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Afina-wow [57]
3 years ago
6

How do the properties of elements change as you move across a period

Chemistry
1 answer:
denpristay [2]3 years ago
3 0
The number of shells increase, the energy level their outer shell have decreases as we go down wards,
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112°C is 385.15 Kelvin
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Do chemical equations with no reaction still have to be balanced
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The same number of atoms of each element must appear on both sides of a chemical equation. However, simply writing down the chemical formulas of reactants and products does not always result in equal numbers of atoms. You have to balance the equation to make the number of atoms equal on each side of an equation.

Explanation:

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Nitrogen dioxide gas is formed by reaction of nitrogen monoxide gas and oxygen gas
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For the formation of nitrogen dioxide, the reactants used are nitrogen monoxide and oxygen gas. 2 moles of nitrogen monoxide gas react with 1 mole of oxygen gas to produce 2 moles of nitrogen dioxide gas.

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Promethium-147 is sometimes used in luminescent paint. It has a half-life of 956.3 days. If 250 grams (g) of promethium-147 is u
Fantom [35]

Answer:

% = 76.75%

Explanation:

To solve this problem, we just need to use the expressions of half life and it's relation with the concentration or mass of a compound. That expression is the following:

A = A₀ e^(-kt)   (1)

Where:

A and A₀: concentrations or mass of the compounds, (final and initial)

k: constant decay of the compound

t: given time

Now to get the value of k, we should use the following expression:

k = ln2 / t₁/₂   (2)

You should note that this expression is valid when the reaction is of order 1 or first order. In this kind of exercises, we can assume it's a first order because we are not using the isotope for a reaction.

Now, let's calculate k:

k = ln2 / 956.3

k = 7.25x10⁻⁴ d⁻¹

With this value, we just replace it in (1) to get the final mass of the isotope. The given time is 1 year or 365 days so:

A = 250 e^(-7.25x10⁻⁴ * 365)

A = 250 e^(-0.7675)

A = 191.87 g

However, the question is the percentage left after 1 year so:

% = (191.87 / 250) * 100

<h2>% = 76.75%</h2><h2>And this is the % of isotope after 1 year</h2>
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3 years ago
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