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Nataly [62]
2 years ago
15

In some reactions, Be behaves like a typical alkaline earth metal; in others, it does not. Complete and balance the following: (

b)BeCl₂(I) + Cl⁻ (l; from moltenNaCl →In which reaction does Be behave like the other Group 2A(2) members?
Chemistry
1 answer:
mina [271]2 years ago
4 0

The correct answer is BeCl_2(l)+2Cl^-(solvated)→BeCl_4^2-.

Evaluating be behavior to see :

how it differs from the other Group 2A (2) members.

In this reaction Be behaves like other alkaline earth metals

The complete equation can be given as

BeCl_2(l)+2Cl^-(solvated)→BeCl_4^2-

BeCl_2 tends to form a chloro bridged dimer in the vapour state, however at high temperatures of the order of 1200K, this dimer dissociates into the linear monomer.

BeCl_2 has a chain structure in its solid form. Each Be atom in this structure is surrounded by chlorine atoms, two of which are connected by conversion bonds and the remaining two by covalent coordinate connections. This chain structure is displayed.

To know more about BeCl₂(I) + Cl⁻ refer the link:

brainly.com/question/5017059

#SPJ4

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Explanation:

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Using the Gibbs free energy equation to solve this question.

∆G = ∆H - T∆S

Where;

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The question states that it is an exothermic reaction with negative entropy. This means that the change in both enthalpy and entropy will be negative. That is;

∆H = >0 ( it's positive)

∆S = < 0 (negative)

Let's remember that an exothermic reaction generally releases energy to it surroundings. This energy is usually released in the form of heat. Therefore, the change in enthalpy H of an exothermic reaction will always be negative. A negative change in entropy S indicates that there is a decrease in disorder, with respect to the reaction.

Using Gibb’s free energy equation at constant temperature and pressure, we have;

∆G = ∆H - T∆S

Now, the change in enthalpy and change in entropy can be written as follows;

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Substitute these values in the above equation;

∆G = ∆H - T - (∆S)

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Change in ∆G will be negative <0 when the value ∆H is greater than T∆S.

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<u>Answer:</u> The molecular formula for the given organic compound is C_2H_{32}O_4

<u>Explanation:</u>

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

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We are given:

Mass of CO_2=0.784g

Mass of H_2O=1.92g

We know that:

Molar mass of carbon dioxide = 44 g/mol

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<u>For calculating the mass of carbon:</u>

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 0.784 g of carbon dioxide, \frac{12}{44}\times 0.784=0.214g of carbon will be contained.

<u>For calculating the mass of hydrogen:</u>

In 18 g of water, 2 g of hydrogen is contained.

So, in 1.92 g of water, \frac{2}{18}\times 1.92=0.213g of hydrogen will be contained.

Mass of oxygen in the compound = (0.844) - (0.214 + 0.213) = 0.417 g

To formulate the empirical formula, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon = \frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.214g}{12g/mole}=0.0134moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.213g}{1g/mole}=0.213moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.417g}{16g/mole}=0.0261moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0134 moles.

For Carbon = \frac{0.0134}{0.0134}=1

For Hydrogen = \frac{0.213}{0.0134}=15.89\approx 16

For Oxygen = \frac{0.0261}{0.0134}=1.95\approx 2

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1 : 16 : 2

The empirical formula for the given compound is CH_{16}O_2

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

n=\frac{\text{Molecular mass}}{\text{Empirical mass}}

We are given:

Mass of molecular formula = 116.2 g/mol

Mass of empirical formula = 60 g/mol

Putting values in above equation, we get:

n=\frac{116.2g/mol}{60g/mol}=1.94\approx 2

Multiplying this valency by the subscript of every element of empirical formula, we get:

C_{(1\times 2)}H_{(16\times 2)}O_{(2\times 2)}=C_2H_{32}O_4

Hence, the molecular formula for the given organic compound is C_2H_{32}O_4

5 0
3 years ago
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