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3 years ago

Periodic table 50 elements with thier valancies

Chemistry

2 answers:

sdas [7]3 years ago

4 0

Answer:

50 elements chemical eq, atomic number, atomic mass, no.of electrons, no.of protons, no.of neutrons, electronic configuration and valancies are given in the above attachment

Explanation:

Hope it helps you

Lubov Fominskaja [6]3 years ago

3 0

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.

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2) Show the calculation of Kc for the following reaction if an initial reaction mixture of 0.800 mole of CO and 2.40 mole of H2

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Answer:

Kc = 3.90

Explanation:

CO reacts with $H_2$ to form $CH_4$ and $H_2O$ . balanced reaction is:

$CO(g) + 3H_2 (g) \leftrightharpoons CH_4(g) + H_2O(g)$

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No. of moles of $H_2$ = 2.40 mol

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Concentration of $H_2$ = $\frac{2.40}{8.00} = 0.300\ mol/L$

$CO(g) + 3H_2 (g) \leftrightharpoons CH_4(g) + H_2O(g)$

Initial 0.100 0.300 0 0

equi. 0.100 -x 0.300 - 3x x x

It is given that,

at equilibrium $H_2O (x)$ = 0.309/8.00 = 0.0386 M

So, at equilibrium CO = 0.100 - 0.0386 = 0.0614 M

At equilibrium $H_2$ = 0.300 - 0.0386 × 3 = 0.184 M

At equilibrium $CH_4$ = 0.0386 M

$Kc=\frac{[H_2O][CH_4]}{[CO][H_2]^3}$

$Kc=\frac{0.0386 \times 0.0386}{(0.184)^3 \times 0.0614} =3.90$

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Periodic table 50 elements with thier valancies​

Periodic table 50 elements with thier valancies