Answer:
The temperature change from the combustion of the glucose is 6.097°C.
Explanation:
Benzoic acid;
Enthaply of combustion of benzoic acid = 3,228 kJ/mol
Mass of benzoic acid = 0.570 g
Moles of benzoic acid =
Energy released by 0.004667 moles of benzoic acid on combustion:
Heat capacity of the calorimeter = C
Change in temperature of the calorimeter = ΔT = 2.053°C
Glucose:
Enthaply of combustion of glucose= 2,780 kJ/mol.
Mass of glucose=2.900 g
Moles of glucose =
Energy released by the 0.016097 moles of calorimeter combustion:
Heat capacity of the calorimeter = C (calculated above)
Change in temperature of the calorimeter on combustion of glucose = ΔT'
The temperature change from the combustion of the glucose is 6.097°C.
Answer:
Boron has a larger radius and the protons in carbon exert more pull.
Explanation:
Remember than elements have greater radius as they are closer to the bottom left corner, so boron would have the larger radius here. Carbon has a smaller radius, which makes it easier for the protons in carbon to exert more pull.
He combustion reaction of hexane:
2 C6H14 + 19 O2 = 12 CO2 + 14 H2O
2 mol C6H14 ---------> 12 mol CO2
x mol C6H14 ---------> 18.4 mol CO2
12 * x = 18.4 * 2
12 x = 36.8
x = 36.8 / 12
x = 3.067 moles of CO2
hope this helps!.
Standar molar entropies from a table:
CH3OH (l): 126.8 J / K*mol
O2 (g): 205.1 J/ K*mol
CO2(g): 213.7 J/K*mol
H2O(g): 188.8 J/J*mol
Now use the coefficients of the reaction and sum the product entropies less the reactant entropies:
4*188.8 + 2*213.7 - 3*205.1 - 2* 126.8 = 313.7 J/mol*K
Answer: option C) +3173.766 J/mol*K