The specific gravity or relative density of a substance is the ratio of its density to the density of a reference material. The relative density of the concentrated salt solution is 1.044.
Mathematically;
Density of the concentrated salt = mass of salt/volume of salt = 5.222 g/5.000 mL = 1.044 g/mL
In the case of specific gravity, the reference material is always water and water has a density of 1 g/mL.
Hence, specific gravity of the concentrated salt solution =
Density of concentrated salt solution/density of equal volume of water
= 1.044 g/mL/1 g/mL
= 1.044
Note that specific gravity is dimensionless.
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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:
1.25 M
Explanation:
Step 1: Given data
Mass of KI (solute): 20.68 g
Volume of the solution: 100 mL (0.100 L)
Step 2: Calculate the moles of solute
The molar mass of KI is 166.00 g/mol.
20.68 g × 1 mol/166.00 g = 0.1246 mol
Step 3: Calculate the molar concentration of KI
Molarity is equal to the moles of solute divided by the liters of solution.
M = 0.1246 mol/0.100 L= 1.25 M
Answer:
The solution is 10^-2 or 0.01M in HCl.
Explanation:
meaning of pH is "power of hydrogen".
what is the molar concentration of a HCl solution with pH=2?
Let say pH=2
[H+]=10^-2M
HCL is a strong acid that dissociates completely:
[H+]=[HCL]
Therefore solution is 10^-2 or 0.01M in HCL.
Answer:
[OH⁻] = 3.34x10⁻³M; Percent ionization = 0.54%; pH = 11.52
Explanation:
Kb of the reaction:
NH3 + H2O(l) ⇄ NH4+ + OH-
Is:
Kb = 1.8x10⁻⁵ = [NH₄⁺] [OH⁻] / [NH₃]
<em>As all NH₄⁺ and OH⁻ comes from the same source we can write: </em>
<em>[NH₄⁺] = [OH⁻] = X</em>
<em>And as </em>[NH₃] = 0.619M
1.8x10⁻⁵ = [X] [X] / [0.619M]
1.11x10⁻⁵ = X²
3.34x10⁻³ = X = [NH₄⁺] = [OH⁻]
<h3>[OH⁻] = 3.34x10⁻³M</h3><h3 />
% ionization:
[NH₄⁺] / [NH₃] * 100 = 3.34x10⁻³M / 0.619M * 100 = 0.54%
pH:
As pOH = -log [OH-]
pOH = 2.48
pH = 14 - pOH
<h3>pH = 11.52</h3>
