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guajiro [1.7K]
1 year ago
12

Calculate the number of Cl atoms that are in 1.1 x 10^-21 moles of Cl atoms.

Chemistry
1 answer:
Margarita [4]1 year ago
6 0

According to Avogadro's number there are 62.53 atoms of chlorine in 1.1x 10^-21 moles of Cl atoms.

<h3>What is Avogadro's number?</h3>

Avogadro's number is defined as a proportionality factor which relates number of constituent particles with the amount of substance which is present in the sample.

According to the definitions, Avogadro's number depend on determined value of mass of one atom of those elements.It bridges the gap between macroscopic and microscopic world by relating amount of substance with number of particles.

Number of atoms can be calculated using Avogadro's number as follows: mass/molar mass×Avogadro's number or number of moles×Avogadro's number.

ON substitution in above formula number of atoms=1.1×10²¹×6.023×10²³=62.53 atoms

Thus, there are 62.53 chlorine  atoms in 1.1x 10^-21 moles of Cl atoms.

Learn more about Avogadro's number,here:

brainly.com/question/11907018

#SPJ1

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Calculate the molar solubility of PbBr2 (Ksp = 4.67x10-6) in 0.10M NaBr solution.
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Explanation:

PbBr_{2} will dissociate into ions as follows.

         PbBr_{2}(s) \rightleftharpoons Pb^{2+}(aq) + 2Br^{-}(aq)

Hence, K_{sp} for this reaction will be as follows.

                   K_{sp} = [Pb^{2+}][Br^{-}]^{2}

We take x as the molar solubility of PbBr_{2} when we dissolve x moles of solution per liter.

Hence, ionic molarities in the saturated solution will be as follows.

               [Pb^{2+}] = [Pb^{2+}]_{o} + x

               [Br^{-}]^{2} = [Br^{-}]_{o} + 2x

So, equilibrium solubility expression will be as follows.

            K_{sp} = ([Pb^{2+}]_{o} + x)([Br^{-}]_{o} + 2x)^{2}

Each sodium bromide molecule is giving one bromide ion to the solution. Therefore, one solution contains [Br^{-}]_{o} = 0.10 and there will be no lead ions. So, [Pb^{2+}]_{o} = 0

So, [Br^{-}]_{o} + 2x will approximately equals to [Br^{-}]_{o}.

Hence, K_{sp} = x[Br^{-}]^{2}_{o}

            4.67 \times 10^{-6} = x \times (0.10)^{2}

                        x = 4.67 \times 10^{-4} M

Thus, we can conclude that molar solubility of PbBr_{2} is 4.67 \times 10^{-4} M.

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