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andrew-mc [135]
4 years ago
11

Can someone give me periodic table till 30?​

Chemistry
1 answer:
inysia [295]4 years ago
6 0

1. H (Hydrogen)

2. He (Helium)

3. Li (Lithium)

4. Be (Beryllium)

5. B (Boron)

6. C (Carbon)

7. N (Nitrogen)

8. O (Oxygen)

9. F (Fluorine)

10. Ne (Neon)

11. Na (Sodium)

12. Mg (Magnesium)

13. Al (Aluminium)

14. Si (Silicon)

15. P (Phosphorus)

16. S (Sulfur)

17. Cl (Chlorine)

18. Ar (Argon)

19. K (Potassium)

20. Ca (Calcium)

21. Sc (Scandium)

22. Ti (Titanium)

23. V (Vanadium)

24. Cr (Chromium)

25. Mn (Manganese)

26. Fe (Iron)

27. Co (Cobalt)

28. Ni (Nickel)

29. Cu (Copper)

30. Zn (Zinc)

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NiS2(s) + O2(g) --> NiO(s) + SO2(g) When 11.2 g of NiS2 react with 5.43 g of O2, 4.86 g of NiO are obtained. The theoretical
makkiz [27]

Answer:

1. The theoretical yield of NiO is 5.09g.

2. O2 is the limiting reactant.

3. The percentage yield of NiO is 95.5%

Explanation:

Step 1:

The balanced equation for the reaction is given below:

2NiS2(s) + 5O2(g) —> 2NiO(s) + 4SO2(g)

Step 2:

Determination of the masses of NiS2 and O2 that reacted and the mass of NiO produced from the balanced equation. This is illustrated below below:

Molar mass of NiS2 = 59 + (32x2) = 123g/mol

Mass of NiS2 from the balanced equation = 2 x 123 = 246g

Molar mass of o3= 16x2 = 32g/mol

Mass of O2 from the balanced equation = 5 x 32 = 160g

Molar mass of NiO = 59 + 16 = 75g/mol

Mass of NiO from the balanced equation = 2 x 75 = 150g

Summary:

From the balanced equation above, 246g of NiS2 reacted with 160g of O2 to produce 150g of NiO

Step 3:

Determination of the limiting reactant. This can be obtain as follow:

From the balanced equation above, 246g of NiS2 reacted with 160g of O2.

Therefore, 11.2g of NiS2 will react with = (11.2 x 160)/246 = 7.28g of O2.

From the above calculation, we can see that it will take a higher mass of O2 i.e 7.28g than what was given i.e 5.43g to react completely with 11.2g of NiS2.

Therefore, O2 is the limiting reactant and NiS2 is the excess reactant.

1. Determination of the theoretical yield of NiO.

In this case, the limiting reactant will be used as all of it is consumed in the reaction. The limiting reactant is O2.

From the balanced equation above, 160g of O2 reacted to produce 150g of NiO.

Therefore, 5.43g of O2 will react to produce = (5.43 x 150)/160 = 5.09g of NiO.

Therefore, the theoretical yield of NiO is 5.09g.

2. The limiting reactant is O2. Please review step 3 above for explanation.

3. Determination of the percentage yield of NiO. This is illustrated below:

Actual yield of NiO = 4.86g

Theoretical yield of NiO = 5.09g

Percentage yield =..?

Percentage yield = Actual yield /Theoretical yield x 100

Percentage yield = 4.86/5.09 x 100

Percentage yield of NiO = 95.5%

3 0
3 years ago
How do we differentiate chemical change from physical change ​
bija089 [108]

Answer:

Chemical change :has change in mass, heat is needed, new element is formed, hard to reverse.......

Physicalchange:does not have change in mass, heat is not necessary, no new element is formed, easy to reverse

5 0
3 years ago
Need help with this question using dimensional analysis
m_a_m_a [10]
The speed limit would be 70.2 mi/h

3 0
3 years ago
Which metal can replace another metal in a reaction?
deff fn [24]

Answer:

O A. A metal higher on the activity series list will replace one that is

lower.

4 0
3 years ago
A solution contains 0.25 M Ni(NO3)2 and 0.25 M Cu(NO3)2. Can the metal ions be separated by slowly adding Na2CO3? Assume that fo
amid [387]

Explanation:

Ksp of NiCO3 = 1.4 x 10^-7

Ksp of CuCO3 = 2.5 x 10^-10

Ionic equations:

NiCO3 --> Ni2+ + CO3^2-

CuCO3 --> Cu2+ + CO3^2-

[Cu2+][CO3^2-]/[Ni2+][CO3^2-]

= (2.5* 10^-10)/(1.4* 10^-7)

= 0.00179.

[Cu2+]/[Ni2+]

= 0.00179

= 0.00179*[Ni2+]

If all of Cu2+ is precipitated before Na2CO3 is added.

= 0.00179 * (0.25)

The amount of Cu2+ not precipitated = 0.000448 M

The percent of Cu2+ precipitated before the NiCO3 precipitates = concentration of Cu2+ unprecipitated/initial concentration of Cu2+ * 100

= 0.000448/0.25 * 100

= 0.18%

Therefore, percentage precipitated = 100 - 0.18

= 99.8%

The two metal ions can be separated by slowly adding Na2CO3. Thus that is the unpptd Cu2+.

8 0
3 years ago
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