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almond37 [142]
2 years ago
5

You have 700,000 atoms of a radioactive substance. After 2 half-lives have past, how

Chemistry
2 answers:
max2010maxim [7]2 years ago
6 0

Answer:

175,000

Explanation:

700,000/4=175,000

2 half-lives = approx 175,000

Ganezh [65]2 years ago
4 0

The number of atoms that remains after 2 half-lives given that it was originally 700,000 atoms is 175000 atoms.

<h3>What is a radioactive substance?</h3>

Radiation in the form of radiant particles or rays, is the result of a nuclear disintegration. Radioactive materials are atoms that have stored energy and may disintegrate in the future, releasing radiation.

Data obtained from the question

Original amount (N₀) = 700,000 atoms

Number of half-lives (n) = 2

Amount remaining (N) =?

How to determine the amount remaining

The amount remaining after 2 half-lives can be obtained as illustrated below:

N = N₀ / 2ⁿ

N = 700,000 / 4

N = 175000 atoms

Hence, 175000 atoms remain.

Learn more about the radioactive substance here:

brainly.com/question/27755919

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<h3>What is balanced nuclear equation?</h3>

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I need help with this, please :00000
mojhsa [17]

theoretical yield of the reaction is 121.38 g of NH₃ (ammonia)

limiting reactant is N₂ (nitrogen)

excess reactant is H₂ (hydrogen)

Explanation:

We have the following chemical reaction:

N₂ + 3 H₂ → 2 NH₃

Now we calculate the number of moles of each reactant:

number of moles = mass / molar weight

number of moles of N₂ = 100 / 28 = 3.57 moles

number of moles of H₂ = 100 / 2 = 50 moles

From the chemical reaction we see that 3 moles of H₂ are reacting with 1 moles of N₂, so 50 moles of H₂ are reacting with 16.66 moles of N₂ but we only have 3.57 moles of  N₂ available, so the limiting reactant will be N₂ and the excess reactant will be H₂.

Knowing the chemical reaction and the limiting reactant we devise the following reasoning:

if          1 mole of N₂ produce 2 moles of NH₃

then    3.57 moles of N₂ produce X moles of NH₃

X = (3.57 × 2) / 1 = 7.14 moles of NH₃

mass = number of moles × molar weight

mass of NH₃ = 7.14 × 17 = 121.38 g

theoretical yield of the reaction is 121.38 g of NH₃

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