Answer: Extracellular [Ca2+]
Explanation:
The sensitivity and density of the alpha receptors serve to <em>enhance the response to the release of</em> <em>norepinephrine (NE)</em> . However, they do not exert a strong influence as the concentration of calcium ions on the amount of <em>norepinephrine (NE)</em> released by sympathic nerve terminals.
The release of neurotransmitters depends more on either an external or internal stimulus.This results in an action potential which on reaching a nerve terminal, results in the opening of Ca²⁺ channels in the neuronal membrane. Because the extracellular concentration of Ca²⁺ is greater than the intracellular Ca²⁺ concentration, Ca²⁺ flows into the nerve terminal. This triggers a series of events that cause the vesicles containing <em>norepinephrine (NE)</em> to fuse with the plasma membrane and release <em>norepinephrine (NE)</em> into the synapse. The higher the action potential, the higher the Ca²⁺ flow into the terminals resulting in higher amount of <em>norepinephrine (NE)</em> into the synapse, and vice versa.
Catechol-O-methyltransferase (COMT) is one of several enzymes that degrade catecholamines such as dopamine, epinephrine, and norepinephrine. It serves a regulatory purpose to lower the concentration of norepinephrine upon its release from nerve terminals.
Answer:
100
Explanation:
Step 1: Subtract 50 by 300 to find the total mass of the marbles
300 - 50 = 250
Not hard, right?
Step 2: Divide 250 by 2.5
250 ÷ 2.5 = 100
And... 100 is your answer :)
Hope this helps :)
-jp524
5 times dilution
0.780M x 1/5 = 0.156M
Hope this help.
Answer:
Percentage mass of copper in the sample = 32%
Explanation:
Equation of the reaction producing Cu(NO₃) is given below:
Cu(s)+ 4HNO₃(aq) ---> Cu(NO₃)(aq) + 2NO₂(g) + 2H₂O(l)
From the equation of reaction, 1 mole of Cu(NO₃) is produced from 1 mole of copper. Therefore, 0.010 moles of Cu(NO₃) will be produced from 0.010 mole of copper.
Molar mass of copper = 64 g/mol
mass of copper = number of moles * molar mass
mass of copper = 0.01 mol * 64 g/mol = 0.64 g
Percentage by mass of copper in the 2.00 g sample = (0.64/2.00) * 100%
Percentage mass of copper in the sample = 32%
