Answer:
0.37 g
Explanation:
The molecular weight for Glycerol = 92
Number of Carbon atoms in glycerol (x) 
= 3
Molecular weight of the Biomass ( Klebsiella aerogenes )
= 
= 
= 26.1
From the molecular weight of the Biomass, we can deduce the Degree of reduction for the substrate(glycerol denoted as 
) as follows:
= (4×1)+(1×1.73)-(2×0.43)-(3×0.24)
= 4.15
Given that the yield of the Biomass = 0.40 g
However;
C = 
C = 
C = 1.41 g
Now , the oxygen requirement can be calculated as:
= 
= 
= 2.1 g/mol
Hence, we can say that the needed oxygen = 2.1 g/mol of the substrate consumed.
Now converting it to mass terms; we have:
= 
= 
= 0.3652 g
≅ 0.37 g
∴ The oxygen requirement for this culture in mass terms = 0.37 g
It means that 22.5×10^5 J of heat is required to change 1 kg of water into steam.
Latent heat of vaporization is amount of energy required to change 1 gram (in this example 1 kilogram) of material from the liquid to the gaseous state at its boiling point.
Boiling point of the water is 100°C.
Joule (J) is the standard unit for energy (in this example heat).
Evaporization is phase change process in which the water changes from a liquid to a gas (water vapor). Fore example, solar radiation can be the source of energy for evaporation.
More about heat of vaporization: brainly.com/question/14679329
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<span>One reaction takes place in less time than the other.</span>
Answer:
Flammability is a material’s ability to burn in the presence of <u><em>oxygen.</em></u>
Explanation:
Flammability can be described as the ability of a substance to get ignited. Flammability will lead to fire or combustion. Some substances are highly flammable like Benzene. Other tend to be just flammable. And there are also compounds which will nor be flammable at all as they won't react with oxygen. Examples of these substances include helium, steel or glass.
The flammability of a substance shall be considered a very important aspect when storing or transporting a substance.
<u>Answer:</u> The final temperature of water is 38.5°C
<u>Explanation:</u>
To calculate the amount of heat released or absorbed, we use the equation:
where,
q = heat absorbed = 1.506 kJ = 1506 J (Conversion factor: 1 kJ = 1000 J)
m = mass of water = 30 g
c = specific heat capacity of water = 4.184 Jl/g.°C
= change in temperature = 
Putting values in above equation, we get:
Hence, the final temperature of water is 38.5°C
