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2 years ago

If ba+2 will reacts with n-3, the formula of the compound formed is:

Chemistry

1 answer:

Ivanshal [37]2 years ago

6 0

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 $Ba_3N_2$ .

Learn more about the ionic compound here:

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3 0

3 years ago

Butane (C4 H10(g), Hf = –125.6 kJ/mol) reacts with oxygen to produce carbon dioxide (CO2 , Hf = –393.5 kJ/mol ) and water (H2 O,

WITCHER [35]

Answer: Enthalpy of combustion (per mole) of $C_4H_{10} (g)$ is -2657.5 kJ

Explanation:

The chemical equation for the combustion of butane follows:

$2C_4H_{10}(g)+4O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(8\times \Delta H^o_f_{CO_2(g)})+(10\times \Delta H^o_f_{H_2O(g)})]-[(1\times \Delta H^o_f_{C_4H_{10}(g)})+(4\times \Delta H^o_f_{O_2(g)})]$

We are given:

$\Delta H^o_f_{(C_4H_{10}(g))}=-125.6kJ/mol\\\Delta H^o_f_{(H_2O(g))}=-241.82kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_{rxn}=?$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(8\times -393.5)+(10\times -241.82)]-[(2\times -125.6)+(4\times 0)]\\\\\Delta H^o_{rxn}=-5315kJ$

Enthalpy of combustion (per mole) of $C_4H_{10} (g)$ is -2657.5 kJ

6 0

3 years ago

Manganese(IV) oxide is reduced to manganese(II) oxide by hydrogen gas. Include all phases in answer

liq [111]

MnO2(s)+H2(g)→MnO(s)+H2O(g)

4 0

3 years ago

Give the number of significant figures indicated 78

Mama L [17]

2 significant figures

6 0

3 years ago

The half-life of tritium (H-3) is 12.3 years. If 48.0mg of tritium is released from a nuclear power plant during the course of a

Rudiy27

Answer:

The amount left after 49.2 years is 3mg.

Explanation:

Given data:

Half life of tritium = 12.3 years

Total mass pf tritium = 48.0 mg

Mass remain after 49.2 years = ?

Solution:

First of all we will calculate the number of half lives.

Number of half lives = T elapsed/ half life

Number of half lives = 49.2 years /12.3 years

Number of half lives = 4

Now we will calculate the amount left after 49.2 years.

At time zero 48.0 mg

At first half life = 48.0mg/2 = 24 mg

At second half life = 24mg/2 = 12 mg

At 3rd half life = 12 mg/2 = 6 mg

At 4th half life = 6mg/2 = 3mg

The amount left after 49.2 years is 3mg.

6 0

4 years ago