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3 years ago

In which compound is the percent compositionby mass of chlorine equal to 42%?

2 answers:

8 0

The answer is (3) HClO3. The percent composition mass of chlorine is equals to the mass of one Cl atom /gram-formula mass of the compound. So after calculation, we can get the right answer.

UkoKoshka [18]3 years ago

3 0

Answer The correct option is, (3) $HClO_3$ (gram-formula mass = 84 g/mol)

Explanation : Given,

Molar mass of chlorine = 35.5 g/mole

Formula used :

$\% \text{ Composition by mass of chlorine}=\frac{\text{Mass of chlorine}}{\text{Mass of given chlorine compound}}\times 100$

Now we have to calculate the percent composition by mass of chlorine.

For option 1 :

$\% \text{ Composition by mass of chlorine}=\frac{35.5g/mole}{52g/mole}\times 100=68.2\%$

For option 2 :

$\% \text{ Composition by mass of chlorine}=\frac{35.5g/mole}{68g/mole}\times 100=52.2\%$

For option 3 :

$\% \text{ Composition by mass of chlorine}=\frac{35.5g/mole}{84g/mole}\times 100=42.2\%$

For option 4 :

$\% \text{ Composition by mass of chlorine}=\frac{35.5g/mole}{100g/mole}\times 100=35.5\%$

From this we conclude that the $HCLO_3$ is the compound in which the percent composition by mass of chlorine equal to 42%.

Hence, the correct option is, (3)

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Calculate the number of sodium ions, perchlorate ions, Cl atoms and O atoms in 17.8 g of sodium perchlorate. Enter your answers

Degger [83]

17.8 g of sodium perchlorate contains 8.73 × 10²² Na⁺ ions, 8.73 × 10²² ClO₄⁻ ions, 8.73 × 10²² Cl atoms and 3.49 × 10²³ O atoms.

First, we will convert 17.8 g of NaClO₄ to moles using its molar mass (122.44 g/mol).

$17.8 g \times \frac{1mol}{122.44g} = 0.145 mol$

Next, we will convert 0.145 moles to molecules of NaClO₄ using Avogadro's number; there are 6.02 × 10²³ molecules in 1 mole of molecules.

$0.145 mol \times \frac{6.02 \times 10^{23}molecules }{mol} = 8.73 \times 10^{22}molecules$

NaClO₄ is a strong electrolyte that dissociates according to the following equation.

NaClO₄ ⇒ Na⁺ + ClO₄⁻

The molar ratio of NaClO₄ to Na⁺ is 1:1. The number of Na⁺ in 8.73 × 10²² molecules of NaClO₄ is:

$8.73 \times 10^{22}moleculeNaClO_4 \times \frac{1 ion Na^{+} }{1moleculeNaClO_4} = 8.73 \times 10^{22}ion Na^{+}$

The molar ratio of NaClO₄ to ClO₄⁻ is 1:1. The number of ClO₄⁻ in 8.73 × 10²² molecules of NaClO₄ is:

$8.73 \times 10^{22}moleculeNaClO_4 \times \frac{1 ion ClO_4^{-} }{1moleculeNaClO_4} = 8.73 \times 10^{22}ion ClO_4^{-}$

The molar ratio of ClO₄⁻ to Cl is 1:1. The number of Cl in 8.73 × 10²² ions of ClO₄⁻ is:

$8.73 \times 10^{22}ion ClO_4^{-} \times \frac{1 atomCl }{1ion ClO_4^{-}} = 8.73 \times 10^{22}atom Cl$

The molar ratio of ClO₄⁻ to O is 1:1. The number of O in 8.73 × 10²² ions of ClO₄⁻ is:

$8.73 \times 10^{22}ion ClO_4^{-} \times \frac{4 atomO }{1ion ClO_4^{-}} = 3.49 \times 10^{23}atom O$

17.8 g of sodium perchlorate contains 8.73 × 10²² Na⁺ ions, 8.73 × 10²² ClO₄⁻ ions, 8.73 × 10²² Cl atoms and 3.49 × 10²³ O atoms.

You can learn more Avogadro's number here: brainly.com/question/13302703

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