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djyliett [7]
4 years ago
5

Naphthalene is soluble in diethyl ether, but it is insoluble in water regardless of the solution pH. Explain why this compound c

annot be readily ionized in aqueous solution.
Chemistry
1 answer:
Mademuasel [1]4 years ago
5 0

Answer:

Naphthalene is a non polar substance

Explanation:

In chemistry, the principle of like dissolves like is the over aching principle that controls the dissolution of one substance in another.

Naphthalene dissolves in diethyl ether because diethyl ether is a non polar solvent just as naphthalene is a nonpolar substance.

Since water is a polar solvent, it can not dissolves naphthalene at any pH because naphthalene is a nonpolar substance.

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¿Cuál es el % m/m de una disolución en que hay disueltos 22 g de soluto en 44 g de disolvente?
Archy [21]

The question is as follows: What is the% m / m of a solution in which 22 g of solute are dissolved in 44 g of solvent?

Answer: The% m/m of a solution in which 22 g of solute are dissolved in 44 g of solvent is 50%.

Explanation:

Given: Mass of solute = 22 g

Mass of solvent = 44 g

The percentage m/m is calculated using the following formula.

Mass percentage = \frac{mass of solute}{mass of solvent} \times 100

Substitute the values into above formula as follows.

Mass percentage = \frac{mass of solute}{mass of solvent} \times 100\\= \frac{22 g}{44 g} \times 100\\= 50 percent

Thus, we can conclude that the% m/m of a solution in which 22 g of solute are dissolved in 44 g of solvent is 50%.

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Using the equations 2 Sr(s) + O₂ (g) → 2 SrO (s) ∆H° = -1184 kJ/mol SrO (s) + CO₂ (g) → SrCO₃ (s) ∆H° = -234 kJ/mol CO₂ (g) → C(
kkurt [141]

<u>Answer:</u> The \Delta H^o_{rxn} for the reaction is 72 kJ.

<u>Explanation:</u>

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction follows:

2SrCO_3(s)\rightarrow 2Sr(s)+2C(s)+3O_2(g)      \Delta H^o_{rxn}=?

The intermediate balanced chemical reaction are:

(1) 2Sr(s)+O_2(g)\rightarrow 2SrO(s)    \Delta H_1=-1184kJ

(2) SrO(s)+CO_2(g)\rightarrow SrCO_3(s)     \Delta H_2=-234kJ      ( × 2)

(3) CO_2(g)\rightarrow C(s)+O_2(g)     \Delta H_3=394kJ    ( × 2)

The expression for enthalpy of the reaction follows:

\Delta H^o_{rxn}=[1\times (\Delta H_1)]+[2\times (-\Delta H_2)]+[2\times (\Delta H_3)]

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(1\times (-1184))+(2\times -(-234))+(2\times (394))]=72kJ

Hence, the \Delta H^o_{rxn} for the reaction is 72 kJ.

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3 years ago
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