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3 years ago

Why can salt and sugar both dissolve in water, even though one is ionic and the other is covalent?

Chemistry

2 answers:

Tema [17]3 years ago

8 0

Answer:

Explanation:

The water molecules, $H_{2} O$ , are polar, with positive partial charges on the hydrogen and a negative partial charge on the oxygen. This makes it possible to interact with ionic compounds such as salt (NaCl). This interaction occurs through the partially positive and negative ends of the water, so that the positive charges attract the negative ones. The NaCl salt in water dissociates (separates) into $Na^{+}$ and $Cl^{-}$ ions. $Na^{+}$ ions are surrounded by negative partial charges of water oxygen, while $Cl^{-}$ ions are surrounded by hydrogen ends with positive partial charge. In this way the salt is dissolved in water.

Sugar is a molecular compound formed by covalent bonds. In a polar covalent bond, electrons are shared unevenly. This creates a slightly negatively charged end and a slightly positively charged end. This is what happens with sugar. These extremes are those that interact with the positive and negative extremes of water, mentioned previously. Then it is possible to say that sugar dissolves in water because both substances are polar substances.

In short, <em>water dissolves most of the substances that are polar or ionic, as in the case of sugar and salt. </em>

polet [3.4K]3 years ago

7 0

Water is polar molecule, ionic compound dissolves in it as like dissolves like.

there is ionic interaction between a salt and water

Sugar has hydroxyl group in it. These hydroxyl group are able to form hydrogen bond with water molecule.

[Hydrogen bond : the interaction between a hydrogen, bonded with a highly electro-negative element (F,O,N) and other highly electro-negative element (F,O,N)]

Thus due to hydrogen bond sugar is soluble in water.

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3 years ago

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3 years ago

Which of the following leads to a higher rate of diffusion?

tiny-mole [99]

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<h3><u>Explanation;</u></h3>

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3 years ago

A sample of CaCO3 (molar mass 100. g) was reported as being 30. percent Ca. Assuming no calcium was present in any impurities, c

natka813 [3]

Answer:

Approximately 75%.

Explanation:

Look up the relative atomic mass of Ca on a modern periodic table:

Ca: 40.078.

There are one mole of Ca atoms in each mole of CaCO₃ formula unit.

The mass of one mole of CaCO₃ is the same as the molar mass of this compound: $\rm 100\; g$ .
The mass of one mole of Ca atoms is (numerically) the same as the relative atomic mass of this element: $\rm 40.078\; g$ .

Calculate the mass ratio of Ca in a pure sample of CaCO₃:

$\displaystyle \frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)} = \frac{40.078}{100} \approx \frac{2}{5}$ .

Let the mass of the sample be 100 g. This sample of CaCO₃ contains 30% Ca by mass. In that 100 grams of this sample, there would be $\rm 30 \% \times 100\; g = 30\; g$ of Ca atoms. Assuming that the impurity does not contain any Ca. In other words, all these Ca atoms belong to CaCO₃. Apply the ratio $\displaystyle \frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)} \approx \frac{2}{5}$ :

$\begin{aligned} m\left(\mathrm{CaCO_3}\right) &= m(\mathrm{Ca})\left/\frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)}\right. \cr &\approx 30\; \rm g \left/ \frac{2}{5}\right. \cr &= 75\; \rm g \end{aligned}$ .

In other words, by these assumptions, 100 grams of this sample would contain 75 grams of CaCO₃. The percentage mass of CaCO₃ in this sample would thus be equal to:

$\displaystyle 100\%\times \frac{m\left(\mathrm{CaCO_3}\right)}{m(\text{sample})} = \frac{75}{100} = 75\%$ .

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