Answer:
28 g/mol, N2
Explanation:
Given data:
Volume of gas = 5.0 L
Mass of gas = 6.3 g
Pressure = 1 atm
Temperature = 273 K
Molar mass of gas = ?
Solution:
We will calculate the density first.
d = mass/ volume
d = 6.3 g/ 5.0 L
d = 1.26 g/L
Molar mass:
d = PM/RT
M = dRT/P
M = 1.26 g/L× 0.0821 atm.L/mol.K × 273 K/ 1 atm
M = 28 g/mol
Molar mass of N₂ is 28 g/mol thus given gas is N₂.
can be split into hydrogen and oxygen chemical property
is liquid at room temperature physical property
has a density of 1.0 g/cm³ physical property
reacts with certain metals chemical property
Explanation:
The physical properties of matter are the properties that occur when no change is occurring to a matter. They are usually observed with our senses or some simple laboratory experiments.
Some of the physical properties are boiling point, density, state of matter, e.t.c
The chemical properties are the properties that occurs when a matter undergoes some chemical changes.
For example rusting, decomposition, reactivity e.t.c
Learn more:
Chemical properties brainly.com/question/2376134
#learnwithBrainly
The skeletal structure of an organic compound is an abbreviated representation of its molecular structure, they are quick and easy to draw.
For example, the following image shows the skeletal structure of a compound:
The peaks represent the carbons. We must remember that carbon can have a maximum of 4 bonds.
Now, I will show you how is the structure of this specific compound:
This is ternary alcohol, called 2-methyl-2-butanol. If you see carefully, you will notice that each carbon has 4 bonds. The functional groups present will be OH. The skeletal structure will be:
The Element that form more compounds than all the rest is Carbon.
Elements of Carbon
<span>A molecule of the compound carbon dioxide contains one atom of the element carbon and two atoms of the element oxygen. Each oxygen atom shares a double bond with the carbon atom. Carbon is the sixth element in the periodic table and occurs in pure form as coal and diamonds.
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Answer:
1.5e+8 atoms of Bismuth.
Explanation:
We need to calculate the <em>ratio</em> of the diameter of a biscuit respect to the diameter of the atom of bismuth (Bi):

For this, it is necessary to know the values in meters for any of these diameters:


Having all this information, we can proceed to calculate the diameters for the biscuit and the atom in meters.
<h3>Diameter of an atom of Bismuth(Bi) in meters</h3>
1 atom of Bismuth = 320pm in diameter.

<h3>Diameter of a biscuit in meters</h3>

<h3>Resulting Ratio</h3>
How many times is the diameter of an atom of Bismuth contained in the diameter of the biscuit? The answer is the ratio described above, that is, the ratio of the diameter of the biscuit respect to the diameter of the atom of Bismuth:





In other words, there are 1.5e+8 diameters of atoms of Bismuth in the diameter of the biscuit in question or simply, it is needed to put 1.5e+8 atoms of Bismuth to span the diameter of a biscuit in a line.