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:
The net energy is 2.196 eV
Explanation:
Basically, the energy of an atom increases when it absorbs a photon. In addition, the wavelength of the emitted photon is longer such that the atom absorbed a net energy in the process.
Using:
ΔE = h*c*(1/λ
- 1/λ
)
where:
ΔE is the net energy in eV (electron-volt). 1 eV is equivalent to 1.602*
J.
h = 4.135*
eVs
c = 3*
m/s
λ = 300 nm = 300*
m
λ = 640 nm = 640* m
Thus:
ΔE = 4.135* eVs*3* m/s*(
)
ΔE = 4.135**3**1.77*
eV = 2.196 eV
Answer:
300.9mL
Explanation:
Given parameters:
V₁ = 280mL
T₁ = 22°C
T₂ = 44°C
Unknown:
V₂ = ?
Solution:
To solve this problem, we apply Charles's law;
it is mathematically expressed as;
We need to convert the temperature to kelvin;
T₁ = 22°C = 22 + 273 = 295K
T₂ = 44°C = 44 + 273 = 317K
Input the parameters and solve;
= 
V₂ x 295 = 280 x 317
V₂ = 300.9mL
Answer: A. the rotational period of the earth is the same as that of the moon
