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I’m not entirely sure
Glucose or fructose. In glycolysis, glucose is oxidized to be pyruvate.
Pepsinogen, an inactive zymogen, is secreted into gastric juice from both mucous cells and chief cells. Once secreted, pepsinogen is activated by stomach acid into the active protease pepsin, which is largely responsible for the stomach's ability to initiate digestion of proteins.
<u>Answer</u>:
Option B. a positive feedback loop
<u>Explanation</u>:
This happens when the product of the reaction is more than that of the reaction. In homeostasis this type of positive feedback loop that can move away. It exacerbates the effect of small disturbance. It also strengthens the change in physiological condition body rather than reversing the change. A slight change in the normal range will result in greater change, and hence the system diverts more from the normal range. It can only be considered normal with definite end point.
Answer:
141C
Explanation:
We need to use the ideal gas law to solve this question. Ideal gas law: PV = nRT
P = pressure = 4.00 atm = 4.00 × 101.325 = 405.3 kPa (pressure needs to be converted to kilopascals)
V = volume = 4.25L
n = number of moles = 0.500
R = gas constant = 8.314 mol/K (found om chemistry data sheet)
T = temperature = ? unknown
Before any other steps are taken, we need to rearrange the the ideal gas law formula to make T the subject.
PV/nR = T
Now we can substitute the values.
(405.3 × 4.25) ÷ (0.500 × 8.314) = 414.3673322107 (answer in kelvin)
Convert 414.3673322107 Kelvin to Celsius.
0C = 273.15K (found on chemistry data sheet)
414.3673322107 - 273.15 = 141.2173322107C
141.2173322107 to nearest C = 141C
