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.
3 first significant figure
6 second significant figure
5 third significant figure
4 cannot exceed 5, so retain 5 instead of increasing it to 6
0.0365
Sagot:
0.1 M NaCl
Paliwanag:
Ang tanong na ito ay nagpapaalala sa atin ng mga patakaran sa solubility. Alalahanin natin na ang lahat ng chlorides ay natutunaw maliban sa mga ng lead, mercury II at silver na hindi natutunaw sa tubig.
Ang sumusunod na reaksyon ay mangyayari na humahantong sa pagbuo ng isang precipitate;
Pb(NO3)2(aq) + 2NaCl(aq) -------> 2NaNO3(aq) + PbCl2(s)
Ang puting precipitate na nabuo ay PbCl2.
The pH level is 1.39. To explain, we start with the given information: the concentration of HClO is 0.15 M, with an acid dissociation constant of 2.9 × 10-8. The objective is to calculate the pH of the solution. Through the process, we find that the equilibrium concentration after applying the formula yields 0.04069 M for H3O⁺, leading us to a pH of 1.39.
Answer:
The volume of calcium hydroxide solution utilized is 0.0235 mL.
Explanation:
Moles of KHP = 
In accordance with the reaction, 2 moles of KHP react with 1 mole of calcium hydroxide, thus 0.0330 moles of KHP will react with;
of calcium hydroxide
The molarity of calcium hydroxide solution = 0.703 M
Volume of calcium hydroxide solution = V
The volume of the calcium hydroxide solution utilized is 0.0235 mL.
