Answer:
3.07 Cal/g
Explanation:
Step 1: Calculate the heat absorbed by the calorimeter
We will use the following expression.
Q = C × ΔT
where,
- C: heat capacity of the calorimeter (37.60 kJ/K = 37.60 kJ/°C)
- ΔT: temperature change (2.29 °C)
Q = 37.60 kJ/°C × 2.29 °C = 86.1 kJ
According to the law of conservation of energy, the heat released by the candy has the same magnitude as the heat absorbed by the calorimeter.
Step 2: Convert 86.1 kJ to Cal
We will use the conversion factor 1 Cal = 4.186 kJ.
86.1 kJ × 1 Cal/4.186 kJ = 20.6 Cal
Step 3: Calculate the number of Cal per gram of candy
20.6 Cal/6.70 g = 3.07 Cal/g
This problem requires a certain equation. That equation is V1/T1=V2/T2, where V1 is your initial volume (535 mL in this case), T1 is your initial temperature in Kelvin(23 degrees C = 296 K), V2 is your final volume (unknown), and T2 is your final temperature (46 degrees C = 319 K). By plugging in these values, the equation looks like this: 535/296=V2/319. Now multiply both sides of the equation by 319, and your final answer is V2= 576.6 mL
Answer : The fuel value and the fuel density of pentane is, 49.09 kJ/g and 
respectively.
Explanation :
Fuel value : It is defined as the amount of energy released from the combustion of hydrocarbon fuels. The fuel value always in positive and in kilojoule per gram (kJ/g).
As we are given that:
Fuel value = 
Molar mass of pentane = 72 g/mol
Fuel value = 
Fuel value = 49.09 kJ/g
Now we have to calculate the fuel density of pentane.
Fuel density = Fuel value × Density
Fuel density = (49.09 kJ/g) × (0.626g/mL)
Fuel density = 30.73 kJ/mL =
Thus, the fuel density of pentane is
This is an application of Boyle's law:
P₁V₁ = P₂V₂. we don't have to convert volume and pressure to standard forms. we can even use the pressure with mmHg
1 atm = 760 mmHg
V₂ = P₁V₁ / P₂ = 745 x 500 / 760 = 490 ml
Note that here we assume constant temperature
