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3 years ago

Explain why anhydrous aluminium chloride is fairly soluble in organic solvent while anhydrous magnesium chloride is insoluble

1 answer:

8 0

Aluminium chloride is covalent hence soluble in organic solvent while magnesium chloride being ionic is insoluble in organic solvent

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In which of the following areas is understanding chemistry NOT necessary?

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The answer is Human behavior.

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3 years ago

Hydrogenated fats are produced by passing hydrogen gas over long saturated carbon chains. A

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4 years ago

If a chlor-alkali cell used a current of 3X10⁴A, how many pounds of Cl₂ would be produced in a typical 8-h operating day?

vova2212 [387]

7 × 10² pounds of Cl₂ would be produced in a typical 8-h operating day.

<h3>What is Stoichiometry ?</h3>

Stoichiometry helps us use the balanced chemical equation to measures quantitative relationships and it is to calculate the amount of products and reactants that are given in a reaction.

<h3>What is Balanced chemical equation ?</h3>

The balanced chemical equation is the equation in which the number of atoms on the reactants side is equal to the number of atoms on the product side in an equation.

2Cl⁻ (aq) → Cl (g) + 2e⁻

According to stoichiometry, moles of Cl₂

$= (3 \times 10^4\ A) \left (\frac{\frac{C}{s}}{A} \right ) \left (\frac{3600\ s}{h} \right ) (8h) \left (\frac{1\ \text{mol}\ e^{-}}{9.65 \times 10^4\ C} \right ) \left (\frac{1\ \text{mol}\ Cl_2}{2\ \text{mol}\ e^-} \right )$

= 4477 moles

Pounds Cl₂

$= (4477\ \text{moles}\ Cl_2) \left (\frac{70.90\ g\ Cl_2}{\text{mol}} \right ) \left (\frac{1\ kg}{1000\ g} \right ) \left (\frac{2.205\ lb}{1\ kg} \right )$

= 7 × 10² lb

Thus from the above conclusion we can say that 7 × 10² pounds of Cl₂ would be produced in a typical 8-h operating day.

Learn more about Stoichiometry here: brainly.com/question/14935523

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5 0

2 years ago

Match each chemical reaction with the type of reaction that best describes it.

r-ruslan [8.4K]

2 Al + 6 HCl → 2 AlCl₃ + 3 H₂ (single displacement)

Ca + Br₂ → CaBr₂ (synthesis)

4 NH₃ + 5 O₂ → 4 NO + 6 H₂O (combustion)

2 NaCl → 2 Na + Cl₂ (decomposition)

FeS + 2 HCl → FeCl₂ + H₂S (double displacement)

single displacement - is a chemical reaction of the following type: A + BC → AC + B

double displacement - is a chemical reaction of the following type: AB + CD → AC + BD

synthesis - the chemical product is obtained by combining in a synthesis the constituent elements

combustion - usually a exothermic reaction of a particular compound with oxygen

decomposition - degradation of a compound in simpler elements

8 0

3 years ago

A volume of 100 mL of 1.00 M HCl solution is titrated with 1.00 M NaOH solution. You added the following quantities of 1.00 M Na

s344n2d4d5 [400]

Answer:

a: before equivalence point

b: equivalence point

c: before equivalence point

d: after the eqivalence point

e: before equivalence point

f: after the eqivalence point

Explanation:

Balanced equation of reaction:

NaOH +HCl =NaCl +H2O;

Volume of HCl is fixed and it 100ml and concentration is 1.0M

N1 and N2 normality of HCl and NaOH respectively;

V1 and V2 volume of HCl and NaOH respectively;

we have given molarity but we need normality;

$Normality=molarity \times n-factor$

<em>but in case of NaOH and HCl n-factor is 1 for each.</em>

hence

normality=molarity;

At equivalence point: $N_1V_1=N_2V_2$

Before equivalence point : $N_1V_1>N_2V_2$

After the equivalence point: $N_1V_1$

$N_1V_1=100\times1=100$

case a: 5.00 mL of 1.00 M NaOH

$N_2V_2=5\times1=5$

$N_1V_1>N_2V_2$ hence it is before equivalence point

case b: 100mL of 1.00 M NaOH

$N_2V_2=100\times1=100$

$N_1V_1=N_2V_2$ hence it is equivalence point

case c: 10.0 mL of 1.00 M NaOH

$N_2V_2=10\times1=10$

$N_1V_1>N_2V_2$ hence it is before equivalence point

case d: 150 mL of 1.00 M NaOH

$N_2V_2=150\times1=150$

$N_1V_1$ hence it is after the eqivalence point

case e: 50.0 mL of 1.00 M NaOH

$N_2V_2=50\times1=50$

$N_1V_1>N_2V_2$ hence it is before equivalence point

case f: 200 mL of 1.00 M NaOH

$N_2V_2=200\times1=200$

$N_1V_1$ hence it is after the eqivalence point

7 0

3 years ago

Explain why anhydrous aluminium chloride is fairly soluble in organic solvent while anhydrous magnesium chloride is insoluble​

Explain why anhydrous aluminium chloride is fairly soluble in organic solvent while anhydrous magnesium chloride is insoluble