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3 years ago

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If cacl2 is dissolved in water, what can be said about the concentration of the ca2+ ion? view available hint(s) if is dissolved

in water, what can be said about the concentration of the ion? it has the same concentration as the cl− ion. its concentration is half that of the cl− ion. its concentration is twice that of the cl− ion. its concentration is one-third that of the cl− ion.

2 answers:

Simora [160]3 years ago

8 0

The concentration of Ca2+ ions is half that of the Cl- ions.

Furkat [3]3 years ago

6 0

<u>Answer:</u> The correct answer is its concentration is half that of the $Cl^-$ ion.

<u>Explanation:</u>

We are given a chemical compound having formula $CaCl_2$

It is an ionic compound because it is formed by the combination of $Ca^{2+}\text{ and }Cl^-$ ion

When an ionic compound is dissolved in water, it breaks into its ions. The chemical equation for the ionization of calcium chloride follows:

$CaCl_2\rightarrow Ca^{2+}+2Cl^-$

By stoichiometry of the reaction:

1 mole of calcium chloride produces 1 mole of calcium ions and 2 moles of chloride ions

From above, it is visible that number of chloride ions is twice that of calcium ion and so will be its concentration.

Thus, the concentration of calcium ions will be half that of chloride ions.

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Given:
<span> 2.1 moles of chlorine gas (Cl2) at standard temperature and pressure (STP)
Required:
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slega [8]

<u>Answer:</u> The mass percentage of carbon element is 63.57 %, hydrogen element is 5.96 %, nitrogen element is 9.28 % and oxygen element is 21.19 %

<u>Explanation:</u>

We are given:

A chemical compound having chemical formula of $C_8H_9NO_2$ . This compound is made by the combination of carbon, hydrogen, nitrogen and oxygen elements.

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To calculate the percentage composition of an element in a compound, we use the equation:

$\%\text{ composition of an element}=\frac{\text{Mass of element}}{\text{Mass of compound}}\times 100$ ........(1)

<u>For carbon element:</u>

Putting values in equation 1, we get:

$\% \text{ composition of carbon element}=\frac{96}{151}\times 100\\\\\% \text{ composition of carbon element}=63.57\%$

<u>For hydrogen element:</u>

Putting values in equation 1, we get:

$\% \text{ composition of hydrogen element}=\frac{9}{151}\times 100\\\\\% \text{ composition of hydrogen element}=5.96\%$

<u>For nitrogen element:</u>

Putting values in equation 1, we get:

$\% \text{ composition of nitrogen element}=\frac{14}{151}\times 100\\\\\% \text{ composition of nitrogen element}=9.28\%$

<u>For oxygen element:</u>

Putting values in equation 1, we get:

$\% \text{ composition of oxygen element}=\frac{32}{151}\times 100\\\\\% \text{ composition of oxygen element}=21.19\%$

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