Answer: -2.79°C
Explanation:
Supposing complete ionization:
Li2SO4 → 2 Li{+} + SO4{2-} [three ions]
(0.50 m Li2SO4) x 3 = 1.50 m ions
(1.50 m) x (1.86 °C/m) = 2.79°C change
0°C - 2.79°C = -2.79°C
A galvanic cell is formed when two metals are immersed in solutions differing in concentration, when two different metals are immersed.
<h3><u>What is a </u><u>
Galvanic</u><u> </u><u>
cell</u><u> ?</u></h3>
- In order to provide a pathway for the flow of electrons along that wire, the galvanic cell makes use of the ability to split the flow of electrons during the oxidation and reduction processes.
- It forces a half-reaction and connects each to the other with a wire.
- A galvanic cell is an electrochemical device that converts chemical redox reaction energy into electrical energy.
- Electrically, it has a potential of 1.1 V. Oxidation takes place at the anode, which is a negative plate in galvanic cells. It is a positive plate where the reduction happens.
- An electrochemical device called a galvanic cell transforms chemical energy's free energy into electrical energy. A photogalvanic cell produces species that are photochemically reactive.
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If it has 2 or more non-metals as well as a metal
Explanation:
Since this is an equilibrium problem, we apply le chatelier principle. This principle states that whenever a system at equilibrium is disturbed due to change in several factors, it would move in a way to annul such change.
C2H4(g) + Cl2 ⇔ 2C2H4Cl2(g)
When the concentration of C2H4 is increased, there is more reactant sin the system. In order to annul this change, the equilibrium position will shift to the right favoring product formation.
When the concentration of C2H4Cl2 is increased, there is more product in the system. To annul this change, the equilibrium position will shift to the left, favoring reactant formation.
