Answer:
Oxygen's mass percent in Fe(OH)3 is 44.92%
Explanation: The mass percentage is a means of indicating the concentration of a specific element within a compound. It is determined through the ratio of the element's mass to the compound's total mass, multiplied by 100.
•First calculate the overall mass of the compound
•Fe's molar mass = 55.85 g/mol
•O's molar mass = 16 g/mol
•H's molar mass = 1 g/mol
Using these values, we can compute the molecular mass of Fe(OH)3 = 55.85 g/mol + (16 g/mol)3 + (1 g/mol)3
=55.85 g/mol + 48 g/mol + 3 g/mol
=106.85 g/mol
Mass percent of an element = mass of element/total mass of compound × 100
In the case of 3 oxygen atoms present within the compound, the mass of oxygen totals 48 g/mol
Mass percent of oxygen= 48 g/mol/106.85 g/mol × 100
= 0.4492×100= 44.92%
[[TAG_31]]Thus, the mass percent of oxygen in Fe(OH)3 amounts to 44.92%[[TAG_32]]
Answer:
81°C.
Justification:
We can arrive at this conclusion using the formula:
Q = m.c.ΔT,
where Q denotes the heat lost by water (Q = - 1200 J).
m represents the mass of water (m = 20.0 g).
c indicates the specific heat of water (c = 4.186 J/g.°C).
ΔT signifies the difference between the starting temperature and the final temperature (ΔT = final T - initial T = final T - 95.0°C).
Given Q = m.c.ΔT
It follows that (- 1200 J) = (20.0 g)(4.186 J/g.°C)(final T - 95.0°C ).
(- 1200 J) = 83.72 final T - 7953.
∴ final T = (- 1200 J + 7953)/83.72 = 80.67°C ≅ 81.0°C.
Consequently, the correct answer is: 81°C.
CxHy + (x+0.25)O₂ → xCO₂ + 0.5yH₂O
m(CO₂)/{xM(CO₂)}=m(H₂O)/{0.5yM(H₂O)}
0.2845/{44.01x}=0.1451/{9.01y}
x/y=0.4=2:5
The empirical formula is C₂H₅.
No, two atoms that have identical mass numbers can never be considered isotopes of one another. This is due to the fact that for atoms to qualify as isotopes, they must consist of the same number of protons while differing in the number of neutrons. If two atoms share the same mass number, their proton count must also be identical, which implies these atoms cannot be isotopes of each other.
Answer:
a) 
b) 1657 €
Explanation:
Hola,
a) En esta cuestión analizaremos el millón de litros de agua anualmente, dado que este dato nos permite calcular el calor necesario para calentar dicha cantidad, considerando que la densidad del agua es de 1 kg/L:
A continuación, utilizamos la entalpía de combustión del metano para determinar la cantidad en kilogramos necesaria, ya que la energía calórica perdida por el metano es equivalente a la energía obtenida por el agua:
b) En este supuesto, tenemos que, bajo condiciones normales de 1 bar y 273 K, el precio de 1 metro cúbico de metano es 0,45 €, lo que nos permite calcular las moles de metano en esas condiciones:
En consecuencia, los kilogramos de metano que se obtienen por 0,45 € son:
Finalmente, usando regla de tres:
0.715 kg ⇒ 0.45 €
2630 kg ⇒ X
X = (2630 kg x 0.45 €) / 0.715 kg
X = 1657 €
Regards.
