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MatroZZZ [7]
1 year ago
14

I2 (s) has a very low solubility in H2O (l) because the weak _____ forces between I2 and H2O are not strong enough to replace th

e _____ between the H2O molecules.
Chemistry
1 answer:
rosijanka [135]1 year ago
7 0

I2 (s) has very low solubility in H2O (l) because the weak <u>dipole-induced dipole</u> forces between I2 and H2O are not strong enough to replace the <u>hydrogen bonds</u> between the H2O molecules.

Dipole interactions are susceptible interactions that get up from the close association of permanent or precipitated dipoles. Collectively these forces are called Van der Waals interactions. Proteins incorporate a massive number of these interactions, which vary substantially in energy.

Solubility depends on the solvent's capacity to overcome the intermolecular forces in a solid. Red blood cells have an excessive concentration of dissolved ions. When positioned into pure water they rupture.

Polar species are soluble in water, whilst nonpolar species are soluble in oils and fats. Covalent solubility uses the like dissolves like a rule. This means that materials with an equal form of polarity might be soluble in one another. moreover, compounds with differing polarities might be insoluble in one another.

Learn more about solubility here brainly.com/question/23946616

#SPJ4

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An enzyme is discovered that catalyzes the chemical reaction SAD ↔ HAPPY A team of motivated researchers sets out to study the e
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Answer:

The K_{m} of a substrate will be "10 μM".

Explanation:

The given values are:

E_{t} = 20 \ nM

[Substract] = 40 \ \mu M

K_{cat}=600 \ s^{-1}

Reaction velocity, Vo=9.6 \ \mu M s^{-1}

As we know,

⇒  Vo=\frac{K_{cat}[E_{t}][S]}{K_{m}+[S]}

On putting the estimated values, we get

⇒  9.6=\frac{600\times 20\times 10^{-3}\times 40}{K_{m}+40}

⇒  K_{m}+40=\frac{600\times 20\times 10^{-3}\times 40}{9.6}

⇒  K_{m}+40=50

On subtracting "40" from both sides, we get

⇒  K_{m}+40-40=50-40

⇒  K_{m}=10 \ \mu M

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How much heat does it take to turn water from -30 degreesC to 120 degreesC?
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Answer:

110 degrees C

Explanation:

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Material A has a small latent heat of fusion. Material B has a large heat of fusion. Which of the following statements is true?
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What must be the molarity of an aqueous solution of trimethylamine, (ch3)3n, if it has a ph = 11.20? (ch3)3n+h2o⇌(ch3)3nh++oh−kb
Stolb23 [73]

0.040 mol / dm³. (2 sig. fig.)

<h3>Explanation</h3>

(\text{CH}_3)_3\text{N} in this question acts as a weak base. As seen in the equation in the question, (\text{CH}_3)_3\text{N} produces \text{OH}^{-} rather than \text{H}^{+} when it dissolves in water. The concentration of \text{OH}^{-} will likely be more useful than that of \text{H}^{+} for the calculations here.

Finding the value of [\text{OH}^{-}] from pH:

Assume that \text{pK}_w = 14,

\begin{array}{ll}\text{pOH} = \text{pK}_w - \text{pH} \\ \phantom{\text{pOH}} = 14 - 11.20 &\text{True only under room temperature where }\text{pK}_w = 14 \\\phantom{\text{pOH}}= 2.80\end{array}.

[\text{OH}^{-}] =10^{-\text{pOH}} =10^{-2.80} = 1.59\;\text{mol}\cdot\text{dm}^{-3}.

Solve for [(\text{CH}_3)_3\text{N}]_\text{initial}:

\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\text{equilibrium}} = \text{K}_b = 1.58\times 10^{-3}

Note that water isn't part of this expression.

The value of Kb is quite small. The change in (\text{CH}_3)_3\text{N} is nearly negligible once it dissolves. In other words,

[(\text{CH}_3)_3\text{N}]_\text{initial} = [(\text{CH}_3)_3\text{N}]_\text{final}.

Also, for each mole of \text{OH}^{-} produced, one mole of (\text{CH}_3)_3\text{NH}^{+} was also produced. The solution started with a small amount of either species. As a result,

[(\text{CH}_3)_3\text{NH}^{+}] = [\text{OH}^{-}] = 10^{-2.80} = 1.58\times 10^{-3}\;\text{mol}\cdot\text{dm}^{-3}.

\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\textbf{initial}} = \text{K}_b = 1.58\times 10^{-3},

[(\text{CH}_3)_3\text{N}]_\textbf{initial} =\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{\text{K}_b},

[(\text{CH}_3)_3\text{N}]_\text{initial} =\dfrac{(1.58\times10^{-3})^{2}}{6.3\times10^{-5}} = 0.040\;\text{mol}\cdot\text{dm}^{-3}.

8 0
3 years ago
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