<span>Enzymes have three main characteristics. First, they increase the rate of a natural chemical reaction. Secondly, they typically only react with one specific substrate or reactant, and thirdly, enzyme activity is regulated and controlled within the cell through several different means, including regulation by inhibitors and activators. It is possible to group enzymes into different categories, including oxidases, transferases, hydrolases, lyaes, isomerases and ligases. In naming enzymes, the "-ase" suffix is often appended to the name of the substrate molecule upon which which the enzyme reacts. For example, the enzyme sucrase catalyzes the transformation of the sugar sucrose in to glucose and fructose. In this case, the "sucr-" suffix represents the molecule upon which the sucrase enzyme reacts. Not all enzymes are named according to this convention.</span>
Answer:-
2328.454 grams
Explanation:-
Volume V = 18.4 litres
Temperature T = 15 C + 273 = 288 K
Pressure P = 1.5 x 10^ 3 KPa
We know universal Gas constant R = 8.314 L KPa K-1 mol-1
Using the relation PV = nRT
Number of moles of oxygen gas n = PV / RT
Plugging in the values
n = (1.5 x 10^3 KPa ) x ( 18.4 litres ) / ( 8.314 L KPa K-1 mol-1 x 288 K)
n = 11.527 mol
Now the balanced chemical equation for this reaction is
2KNO3 --> 2KNO2 + O2
From the equation we can see that
1 mol of O2 is produced from 2 mol of KNO3.
∴ 11.527 mol of O2 is produced from 2 x 11.527 mol of KNO3.
= 23.054 mol of KNO3
Molar mass of KNO3 = 39 x 1 + 14 x 1 + 16 x 3 = 101 grams / mol
Mass of KNO3 = 23.054 mol x 101 gram / mol
= 2328.454 grams
Explanation:
The given data for case (1) is as follows.
h = 20 cm = 0.2 m
Assuming that a rectangular slab is placed above the pipe and we will calculate the heat transfer as follows.
Q = 
where, A = area
L = length
k = thermal conductivity = 0.8 W/m
= change in temperature.
Therefore, putting the given values into the above formula as follows.
Q =
= 
= 168 W
For case (2), h = 180 cm = 1.8 m
Therefore, heat lost will be calculated as follows.
Q =
= 
= 18.67 W
Thus, we can conclude that 18.67 W heat lost if the pipe was buried at a depth of 180 cm.
I think D?? I apologize if not-check in other answers to be sure ^^
