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3 years ago

What occurs when a magnesium atom becomes a magnesium ion

Chemistry

2 answers:

Law Incorporation [45]3 years ago

7 0

Answer is: Electrons are lost and the oxidation number increases.

Magnesium (Mg) is metal from 2. group of Periodic table of elements and has low ionisation energy and electronegativity, which means it easily lose valence electons (two valence electrons).

Magnesium has atomic number 12, which means it has 12 protons and 12 electrons. It lost two electrons to form magnesium cation (Mg²⁺) with stable electron configuration like closest noble gas neon (Ne) with 10 electrons.

Electron configuration of magnesium atom: ₁₂Mg 1s² 2s² 2p⁶ 3s².

Electron configuration of magnesium ion: ₁₂Mg²⁺ 1s² 2s² 2p⁶.

Karo-lina-s [1.5K]3 years ago

4 0

The electrons are lost.

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There is an error in the rate-determining step of the following proposed mechanism.

yulyashka [42]

Answer:

The steps with correct mechanism are given below:

1) CH₄(g) + Cl(g) → CH₃(g) + HCl(g) : This is a slow step.

The rate is given as: R1 = k₁[CH₄][Cl]

2) CH₃(g) + Cl₂(g) → CH₃Cl(g) + Cl(g): This is a fast step.

The rate is given as: Rate = k₂[CH₃][Cl₂]

∴ CH₄(g) + Cl₂(g) → CH₃Cl(g) + HCl(g)

Here, the slowest step will be the rate-determining step.

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3 years ago

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Luda [366]

Answer:

118

Explanation:

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3 years ago

How does a heat affect the chemical reaction?

patriot [66]

Answer:

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Explanation:

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3 years ago

Read 2 more answers

Can someone help me with this I've been stuck on it for a few days

Fiesta28 [93]

So I haven’t got time to answer all of it for you but the id you look at the picture of the periodic table I’ve added the top number in the red boxes are the groups and the period is how many elements down from the top it is (remember that the hydrogen and helium make up period ONE) so remember to include them when counting the elements as you go down the table

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2 years ago

5. How much heat (in calories) is absorbed by a reaction when

Wewaii [24]

Answer:

1.2 × 10⁴ cal

Explanation:

Given data

Mass of water (m): 300 g

Initial temperature: 80 °C

Final temperature: 40°C

We can calculate the heat released by the water ( $Q_w$ ) when it cools using the following expression.

$Q_w = c \times m \times (T_f - T_i)$

where

c is the specific heat capacity of water (1 cal/g.°C)

$Q_w = \frac{1cal}{g.\°C} \times 300g \times (40\°C - 80\°C) = -1.2 \times 10^{4} cal$

According to the law of conservation of energy, the sum of the heat released by the water ( $Q_w$ ) and the heat absorbed by the reaction ( $Q_r$ ) is zero.

$Q_w + Q_r = 0\\Q_r = -Q_w = 1.2 \times 10^{4} cal$

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3 years ago