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melisa1 [442]
3 years ago
9

If we start with 8000 atoms of radium-226 how much would remain after 3200 years?

Chemistry
1 answer:
olya-2409 [2.1K]3 years ago
6 0
<h3>Answer:</h3>

2000 atoms

<h3>Explanation:</h3>

We are given the following;

Initial number of atoms of radium-226 as 8000 atoms

Time taken for the decay 3200 years

We are required to determine the number of atoms that will remain after 3200 years.

We need to know the half life of Radium

  • Half life is the time taken by a radio active material to decay by half of its initial amount.
  • Half life of Radium-226 is 1600 years
  • Therefore, using the formula;

Remaining amount = Original amount × 0.5^n

where n is the number of half lives

n = 3200 years ÷ 1600 years

 = 2

Therefore;

Remaining amount = 8000 atoms × 0.5^2

                                = 8000 × 0.25

                                 = 2000 atoms

Thus, the number of radium-226 that will remain after 3200 years is 2000 atoms.

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Avocados and tomatoes were mainly eaten by the Aztecs and Maya.

<h3>How did Mayans cook their food?</h3>

"Mayans cooking method includes digging a shallow pit, lining it with stones or clay balls.

Avocados and tomatoes were mainly eaten by the Aztecs and Maya, with different types of fruit. Corn was made into a kind of porridge, called atole in Mesoamerica and capita in Inca territory.

Avocados and tomatoes were mainly eaten by the Aztecs and Maya.

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Acteria and archaea fix nitrogen (n2) by reducing it to ammonia (nh3), the form of nitrogen assimilated by plants. the main rese
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The correct answer for this question is the atmosphere.

The transformation of atmospheric N2 into NH3, which is a form that plants may use, is known as biological nitrogen fixation. However, eukaryotes do not engage in the process; it is exclusive to bacteria and archaea.

It has long been recognized that some archaea can contribute to the nitrogen cycling processes (albeit some of these organisms were not initially recognized as archaea). These included both assimilatory and dissimilatory activities, such as the fixing of atmospheric nitrogen. However, rather than large terrestrial or aquatic environments, these reactions were linked to extremophilic archaea that are generally found in 'exotic' habitats like hot springs or salt-saturated lakes.

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2 years ago
Calculate the molar solubility of PbBr2 (Ksp = 4.67x10-6) in 0.10M NaBr solution.
Softa [21]

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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