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yuradex [85]
2 years ago
15

A solution of 20.0 g of which hydrated salt dissolved in 200 g H2O will have the lowest freezing point? (A) CuSO4 • 5 H2O (M = 2

50) (B) NiSO4 • 6 H2O (M = 263) (C) MgSO4 • 7 H2O (M = 246) (D) Na2SO4 • 10 H2O (M = 286) The correct answer is D, but I really don't know WHY, so I ask... Thank you very much for your help and explanation ! :)
Chemistry
1 answer:
Andrews [41]2 years ago
4 0

Answer:

(D) Na₂SO₄•10H₂O (M = 286).

Explanation:

  • The depression in freezing point of water by adding a solute is determined using the relation:

ΔTf = i.Kf.m,

Where, <em>ΔTf </em>is the depression in freezing point of water.

<em>i</em> is van't Hoff factor.

<em>Kf </em>is the molal depression constant.

<em>m</em> is the molality of the solute.

  • Since, Kf and m is constant for all the mentioned salts. So, the depression in freezing point depends strongly on the van't Hoff factor (i).
  • van't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass.

(A) CuSO₄•5H₂O:

CuSO₄ is dissociated to Cu⁺² and SO₄²⁻.

So, i = dissociated ions/no. of particles = 2/1 = 2.

(B) NiSO₄•6H₂O:

NiSO₄ is dissociated to Ni⁺² and SO₄²⁻.

So, i = dissociated ions/no. of particles = 2/1 = 2.

(C) MgSO₄•7H₂O:

MgSO₄ is dissociated to Mg⁺² and SO₄²⁻.

So, i = dissociated ions/no. of particles = 2/1 = 2.

(D) Na₂SO₄•10H₂O:

Na₂SO₄ is dissociated to 2 Na⁺ and SO₄²⁻.

So, i = dissociated ions/no. of particles = 3/1 = 3.

∴ The salt with the high (i) value is Na₂SO₄•10H₂O.

So, the highest ΔTf resulted by adding Na₂SO₄•10H₂O salt.

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Which balanced redox reaction is occurring in the voltaic cell represented by the notation of A l ( s ) | A l 3 ( a q ) | | P b
frez [133]

The question is missing. Here is the complete question.

Which balanced redox reaction is ocurring in the voltaic cell represented by the notation of Al_{(s)}|Al^{3+}_{(aq)}||Pb^{2+}_{(aq)}|Pb_{(s)}?

(a) Al_{(s)}+Pb^{2+}_{(aq)} ->Al^{3+}_{(aq)}+Pb_{(s)}

(b) 2Al^{3+}_{(aq)}+3Pb_{(s)} -> 2Al_{(s)}+3Pb^{2+}_{(aq)}

(c)Al^{3+}_{(aq)}+Pb_{(s)} ->Al_{(s)}+Pb^{2+}_{(aq)}

(d) 2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}

Answer: (d) 2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}

Explanation: <u>Redox</u> <u>Reaction</u> is an oxidation-reduction reaction that happens in the reagents. In this type of reaction, reagent changes its oxidation state: when it loses an electron, oxidation state increases, so it is oxidized; when receives an electron, oxidation state decreases, then it is reduced.

Redox reactions can be represented in shorthand form called <u>cell</u> <u>notation,</u> formed by: <em><u>left side</u></em> of the salt bridge (||), which is always the <em><u>anode</u></em>, i.e., its half-equation is as an <em><u>oxidation</u></em> and <em><u>right side</u></em>, which is always <em><u>the cathode</u></em>, i.e., its half-equation is always a <em><u>reduction</u></em>.

