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.
Answer:
1.9 × 10² g NaN₃
1.5 g/L
Explanation:
Step 1: Write the balanced decomposition equation
2 NaN₃(s) ⇒ 2 Na(s) + 3 N₂(g)
Step 2: Calculate the moles of N₂ formed
N₂ occupies a 80.0 L bag at 1.3 atm and 27 °C (300 K). We will calculate the moles of N₂ using the ideal gas equation.
P × V = n × R × T
n = P × V / R × T
n = 1.3 atm × 80.0 L / (0.0821 atm.L/mol.K) × 300 K = 4.2 mol
We can also calculate the mass of nitrogen using the molar mass (M) 28.01 g/mol.
4.2 mol × 28.01 g/mol = 1.2 × 10² g
Step 3: Calculate the mass of NaN₃ needed to form 1.2 × 10² g of N₂
The mass ratio of NaN₃ to N₂ is 130.02:84.03.
1.2 × 10² g N₂ × 130.02 g NaN₃/84.03 g N₂ = 1.9 × 10² g NaN₃
Step 4: Calculate the density of N₂
We will use the following expression.
ρ = P × M / R × T
ρ = 1.3 atm × 28.01 g/mol / (0.0821 atm.L/mol.K) × 300 K = 1.5 g/L
Answer:
<em>An object in a fluid medium displaces a set amount of fluid upon immersion. Archimedes' principle states that the weight of the displaced fluid is equal to the buoyant force exerted on the object</em>
Explanation:
The reaction is CaCO
3
+2HCl→CaCl
2
+H
2
O+CO
2
.
Thus, 2 moles of HCl reacts with one mole of calcium carbonate to produce one mole each of calcium chloride, water and carbon dioxide respectively.
Hence, 3 moles of HCl will react with excess of calcium carbonate to produce 3×
2
1
=1.5 mol of carbon dioxide
Radio waves has the longest wavelength
