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MakcuM [25]
3 years ago
10

A sample of sodium sulfite has a mass of 2.80 g.

Chemistry
1 answer:
Gnom [1K]3 years ago
7 0

<u>Answer:</u>

<u>For a:</u> The number of sodium ions in given amount of sodium sulfite are 2.65\times 10^{22}

<u>For b:</u> The number of sulfite ions in given amount of sodium sulfite are 1.325\times 10^{22}

<u>For c:</u> The mass of one formula unit of sodium ions is 2.09\times 10^{-22} grams

<u>Explanation:</u>

The chemical formula of sodium sulfite is  Na_2SO_3. It is formed by the combination of 2 sodium (Na^+) ions and 1 sulfite (SO_3^{2-}) ions

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass of sodium sulfite = 2.80 g

Molar mass of sodium sulfite = 126 g/mol

Putting values in above equation, we get:

\text{Moles of sodium sulfite}=\frac{2.80g}{126g/mol}=0.022mol

  • <u>For a:</u>

Moles of sodium ions in sodium sulfite = (2 × 0.022) moles

According to mole concept:

1 mole of a compound contains 6.022\times 10^{23} number of particles

So, 0.022 moles of sodium sulfite will contain = (2\times 0.022\times 6.022\times 10^{23})=2.65\times 10^{22} number of sodium ions

Hence, the number of sodium ions in given amount of sodium sulfite are 2.65\times 10^{22}

  • <u>For b:</u>

Moles of sulfite ions in sodium sulfite = (1 × 0.022) moles

According to mole concept:

1 mole of a compound contains 6.022\times 10^{23} number of particles

So, 0.022 moles of sodium sulfite will contain = (1\times 0.022\times 6.022\times 10^{23})=1.325\times 10^{22} number of sulfite ions

Hence, the number of sulfite ions in given amount of sodium sulfite are 1.325\times 10^{22}

  • <u>For c:</u>

Molar mass of sodium sulfite = 126 g/mol

According to mole concept:

6.022\times 10^{23} number of formula units are present in 1 mole of a compound

Or, 6.022\times 10^{23} number of formula units of sodium sulfite have a mass of 126 grams

So, 1 formula unit of sodium sulfite will have a mass of = \frac{126}{6.022\times 10^{23}}\times 1=2.09\times 10^{-22}g

Hence, the mass of one formula unit of sodium ions is 2.09\times 10^{-22} grams

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<h3>Which are diatomic molecules?</h3>

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Which is an example of ionization?
strojnjashka [21]
Answer: SO₂ + H₂O → HSO₃ ⁻ + H⁺

Justification:

1) Ionization means formation of ions.

2) Ions are species that are not neutral, they are charged, in virtue of having less or more electrons than protons.

3) Ionization may happen in different environments.

4) Ionic compunds, like Mg(OH)₂ dissociate into ions (ionize) in water. That is the example shown in the fourth option:

Mg(OH)₂ → Mg ²⁺ + 2OH⁻

5) How much a ionic compound dissociates in water (ionize) depends on the Ksp (product solubility constant) which measures the concentrations of the ions that can be in the solution.


6) The Ksp for Mg(OH)₂ is very low, meaning that it will slightly ionize.

7) SO₂ + H₂O forms H₂SO₄, which is a strong acid, meaning that it will ionize fully in water, into the ions HSO₃ ⁻ and H⁺, so the third option is a good example of ionization.

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Which of the following statements is NOT a part of John Dalton's atomic theory?
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The statement that is NOT a part of John Dalton's atomic theory Electrons move in specific orbits around the nucleus of an atom. The statements ‘All elements are composed of atoms that cannot be divided’, ‘All atoms of the same element are exactly alike and have the same mass’ and ‘Every compound is composed of atoms of different elements, combined in a specific ratio’ are not his theory.

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3 years ago
Calculate either [ H 3 O + ] or [ OH − ] for each of the solutions.
STALIN [3.7K]

Answer: Solution A : [H_3O^+]=0.300\times 10^{-7}M

Solution B : [OH^-]=0.107\times 10^{-5}M

Solution C : [OH^-]=0.177\times 10^{-10}M

Explanation:

pH or pOH is the measure of acidity or alkalinity of a solution.

pH is calculated by taking negative logarithm of hydrogen ion concentration and pOH is calculated by taking negative logarithm of hydroxide ion concentration.

pH=-\log[H_3O^+]

pOH=-log[OH^-}

pH+pOH=14

[H_3O^+][OH^-]=10^{-14}

a. Solution A: [OH^-]=3.33\times 10^{-7}M

[H_3O^+]=\frac{10^{-14}}{3.33\times 10^{-7}}=0.300\times 10^{-7}M

b. Solution B : [H_3O^+]=9.33\times 10^{-9}M

[OH^-]=\frac{10^{-14}}{9.33\times 10^{-9}}=0.107\times 10^{-5}M

c. Solution C : [H_3O^+]=5.65\times 10^{-4}M

[OH^-]=\frac{10^{-14}}{5.65\times 10^{-4}}=0.177\times 10^{-10}M

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3 years ago
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