For this problem, we use the Beer Lambert's Law. Its usual equation is:
A = ∈LC
where
A is the absorbance
∈ is the molar absorptivity
L is the path length
C is the concentration of the sample solution
As you notice, we only have to find the absorbance. But since we are not given with the molar absorptivity, we will have to use the modified equation that relates % transmittance to absorbance:
A = 2 - log(%T)
A = 2 - log(27.3)
A = 0.5638
Answer:
True
Explanation:
This means the fuel is highly flammable hence it has a low ignition temperature. This, therefore, means it burns fast without putting in a lot of energy to ignite it. Even in cars, therefore, the compression ratios, in engines using these of these fuels, is lesser.
Answer:
Tc
Explanation:
You just have to follow the rows with the exponents. Just remember that when we get to d, the number in the front is a period lower. Hope this helps!
Answer:
330.95K
Explanation:
V₁ = 1.2L
T₁ = 25°C = (25 + 273.15)K = 298.15K
P₁ = 1.0 atm
P₂ = 0.74 atm
V₂ = 1.8L
T₂ =?
From combined gas equation,
(P₁ * V₁) / T₁ = (P₂ * V₂) / T₂
Solve for T₂
T₂ = (P₂ * V₂ * T₁) / (P₁ * V₁)
T₂ = (0.74 * 1.8 * 298.15) / (1.0 * 1.2)
T₂ = 397.1358 / 1.2
T₂ = 330.9465K
T₂ = 330.95K or T₂ = (330.95 - 273.15)°C = 57.8°C
