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Sati [7]
4 years ago
11

Which shows the electron configuration for zinc (zn)?

Chemistry
2 answers:
mina [271]4 years ago
6 0

Answer:

The atomic number of zinc is 30 therefore The electronic configuration of zinc is expressed as 1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 or in simpler form expressed as [Ar] 3d^10 4s^2. The electronic configuration shows the distribution of the electrons of an atom or a molecule.

Explanation:

what they said, i just need points

Lyrx [107]4 years ago
3 0
The atomic number of zinc is 30 therefore The electronic configuration of zinc is expressed as 1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 or in simpler form expressed as [Ar] 3d^10 4s^2. The electronic configuration shows the distribution of the electrons of an atom or a molecule.
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Identify the limiting reactant in the reaction of nitrogen and hydrogen to form NH3 if 5.23 g of N2 and 5.52 g of H2 are combine
Marrrta [24]

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:

  • Firstly, we should write the balanced equation of the reaction:

<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.

4 0
3 years ago
Use the following information to calculate the concentration, Ka and pka for an unknown monoprotic weak acid. (8 pts.) 20.00 mL
lisov135 [29]

Answer:

Concentration: 0.185M HX

Ka = 9.836x10⁻⁶

pKa = 5.01

Explanation:

A weak acid, HX, reacts with NaOH as follows:

HX + NaOH → NaX + H2O

<em>Where 1 mole of HX reacts with 1 mole of NaOH</em>

To solve this question we need to find the moles of NaOH at equivalence point (Were moles HX = Moles NaOH).

18.50mL = 0.01850L * (0.20mol / L) = 0.00370 moles NaOH = Moles HX

In 20.0mL = 0.0200L =

0.00370 moles HX / 0.0200L = 0.185M HX

The equilibrium of HX is:

HX(aq) ⇄ H⁺(aq) + X⁻(aq)

And Ka is defined as:

Ka = [H⁺] [X⁻] / [HX]

<em>Where [H⁺] = [X⁻] because comes from the same equilibrium</em>

As pH = 2.87, [H+] = 10^-pH = 1.349x10⁻³M

Replacing:

Ka = [H⁺] [H⁺] / [HX]

Ka = [1.349x10⁻³M]² / [0.185M]

Ka = 9.836x10⁻⁶

pKa = -log Ka

<h3>pKa = 5.01</h3>
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