Question

A 0.0447−mol sample of a nutrient substance, with a formula weight of 114 g/mol, is burned in a bomb calorimeter containing 6.19 × 102 g H2O. Given that the fuel value is 6.13 × 10−1 in nutritional Cal when the temperature of the water is increased by 5.05°C, what is the fuel value in kJ in scientific notation?

Answer 1

Answer:

The value is  $x = 2.565 *10^{3} \ kJ/kg$

Explanation:

From the question we are told that

  The  no of moles of the sample is  n = 0.0447 mole

  The formula weight is  $M = 114 \ g/mol$

   The  mass of water is  $m = 6.19 *10^{2}\ g$

   The amount of the fuel is  $f= 6.13*10^{-1} \ nutritional \ Cal$

   The temperature rise is  $\Delta T = 5.05^o$

Generally

      $1 \ nutritional \ Cal => 4.184*10^{3} \ kJ/kg$

=>  $f= 6.13*10^{-1} \ nutritional \ Cal \to x$

=>     $x = \frac{6.13 *10^{-1} * 4.184 *10^{3}}{1}$

=>     $x = 2.565 *10^{3} \ kJ/kg$