V1/T1=V2/T2
524 cm^3/491 K =V2/297K
V2=(524 cm^3*297K)/491 K= 317K
<h2>
1.25 g of
would be produced from the complete reaction of 25 mL of 0.833 mol/L
with excess
</h2>
Explanation:
To calculate the number of moles for given molarity, we use the equation:


According to stoichiometry:
1 mole of
will give = 1 mole of 
0.0208 moles of
will give =
of 
Mass of 
Thus 1.25 g of
would be produced from the complete reaction of 25 mL of 0.833 mol/L
with excess
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When a specific amount of energy is emitted when excited electrons in an atom in a sample of an element return to the ground state, this emitted energy can
<span>be used to determine the "identity of the element".</span>
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.