if during the interaction of 36.5 g of hydrochloric acid with magnesium 137 kg of joules energy are released, how much energy wo
1 answer:
You might be interested in
Answer:
-1.05 V
Explanation:
A detailed diagram of the setup as required in the question is shown in the image attached to this answer. The electrolytes chosen are SnCl2 for the anode half cell and MnCl2 for the cathode half cell. Tin rod and manganese rod are used as the anode and cathode materials respectively. Electrons flow from anode to cathode as indicated. The battery connected to the set up drives this non spontaneous electrolytic process.
Oxidation half equation;
Sn(s) ------> Sn^2+(aq) + 2e
Reduction half equation:
Mn^2+(aq) + 2e ----> Mn(s)
Cell voltage= E°cathode - E°anode
E°cathode= -1.19V
E°anode= -0.14 V
Cell voltage= -1.19 V - (-0.14V)
Cell voltage= -1.05 V
Answer:
The concentration of the copper sulfate solution is 83 mM.
Explanation:
The absorbance of a copper sulfate solution can be calculated using Beer-Lambert Law:
A = ε . c . <em>l</em>
where
ε is the extinction coefficient of copper sulfate (ε = 12 M⁻¹.cm⁻¹)
c is its molar concentration (what we are looking for)
l is the pathlength (0.50 cm)
We can use this expression to find the molarity of this solution: