Given the connection between Aw and K (Aw=2k) could you use the ideal gas law and derive the Boltzmann constant. Water freezes a

The work done is 19.26 kJ.

Answer:

La masa molar del compuesto es: 168.82 g/mol

La masa molar del gas es: 16.38 g/mol

Explanation:

(a)

Utilizando la ecuación de gases ideales:

donde,

P es la presión

V es el volumen

n es el número de moles

T es la temperatura

R es la constante de los gases, cuyo valor es = 0.0821 L.atm/K.mol

Además,

Moles = masa (m) / Masa molar (M)

La densidad (d) = Masa (m) / Volumen (V)

Así, la ecuación de gases ideales se puede expresar como:

Dado que:-

Presión = 20 kPa = 20000 Pa

La expresión para la conversión de presión en Pascal a presión en atm se muestra a continuación:

P (Pa) = P (atm)

20000 Pa = atm

Presión = 0.1974 atm

Temperatura = 330 K

d = 1.23 kg/m³ = 1.23 g/L

Masa molar =?

Aplica la fórmula:

0.1974 atm × M = 1.23 g/L × 0.0821 L.atm/K.mol × 330 K

⇒M = 168.82 g/mol

La masa molar del compuesto es: 168.82 g/mol

(b)

Dado que:

Presión = 152 Torr

Temperatura = 298 K

Volumen = 250 cm³ = 0.25 L

Utilizando la ecuación de gases ideales:

R =

Aplicando la fórmula:

152 Torr × 0.25 L = n × 62.3637 L.torr/K.mol × 298 K

⇒n = 0.002045 moles

Dado que:

Masa del gas = 33.5 mg = 0.0335 g

Masa molar =?

La fórmula para calcular los moles se muestra a continuación:

Así,

La masa molar del gas es: 16.38 g/mol

Answer:

Explanation:

0.3(125)+0.7(126)=37.5+88.2=125.7

2. The most prevalent isotope is 32, since the average is very close to this value.

Janice's teacher recommended using temperature to differentiate between the alcohol and water mixture. The relevant property illustrated through this experiment is D. boiling. The boiling point is the temperature at which a liquid transitions into vapor. Water boils at 100°C under atmospheric pressure, while most alcohols have lower boiling points. Recognizing the significant boiling temperature disparity between the two substances was essential for their separation.

Response:

a. 3 Br₂(l) + 6 OH⁻(aq) → 5 Br⁻(aq) + BrO₃⁻(aq) + 3 H₂O

b. Br₂

c. Br₂

Clarification:

Balancing a redox reactionis performed using the ion-electron method.

Step 1: Identify both half-reactions.

Reduction: Br₂(l) → Br⁻(aq)

Oxidation: Br₂(l) → BrO₃⁻(aq)

Step 2: Perform mass balance. This reaction occurs in basic conditions, thus we must add OH⁻ and H₂O as needed.

0.5 Br₂(l) → Br⁻(aq)

6 OH⁻(aq) + 0.5 Br₂(l) → BrO₃⁻(aq) + 3 H₂O

Step 3: Ensure electrical balance by incorporating electrons when necessary.

1 e⁻ + 0.5 Br₂(l) → Br⁻(aq)

6 OH⁻(aq) + 0.5 Br₂(l) → BrO₃⁻(aq) + 3 H₂O + 5 e⁻

Step 4: Scale both half-reactions to ensure the electron counts balance out.

5 × (1 e⁻ + 0.5 Br₂(l) → Br⁻(aq))

1 × (6 OH⁻(aq) + 0.5 Br₂(l) → BrO₃⁻(aq) + 3 H₂O + 5 e⁻)

Step 5: Combine both half-reactions and simplify as appropriate.

5 e⁻ + 3 Br₂(l) + 6 OH⁻(aq) → 5 Br⁻(aq) + BrO₃⁻(aq) + 3 H₂O + 5 e⁻

3 Br₂(l) + 6 OH⁻(aq) → 5 Br⁻(aq) + BrO₃⁻(aq) + 3 H₂O

The species that undergoes reduction is identified as the oxidizer. The species that undergoes oxidation is termed the reducer. In this situation, Br₂ qualifies as both.