\left|\frac{-52}{-603}\right|\times100%$
Step3: Calculate the fraction and percentage
$\frac{52}{603}\approx0.08624$, and $0.08624\times100% = 8.624%$ Rounding to the nearest hundredth, we get $8.62%$
Answer:
$8.62%$