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Nina [5.8K]
3 years ago
8

You have 300,000 atoms of a radioactive substance. After 3 half-lives have past, how many atoms remain? Remember that you cannot

have a fraction of an atom, so round the answer to the nearest whole number.
Chemistry
1 answer:
padilas [110]3 years ago
3 0

The number of atoms that remains after 3 half-lives given that it was originally 300000 atoms is 37500 atoms

<h3>Data obtained from the question</h3>
  • Original amount (N₀) = 300000 atoms
  • Number of half-lives (n) = 3
  • Amount remaining (N) =?

<h3>How to determine the amount remaining</h3>

The amount remaining after 3 half-lives can be obtained as illustrated below:

N = N₀ / 2ⁿ

N = 300000 / 2³

N = 300000 / 8

N = 37500 atoms

Learn more about half life:

brainly.com/question/26374513

#SPJ1

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If you want 2 miles of Silver (Ag) from the reaction below, how many grams of copper should you start with? Reaction: Cu + 2AgNO
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Answer:

You should start with 63.54 grams of copper.

Explanation:

The chemical reactions are processes in which the nature of the substances changes, that is, from some initial substances called reactants, totally different ones called products are obtained.

In the chemical reaction, the formulas of reagents and products appear preceded by numbers (the stoichiometric coefficients) that indicate the proportions according to which the transformation occurs. So you can say that stoichiometry establishes relationships between the molecules or elements that make up the reactants of a chemical equation with the products of said reaction. The relationships that are established are MOLAR relationships between the compounds or elements that make up the chemical equation: always in MOLES.

The stoichiometric coefficients of a chemical equation are due to the fact that the atoms present before the reaction must be the same after the reaction, although they will have been rearranged to produce new substances.

If you want 2 moles of silver (Ag), for stoichiometry of the reaction you need a moles of copper Cu. Being the molar mass of copper Cu 63.54 g / mole, then:

1 mole*63.54 g/mole= 63.54 g

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3 years ago
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In the Haber process for ammonia synthesis, K " 0.036 for N 2 (g) ! 3 H 2 (g) ∆ 2 NH 3 (g) at 500. K. If a 2.0-L reactor is char
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Answer : The partial pressure of N_2,H_2\text{ and }NH_3 at equilibrium are, 1.133, 2.009, 0.574 bar respectively. The total pressure at equilibrium is, 3.716 bar

Solution :  Given,

Initial pressure of N_2 = 1.42 bar

Initial pressure of H_2 = 2.87 bar

K_p = 0.036

The given equilibrium reaction is,

                              N_2(g)+H_2(g)\rightleftharpoons 2NH_3(g)

Initially                   1.42      2.87             0

At equilibrium    (1.42-x)  (2.87-3x)     2x

The expression of K_p will be,

K_p=\frac{(p_{NH_3})^2}{(p_{N_2})(p_{H_2})^3}

Now put all the values of partial pressure, we get

0.036=\frac{(2x)^2}{(1.42-x)\times (2.87-3x)^3}

By solving the term x, we get

x=0.287\text{ and }3.889

From the values of 'x' we conclude that, x = 3.889 can not more than initial partial pressures. So, the value of 'x' which is equal to 3.889 is not consider.

Thus, the partial pressure of NH_3 at equilibrium = 2x = 2 × 0.287 = 0.574 bar

The partial pressure of N_2 at equilibrium = (1.42-x) = (1.42-0.287) = 1.133 bar

The partial pressure of H_2 at equilibrium = (2.87-3x) = [2.87-3(0.287)] = 2.009 bar

The total pressure at equilibrium = Partial pressure of N_2 + Partial pressure of H_2 + Partial pressure of NH_3

The total pressure at equilibrium = 1.133 + 2.009 + 0.574 = 3.716 bar

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