Answer:-
5.554 litres
Explanation:-
Initial Volume V 1 = 5.00 litres
Initial temperature T 1 = 34 C + 273 = 307 K
Final temperature T 2 = 68 C + 273 = 341 K
Using the relation of Charles Law (V 1 / T 1) = (V 2 / T 2)
New Volume V 2 = V1 x T2 / T1
Plugging in the values
V2 = 5.00 litres x 341 K / 307 K
= 5.554 litres.
Answer:
moles Na = 0.1114 g / 22.9898 g/mol=0.004846
moles Tc = 0.4562g /98.9063 g/mol=0.004612
mass O = 0.8961 - ( 0.1114 + 0.4562)=03285 g
moles O = 0.3285 g/ 15.999 g/mol=0.02053
divide by the smallest
0.02053/ 0.004612 =4.45 => O
0.004846/ 0.004612 = 1.0 => Tc
to get whole numbers multiply by 2
Na2Tc2O 9
Explanation:
Hope it right hope it helps
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:
Theoretical maximum moles of hydroquinone: 0.2167 mol.
Explanation:
Hello,
In this case, the undergoing chemical reaction is like:
In such a way, since the molar mass of quinone is 108.1 g/mol and it is in a 1:1 molar ratio with hydroquinone, we can easily compute the theoretical maximum moles of hydroquinone by stoichiometry:
Clearly, this is the theoretical yield which in grams is:
Which allows us to compute the percent yield as well since the obtained mass of the product is 13.0 g:
Best regards.
