Molar mass of MgCO3 is 84.313 g/mol
You can calculate this from data on the periodic table:
Molar mass Mg = 24.305g/mol
molar mass C = 12.011g/mol
molar mass O = 15.999g/mol mass 3 mol = 47.997g
Total = 84.313g/mol
Mass to be used in 1.2L of 1.5M solution = 84.313g * 1.2L * 1.5mol /L = 151.763g
I have not taken significant figures into account
The balanced equation you provide is not necessary in this calculation
Answer:
used=Evaporation and sublimation
released=condensation and freezing
Explanation:
2.23 moles of propane react when 294 g of CO₂ is formed .
<h3>What is moles ?</h3>
Moles is a unit which is equal to the molar mass of an element.
A reaction is given
C₃H₈ +50₂ → 3CO₂ + 4H₂O
Grams of CO₂ formed = 294 gm
In moles = 294 /44 = 6.68 moles.
Let x be the moles of C₃H₈ is x
Mole ratio of CO₂ to C₃H₈ = 3 : 1
so
6.68 /x = 3/1
x = 6.68 /3 = 2.23 moles
Therefore 2.23 moles of propane react when 294 g of CO₂ is formed .
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b) It is based on atomic properties as alkali metals requires 7 more electrons to complete their outer orbit. And they try to give those electrons to other elements to obtain noble gas configuration.
Noble gases are the gases which do not react easily with anything. They are also called as Inert gases, and belongs to group 18 of the periodic table.
Alkali metals are the substances which are found in Group I of a periodic table. Mostly the elements which are present are:
Properties of alkali metals are: Soft, shiny reactive metals. They are soft enough to cut with knife. Metals react with water and air quickly and gets tarnish, so pure metals are stored in container by dipping them in oil to prevent oxidation.
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The answer would be volume
