Answer:
0.35 V
Explanation:
(a) Standard reduction potentials
<u> E°/V</u>
Fe²⁺ + 2e- ⇌ Fe; -0.41
Cr³⁺ + 3e⁻ ⇌ Cr; -0.74
(b) Standard cell potential
<u> E°/V</u>
2Cr³⁺ + 6e⁻ ⇌ 2Cr; +0.74
<u>3Fe ⇌ 3Fe²⁺ + 6e-; </u> <u>-0.41
</u>
2Cr³⁺ + 3Fe ⇌ 2Cr + 3Fe²⁺; +0.33
3. Cell potential
2Cr³⁺(0.75 mol·L⁻¹) + 6e⁻ ⇌ 2Cr
<u>3Fe ⇌ 3Fe²⁺(0.25 mol·L⁻¹) + 6e-
</u>
2Cr³⁺(0.75 mol·L⁻¹) + 3Fe ⇌ 2Cr + 3Fe²⁺(0.25 mol·L⁻¹)
The concentrations are not 1 mol·L⁻¹, so we must use the Nernst equation

(a) Data
E° = 0.33 V
R = 8.314 J·K⁻¹mol⁻¹
T = 298 K
z = 6
F = 96 485 C/mol
(b) Calculations:
![Q = \dfrac{\text{[Fe}^{2+}]^{3}}{ \text{[Cr}^{3+}]^{2}} = \dfrac{0.25^{3}}{ 0.75^{2}} =\dfrac{0.0156}{0.562} = 0.0278\\\\E = 0.33 - \left (\dfrac{8.314 \times 298}{6 \times 96485}\right ) \ln(0.0278)\\\\=0.33 -0.00428 \times (-3.58) = 0.33 + 0.0153 = \textbf{0.35 V}\\\text{The cell potential is }\large\boxed{\textbf{0.35 V}}](https://tex.z-dn.net/?f=Q%20%3D%20%5Cdfrac%7B%5Ctext%7B%5BFe%7D%5E%7B2%2B%7D%5D%5E%7B3%7D%7D%7B%20%5Ctext%7B%5BCr%7D%5E%7B3%2B%7D%5D%5E%7B2%7D%7D%20%3D%20%5Cdfrac%7B0.25%5E%7B3%7D%7D%7B%200.75%5E%7B2%7D%7D%20%3D%5Cdfrac%7B0.0156%7D%7B0.562%7D%20%3D%200.0278%5C%5C%5C%5CE%20%3D%200.33%20-%20%5Cleft%20%28%5Cdfrac%7B8.314%20%5Ctimes%20298%7D%7B6%20%5Ctimes%2096485%7D%5Cright%20%29%20%5Cln%280.0278%29%5C%5C%5C%5C%3D0.33%20-0.00428%20%5Ctimes%20%28-3.58%29%20%3D%200.33%20%2B%200.0153%20%3D%20%5Ctextbf%7B0.35%20V%7D%5C%5C%5Ctext%7BThe%20cell%20potential%20is%20%7D%5Clarge%5Cboxed%7B%5Ctextbf%7B0.35%20V%7D%7D)
Answer:
Final volume=V₂ = 216.3 mL
Explanation:
Given data:
Initial volume = 120.0 mL
Initial temperature = -12.3 °C (-12.3 +273 = 260.7 K)
Final volume = ?
Final temperature = 197.0 °C (197+273 = 470 K)
Solution:
We will apply Charles Law to solve the problem.
According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.
Mathematical expression:
V₁/T₁ = V₂/T₂
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
Now we will put the values in formula.
V₁/T₁ = V₂/T₂
V₂ = V₁T₂/T₁
V₂ = 120 mL × 470 K /260.7K
V₂ = 56400 mL.K /260.7K
V₂ = 216.3 mL
B hibernation but it deppends on the animal
Answer:
one mole of atom of any element contains6.022×1033 atoms regardless of the type of elements the mass of one mole of an element depend on what that element is and is equal to atom mass of that element in gram
is 7 cus the valance electron for sodium is Ne 3s1 and for chlorine is Ne 3s13p5