The answer is (2) Copper. Because copper is consist of atom. If it is decomposed, the atom will be changed. While chemical change is just the break down and form of chemical bonds between atoms.
Answer:
The first 50 elements along with their valences are given below :
1. Hydrogen = 1
2. Helium = 0
3. Lithium = 1
4. Beryllium = 2
5. Boron = 3
6. Carbon = 4
7. Nitrogen = 3
8. Oxygen = 2
9. Fluorine = 1
10. Neon = 0
11. Sodium = 1
12. Magnesium = 2
13. Aluminium = 3
14. Silicon = 4
15. Phosphorus = 3
16. Sulphur = 2
17. Chlorine = 1
18. Argon = 0
19. Potassium = 1
20. Calcium = 2
21. Scandiun = 3
22. Titanium = 3
23. Vanadium = 4
24. Chromium = 3
25. Manganese = 4
26. Iron = 2
27. Cobalt = 2
28. Nickel = 2
29. Copper = 2
30. Zinc = 2
31. Gallium = 3
32. Germanium = 4
33. Arsenic = 3
34. Selenium = 2
35. Bromine = 1
36. Krypton = 0
37. Rubidium = 1
38. Strontium = 2
39. Yttrium = 3
40. Zirconium = 4
41. Niobium = 3
42. Molybdenum = 3
43. Technetium = 7
44. Ruthenium = 4
45. Rhodium = 3
46. Palladium = 4
47. Sliver = 1
48. Cadmium = 2
49. Indium = 3
50. Tin = 4
<u>Note</u> :
An element like Iron, copper can have more than one valencies.
Mn metal can be used as a sacrificial electrode to prevent the rusting of an iron pipe. So, the correct option is (c) Mn.
Commonly, sacrificial electrodes are employed to stop another metal from corroding or oxidising. A metal that is more reactive than the metal being shielded must serve as the sacrificial electrode. Magnesium, aluminium, and zinc are the three metals most frequently used in sacrificial anodes.
Manganese-Magnesium (Mn-Mg) electrode is more suited for on-shore pipelines where the electrolyte (soil or water) resistivity is higher since it has the highest negative electropotential of the three. In order to replenish any electrons that could have been lost during the oxidation of the shielded metal, the highly active metal offers its electrons.
Therefore, Mn metal can be used as a sacrificial electrode to prevent the rusting of an iron pipe. So, the correct option is (c) Mn.
Learn more about electrode here:
brainly.com/question/17060277
#SPJ4
Answer:
2.33g of iron (iii) chloride
50.0 mL of 5.00 M of sodium phosphate
FeCl3 + Na3PO4 > Fe(PO4) + 3NaCl
mol = conc × vol = 0.5 × 50/1000 = 0.025 mol Na3PO4
from the equation:
1 mol of Na3PO4 reacts with 1 mol FeCl3 = 3 mol of NaCl
0.025 mol = x
x = 0.0025 × 3 = 0.075 mol NaCl
mass = 0.075 g × 59 g/mol = 4.425 g NaCl
i guessed all of this so i dont know i it is correct
