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:
1.02 × 10⁶ g
Explanation:
Step 1: Given data
- Volume of the balloon (V): 5400 m³
- Absolute pressure (P): 1.10 × 10⁵ Pa
- Molar mass of He (M): 4.002 g/mol
Step 2: Convert "V" to L
We will use the conversion factor 1 m³ = 1000 L.
5400 m³ × 1000 L/1 m³ = 5.400 × 10⁶ L
Step 3: Convert "P" to atm
We will use the conversion factor 1 atm = 101325 Pa.
1.10 × 10⁵ Pa × 1 atm / 101325 Pa = 1.09 atm
Step 4: Calculate the moles of He (n)
We will use the ideal gas equation.
P × V = n × R × T
n = P × V / R × T
n = 1.09 atm × 5.400 × 10⁶ L / 0.08206 atm.L/mol.K × 280 K
n = 2.56 × 10⁵ mol
Step 5: Calculate the mass of He (m)
We will use the following expression.
m = n × M
m = 2.56 × 10⁵ mol × 4.002 g/mol
m = 1.02 × 10⁶ g
The molar mass of the gene fragment is 19182 g/mol.
What is osmotic pressure ?
Osmotic pressure is the minimum pressure which needs to be applied to a solution to prevent the inward flow of its pure solvent across a semipermeable membrane. It is also defined as the measure of the tendency of a solution to take in a pure solvent by osmosis. Potential osmotic pressure is the maximum osmotic pressure that could develop in a solution if it were separated from its pure solvent by a semipermeable membrane.
We employ the osmotic pressure equation to determine the solute's concentration, which is:
π = iMRT
Using the values in the equation above, we obtain: 19182 g/mol.
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The current divides according to the resistance; more current in the lower resistance, less in the higher resistance. This is called “parallel branches” or paths. The voltage is the same for both (all) branches.
