Answer:
C.0.28 V
Explanation:
Using the standard cell potential we can find the standard cell potential for a voltaic cell as follows:
The most positive potential is the potential that will be more easily reduced. The other reaction will be the oxidized one. That means for the reactions:
Cu²⁺ + 2e⁻ → Cu E° = 0.52V
Ag⁺ + 1e⁻ → Ag E° = 0.80V
As the Cu will be oxidized:
Cu → Cu²⁺ + 2e⁻
The cell potential is:
E°Cell = E°cathode(reduced) - E°cathode(oxidized)
E°cell = 0.80V - (0.52V)
E°cell = 1.32V
Right answer is:
<h3>C.0.28 V
</h3>
<h3 />
<h3>Given:</h3>
M₁ = 2.0 mol/L
V₁ = 1 L
M₂ = 0.1 mol/L
<h3>Required:</h3>
V₂
<h3>Solution:</h3>
M₁V₁ = M₂V₂
V₂ = M₁V₁ / M₂
V₂ = (2.0 mol/L)(1 L) / (0.1 L)
<u>V₂ = 20 L</u>
Therefore, the volume of the new solution will be 20 L.
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Answer:
212.8 dm^3 or L
Explanation:
1 mole of any sub=6.02×10^23 molecules
X mole of O2=5.7×10^24 molecules
X mole=5.7×10^24/6.02×10^23
=9.5 mole
1 mole of any gas at stp=22.4 dm^3
Therefore, 9.5 mole of O2 will be 22.4×9.5
=212.8 dm^3
<u>61.25 grams</u> of CO can be formed from 35 grams of oxygen.
The molecular mass of oxygen is <u>16 gmol⁻¹</u>
The molecular mass of carbon monoxide is<u> 28 gmol⁻¹</u>
Explanation:
The molar mass of carbon monoxide is molar mass of C added to that of O;
12 + 16 = 28
= 28g/mol
The molar mass of oxygen is 16 g/mol while that of oxygen gas (O₂) is 32 g/mol
Since the ration oxygen to carbon monoxide is 1: 2 moles, we begin to find out how many moles of carbon monoxide are formed by 35 g of oxygen;
35/32 * 2
= 70/32 moles
Then multiply by the molar mass of carbon monoxide;
70/32 * 28
= 61.25 g
