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.
The answer is C. because it gives up two electrons.
KE = 0
<h3>Further explanation </h3>
Energy is the ability to do work
Energy because its motion is expressed as Kinetic energy (KE) which can be formulated as:

So for two objects that have the same speed, the greater the mass of the object, the greater the kinetic energy
The stone in hand is in a motionless state (at rest) so that its velocity (v) = 0, so it has no kinetic energy
But this stone can have <em>potential energy that is gained due to its height</em>
Answer:
6.23 KOH 90% son necesarios
Explanation:
Una solución 1N de KOH requiere 1equivalente (En KOH, 1eq = 1mol) por cada litro de solución.
Para responder esta pregunta se requiere hallar los equivalentes = Moles de KOH para preparar 100mL = 0.100L de una solución 1N. Haciendo uso de la masa molar de KOH y del porcentaje de pureza del KOH se pueden calcular los gramos requeridos para preparar la solución así:
<em>Equivalentes KOH:</em>
0.100L * (1eq / L) = 0.100eq = 0.100moles
<em>Gramos KOH -Masa molar: 56.1056g/mol-:</em>
0.100moles * (56.1056g/mol) = 5.61 KOH se requieren
<em>KOH 90%:</em>
5.61g KOH * (100g KOH 90% / 90g KOH) =
<h3>6.23 KOH 90% son necesarios</h3>