A solid is cooled to a very low temperature. Assuming the mass remains constant, how, if at all, does this affect the density of
2 answers:
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Explanation:
According to the given data, we will calculate the following.
Half life of lipase = 8 min x 60 s/min
= 480 s
Rate constant for first order reaction is as follows.
=
Initial fat concentration = 45
= 45 mmol/L
Rate of hydrolysis = 0.07 mmol/L/s
Conversion X = 0.80
Final concentration (S) =
= 45 (1 - 0.80)
= 9
or, = 9 mmol/L
It is given that = 5mmol/L
Therefore, time taken will be calculated as follows.
t =
Now, putting the given values into the above formula as follows.
t =
=
=
= 27.38 min
Therefore, we can conclude that time taken by the enzyme to hydrolyse 80% of the fat present is 27.38 min.
Answer:
Q = 114349.5 J
Explanation:
Hello there!
In this case, since this a problem in which we need to calculate the total heat of the described process, it turns out convenient to calculate it in three steps; the first one, associated to the heating of the liquid water from 40 °C to 100 °C, next the vaporization of liquid water to steam at constant 100 °C and finally the heating of steam from 100 °C to 115 °C. In such a way, we calculate each heat as shown below:
Thus, the total energy turns out to be:
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