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Vlada [557]
2 years ago
15

Many chemical names are similar at first glance. Give the formulas of the species in each set: (a) ammonium ion and ammonia; (b)

magnesium sulfide, magnesium sulfite, and magnesium sulfate; (c) hydrochloric acid, chloric acid, and chlorous acid; (d) cuprous bromide and cupric bromide.
Chemistry
2 answers:
ohaa [14]2 years ago
5 0

Answer:

a) Ammonium ion: NH₄⁺, ammonia: NH₃.

b)Magnesium sulfide: MgS, magnesium sulfite: MgSO₃, magnesium sulfate: MgSO₄.

c) Chloric acid: HClO₃, chlorous acid HClO₂.

d) Cupric bromide: CuBr₂, cuprous bromide: CuBr.

Explanation:

a) Ammonia is the neutral substance, which as molecular formula NH₃ when it gains a proton, it becomes the ammonium ion NH₄⁺.

b) In the name of salts, first, it comes the cation name, in this case, all the cations are the magnesium (Mg²⁺). The ions of sulfur can be oxidized or not, the termination "ide" indicates the non-oxidized ion (S⁻²), "ite" the less oxidized (SO₃⁻²), and "ate" the more oxidized (SO₄⁻²). To do the molecular formula, the charges are replaced without the signals and put down. If they are equal, the number is not put.

Magnesium sulfide: MgS

Magnesium sulfite: MgSO₃

Magnesium sulfate: MgSO₄

c) The acids can be oxidized or not, when it has no oxygens, the name is given "hydro" + the anion name, as hydrochloric acid (HCl). When there's oxygen in the molecule, the more oxidized is name as "'per" + anion name + "ic", the second more oxidized as anion name + "ic", the second less oxidized as anion name + "us", and the less oxidized as "hypo" + anion name + "us". In this case:

chloric acid: HClO₃

chlorous acid HClO₂

d) The salts of copper are named depends on the oxidation number of the cupper. The one with a higher oxidation number( +2) is called cupric, and the other (+1), cuprous. Thus:

cupric bromide: CuBr₂ (oxidation number = +2)

cuprous bromide: CuBr (oxidation number = +1)

g100num [7]2 years ago
5 0

Answer:

The formulae are

a) ammonia : NH₃

ammonium ion: NH₄⁺

b) Magnesium sulfide: Mgs

Magnesium sulfite: MgSO₃

c) Hydrochloric acid: HCl

chloric acid: HClO₃

chlorous acid: HClO₂

d) cuprous bromide: CuBr

cupric bromide: CuBr₂

Explanation:

The formula are

a) ammonia : NH₃ and when ammonia gains a proton it becomes

ammonium ion: NH₄⁺

b) Magnesium sulfide: MgS : the sulfide group is S⁻²

Magnesium sulfite: MgSO₃ : the sulfite group is SO₃⁻²

c) Hydrochloric acid: HCl :

chloric acid: HClO₃ These are oxy acids

chlorous acid: HClO₂

d) cuprous bromide: CuBr : cuprous word is used for copper with one positive charge

cupric bromide: CuBr₂: cupric is used for di-positive copper

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inysia [295]

Gunpowder is a heterogenous mixture,

Reason : the composiotion of the mixture is not uniform throughout, the mixture can be separated by a suitable method , and finally , it composes of charcoal, sulfur and potassium nitrate which is not homogeneous mixture.

4 0
1 year ago
If a reaction mixture initially contains 0.168 MSO2Cl2, what is the equilibrium concentration of Cl2 at 227 ∘C?
Semenov [28]

Answer:

see explanation below

Explanation:

The question is incomplete. Here's the complete question:

<em>Consider the following reaction: </em>

<em>SO2Cl2 -----> SO2(g) + Cl2(g) </em>

<em>Kc= 2.99 x 10^-7 at 227 degrees celcius </em>

<em>If a reaction mixture initially contains 0.168 MSO2Cl2, what is the equilibrium concentration of Cl2 at 227 ∘C?</em>

