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Fynjy0 [20]
2 years ago
12

14.Sorbitol is used in sugar free gums as a sweetener. After analysis, a sample of

Chemistry
1 answer:
evablogger [386]2 years ago
3 0
<h3>Answer:</h3>

C6H12O6

Explanation:

To find empirical formular you write out the element

Carbon Hydrogen Oxygen

3.96 769 5.28

then u althrough divide by their atomic numbers

3.96÷ 12 769÷1 5.28÷16

=0.33 =769 =0.33

also divide althrough by the lowest number

the lowest number is 0.33 so we use it to divide

0.33÷0.33. 769÷0.33. 0.33÷0.33

= 1 = 2 = 1

empirical formular= CH2O

To find the Molecular formular

Molecular number = molar mass

(CH2O)n = 182

(12+1×2+16)n = 182

30n = 182

n = 6

therefore the Molecular formular is C6H12O6

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5.943x10^24 molecules of H3PO4 will need how many grams of Mg(OH)2 in the reaction below? 3 Mg(OH)2 + 2 H3PO4 -------&gt; 1 Mg3(
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Answer:

Mass of Mg(OH)₂ required for the reaction = 863.13 g

Explanation:

3Mg(OH)₂ + 2H₃PO₄ -------> Mg₃(PO₄)₂ + 6H₂O

(5.943 x 10²⁴) molecules of H₃PO₄ is available fore reaction. Mass of Mg(OH)₂ required for reaction.

According to Avogadro's theory, 1 mole of all substances contain (6.022 × 10²³) molecules.

This can allow us find the number of moles that (5.943 x 10²⁴) molecules of H₃PO₄ represents.

1 mole = (6.022 × 10²³) molecules.

x mole = (5.943 x 10²⁴) molecules

x = (5.943 x 10²⁴) ÷ (6.022 × 10²³)

x = 9.87 moles

From the stoichiometric balance of the reaction,

2 moles of H₃PO₄ reacts with 3 moles of Mg(OH)₂

9.87 moles of H₃PO₄ will react with y moles of Mg(OH)₂

y = (3×9.87)/2 = 14.80 moles

So, 14.8 moles of Mg(OH)₂ is required for this reaction. We them convert this to mass

Mass = (number of moles) × Molar mass

Molar mass of Mg(OH)₂ = 58.3197 g/mol

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= 14.8 × 58.3197 = 863.13 g

Hope this Helps!!!

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3 years ago
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<h3>1</h3>

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<h3>2</h3>

A double replacement reaction takes place only if it reduces in the concentration of ions in the solution. For example, the reaction between Ca(NO₃)₂ and KOH produces Ca(OH)₂. Ca(OH)₂ barely dissolves. The reaction has removed Ca²⁺ and OH⁻ ions from the solution.

Some of the reactions lead to neither precipitates nor gases. They will not take place since they are not energetically favored.


<h3>3</h3>

Compare the first and last row:

Both Ca(NO₃)₂ and Zn(NO₃)₂ react with KOH. However, between the two precipitates formed, Ca(OH)₂ is more soluble than Zn(OH)₂.

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<h3>4</h3>

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