We know the formula of the density:
ρ =

;
So the mass will be equal to:
m = ρ * V = 2.70 * 21.3 = 57,51 g =
57510 mg of substance.
So the answer is 57510.
This is a true statement if it is density you are looking for... Density problem.....
Density is the ratio of the mass of an object to its volume.
D = m / V
D = 104g / 14.3 cm³ = 7.27 g/cm³ .............. to three significant digits
The conventions for the units of density is that grams per cubic centimeter (g/cm³) are usually used for solids, but will work for anything. Grams per milliliter (g/mL) are usually used for liquids and grams per liter (g/L) are for gases. Therefore, by convention, the units for tin (a solid) should be in grams per cubic centimeter.
Since 1 mL is equivalent to 1 cm³, then the density could be expressed as 7.27 g/mL.
The accepted value for the density of tin is 7.31 g/cm³
Answer:
There are three ways that scientists have proved that these sub-atomic particles exist. They are direct observation, indirect observation or inferred presence and predictions from theory or conjecture. Scientists in the 1800's were able to infer a lot about the sub-atomic world from chemistry.
Explanation:
Hope this helps
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.
I hope you find this informatiou useful and interesting! Good luck!
Excess reactant : Na
NaCl produced : = 16.497 g
<h3>Further explanation</h3>
Given
Reaction(balanced)
2Na + Cl₂⇒ 2NaCl
20 g Na
10 g Cl₂
Required
Excess reactant
NaCl produced
Solution
mol Na(Ar = 23 g/mol) :
= 20 : 23 = 0.87
mol Cl₂(MW=71 g/mol):
= 10 : 71 g/mol = 0.141
mol : coefficient :
Na = 0.87 : 2 = 0.435
Cl₂ = 0.141 : 1 = 0.141
Limiting reactant : Cl₂(smaller ratio)
Excess reactant : Na
Mol NaCl based on mol Cl₂, so mol NaCl :
= 2/1 x mol Cl₂
= 2/1 x 0.141
= 0.282
Mass NaCl :
= 0.282 x 58.5 g/mol
= 16.497 g