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

The value is
`x = 2.565 *10^(3) \ kJ/kg`

Step-by-step 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 = (6.13 *10^(-1) * 4.184 *10^(3))/(1)`

=>
`x = 2.565 *10^(3) \ kJ/kg`