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e-lub [12.9K]
3 years ago
15

From his experiments, J. J. Thomson concluded that cathode ray particles can move at very fast speeds. cathode ray particles can

be moved by electric current. atoms contain small positively charged particles that are called protons. atoms contain small negatively charged particles that are called electrons. Mark this and return
Chemistry
1 answer:
Semenov [28]3 years ago
4 0

Answer:atoms contain small negatively charged particles that are called electrons

Explanation:

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Origin of hard water​
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What will the sun become when it dies
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It will become a red giant

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5 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
3 years ago
Two different compounds are formed by the elements nitrogen and oxygen. The first compound, compound P1, contains 64.17% by mass
xz_007 [3.2K]

Answer:

Explanation:

To solve the problem, we must know the kind of compounds we are dealing with.

For the first compound, P1 and second compound P2:

                                N                       O                         N                     O

Mass percent       64.17                 35.73                  47.23               52.79

Atomic mass          14                      16                         14                    16

Number of

moles            64.17/14            35.73/16            47.23/14    52.79/16      

                            4.58                  2.23                     3.37                  3.30

Simplest

ratio                 4.58/2.23            2.23/2.23             3.37/3.30         3.3/3.3

                              2                           1                             1                        1

                             

P1 compound is N₂O

P2 compound is NO

These are the compounds,

   In N₂O = 28:16

          NO = 14:16

This is the ratio of nitrogen to a fixed mass of oxygen for the two compounds.

7 0
2 years ago
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