Answer:
2.01V ( To three significant digits)
Explanation:
First we show the standard reduction potentials of Cu2+(aq)/Cu(s) system and Al3+(aq)/Al(s) system. We can clearly see from the balanced redox reaction equation that aluminium is the anode and was the oxidized specie while copper is the cathode and was the reduced specie. This observation is necessary when substituting values of concentration into the Nernst equation.
The next thing to do is to obtain the standard cell potential as shown in the image attached and subsequently substitute values of concentration and standard cell potential into the Nernst equation as shown. This gives the cell potential under the given conditions.
Answer:
i think d maybe correct me if im wrong
Explanation:
Answer:
Intermolecular forces are much weaker than the strong covalent bonds within the molecules. ... Very little energy is needed to overcome the intermolecular forces, so simple molecular substances usually have low melting and boiling points. They are often liquids or gases at room temperature
Answer:
HI.
Explanation:
- Thomas Graham found that, at a constant temperature and pressure the rates of effusion of various gases are inversely proportional to the square root of their masses.
Rate of effusion ∝ 1/√molar mass.
- <em>(Rate of effusion of O₂) / (Rate of effusion of unknown gas) = (√molar mass of unknown gas) / (√molar mass of O₂).</em>
- An unknown gas effuses at one half the speed of that of oxygen.
∵ Rate of effusion of unknown gas = 1/2 (Rate of effusion of O₂)
∴ (Rate of effusion of O₂) / (Rate of effusion of unknown gas) = 2.
Molar mass of O₂ = 32.0 g/mol.
∵ (Rate of effusion of O₂) / (Rate of effusion of unknown gas) = (√molar mass of unknown gas) / (√molar mass of O₂).
∴ 2.0 = (√molar mass of unknown gas) / √32.0.
(
√molar mass of unknown gas) = 2.0 x √32.0
By squaring the both sides:
∴ molar mass of unknown gas = (2.0 x √32.0)² = 128 g/mol.
∴ The molar mass of sulfur dioxide = 80.91 g/mol and the molar mass of HI = 127.911 g/mol.
<em>So, the unknown gas is HI.</em>
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