Answer:
A solution in which the maximum amount of solvent has been dissolved. Any more solute added will sit as crystals on the bottom of the container.
Explanation:
A solution in which the maximum amount of solvent has been dissolved. Any more solute added will sit as crystals on the bottom of the contain
er.
Answer : The moles of 
are, 108.0948 mole
Explanation : Given,
Mass of aspirin = 0.600 g
Molar mass of aspirin = 180.158 g/mole
The molecular formula of aspirin is,
Formula used :
Now put all the given values in this formula, we get the moles of .
Therefore, the moles of are, 108.0948 mole
Answer:
97.0%m/m es la concentración de la solución
Explanation:
El porcentaje masa/masa (%m/m) se define como 100 veces el radio entre la masa de soluto (300g de HCl) y la masa de la solución. Para hallar la masa de la solución debemos hallar la masa del agua (Solvente) haciendo uso de la ecuación del volumen de un cono. Con el volumen del cono podemos hallar la masa del agua haciendo uso de la densidad, así:
<em>Volumen:</em>
Volumen Cono = π*r²*h / 3
Donde r es el radio = 0.300m
h la altura = 5m
Volumen = π*(5m)²*0.300m / 3
7.85m³
<em>Masa Agua:</em>
7.85m³ * (1.2g / m³) = 9.42g Agua
<em>Masa solución:</em>
300g HCl + 9.42g Agua = 309.42g Solución
<em>%m/m:</em>
300g HCl / 309.42g * 100 =
<h3>97.0%m/m es la concentración de la solución</h3>
The balanced equation for the reaction is as follows;
2H₂S + SO₂ —> 2H₂O + 3S
Stoichiometry of H₂S to SO₂ is 2:1
Limiting reactant is fully used up in the reaction and amount of product formed depends on amount of limiting reactant present.
Number of H₂S moles - 8.0 g / 34 g/mol = 0.24 mol of H₂S
Number of SO₂ moles = 12.0 g / 64 g/mol = 0.188 mol of SO₂
According to molar ratio of 2:1
If we assume H₂S to be the limiting reactant
2 mol of H₂S reacts with 1 mol of SO₂
Therefore 0.24 mol of H₂S requires - 1/2 x 0.24 = 0.12 mol of SO₂
But 0.188 mol of SO₂ is present therefore SO₂ is in excess and H₂S is the limiting reactant.
H₂S is the limiting reactant
Amount of S produced depends on amount of H₂S present
Stoichiometry of H₂S to S is 2:3
2 mol of H₂S forms 3 mol of S
Therefore 0.24 mol of H₂S forms - 3/2 x 0.24 mol = 0.36 mol of S
Mass of S produced = 0.36 mol x 32 g/mol = 11.5 g of S is produced
