Hi!
The correct options would be:
1. Cathode - <em>reduction</em>
The cathode is the negatively charged electrode, and so has an excess of electrons. Cations (positively charged ions) are attracted to the cathode, and gain electrons to acquire a neutral charge. The process in which a gain of electron occurs is called reduction.
2. Anode - <em>oxidation</em>
The opposite occurs at the anode which is positively charged and attracts negatively charged ions, anions. These anions lose their electrons at the anode to acquire a neutral charge, and the process involving loss of electrons is known as oxidation.
3. Salt Bridge - <em>ion transport </em>
Salt bridge is a physical connection between the the anodic and cathodic half cells in an electrochemical cell and is a pathway that facilitates the flow of ions back and forth these half cells. Salt bridge is involved in maintaining a neutral condition in the electrochemical cells, and its absence would result in the accumulation of positive charge in the anodic cell, and negative charge in the cathodic cell.
4. Wire - <em>electron transport </em>
Wires have a universal role of being a pathway for the transport of electrons in circuit. This role is also the same in the wires involved in an electrochemical cells where they are used to transport electrons from the anodic half cell, and this electron transport results in the generation of electricity in the internal circuit of the electrochemical cell.
Hope this helps!
Answer:
3. 116.5 V
4. 119.6 V
Explanation:
3. Determination of the voltage.
Resistance (R) = 25 Ω
Current (I) = 4.66 A
Voltage (V) =?
V = IR
V = 4.66 × 25
V = 116.5 V
Thus, the voltage is 116.5 V
4. Determination of the voltage.
Current (I) = 9.80 A
Resistance (R) = 12.2 Ω
Voltage (V) =?
V = IR
V = 9.80 × 12.2
V = 119.6 V
Thus, the voltage is 119.6 V
Answer:
the 3rd one (0.01 cm the one selected already)
Explanation:
copper wire isn't excessively big, and it wraps around the pencil because its malleable. I think that the most accurate would be 0.01 cm
Balanced equation: 2Na(s) + Cl₂(g) ---> 2NaCl(s)
when we have STP conditions, we can use this conversion: 1 mol = 22.4 L
first, we have to convert grams to molecules using the molar mass, and then use mole to mole ratio from the balanced equation.
molar mass of Na= 23.0 g/mol
ratio: 2 mol Na= 1 mol Cl₂ (based on coefficients of balanced equation)
calculations:
