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ivanzaharov [21]
3 years ago
12

Given the following chemical formula:

Chemistry
1 answer:
aliina [53]3 years ago
3 0

Answers:

Explanation:

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The nuclei of atoms also contain neutrons, which help hold the nucleus together. ... The total weight of an atom is called the atomic weight. It is approximately equal to the number of protons and neutrons, with a little extra added by the electrons.


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Would the ph at the equivalence point be acidic, basic, or neutral for each given titration?.
Lesechka [4]

Answer:

At the equivalence point, equal amounts of H+ and OH– ions will combine to form H2O, resulting in a pH of 7.0 (neutral). The pH at the equivalence point for this titration will always be 7.0, note that this is true only for titrations of strong acid with strong base.

Explanation:

3 0
2 years ago
A chemist wants to extract copper metal from copper chloride solution. The chemist places 0.50 grams of aluminum foil in a solut
Irina-Kira [14]

Answer:

Approximately 0.36 grams, because copper (II) chloride acts as a limiting reactant.

Explanation:

  • It is a stichiometry problem.
  • We should write the balance equation of the mentioned chemical reaction:

<em>2Al + 3CuCl₂ → 3Cu + 2AlCl₃.</em>

  • It is clear that 2.0 moles of Al foil reacts with 3.0 moles of CuCl₂ to produce 3.0 moles of Cu metal and 2.0 moles of AlCl₃.
  • Also, we need to calculate the number of moles of the reported masses of Al foil (0.50 g) and CuCl₂ (0.75 g) using the relation:

<em>n = mass / molar mass</em>

  • The no. of moles of Al foil = mass / atomic mass = (0.50 g) / (26.98 g/mol) = 0.0185 mol.
  • The no. of moles of CuCl₂ = mass / molar mass = (0.75 g) / (134.45 g/mol) = 5.578 x 10⁻³  mol.
  • <em>From the stichiometry Al foil reacts with CuCl₂ with a ratio of 2:3.</em>

∴ 3.85 x 10⁻³  mol of Al foil reacts completely with 5.578 x 10⁻³  mol of CuCl₂ with <em>(2:3)</em> ratio and CuCl₂ is the limiting reactant while Al foil is in excess.

  • From the stichiometry 3.0 moles of  CuCl₂ will produce the same no. of moles of copper metal (3.0 moles).
  • So, this reaction will produce 5.578 x 10⁻³ mol of copper metal.
  • Finally, we can calculate the mass of copper produced using:

mass of Cu = no. of moles x Atomic mass of Cu = (5.578 x 10⁻³  mol)(63.546 g/mol) = 0.354459 g ≅ 0.36 g.

  • <u><em>So, the answer is:</em></u>

<em>Approximately 0.36 grams, because copper (II) chloride acts as a limiting reactant.</em>

5 0
4 years ago
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