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3 years ago

Use IUPAC rules to name this compound. A) nitrogen iodide B) nitrogen triiodide C) mononitrogen iodide D) mononitrogen triiodide

Chemistry

2 answers:

Karolina [17]3 years ago

7 0

Answer:

its nitrogen triiodide?

Explanation:

just took it on usa test prep

Mariulka [41]3 years ago

3 0

Answer:

Name of the compound is Nitrogen triiodide.

Explanation:

According to the IUPAC rules, to naming of the compound the following formula can be applied.

Prefix + Name of first element + Base name element of second element + Suffix.

The given compound - $NI_{3}$

Name of first element- Nitrogen

Base name element of second element - Iodine

Suffix = 3 = tri

Here, iodine is in ionic form therefore, it becomes iodide. and then suffix will be added in front of the halogen.

Therefore, name of the compound will be Nitrogen triiodide..

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Which compound are ionic and which are covalent? (N2) (CCl4) (SiO2) (AlCl3) (CaCl2) (LiBr)

muminat

Covalent compounds: N2, CCl4, SiO2 and AlCl3.

Ionic compounds: CaCl2 and LiBr.

Hope this helps!

3 0

3 years ago

If you were to react 4.3 miles of magnesium phosphide how many moles of potassium would you need?

bixtya [17]

Answer:

25.8

Explanation:

Let's write the reaction between magnesium-phosphide and potassium:

Mg3P2 + K = Mg + K3P

And now let's balance this equation:

Mg3P2+6K=3Mg+2K3P

We see that the ratio of magnesium-phosphide and potassium is 1:6, which means that for every mole of magnesium-phosphide there need to be 6 moles of potassium.

Since we have 4.3 moles of Mg3P2, there need to be 6 • 4.3 = 25.8 moles of potassium.

5 0

2 years ago

A chemist fills a reaction vessel with mercurous chloride solid, mercury (I) aqueous solution, and chloride aqueous solution at

makvit [3.9K]

Answer:

ΔG° = -533.64 kJ

Explanation:

Let's consider the following reaction.

Hg₂Cl₂(s) ⇄ Hg₂²⁺(aq) + 2 Cl⁻(aq)

The standard Gibbs free energy (ΔG°) can be calculated using the following expression:

ΔG° = ∑np × ΔG°f(products) - ∑nr × ΔG°f(reactants)

where,

ni are the moles of reactants and products

ΔG°f(i) are the standard Gibbs free energies of formation of reactants and products

ΔG° = 1 mol × ΔG°f(Hg₂²⁺) + 2 mol × ΔG°f(Cl⁻) - 1 mol × ΔG°f(Hg₂Cl₂)

ΔG° = 1 mol × 148.85 kJ/mol + 2 mol × (-182.43 kJ/mol) - 1 mol × (-317.63 kJ/mol)

ΔG° = -533.64 kJ

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3 years ago

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