The mass of atoms of carbon and 3 molecules of hydrogen : 18 g/mol
<h3>Further explanation
</h3>
An atomic mass unit ( amu or "u") is a relative atomic mass of 1/12 the mass of an atom of carbon-12.
The molar mass(molecular mass-formula mass-molecular weight(MW)) of a compound is the sum of the relative atomic mass (Ar) of the constituent elements of the compound
Can be formulated :
M AxBy = (x.Ar A + y. Ar B)
The mass of atom of Carbon(C)⇒Ar = 12 g/mol
The mass of 1 molecule of Hydrogen - H₂(MW) : 2 g/mol
The mass of 3 molecules of Hydrogen : 3 x 2 = 6 g/mol
So the mass of atoms of carbon and 3 molecules of hydrogen :
The chemical equation is said to be balanced if the number of atoms in the reactants and products is the same
<h3>Further explanation</h3>
Equation balanced ⇒ total number of atoms in reactants(on the left)= total number of atoms in products(on the right)
H₂+O₂---> H₂O
Reactants : H₂, O₂
Products : H₂O
not balanced
H₂O₂ ---> H₂O+O₂
Reactants : H₂O₂
Products : H₂O, O₂
not balanced
Na+O₂ ---> Na₂O
Reactants : Na, O₂
Products : Na₂O
not balanced
N₂+H₂ ---> NH₃
Reactants : N₂, H₂
Products : NH₃
not balanced
P₄+O₂---> P₄O₁₀
Reactants : P₄, O₂
Products : P₄O₁₀
not balanced
Fe+H₂O ----> Fe₃O₄ + H₂
Reactants : Fe, H₂O
Products : Fe₃O₄
not balanced
As in math or
?
/.........
Answer:
The volume of the air is 0.662 L
Explanation:
Charles's Law is a gas law that relates the volume and temperature of a certain amount of gas at constant pressure. This law says that for a given sum of gas at a constant pressure, as the temperature increases, the volume of the gas increases and as the temperature decreases, the volume of the gas decreases because the temperature is directly related to the energy of the movement they have. the gas molecules. This is represented by the quotient that exists between volume and temperature will always have the same value:
If you have a certain volume of gas V1 that is at a temperature T1 at the beginning of the experiment and several the volume of gas to a new value V2, then the temperature will change to T2, and it will be true:
In this case:
- V1= 0.730 L
- T1= 28 °C= 301 °K (0°C= 273°K)
- V2= ?
- T2= 0°C= 273 °K
Replacing:
Solving:
V2=0.662 L
<u><em>The volume of the air is 0.662 L</em></u>
