Answer:Estructura de la materia, átomo de hidrógeno, modelo de Bohr. ... Esto significaba que el electrón no puede tener cualquier velocidad, y por lo ... Despejando el valor del radio de la órbita del electrón en el primer postulado de ... ecuación (10) cuando n = 1; el segundo r2 para n = 2; el tercero r3 para n = 3
Explanation:
Answer:
Percent yield = 50%
Explanation:
Given data:
Mass of CH₄ = 16 g
Mass of O₂ = 32 g
Mass of CO₂ = 11 g
Percent yield of CO₂ = ?
Solution:
Chemical equation:
CH₄ + 2O₂ → CO₂ + 2H₂O
Number of moles of CH₄:
Number of moles = mass/ molar mass
Number of moles = 16 g /16 g/mol
Number of moles = 1 mol
Number of moles of O₂:
Number of moles = mass/ molar mass
Number of moles = 32 g /32 g/mol
Number of moles = 1 mol
Now we will compare the moles of CO₂ with both reactant.
O₂ : CO₂
2 : 1
1 : 1/2×1= 0.5 mol
CH₄ : CO₂
1 : 1
Number of moles of CO₂ produced by oxygen are less so it will limiting reactant.
Theoretical yield:
Mass of CO₂:
Mass = number of moles × molar mass
Mass = 0.5 mol × 44 g/mol
Mass = 22 g
Percent yield:
Percent yield = actual yield / theoretical yield × 100
Percent yield = 11 g/ 22 g × 100
Percent yield = 50%
Answer:
3.89 kg P2O5 must be used to supply 1.69 kg Phosphorus to the soil.
Explanation:
The molecular mass of P2O5 is
P2 = 2* 31 = 62
O5 = 5 *<u> 16 = 80</u>
Molecular Mass = 142
Set up a Proportion
142 grams P2O5 supplies 62 grams of phosphorus
x kg P2O5 supplies 1.69 kg of phosphorus
Though this might be a bit anti intuitive, you don't have to convert the units for this question. The ratio is all that is important.
142/x = 62/1.69 Cross multiply
142 * 1.69 = 62x combine the left
239.98 = 62x Divide by 62
239.98/62 = x
3.89 kg of P2O5 must be used.
Answer:
why are you trying to find a date on a homework help website?
Answer:
A. air pressure,this is answer.
