Answer : The volume of the cube submerged in the liquid is, 29.8 mL
Explanation :
First we have to determine the mass of ice.
Formula used :
Given:
Density of ice = 
Volume of ice = 45.0 mL
The cube will float when 40.5 g of liquid is displaced.
Now we have to determine the volume of the cube is submerged in the liquid.
Thus, the volume of the cube submerged in the liquid is, 29.8 mL
The answer is (4) Add enough solvent to 30.0 g of solute to make 1.0 L solution. The molarity is calculated using volume of the solution. When solute dissolving, the total volume will change. So the final volume of solution need to be 1.0 L.
Answer:
1. Galvanic oxidation. Example is the corrosion of aluminium wires when in contact with copper wires under wet conditions.
2. Rainwater or Damp/moist air
3. Chromium-plated steel screws or stainless steel screws or galvanized steel screws
Explanation:
1. Galvanic oxidation or corrosion occurs when two different metals with different electrode potentials are brought into contact with each other by means of an electrolyte (usually a aqueous solution), such that a redox reaction occurs leading to one metal with the more negative electrode potential (the anode) becoming oxidized, while the other less negative potential (the cathode) is reduced.
In order for galvanic corrosion to occur, three elements are required.
i. Two metals with different corrosion potentials (anode and cathode)
ii. Direct metal-to-metal electrical contact
iii. A conductive electrolyte solution (e.g. water) must connect the two metals on a regular basis.
For example oxidation (corrosion) of aluminium wires when in contact with copper wire under wet conditions.
2. The most likely electrolyte will be rainwater containing dissoved solutes (if the panel is in an exposed part of the house) or damp/moist air.
3. From the table, the most likely screw will be chromium-plated steel screws or stainless steel (made of iron and nickel) screws or galvanized steel (zinc-plated) screws.
All these possible screw components have a more negative electrode potential than copper. Thus they will serve as the anode in a galvanic oxidation with copper.
The formula to be used for this problem is as follows:
E = hc/λ, where h is the Planck's constant, c is the speed of light and λ is the wavelength. Also 1 aJ = 10⁻¹⁸ J
0.696×10⁻¹⁸ = (6.62607004×10⁻³⁴ m²·kg/s)(3×10⁸ m/s)/λ
Solving for λ,
λ = 2.656×10⁻⁷ m or <em>0.022656 nm</em>
