Convert 0.60 moles of Fe to moles of Fe2O3

What is the last energy level of N? Mg?

Alexandra [31]

Answer:

2

3

Explanation:

To infer the last energy of the given atoms, we need to write their electronic configuration:

For N = 1S² 2S² 2P³

Mg = 1S² 2S² 2P⁶ 3S²

The energy levels are usually designated as;

n = 1

n = 2

n = 3

n =4

For N, the last energy level is 2

Mg, the last energy level is 3

We can also determine this number by the periods the atoms can be found.

3 0

3 years ago

When the reaction shown is correctly balanced, the coefficients are HNO3 + KOH → KNO3 + H2O

GuDViN [60]

Coefficients are 1,1,1,1

4 0

3 years ago

Based on the chemical bonds present in this molecule, we would classify it as a(n)

MrMuchimi

Answer is C – Unsaturated fat.

Fats are large molecules basically made from a combination of glycerol and three chains of fatty acids. Usually fats consist of C, H and O. According to the C – C bond, there are two types of fats as saturated fats and unsaturated fats. Saturated fats have only C – C single bonds and all C atoms are sp³ hybridized atoms. But in unsaturated fats other than C – C single bonds there are C – C double bonds. Hence, both sp³ and sp² hybridized C atoms can be seen.

8 0

3 years ago

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A solution may contain Ag+, Pb2+, and/or Hg22+. A white precipitate forms when 6 M HCl is added. The precipitate is partially so

omeli [17]

Answer:

All three are present

Explanation:

Addition of 6 M HCl would form precipitates of all the three cations, since the chlorides of these cations are insoluble: $AgCl (s), PbCl_2 (s), Hg_2Cl_2 (s)$ .

Firstly, the solid produced is partially soluble in hot water. Remember that out of all the three solids, lead(II) choride is the most soluble. It would easily completely dissolve in hot water. This is how we separate it from the remaining precipitate. Therefore, we know that we have lead(II) cations present, as the two remaining chlorides are insoluble even at high temperatures.
Secondly, addition of liquid ammonia would form a precipitate with silver: $AgCl (s) + 2 NH_3 (aq) + H_2O (l)\rightarrow [Ag(NH_3)_2]OH (s) + HCl (aq)$ ; Silver hydroxide at higher temperatures decomposes into black silver oxide: $2 [Ag(NH_3)_2]OH (s)\rightarrow Ag_2O (s) + H_2O (l) + 4 NH_3 (g)$ .
Thirdly, we also know we have $Hg_2^{2+}$ in the mixture, since addition of potassium chromate produces a yellow precipitate: $Hg_2^{2+} (aq) + CrO_4^{2-} (aq)\rightarrow Hg_2CrO_4 (s)$ . The latter precipitate is yellow.