Answer:
C. connecting an active metal to designate the pipe as the cathode in an electrochemical cell.
Explanation:
Cathodic protection involves a method to manage the accelerated corrosion of a metal surface by designating it as the cathode within an electrochemical cell. This is accomplished by attaching the protected metal to a more sacrificial metal, which acts as the anode.
This method helps to preserve the metal by introducing a highly reactive metal that serves as the anode, supplying free electrons. By adding these free electrons, the active metal gives up its ions, protecting the less reactive steel from corrosion.
The visual representation is displayed in the following image.
For calculations, consider 100 grams of the compound:
ω(Cl) = 85.5% ÷ 100%.
ω(Cl) = 0.855; signifying the mass percentage of chlorine in the compound.
m(Cl) = 0.855 · 100 g.
m(Cl) = 85.5 g; this represents the mass of chlorine.
m(C) = 100 g - 85.5 g.
m(C) = 14.5 g; indicating the mass of carbon.
n(Cl) = m(Cl) ÷ M(Cl).
n(Cl) = 85.5 g ÷ 35.45 g/mol.
n(Cl) = 2.41 mol; this is the quantity of chlorine.
n(C) = 14.5 g ÷ 12 g/mol.
n(C) = 1.21 mol; this is the quantity of carbon.
n(Cl): n(C) = 2.41 mol: 1.21 mol = 2: 1.
The compound in question is identified as dichlorocarbene CCl₂.
Response:
CRYSTAL
A LARGE NUMBER OF ATOMS ORGANIZED IN A REGULAR STRUCTURE
1:1
Reasoning:
I predict that there will be an increase in the seconds recorded in the time column. This is because, as more water is mixed with sodium thiosulfate, its concentration diminishes in each flask. Additionally, a lower concentration results in a slower reaction rate since fewer molecules of sodium thiosulfate means there are less frequent collisions with sulfuric acid. With fewer collisions occurring in the reaction, it takes a longer time for the reaction to complete, leading to increased time when sodium thiosulfate is diluted.
Explanation:
I can confirm that this explanation is accurate.
Answer:
vHe / vNe = 2.24
Explanation:
To determine the velocity of an ideal gas, one should apply the formula:
v = √3RT / √M
In this equation, R represents the gas constant (8.314 kgm²/s²molK); T refers to temperature, and M indicates the molar mass of the gas (4x10⁻³kg/mol for helium and 20.18x10⁻³ kg/mol for neon). Hence:
vHe = √3×8.314 kgm²/s²molK×T / √4x10⁻³kg/mol
vNe = √3×8.314 kgm²/s²molK×T / √20.18x10⁻³kg/mol
The ratio simplifies to:
vHe / vNe = √3×8.314 kgm²/s²molK×T / √4x10⁻³kg/mol / √3×8.314 kgm²/s²molK×T / √20.18x10⁻³kg/mol
vHe / vNe = √20.18x10⁻³kg/mol / √4x10⁻³kg/mol
vHe / vNe = 2.24
I hope it assists you!
