Answer:
Mescarinic and Nicotinic
Explanation:
Postganglionic fibers can be present in both sympathetic and parasympathetic divisions, their main difference resides in how in the sympathetic division the postganglionic fibers are adrenergic and use norepinephrine (noradrenalin) as a neurotransmitter, in the parasympathetic division, on the other hand, fibers are cholinergic and use acetylcholine as a neurotransmitter, the<em> postganglionic neurons of sweat glands release acetylcholine for the activation of muscarinic receptors, another kind of receptor for acetylcholine are nicotinic receptors </em>that act as transmembrane sodium/potassium channels, while muscarinic receptors need to act through intracellular proteins.
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Answer:
The difference in the electronegativities of chlorine and boron is 3.0 - 2.0 = 1.0 ; the difference in between chlorine and carbon is 3.0 = 2.5 = 0.5 . Consequently, the B-Cl bond is more polar ; the chlorine atom asrries the partial negative charge because it has higher electronegativity .
Explanation:
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Answer:
802.69 g
Explanation:
The molar mass of Barium nitrite is 229.34 g/mol, so 3.5 moles of it will have a mass of ...
3.5 mol × 229.34 g/mol = 802.69 g
Answer:
Those two horizontal lines.
Explanation:
Hello there!
In this case, when focusing on these heating curves, it is important to say they tend to have two constant-temperature sections and three variable-temperature sections. Thus, from lower to higher temperature, the first constant-temperature section corresponds to melting and the second one vaporization, whereas the three variable-temperature sections correspond to the heating of the solid until melting, the liquid until vaporization and the gas until the critical point.
In such a way, we infer that the boxes referred to constant temperature are referred to a gain in potential energy, that is, the two horizontal lines.
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Answer:
-3.7771 × 10² kJ/mol
Explanation:
Let's consider the following equation.
3 Mg(s) + 2 Al³⁺(aq) ⇌ 3 Mg²⁺(aq) + 2 Al(s)
We can calculate the standard Gibbs free energy (ΔG°) using the following expression.
ΔG° = ∑np . ΔG°f(p) - ∑nr . ΔG°f(r)
where,
n: moles
ΔG°f(): standard Gibbs free energy of formation
p: products
r: reactants
ΔG° = 3 mol × ΔG°f(Mg²⁺(aq)) + 2 mol × ΔG°f(Al(s)) - 3 mol × ΔG°f(Mg(s)) - 2 mol × ΔG°f(Al³⁺(aq))
ΔG° = 3 mol × (-456.35 kJ/mol) + 2 mol × 0 kJ/mol - 3 mol × 0 kJ/mol - 2 mol × (-495.67 kJ/mol)
ΔG° = -377.71 kJ = -3.7771 × 10² kJ
This is the standard Gibbs free energy per mole of reaction.
