Carbon cycle is returned to the atmosphere primarily by the respiration of consumer.

Naming Compounds Test Review Practice Name the following ionic compounds: NH4Cl _____________________________________ Fe(NO3)3 _

Mariulka [41]

Answer :

The name of $NH_4Cl$ is ammonium chloride.

The name of $Fe(NO_3)_3$ is iron(III) nitrate.

The name of $TiBr_3$ is titanium(III) bromide.

The name of $Cu_3P$ is copper(I) phosphide.

The name of $SnSe_2$ is tin(IV) selenide.

The name of $GaAs$ is gallium arsenide.

The name of $Pb(SO_4)_2$ is lead(IV) sulfate.

The name of $Be(HCO_3)_2$ is beryllium hydrogen carbonate.

The name of $Mn_2(SO_3)_3$ is manganese(III) sulfite.

The name of $Al(CN)_3$ is aluminium cyanide.

Explanation :

Ionic compound : It is defined as the compound which is formed when electron gets transferred from one atom to another atom.

Ionic compound are usually formed when a metal reacts with a non-metal.

The nomenclature of ionic compounds is given by:

1. Positive is written first.

2. The negative ion is written next and a suffix is added at the end of the negative ion. The suffix written is '-ide'.

3. In case of transition metals, the oxidation state are written in roman numerals in bracket in-front of positive ions.

(a) $NH_4Cl$

It is an ionic compound because it is formed by the combination of one polyatomic cation (ammonium ion) and anion (chloride ion). The charge on ammonium ion is (+1) and the the charge on chloride ion is (-1).

Thus, the name of $NH_4Cl$ is ammonium chloride.

(b) $Fe(NO_3)_3$

It is an ionic compound because it is formed by the combination of one polyatomic anion (nitrate ion) and cation (iron ion). The charge on iron ion is (+3) and the the charge on nitrate ion is (-1).

Thus, the name of $Fe(NO_3)_3$ is iron(III) nitrate.

(c) $TiBr_3$

It is an ionic compound because it is formed by the combination of titanium cation and bromide anion. The charge on titanium is (+3) and the the charge on bromide ion is (-1).

Thus, the name of $TiBr_3$ is titanium(III) bromide.

(d) $Cu_3P$

It is an ionic compound because it is formed by the combination of copper cation and phosphorous anion. The charge on copper is (+1) and the the charge on phosphorous ion is (-3).

Thus, the name of $Cu_3P$ is copper(I) phosphide.

(e) $SnSe_2$

It is an ionic compound because it is formed by the combination of tin cation and selenium anion. The charge on tin is (+4) and the the charge on selenium ion is (-2).

Thus, the name of $SnSe_2$ is tin(IV) selenide.

(f) $GaAs$

It is an ionic compound because it is formed by the combination of gallium cation and arsenic anion. The charge on gallium is (+3) and the the charge on arsenic ion is (-3).

Thus, the name of $GaAs$ is gallium arsenide.

(g) $Pb(SO_4)_2$

It is an ionic compound because it is formed by the combination of one polyatomic anion (sulfate ion) and cation (lead ion). The charge on lead ion is (+4) and the the charge on sulfate ion is (-2).

Thus, the name of $Pb(SO_4)_2$ is lead(IV) sulfate.

(h) $Be(HCO_3)_2$

It is an ionic compound because it is formed by the combination of one polyatomic anion (hydrogen carbonate ion) and cation (beryllium ion). The charge on beryllium ion is (+2) and the the charge on hydrogen carbonate ion is (-1).

Thus, the name of $Be(HCO_3)_2$ is beryllium hydrogen carbonate.

(i) $Mn_2(SO_3)_3$

It is an ionic compound because it is formed by the combination of one polyatomic anion (sulfite ion) and cation (manganese ion). The charge on manganese ion is (+3) and the the charge on sulfite ion is (-2).

Thus, the name of $Mn_2(SO_3)_3$ is manganese(III) sulfite.

(j) $Al(CN)_3$

It is an ionic compound because it is formed by the combination of one polyatomic anion (cyanide ion) and cation (aluminium ion). The charge on aluminium ion is (+3) and the the charge on cyanide ion is (-1).

Thus, the name of $Al(CN)_3$ is aluminium cyanide.

3 0

3 years ago

Consider the reaction between reactants s and o2: 2s(s)+3o2(g)→2so3(g)if a reaction vessel initially contains 7 mols and 9 mol o

nadya68 [22]

Answer: -

1 mol

Explanation: -

Number of moles of Sulphur S = 7

Number of moles of O2 = 9

The balanced chemical equation for the reaction is

2S (s)+3 O2 (g)→2SO3(g)

From the above reaction we can see that

3 mol of O2 react with 2 mol of S

9 mol of O2 will react with $\frac{2 mol S x 9 mol O2}{3 mol O2}$

= 6 mol of S

Unreacted S = 7 - = 1 mol.

If a reaction vessel initially contains 7 mol S and 9 mol O2

1 mole of s will be in the reaction vessel once the reactants have reacted as much as possible

6 0

4 years ago

Please help, only a couple of days left!!!

s344n2d4d5 [400]

Answer:

the heat from the universe during the big bang

a.k.a.

ITS A

Explanation:

the universe is still expanding, so what is gonna be used to create that new matter that's expanding the universe? energy and heat from the beginning of the universe! the universe started off as this point that was probably smaller than a quantum foam and infinite heat and started expanding from there

3 0

3 years ago

Read 2 more answers

An element consists of 1.40% of an isotope

Dmitriy789 [7]

Answer:

I think it's lead (pb). Hope this helps.

6 0

3 years ago

Read 2 more answers

A mixture of helium and neon gases has a density of 0.2460 g/L at 35.2°C and 288 torr. What is the mole fraction of neon in this

pshichka [43]

We know the volume of one mole of gas at 273 K and 760 Torr is 22.4 L. Using
(PV)/T = constant
We can calculate the volume of the gasses at the given conditions:
(P₁V₁)/T₁ = (P₂V₂)/T₂
(760 * 22.4) / 273 = (288 V₂) / 308.2
V₂ = 66.7 L
Mass of He: 4
Mass of Ne: 20
Fraction of Ne: x
Fraction of He: 1 - x
avg density = (∑(component fraction × component mass))/volume
0.2460 = (20x + 4(1 - x))/ 66.7
x = 0.775

6 0

4 years ago