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balandron [24]
3 years ago
8

How many moles are contained in 3.131 × 1024 particles?

Chemistry
1 answer:
Mnenie [13.5K]3 years ago
5 0
One mole contains 6.02×10²³
If x mole contains 3.131×10²⁴

x = 3.131×10²⁴
-------------
6.02×10²³

x = 0.52×10²⁴¯²³

x = 0.52×10¹

x = 5.2moles....

Hope this helped...?
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Un mol de amoniaco Tiene una masa molar de 17 g y ocupa un volumen de 22.4 l qué volumen ocupa el 50 g amoniaco en condiciones n
mr_godi [17]

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>
4 0
3 years ago
Water has the following thermodynamic values: ΔH°fus of H2O = 6.02 kJ/mol ΔH°vap of H2O = 40.7 kJ/mol heat capacity of solid H2O
Brrunno [24]

Answer:

Qtotal = 90.004 kJ

Explanation:

To start resolving the problem we need to first convert the kJ/mol units from the thermodynamic values to J/g, so that we can work with the units of the heat capacity values. We know that the molar mass of water is 18.015 g/mol, so with this we do the respective conversion:

ΔH°fus of H2O = (6.02 kJ/mol) (1 mol/18.015g) (1000J/kJ) = 334.165 J/g

ΔH°vap of H2O = (40.7 kJ/mol) (1mol/18.015g) (1000J/kJ) = 2259.228 J/g

Now we need to find out the heat energy required to rise the temperature (specific heat capacity) and the energy required for each change of phase (specific latent heat), and add everything up. For this we will require the specific heat capacity and latent heat equations:

Q = mCΔT ; where m = mass, C = Hear capacity, ΔT = change of temperature

Q = mL ; where m = mass, L = specific latent heat

<u />

<u>First change of phase (solid to liquid - fusion)</u>

Q1 = (25g) (2.09 J/g°C) (0°C - (-129°C) = 6740.25 J

Q2 = (25g) (334.165 J/g) = 8354.125 J

<u>Second change of phase (liquid to gas - vaporization)</u>

Q3 = (25g) (4.18 J/g°C) (100°C - 0°C = 10450 J

Q4 = (25g) (2259.228 J/g) = 56480.7 J

<u>Rise of temperature of the gaseous water</u>

Q5 = (25g) (1.97 J/g°C) (262°C - 100°C = 7978.5 J

Finally we add everything up:

Qtotal = Q1 + Q2 + Q3 + Q4 + Q5 = 6740.25 J + 8354.125 J + 10450 J + 56480.7 J + 7978.5 J = 90003.575 J = 90.004 kJ

8 0
3 years ago
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