Answer:
b. Na+ leaves the salt bridge and enters enters the cathode
Explanation:
A galvanic cell or electrochemical cell depicts an oxidation -reduction half reactions (redox) reaction. it consists of two half cells ; one for the reduction reaction which involves the gain of electrons and the other for the oxidation reaction which involves the loss of electrons. One half cell contains the anode and oxidation occurs at the anode while the other half cell contains the cathode and reduction occurs at the cathode. The anode is usually connected to the cathode, a salt bridge is added to complete the circuit and allow current to flow. The salt bridge serves as a counter ions, they do not interfere with the electrochemical reaction but provides a passage for the migration of ions thereby preventing the cells from reaching equilibrium too quickly and thus the electrons in the salt are able to move along with any electrons.
In this galvanic cell, Cu at the anode losses two electrons to become Cu2+, and the electrons moves from the anode to the cathode where Mg2+ gain these two electrons to become negatively charged. Positively charged ions in the salt brigde Na+ will move to the cathode to pick negatively charged ions from the cathode solution. this helps to remove the strong negative charge from the cathode and allows the electrons to continue to move to the cathode.
The answer is B charged particles
I believe the answer would be D. Metallic bonding forms a “sea” of free electrons. These free electrons are able to move around much more freely than electrons in a regular lattice formation. This ability of the electrons to move around easily means that metal can be hammered into sheets or pulled into wires and still be stable. So the free electrons allow the metal to have the property if malleability.
The total percent yield:
After the combustion reaction with methane, the percent yield was 66.7%.
Combustion of Methane:
- Methane produces a blue flame as it burns in the atmosphere.
- Methane burns in the presence of enough oxygen to produce carbon dioxide (CO₂) and water (H₂O).
- It creates a significant quantity of heat during combustion, making it an excellent fuel source.
The other reactant, air's excess oxygen, is always present, making methane the limiting reactant. As a result, the amount of CH₄ burned will determine how much CO₂ and H₂O are produced.
The following chemical process produces carbon dioxide from methane:
CH₄ + 2O₂ ⇒ CO₂ + 2H₂O
Calculations:
1. <u><em>Theoretical quantity of carbon dioxide:</em></u>
All calculations will be based on the amount of methane because the problem specifies that it is the limiting reagent:
12.0g of CH₄ × (1 mol of CH₄/16g CH₄) × (1 mole of CO₂/1 mole of CH₄) × (44g CO₂/1 mole of CO₂)
= 33g of CO₂
2. <u><em>Percent yield:</em></u>
= Actual yield/Theoretical yield × 100
= 22.0g/33g × 100
= 66.7%
Learn more about the percent yield here,
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