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3 years ago

It is logical to conclude that water cannot dissolve vegetable oil because the oil is

Chemistry

2 answers:

marin [14]3 years ago

6 0

Answer:

The oil is insoluble with water.

Explanation:

The water and oil do not mix, they are illustrated as immiscible. The molecules of water are polar, that is, they exhibit a small positive charge at one terminal and a small negative charge at the other terminal, and they attach with each other. The molecules of oil are non-polar, and they possess no charge. Due to this, the molecules of oil are more fascinated with each other than to the molecules of water, and the molecules of water are more fascinated towards each other than to the molecules of oil.

Gekata [30.6K]3 years ago

3 0

In chemistry, there is a common note that says, "Like dissolves like".

This pertains to the concept that polar substances can dissolve only other polar substances. Also, nonpolar substances are also only able to dissolve nonpolar substances.

Polarity of the substance depends primarily on the type of bond and the difference in electronegativity.

Water is a polar substance while vegetable oil is not. From the concept presented above, it may be concluded that water will not be able to dissolve the vegetable oil and the assumption is logical.

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Answer:

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2 years ago

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Which of the following explains why the lab safety policy specifically addresses loose clothing, contact lenses, and large, loos

valentina_108 [34]

Answer:

b. Loose clothing and jewelry is more likely to impede movement, or accidentally knock materials over, causing a spill or other accident.

Explanation:

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3 years ago

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Jobisdone [24]

Answer:

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Explanation:

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3 years ago

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2 years ago

If the concentration of acetic acid is 0.10 M, what is the pH of the resulting solution?

Arisa [49]

Answer:

Option D) 2.89.

Explanation:

Look up the acid dissociation constant of acetic acid:

$K_{\rm a} \approx \rm 1.75\times 10^{-5}$

(CRC Handbook of Chemistry and Physics, 84th Edition (2004).)

Acetic acid partially dissociate to produce acetate ions and hydrogen ions:

$\rm CH_3COOH \rightleftharpoons CH_3COO^{-} + H^{+}$ .

Let the final concentration of $\rm H^{+}$ in the solution be $x\; \rm M$ . The concentration of acetic acid would have dropped by $x\; \rm M$ and the concentration of acetate ions would have increased by $x\; \rm M$ . The initial concentration of $\rm H^{+}$ in pure water is $1\times 10^{-7}\;\rm M$ and will barely influence the outcome.

Construct a RICE table for this reaction: (all values here are in $M$ , which stands for concentrations in moles per liter.)

$\begin{array}{c|ccccc}\textbf{R}& \mathrm{CH_3COOH} & \rightleftharpoons & \mathrm{CH_3COO^{-}} & + & \mathrm{H^{+}}\\\textbf{I} & 0.10\\ \textbf{C} & -x & & +x & & +x \\\textbf{E} & 0.10 - x & & x & & x \end{array}$ .

At equilibrium:

$[\mathrm{CH_3COOH}] = (0.10 - x) \; \rm M$ ;
$[\mathrm{CH_3COO^{-}}] = x\; \rm M$ ;
$[\mathrm{H^{+}}] = x\; \rm M$ .

By the definition of the acid dissociation constant, $K_{\rm a}$ :

$\displaystyle K_{\rm a}(\mathrm{CH_3COOH}) = \frac{[\mathrm{CH_3COO^{-}}]\cdot [\mathrm{H^{+}}]}{[\mathrm{CH_3COOH}]}$ .

That is:

$\displaystyle \frac{x^{2}}{0.10 - x} = \rm 1.75\times 10^{-5}$ .

Rearrange and solve for $x$ :

$x^{2} + 1.75\times 10^{-5} ~x - 1.75 \times 10^{-6} = 0$ .

$\displaystyle x = \frac{-1.75\times 10^{-5} \pm \sqrt{{\left(1.75\times 10^{-5}\right)^{2}}- 4\times \left(-1.75\times 10^{-6}\right)}}{2}$ .

There might be more than one solution to this equation. However, keep in mind that all concentration should be positive (at least non-negative.) The only possible value of $x$ will thus be approximately $0.00131$ .

In other words, at equilibrium $[\mathrm{H^{+}}] \approx 0.00131 \; \rm M$ . By the definition of pH,

$\begin{aligned} \rm pH &= -\log_{10}{[\mathrm{H^{+}]}\\&= - \log_{10}{0.00131} \\&\approx 2.9\end{aligned}$ .

Note that depending on the $K_{\rm a}$ value, the final result might slightly vary.

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3 years ago