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vodomira [7]
3 years ago
6

Use the portion of the periodic table shown below to answer the questions. A portion of the first three columns of the periodic

table is shown. Column one from top to bottom reads 11 sodium 22.990, 19 potassium 39.098, and 37 rubidium 85.468. Column two reads 12 magnesium 24.305, 20 calcium 40.078, and 38 strontium 87.62. Column three reads 21 scandium 44.956 and 39 yttrium 88.906. Part 1: Name two elements that have the same properties as calcium (Ca). (4 points) Part 2: Determine the number of protons, electrons, and neutrons present in an atom of sodium (Na). Explain how you determined your answer using complete sentences. (6 points)
Chemistry
2 answers:
Anna71 [15]3 years ago
7 0

I'm not sure about part 1, you may need to google it, but part two is 11 protons, 11 electrons and 12 neutrons. You can find protons and electrons by just looking at the atomic number, and you can find neutrons by subtracting the atomic number from the atomic mass.

Oxana [17]3 years ago
3 0

Answer:

1) magnesium and strontium

2) number of protons = 11

number of electrons = 11

number of neutrons = 12

Explanation:

Part 1:  

Two elements that have the same properties as calcium are magnesium and strontium. Elements in the same group (represented as columns in periodic table) share the same properties because they have the same number of electrons in the outermost electron shell.

Part 2:

For sodium, from the periodic table we know:

atomic number: 11

atomic mass: 22.990 ≈ 23

The number of protons and electrons is indicated by atomic number, in this case sodium has 11 protons and electrons (remember that atoms of the elements are electrically neutral, then they have the same amount of protons and electrons).

The mass of the atom is almost concentrated in the atomic nucleus, then the atomic mass represents the amount of protons and neutrons combined. So,

number of neutrons = atomic mass - atomic number = 23 - 11 = 12

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Answer:

1) The limiting reactant is N₂ because it is present with the lower no. of moles than H₂.

2) The amount (in grams) of excess reactant H₂ = 4.39 g.

Explanation:

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<em>N₂ + 3H₂ → 2NH₃.</em>

<em>1) To determine the limiting reactant of the reaction:</em>

  • From the stichiometry of the balanced equation, 1.0 mole of N₂ reacts with 3.0 moles of H₂ to produce 2.0 moles of NH₃.
  • This means that <em>N₂ reacts with H₂ with a ratio of (1:3).</em>
  • We need to calculate the no. of moles (n) of N₂ (5.23 g) and H₂ (5.52 g) using the relation:<em> n = mass / molar mass.</em>

The no. of moles of N₂ in (5.23 g) = mass / molar mass = (5.23 g) / (28.00 g/mol) = 0.1868 mol.

The no. of moles of H₂ (5.52 g) = mass / molar mass = (5.52 g) / (2.015 g/mol) = 2.74 mol.

  • From the stichiometry, N₂ reacts with H₂ with a ratio of (1:3).

The ratio of the reactants of N₂ (5.23 g, 0.1868 mol) to H₂ (5.52 g, 2.74 mol) is (1:14.67).

∴ The limiting reactant is N₂ because it is present with the lower no. of moles than H₂.

0.1868 mol of N₂ react completely with 0.5604 mol of H₂ and the remaining of H₂ is in excess.

<em>2) To determine the amount (in grams) of excess reactant of the reaction:</em>

  • As showed in the part 1, The limiting reactant is N₂ because it is present with the lower no. of moles than H₂.
  • Also, 0.1868 mol of N₂ react completely with 0.5604 mol of H₂ and the remaining of H₂ is in excess.
  • The no. of moles are in excess of H₂ = 2.74 mol - 0.5604 mol (reacted with N₂) = 2.1796 mol.
  • ∴ The amount (in grams) of excess reactant H₂ = n (excess moles) x molar mass = (2.1796 mol)((2.015 g/mol) = 4.39 g.

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