Which condition applies when water boils at 100oC

The standard cell potential, E°cell, for a reaction in which two electrons are transferred between the reactants is +1.33 V. Cal

Alinara [238K]

Answer:

ΔG° = 2.57 × 10² kJ

The reaction is spontaneous.

Explanation:

<em>The standard cell potential, E°cell, for a reaction in which two electrons are transferred between the reactants is +1.33 V. Calculate the standard free energy change, ΔG°, in kJ for this reaction and determine if it is spontaneous or nonspontaneous at 25°C.</em>

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We can calculate the standard Gibbs free energy (ΔG°) using the following expression.

ΔG° = -n × F × E°cell

where,

n: moles of electrons transferred

F: Faraday's constant

E°cell: standard cell potential

ΔG° = - (2 mol) × (96468 J/V . mol) × 1.33 V

ΔG° = -2.57 × 10⁵ J = 2.57 × 10² kJ

ΔG° < 0 means that the reaction is spontaneous.

4 0

3 years ago

A sample of neon occupies a volume of 752 ml at 25 0

lyudmila [28]

$V_{initial} = 752\:mL$
$T_{initial} = 25.0^0C$
converting to Kelvin
TK = TC + 273
TK = 25.0 + 273 → TK = 298.0 → $T_{initial} = 298.0\:K$
$V_{final} = ? (in\:milliliters)$
$T_{final} = 50.0^0C$
TK = TC + 273
TK = 50.0 + 273 → TK = 323.0 → $T_{final} = 323.0\:K$

By the first Law of Charles and Gay-Lussac, we have:
$\frac{ V_{i} }{ T_{i} } = \frac{ V_{f} }{ T_{f} }$

Solving:
$\frac{ V_{i} }{ T_{i} } = \frac{ V_{f} }{ T_{f} }$
$\frac{ 752 }{ 298.0 } = \frac{ V_{f} }{ 323.0 }$
Product of extremes equals product of means:
$298.0* V_{f} = 752*323.0$
$298.0 V_{f} = 242896$
$V_{f} = \frac{242896}{298.0}$
$\boxed{\boxed{V_{f} \approx 815.08\:mL}}\end{array}}\qquad\quad\checkmark$

5 0

4 years ago

A sample of gas occupies a volume of 50.0 milliliters in a cylinder with a movable piston. The pressure of the sample is 0.90 at

alina1380 [7]

P₁ = 0.90 atm

V₁ = 50.0 mL

T₁ = 298 K

P₂ = 1 atm

T₂ = 273 K

V₂ = P₁ x V₁ x T₂ / T₁ x P₂

V₂ = 0.90 x 50.0 x 273 / 298 x 1

V₂ = 12285 / 298

V₂ = 41 mL

Answer (1)

hope this helps!

7 0

3 years ago

The spongy layer of a leaf<br> HELLPROPRPTPR

Ulleksa [173]

The spongy layer of the leaves makes up the Mesophyll

8 0

4 years ago

A heliox tank contains 32% helium and 68% oxygen. The total pressure in the tank is 475 kPa. What is the partial pressure of hel

Paladinen [302]

Answer:

152 kPa = Partial pressure O₂

Explanation:

Data by percent is the molar fraction . 100.

Molar fraction of Helium = 32/ 100 → 0.32

Molar fraction of O₂ = 68/100 → 0.68

Sum of molar fractions in a mixture = 1

0.68 + 0.32 = 1

If we apply the molar fraction, we can determine the partial pressure.

Mole fraction = Partial pressure / Total pressure

0.32 = Partial pressure O₂ / 475kPa → 0.32 . 475 kPa = Partial pressure O₂

152 kPa = Partial pressure O₂

8 0

3 years ago