This is a problem of equilibrium, therefore, we need to solve this using the expression of equilibrium constant. To do that, we need to wirte an ICE chart and solve from there:

       SOCl2 ---------> SO2 + Cl2     Kc = 2.99x10⁻⁷

i)       0.168                  0        0

c)         -x                    +x       +x

e)     0.168-x                x         x

Writting the Kc expression:

Kc = [SO2] [Cl2] / [SOCl2]

Replacing the values from the chart:

2.99x10⁻⁷ = x² / 0.168 - x

However, Kc is a very very small value, therefore, we can assume that the value of "x" would be very small too, and we can neglect the 0.168-x and just round it to 0.168:

2.99x10⁻⁷ = x²/0.168

2.99x10⁻⁷ * 0.168 = x²

√5.02x10⁻⁸ = x

x = 2.24x10⁻⁴ M

This means then, that the concentration of Cl2 in equilibrium would be:

<em>[Cl₂] = 2.24x10⁻⁴ M</em>

5 0
3 years ago
Answer the following questions about the solubility of AgCl(s). The value of Ksp for AgCl(s) is 1.8 × 10−10.
Firlakuza [10]

Answer:

  • [Ag⁺] = 1.3 × 10⁻⁵M
  • s = 3.3 × 10⁻¹⁰ M
  • Because the common ion effect.

Explanation:

<u></u>

<u>1. Value of [Ag⁺]  in a saturated solution of AgCl in distilled water.</u>

The value of [Ag⁺]  in a saturated solution of AgCl in distilled water is calculated by the dissolution reaction:

  • AgCl(s)    ⇄    Ag⁺ (aq)    +    Cl⁻ (aq)

The ICE (initial, change, equilibrium) table is:

  • AgCl(s)    ⇄    Ag⁺ (aq)    +    Cl⁻ (aq)

I            X                      0                    0

C          -s                      +s                  +s

E         X - s                   s                     s

Since s is very small, X - s is practically equal to X and is a constant, due to which the concentration of the solids do not appear in the Ksp equation.

Thus, the Ksp equation is:

  • Ksp = [Ag⁺] [Cl⁻]
  • Ksp = s × s
  • Ksp = s²

By substitution:

  • 1.8 × 10⁻¹⁰ = s²
  • s = 1.34 × 10⁻⁵M

Rounding to two significant figures:

  • [Ag⁺] = 1.3 × 10⁻⁵M ← answer

<u></u>

<u>2. Molar solubility of AgCl(s) in seawater</u>

Since, the conentration of Cl⁻ in seawater is 0.54 M you must introduce this as the initial concentration in the ECE table.

The new ICE table will be:

  • AgCl(s)    ⇄    Ag⁺ (aq)    +    Cl⁻ (aq)

I            X                      0                  0.54

C          -s                      +s                  +s

E         X - s                   s                     s + 0.54

The new equation for the Ksp equation will be:

  • Ksp = [Ag⁺] [Cl⁻]
  • Ksp = s × ( s + 0.54)
  • Ksp = s² + 0.54s

By substitution:

  • 1.8 × 10⁻¹⁰ = s² + 0.54s
  • s² + 0.54s - 1.8 × 10⁻¹⁰ = 0

Now you must solve a quadratic equation.

Use the quadratic formula:

     

     s=\dfrac{-0.54\pm\sqrt{0.54^2-4(1)(-1.8\times 10^{-10})}}{2(1)}

The positive and valid solution is s = 3.3×10⁻¹⁰ M ← answer

<u>3. Why is AgCl(s) less soluble in seawater than in distilled water.</u>

AgCl(s) is less soluble in seawater than in distilled water because there are some Cl⁻ ions is seawater which shift the equilibrium to the left.

This is known as the common ion effect.

By LeChatelier's principle, you know that an increase in the concentrations of one of the substances that participate in the equilibrium displaces the reaction to the direction that minimizes this efect.

In the case of solubility reactions, this is known as the common ion effect: when the solution contains one of the ions that is formed by the solid reactant, the reaction will proceed in less proportion, i.e. less reactant can be dissolved.

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