Answer:
Your questions requires diagrams of the cell to get which one is on the left or right. However, see the attached file below
The correct answer is (d) the left half-cell will decrease in concentration; and the right half-cell will increase in concentration.
Explanation:
The concentration of the Pb2+ increases in the oxidation half cell while the concentration of the Pb2+ decreases in the reduction half cell during the reaction.
In the Left Beaker (Left half cell), their is less concentration
Pb(s) ---> Pb2+(aq) + 2 e- Concentration of Pb2+(aq) increase ; Electrons going out from this side
In the Right Beaker (right half cell), their is more concentration
Pb2+(aq) + 2 e- ---> Pb(s) Concentration of Pb2+(aq) decrease ; Electrons coming in to this side
Electrons will flow from Left to Right direction.
Answer:
T₂ = 43.46 °C
Explanation:
Given that:
The heat of the formation of carbon dioxide = - 393.5 kJ/mol (Negative sign suggests heat loss)
It means that energy released when 1 mole of carbon undergoes combustion = 393.5 kJ = 393500 J
Heat gain by water = Heat lost by the reaction
Thus,
For water:
Mass of water = 5100 g
Specific heat of water = 4.18 J/g°C
T₁ = 25 °C
T₂ = ?
Q = 393500 J
So,
T₂ = 43.46 °C
<span>1. Which variable is the independent variable and which is the dependent variable? Density vs. ethylene glycol
The independent variable would be ethylene glycol and dependent variable would be density.
A. A 25-mL volumetric flask with its stopper has a mass of 32.6341 g. The same flask filled to the line with ethylene glycol (C2H6O2, automotive antifreeze) solution has a mass of 58.0091 g. What is the density of the ethylene glycol solution?
Density = 58.0091 - 32.6341 / .025 = 1015 g/L
B. What is the molarity of the ethylene glycol solution, if the mass of ethylene glycol in the solution is 12.0439 g?
Molarity = 12.0439 ( 1 mol / 62.07 g) / 0.025 = 7.8 M</span>
Answer:
In oxidation reduction reactions, one species gets reduced by taking on electron(s) and another species gets oxidized by losing electrons. The movement of electrons can be used to do work. ... The electron flow can be run through a wire and these electrons can be used to do work (like run a battery). Hope this helps.
