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Rudiy27
3 years ago
10

Mr. Smith is hyperglycemic with a blood glucose level of 300mg/ml of blood. Explain how homeostasis would regulate his glucose l

evels through a negative feedback system.
Chemistry
1 answer:
aleksley [76]3 years ago
6 0

Answer:

When the animal has eaten food and the blood glucose level in the body increases. The pancreas cells in the body detects the increase in the blood glucose which leads to increase the insulin hormone.

This decreases the blood glucose level in the level. This is how the negative feedback works in the body if the level of glucose increases.

Negative feedback is the way by which the body maintains homeostasis and maintains equilibrium in the body.

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Consider the following reaction: CO(g)+2H2(g)⇌CH3OH(g) This reaction is carried out at a different temperature with initial conc
Anni [7]

Answer:

Ka = 4.76108

Explanation:

  • CO(g) + 2H2(g) ↔ CH3OH(g)

∴ Keq = [CH3OH(g)] / [H2(g)]²[CO(g)]

                      [ ]initial         change         [ ]eq

CO(g)              0.27 M       0.27 - x        0.27 - x

H2(g)              0.49 M       0.49 - x        0.49 - x

CH3OH(g)          0                0 + x               x = 0.11 M

replacing in Ka:

⇒ Ka = ( x ) / (0.49 - x)²(0.27 - x)

⇒ Ka = (0.11) / (0.49 - 0.11)² (0.27 - 0.11)

⇒ Ka = (0.11) / (0.38)²(0.16)

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7 0
3 years ago
Calculate the pOH of a solution if the concentration of hydroxide ions (OH-) is 1.9 x 10-5M?
OLEGan [10]

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Concentration of the hydroxide ions = 1.9 x 10-5 M

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6 0
3 years ago
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  <span>C6H12 = 6x12 + 6x1 = 78. 
The equation indicates that 2x78 = 156g benzene will produce 6542kJ. 
Using proportions you can then calculate that 
x/6542kJ = 7.9g / 156g 
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heat = mass x ΔT x 4.18J/g° 
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final temp = 21 + 14° = 35°C</span>
4 0
3 years ago
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