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3 years ago

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What is the theoretical yield of cyclohexene (in grams) that could be formed from 0.105 moles of cyclohexanol and 0.0444 moles o

f 85% phosphoric acid

1 answer:

zubka84 [21]3 years ago

6 0

Answer:

3.65 g of cyclohexene

Explanation:

Cyclohexanol + phosphoric acid ----> cyclohexene

The reaction is 1:1 hence the limiting reactant is phosphoric acid.

Hence,

1 mole of phosphoric acid yields 1 mole of cyclohexene

0.0444 moles of phosphoric acid yields 0.0444 moles of cyclohexene

Theoretical yield = number of moles of cyclohexene × molar mass of cyclohexene

Theoretical yield = 0.0444 moles of cyclohexene × 82.143 g/mol

Theoretical yield = 3.65 g of cyclohexene

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Answer:

1. The concentration of N₂O₄ decrease as the temperature of the system increased.

2. The formation of products was favored by the addition of heat.

3. The reaction going from right to left is exothermic.

4. N₂O₄ → 2NO₂; HR = +14 kcal.

Explanation:

Equation of the reaction is as follows: N₂O₄ ⇄ 2NO₂

The forward reaction proceeds with an increase in temperature. As the temperature of the reaction was increased, more of the N₂O₄ decomposed to form NO₂. Therefore, the concentration of N₂O₄ deceased.

2. The formation of products, that is the forward reaction leading to formation of NO₂ was favored by the addition of heat. Thus, with increase in temperature, the concentration of NO₂ increased.

3. An endothermic reaction is a type of reaction which requires energy input in the form of heat in order to proceed.

From the observations during the reaction, an increase in temperature by the addition of heat resulted in an increase in the forward reaction, therefore, the forward reaction is endothermic and the backward reaction is exothermic. Thus, the reaction in which colorless N₂O₄ is produced, is an exothermic reaction.

4. The change in enthalpy of a reaction is the difference in the heat content of reactants and products. For exothermic reactions, enthalpy change is negative, whereas for endothermic reactions, enthalpy change is positive.

The decomposition of N₂O₄ to NO₂ is an endothermic reaction. Hence, the correct chemical equation is: N₂O₄ → 2NO₂; HR = +14 kcal.

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Counting atoms worksheet

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1. The solubility of lead(II) chloride at some high temperature is 3.1 x 10-2 M. Find the Ksp of PbCl2 at this temperature.

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Answer:

1) The solubility product of the lead(II) chloride is $1.2\times 10^{-4}$ .

2) The solubility of the aluminium hydroxide is $1.6\times 10^{-10} M$ .

3)The given statement is false.

Explanation:

1)

Solubility of lead chloride = $S=3.1\times 10^-2M$

$PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)$

S 2S

The solubility product of the lead(II) chloride = $K_{sp}$

$K_{sp}=[Pb^{2+}][Cl^-]^2$

$K_{sp}=S\times (2S)^2=4S^3=4\times (3.1\times 10^{-2})^3=1.2\times 10^{-4}$

The solubility product of the lead(II) chloride is $1.2\times 10^{-4}$ .

2)

Concentration of aluminium nitrate = 0.000010 M

Concentration of aluminum ion = $1\timed 0.000010 M=0.000010 M$

Solubility of aluminium hydroxide in aluminum nitrate solution = $S$

$Al(OH)_3(aq)\rightleftharpoons Al^{3+}(aq)+3OH^-(aq)$

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The solubility product of the aluminium nitrate = $K_{sp}=1.0\times 10^{-33}$

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The solubility of the aluminium hydroxide is $1.6\times 10^{-10} M$ .

3.

$Molarity=\frac{Moles}{Volume (L)}$

Mass of NaCl= 3.5 mg = 0.0035 g

1 mg = 0.001 g

Moles of NaCl = $\frac{0.0035 g}{58.5 g/mol}=6.0\times 10^{-5} mol$

Volume of the solution = 0.250 L

$[NaCl]=\frac{6.0\times 10^{-5} mol}{0.250 L}=0.00024 M$

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$[Cl^-]=[NaCl]=0.00024 M$

Moles of lead (II) nitrate = n

Volume of the solution = 0.250 L

Molarity lead(II) nitrate = 0.12 M

$n=0.12 M]\times 0.250 L=0.030 mol$

1 mole of lead nitrate gives 1 mole of lead (II) ion and 2 moles of nitrate ions.

$[Pb^{2+}]=[Pb(NO_2)_3]=0.030 M$

$PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)$

Solubility of lead(II) chloride = $K_{sp}=1.2\times 10^{-4}$

Ionic product of the lead chloride in solution :

$Q_i=[Pb^{2+}][Cl^-]^2=0.030 M\times (0.00024 M)^2=1.7\times 10^{-9}$

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