Answer:
282.7KPa
Explanation:
Step 1:
Data obtained from the question.
Number of mole of (n) = 1.5 mole
Volume (V) = 13L
Temperature (T) = 22°C = 22 + 273°C = 295K
Pressure (P) =..?
Gas constant (R) = 0.082atm.L/Kmol
Step 2:
Determination of the pressure exerted by the gas.
This can be obtained by using the ideal gas equation as follow:
PV = nRT
P = nRT /V
P = 1.5 x 0.082 x 295 / 13
P = 2.79atm.
Step 3:
Conversion of 2.79atm to KPa.
This is illustrated below:
1 atm = 101.325KPa
Therefore, 2.79atm = 2.79 x 101.325 = 282.7KPa
Therefore, the pressure exerted by the gas in KPa is 282.7KPa
Answer:
Explanation:
In this question, we wish to find the missing nuclei for the equation:
In order to find the missing species, we need to use the charge and mass balance law. That is, the mass should be conserved: the total mass on the left-hand side with respect to the arrow should be equal to the total mass on the right-hand side with respect to the arrow:
Notice from here that:
So far we know that the mass of X is 4. Similarly, we apply the law of charge conservation. The total charge should be conserved:
From here:
We have a particle:
Looking at the periodic table, an atom with Z = 2 corresponds to helium. This can also be written as an alpha particle:
Answer: An increase in the ratio of insulin to glucagon will increase the activity of --
- Acetyl-CoA carboxylase(+)
-Phosphofructokinase PFK2(+)
-Glycogen synthase(+)
- Hormone sensitive lipase (-). The hormone sensitive lipase activity is not increased with increased insulin activity.
Explanation: increased insulin - glucagon ratio is usually high in fed state.Insulin helps the cells absorb glucose, reducing blood sugar and providing the cells with glucose for energy. When blood sugar levels are too low, the pancreas releases glucagon. Glucagon instructs the liver to release stored glucose, which causes blood sugar to rise.
