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2 years ago

Is nitrogen triiodide an ionic compound or covalent compound?

Chemistry

2 answers:

DerKrebs [107]2 years ago

8 0

Answer: The given compound is a covalent compound.

Explanation:

Covalent compound is defined as the compound which is formed by the sharing of electrons between the atoms forming a compound. These are usually formed when two non-metals react.

An ionic compound is defined as the compound which is formed when electron gets transferred from one atom to another atom. These are usually formed when a metal reacts with a non-metal.

We are given:

A chemical compound having chemical name as nitrogen triiodide.

This compound is formed by the combination of nitrogen and iodine atoms. Both these elements are non-metals and thus form covalent compound.

The chemical formula for the given compound is $NI_3$

Hence, the given compound is a covalent compound.

notka56 [123]2 years ago

5 0

Nitrogen triiodide is the inorganic compound with the formula NI3. It is an extremely sensitive contact explosive: small quantities explode with a loud, sharp snap when touched even lightly, releasing a purple cloud of iodine vapor; it can even be detonated by alpha radiation. NI3 has a complex structural chemistry that is difficult to study because of the instability of the derivatives.

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7. A 2.0 L container had 0.40 mol of He(g) and 0.60 mol of Ar(g) at 25°C.

Finger [1]

Answer:

a) Ek Ar > Ek He

b) v Ar < v He

c) If v Ar = 431 m/s ⇒ v He = 1710.44 m/s

d) Pt = 12.218 atm

e) P He = 4.887 atm and P Ar = 7.33 atm

Explanation:

container:

∴ V = 2.0 L

∴ n He = 0.4 mol

∴ n Ar = 0.6 mol

∴ T = 25°C ≅ 298 K

a) Internal energy (U) :

∴ U = Ek + Ep = kinetic energy + potential energy

∴ Ep: the potential interaction energy is neglected, assuming ideal gas mixture

⇒ U = Ek = N(1/2mv²)= 3/2 NKT

∴ N = nNo ....number of moleculas

∴ K = 1.380 E-23 J/K....Boltzmann's constant

∴ No = 6.022 E23 molec/mol....Avogadro's number

for He:

⇒ N = (0.4)(6.022 E23) = 2.4088 E23 molec

⇒ Ek = (3/2)(2.4088 E23)(1.380 E-23 J/K)(298) = 1485.892 J

for Ar:

⇒ N = (0.6)(6.022 E23) = 3.6132 E 23 molec

⇒ Ek = (3/2)(3.6132 E23)(1.380 E-23 J/K)(298) = 2228.838 J

** Ar gas has a greater average kinetic energy

b) He:

∴ N(1/2)mv² = (3/2)NKT

⇒ mv² = 3KT

⇒ v² = 3KT/m

⇒ v = √3KT/m

∴ m He = (0.4 mol)(4.0026 g/mol) = 1.601 g He = 1.601 E-3 Kg He

⇒ v = √(3(1.380 E-23)(298)/(1.601 E-3)) = 2.776 E-9 m/s He

Ar:

∴ m Ar = (0.6)(39.948 g/mol) = 23.969 g = 0.0239 Kg Ar

⇒ v = 6.99 E-10 m/s

** v Ar < v He

c) r = V Ar / v He = (6.99 E-10 m/s)/(2.776 E-9 m/s) = 0.252

∴ If v Ar = 431 m/s

⇒ v He = v Ar/0.252 = 431 m/s / 0.252 = 1710.44 m/s

d) Pt = ntRT / V

∴ nt = 0.4 + 0.6 = 1 mol

⇒ Pt = (1mol)(0.082 atm.L/K.mol)(298 K)/(2.00 L) = 12.218 atm

e) P He = nRT/V = (0.4)(0.082)(298)/2 = 4.8872 atm

⇒ P Ar = Pt - PHe = 12.218 - 4.8872 = 7.33 atm

3 0

3 years ago

For an organism’s scientific names, the first part is the ________ and the second is the ________.

kari74 [83]

Answer:

Explanation:

For an organism scientific names the first part is the genus and the second is the species

3 0

2 years ago

Read 2 more answers

If mercury (Hg) and oxygen (O2) were reacted to form mercury oxide how many molecules of each reactant and product would be need

dimulka [17.4K]

In order to get HgO you would need 2Hg+1O2=2HgO. Since oxygen is diatomic you need two when it stands alone causing you to need two mercuries to balance out the reactants and the product I hope this helps

5 0

3 years ago

Describe uses of H2S as analytical regent

konstantin123 [22]

Answer:

Hydrogen sulfide is used primarily to produce sulfuric acid and sulfur. It is also used to create a variety of inorganic sulfides used to create pesticides, leather, dyes, and pharmaceuticals. Hydrogen sulfide is used to produce heavy water for nuclear power plants (like CANDU reactors specifically).

Explanation:

Sana Po it's help

8 0

3 years ago

The equilibrium constant for the dissolution of silver chloride (AgCl(s) Ag+(aq) + Cl–(aq)) has a value of 1.79 × 10–10. Which s

ryzh [129]

"Silver chloride is essentially insoluble in water" this statement is true for the equilibrium constant for the dissolution of silver chloride.

Option: b

Explanation:

As silver chloride is essentially insoluble in water but also show sparing solubility, its reason is explained through Fajan's rule. Therefore when AgCl added in water, equilibrium take place between undissolved and dissolved ions. While solubility product constant $\left(\boldsymbol{K}_{s p}\right)$ for silver chloride is determined by equilibrium concentrations of dissolved ions. But solubility may vary also at different temperatures. Complete solubility is possible in ammonia solution as it form stable complex as water is not good ligand for Ag+.

To calculate $\left(\boldsymbol{K}_{s p}\right)$ firstly molarity of ions are needed to be found with formula: $\text { Molarity of ions }=\frac{\text { number of moles of solute }}{\text { Volume of solution in litres }}$

Then at equilibrium cations and anions concentration is considered same hence:

$\left[\mathbf{A} \mathbf{g}^{+}\right]=[\mathbf{C} \mathbf{I}]=\text { molarity of ions }$

Hence from above data $\left(\boldsymbol{K}_{s p}\right)$ can be calculated by: $\left(\boldsymbol{K}_{s p}\right)$ = $\left[\mathbf{A} \mathbf{g}^{+}\right] \cdot[\mathbf{C} \mathbf{I}]$

6 0

3 years ago