For the cell notation: Al_{(s)}|Al^{3+}_{(aq)}||Pb^{2+}_{(aq)}|Pb_{(s)}

Aluminum's half-equation is oxidation:

Al_{(s)} -> Al^{3+}_{(aq)}+3e^{-}

For Lead, half-equation is reduction:

Pb^{2+}_{(aq)}+2e^{-} -> Pb_{(s)}

Multiply first half-equation for 2 and second half-equation by 3:

2Al_{(s)} -> 2Al^{3+}_{(aq)}+6e^{-}

3Pb^{2+}_{(aq)}+6e^{-} -> 3Pb_{(s)}

Adding them:

2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}

The balanced redox reaction with cell notation Al_{(s)}|Al^{3+}_{(aq)}||Pb^{2+}_{(aq)}|Pb_{(s)} is

2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}

6 0
2 years ago
Determine whether each description applies to electrophilic aromatic substitution or nucleophilic aromatic substitution.
Alborosie

Answer:

a. electrophilic aromatic substitution

b. nucleophilic aromatic substitution

c. nucleophilic aromatic substitution

d. electrophilic aromatic substitution

e. nucleophilic aromatic substitution

f. electrophilic aromatic substitution

Explanation:

Electrophilic aromatic substitution is a type of chemical reaction where a hydrogen atom or a functional group that is attached to the aromatic ring is replaced by an electrophile. Electrophilic aromatic substitutions can be classified into five classes: 1-Halogenation: is the replacement of one or more hydrogen (H) atoms in an organic compound by a halogen such as, for example, bromine (bromination), chlorine (chlorination), etc; 2- Nitration: the replacement of H with a nitrate group (NO2); 3-Sulfonation: the replacement of H with a bisulfite (SO3H); 4-Friedel-CraftsAlkylation: the replacement of H with an alkyl group (R), and 5-Friedel-Crafts Acylation: the replacement of H with an acyl group (RCO). For example, the Benzene undergoes electrophilic substitution to produce a wide range of chemical compounds (chlorobenzene, nitrobenzene, benzene sulfonic acid, etc).

A nucleophilic aromatic substitution is a type of chemical reaction where an electron-rich nucleophile displaces a leaving group (for example, a halide on the aromatic ring). There are six types of nucleophilic substitution mechanisms: 1-the SNAr (addition-elimination) mechanism, whose name is due to the Hughes-Ingold symbol ''SN' and a unimolecular mechanism; 2-the SN1 reaction that produces diazonium salts 3-the benzyne mechanism that produce highly reactive species (including benzyne) derived from the aromatic ring by the replacement of two substituents; 4-the free radical SRN1 mechanism where a substituent on the aromatic ring is displaced by a nucleophile with the formation of intermediary free radical species; 5-the ANRORC (Addition of the Nucleophile, Ring Opening, and Ring Closure) mechanism, involved in reactions of metal amide nucleophiles and substituted pyrimidines; and 6-the Vicarious nucleophilic substitution, where a nucleophile displaces an H atom on the aromatic ring but without leaving groups (such as, for example, halogen substituents).

3 0
2 years ago
The ksp of agcl is 1.6 x 10-10. what is the solubility of agcl in 0.010 m fecl3? give your answer using scientific notation (1.2
bija089 [108]

Ksp of AgCl= 1.6×10⁻¹⁰

AgCl=Ag⁺ +Cl⁻

Ksp=[Ag⁺][Cl⁻]

Assume [Ag⁺]=[Cl⁻]=x

Ksp=x²

1.6×10⁻¹⁰=x²

x=0.000012

In FeCl₃:

FeCl₃------>Fe⁺³+ 3Cl⁻

as there is 0.010 M FeCl₃

So there will be ,

[Cl⁻]= 0.030

So

[Ag⁺]=Ksp/[Cl⁻]

=1.6×10⁻¹⁰/0.030

=5.3×10⁻⁹

so solubility of AgCl in FeCl₃ will be 5.3×10⁻⁹.

7 0
2 years ago
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Lina20 [59]

Answer:

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Explanation:

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We can calculate the standard Gibbs free energy of the reaction (ΔG°rxn) using the following expression:

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where,

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T is the absolute temperature

ΔS° is the standard entropy of the reaction

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Answer:

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Explanation:

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