Answer:
V = 65.81 L
Explanation:
En este caso, debemos usar la expresión para los gases ideales, la cual es la siguiente:
PV = nRT (1)
Donde:
P: Presion (atm)
V: Volumen (L)
n: moles
R: constante de gases (0.082 L atm / mol K)
T: Temperatura (K)
De ahí, despejando el volumen tenemos:
V = nRT / P (2)
Sin embargo como estamos hablando de condiciones normales de temperatura y presión, significa que estamos trabajando a 0° C (o 273 K) y 1 atm de presión. Lo que debemos hacer primero, es calcular los moles que hay en 50 g de amoníaco, usando su masa molar de 17 g/mol:
n = 50 / 17 = 2.94 moles
Con estos moles, reemplazamos en la expresión (2) y calculamos el volumen:
V = 2.94 * 0.082 * 273 / 1
<h2>
V = 65.81 L</h2>
Answer:
Sn₃(PO₄)₄ - tin(IV) phosphate.
Explanation:
Hope it helps! :)
Answer:
The amount of work done on the system is 18234 J and the final positive sign means that this work corresponds to an increase in internal energy of the gas.
Explanation:
Thermodynamic work is called the transfer of energy between the system and the environment by methods that do not depend on the difference in temperatures between the two. When a system is compressed or expanded, a thermodynamic work is produced which is called pressure-volume work (p - v).
The pressure-volume work done by a system that compresses or expands at constant pressure is given by the expression:
W system= -p*∆V
Where:
- W system: Work exchanged by the system with the environment. Its unit of measure in the International System is the joule (J)
- p: Pressure. Its unit of measurement in the International System is the pascal (Pa)
- ∆V: Volume variation (∆V = Vf - Vi). Its unit of measurement in the International System is cubic meter (m³)
In this case:
- p= 10 atm= 1.013*10⁶ Pa (being 1 atm= 101325 Pa)
- ΔV= 2 L- 20 L= -18 L= -0.018 m³ (being 1 L=0.001 m³)
Replacing:
W system= -1.013*10⁶ Pa* (-0.018 m³)
Solving:
W system= 18234 J
<u><em>The amount of work done on the system is 18234 J and the final positive sign means that this work corresponds to an increase in internal energy of the gas.</em></u>
