Answer:
describes properties characteristic of no more than two electrons in the vicinity of an atomic nucleus or of a system of nuclei as in a molecule
Let's divide the three experiments: The experiment with 10.00 mL of water is A), the experiment with 15.00 mL is B), and the experiment with 25.00 mL is C).
- (1) Now let's calculate the experimental density of each experiment. Density (ρ) is equal to the mass divided by the volume, thus:
- (2)To calculate the average density, we add each density and divide the result by the number of experiments (in this case 3):
- (3) The percent error is calculated by dividing the absolute value of the substraction of the theorethical and experimental values, by the theoretical value, times 100:
%error=
%error=
%error=2.44 %
Answer:
Check explanation
Explanation:
From the question, the parameters given are 64.7g of benzene,C6H6; a starting temperature of 41.9°C and bringing it to 33.2°C.
Molar mass of benzene,C6H6= 78.11236 g/mol.
Things to know: heat capacity of benzene, C6H6= 1.63 J/g.K, the heat of fusion = 9.87 kj/mol.
STEP ONE(1): ENERGY USED IN MELTING BENZENE SOLID.
Using the formula below;
Energy used in melting the solid(in JOULES) = (mass of benzene/molar mass of benzene) × heat of Fusion.
=(64.7 g of C6H6/ 78.11236(g per mol) of C6H6) × 9.87 kJ per mol.
= 8.175 J.
= 0.008175 kJ.
STEP TWO (2): ENERGY OF HEATING THE LIQUID.
It can be calculated from the formula below;
Energy= heat capacity (J/g.K) × mass of benzene× (∆T).
= 1.63 J/g.K × 64.7 × (41.9-33.2).
= 917.5J.
= 0.9175 kJ.
Energy required to boil benzene= Energy required to melt the bezene + energy required for boiling.
= 0.008175+ 0.9175.
= 0.93kJ
Approximately, 1 kJ