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Alik [6]
3 years ago
15

The oxidation number of magnesium in magnesium chloride is _____.

Chemistry
1 answer:
LenaWriter [7]3 years ago
3 0
I hope this would be able to help u
dont forget the symbol its +2

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If ba+2 will reacts with n-3, the formula of the compound formed is:
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If Ba^{+2} will reacts with N^{-3}, the formula of the compound formed is Ba_3N_2.

<h3>What is an ionic compound?</h3>

Ionic compounds are neutral compounds made up of positively charged ions called cations and negatively charged ions called anions.

Barium is a metal element and Nitrogen is a non-metal element, so they form an ionic compound. To make the formula of an ionic compound, first, we need to see the valence of the atoms, that is, how many electrons it can donate or it has to gain to have 8 electrons in the valence layer (octet rule).

Barium is in family 2, so it has two electrons in its valence layer, so it must donate these two electrons; Nitrogen is in family 15, so it has 5 electrons ins its valence layer and must gain 3 electrons.

Writing the ions they must form:

Ba^{+2}N^{-3}

First came the cation (positive charge), then the anion (negative charge), and the charges must change between them, without the signal, and will inform how many atoms are necessary for which element:

Ba_3N_2

Hence, If Ba^{+2} will reacts with N^{-3}, the formula of the compound formed is Ba_3N_2.

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What type pf asexual reproduction produces the most offspring?
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The rate constants of some reactions double with every 10-degree rise in temperature. Assume that a reaction takes place at 295
LUCKY_DIMON [66]

Answer : The activation energy for the reaction is, 51.9 kJ

Explanation :

According to the Arrhenius equation,

K=A\times e^{\frac{-Ea}{RT}}

or,

\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_1 = rate constant at 295 K

K_2 = rate constant at 305 K = 2K_1

Ea = activation energy for the reaction = ?

R = gas constant = 8.314 J/mole.K

T_1 = initial temperature = 295 K

T_2 = final temperature = 305 K

Now put all the given values in this formula, we get:

\log (\frac{2K_1}{K_1})=\frac{Ea}{2.303\times 8.314J/mole.K}[\frac{1}{295K}-\frac{1}{305K}]

Ea=51879.96J=51.9kJ

Therefore, the activation energy for the reaction is, 51.9 kJ

8 0
4 years ago
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