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ELEN [110]
3 years ago
13

Consider a galvanic cell consisting of the following two redox couplesAlB+ 3e Al , Mga + 2e Mg, (ag. E® = 1.676 V ) E° = 2.356 V

A) Write the equation for the half-reaction occurring at the cathode.B) Write the equation for the half-reaction occurring at th anodeC) Write the overall equation for the cell reation.D) what is the standard cell potential EoCell, for the Cell?
Chemistry
1 answer:
malfutka [58]3 years ago
4 0

Answer:

0.68 V

Explanation:

For anode;

3Mg(s) ---->3Mg^2+(aq) + 6e

For cathode;

2Al^3+(aq) + 6e -----> 2Al(s)

Overall balanced reaction equation;

3Mg(s) + 2Al^3+(aq) ----> 3Mg^2+(aq) + 2Al(s)

Since

E°anode = -2.356 V

E°cathode = -1.676 V

E°cell=-1.676 -(-2.356)

E°cell= 0.68 V

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How does the emission spectrum support the idea of quantized energy levels
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Explanation:

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3 years ago
25.88 grams of tin (ll) phosphate reacts with 31.73 grams of zinc.
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Explanation:

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is the answer

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6 0
2 years ago
A sample of nitrogen occupies 10.0 liters at 25°C what would be the new volume at 20°C? 7.9 L Ob 9.8 L 10.2 L 10.6 L
PtichkaEL [24]

Answer:

  9.4 liter

Explanation:

1) Data:

V₁ = 10.0 L

T₁ = 25°C = 25 + 273.15 K = 298.15 K

 P₁ = 98.7 Kpa

 T₂ = 20°C = 20 + 273.15 K = 293.15 K

  P₂ = 102.7 KPa

  V₂ = ?

2) Formula:

Used combined law of gases:

  PV / T = constant

  P₁V₁ / T₁ = P₂V₂ / T₂

3) Solution:

Solve the equation for V₂:

  V₂ = P₁V₁ T₂ / (P₂ T₁)

Substitute and compuite:

V₂ = P₁V₁ T₂ / (P₂ T₁)

V₂ = 98.7 KPa × 10.0 L × 293.15 K / (102.7 KPa × 298.15 K)

V₂ =  9.4 liter ← answer

You can learn more about gas law problems reading this other answer on

Explanation:

7 0
2 years ago
What volume of 0.350 m koh is required to react completely with 24.0 ml of 0.650 m h3po4?
BartSMP [9]

The complete balanced chemical equation for this is:

<span>3KOH  +  H3PO4  -->  K3PO4  +  3H2O</span>

 

First we calculate the number of moles of H3PO4:

moles H3PO4 = 0.650 moles / L * 0.024 L = 0.0156 mol

 

From stoichiometry, 3 moles of KOH is required for every mole of H3PO4, therefore:

moles KOH = 0.0156 mol H3PO4 * (3 moles KOH / 1 mole H3PO4) = 0.0468 mol

 

Calculating for volume given molarity of 0.350 M KOH:

Volume = 0.0468 mol / (0.350 mol / L) = 0.1337 L = 133.7 mL

 

Answer:

<span>133.7 mL KOH</span>

7 0
3 years ago
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