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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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Suppose of hcl and of o2 are added to an empty flask. How much will be in the flask at equilibrium?A. None. B. Some, but less th
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This is an incomplete question, here is a complete question.

Hydrogen chloride and oxygen react to form water and chlorine, like this:

4HCI(g)+O_2(g)\rightarrow 2H_2(g)+2CI_2(g)

Use this chemical equation to answer the questions in the table below.

Suppose 155. mmol of HCl and 38.8 mmol of O₂ are added to an empty flask. How much HCl will be in the flask at equilibrium?

(1) None.

(2) Some, but less than 155. mmol,

(3) 155. mmol,

(4) More than 155. mmol,

Answer : The correct option is, (1) None

Explanation :

The given balanced chemical reaction is:

4HCI(g)+O_2(g)\rightarrow 2H_2(g)+2CI_2(g)

From the balanced chemical reaction, we conclude that

As, 4 mmole of HCl react with 1 mmoles of O_2

So, 155. mmole of HCl react with \frac{155.}{4}\times 1=38.75 mmoles of O_2

That means, O_2 is excess reagent and HCl is a limiting reagent. All the moles of HCl will consume.

Hence, the correct option is, None.

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3 years ago
A chemist prepares a solution of barium chloride by measuring out of barium chloride into a volumetric flask and filling the fla
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Answer:

30 μmol/L

Explanation:

<em>A chemist prepares a solution of barium chloride by measuring out 8.9 μmol of barium chloride into a 300mL volumetric flask and filling the flask to the mark with water. </em>

<em> Calculate the concentration in μmol L of the chemist's barium chloride solution. Round your answer to 2 significant digits.</em>

<em />

The chemist has 8.9 μmol of solute (barium chloride) and he adds water until the mark of 300 mL in the container, which is the volume of the solution. We will need the conversion factor 1 L = 1000 mL. The concentration of barium chloride in μmol/L is:

\frac{8.9\mu mol}{300mL} .\frac{1000mL}{1L} =30\mu mol/L

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Answer:

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Explanation:

Step 1: Write the balanced neutralization reaction

Al(OH)₃ + 3 HCl ⇒ AlCl₃ + 3 H₂O

Step 2: Calculate the theoretical yield of AlCl₃

According to the balanced equation, the mass ratio of Al(OH)₃ to AlCl₃ is 81.03:133.34.

28 g Al(OH)₃ × 133.34 g AlCl₃/81.03 g Al(OH)₃ = 46 g AlCl₃

Step 3: Calculate the percent yield of AlCl₃

The real yield of AlCl₃ is 44 g. We can calculate the percent yield using the following expression.

%yield = real yield / theoretical yield × 100%

%yield = 44 g / 46 g × 100% = 96%

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