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Alex17521 [72]
3 years ago
13

Oday, you will observe drops of liquid water and compare them to drops of liquid isopropanol, noting whether either liquid evapo

rates. Given that the energy transferred in from the air will be the same for both liquids, make a prediction about whether both drops or only one drop will evaporate within five minutes. Explain your prediction below.
Chemistry
1 answer:
elena-s [515]3 years ago
8 0

Answer:

Liquid isopropanol.

Explanation:

Liquid isopropanol will evaporate earlier than water because the boiling point of isopropanol is less than water. That liquid which have high boiling point require more heat energy for change into vapor state as compared to those liquids that has low boiling point. The isopropanol has boiling point i.e. 82.5 °C while on the other hand, water has boiling point i.e. 100 °C so we can say that isopropanol will evaporate first.

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3 years ago
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What would be the voltage (Ecell) of a voltaic cell comprised of Cr (s)/Cr3+(aq) and Fe (s)/Fe2+(aq) if the concentrations of th
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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

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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

E = E^{\circ} - \dfrac{RT}{zF}\ln Q

(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}}

 

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