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Readme [11.4K]
2 years ago
14

A compound contains only carbon, hydrogen, and oxygen. Combustion of 10.68 mg of the compound yields 16.01 mg and 4.37 mg . The

molar mass of the compound is 176.1 g/mol. What are the empirical and molecular formulas of the compound
Chemistry
1 answer:
Verizon [17]2 years ago
5 0

Answer: See below

Explanation:

n of CO2​ = 0.364mmol

Mass of C = 0.364*12 = 4.368 mg

n of H2​O = 184.37​ = 0.243 mol

The compound has 2*0.243mmol of H

Mass of H = 0.486 mg

Mass of O = 10.68 − (4.368+0.486) = 5.826mg

Moles of O = 0.364

<u>C:H:O Ratios</u>

0.364 : 0.486 : 0.364

= 1 : 1.34 : 1

= 3 : 4 : 3

<u>So the empirical formula is C3​H4​O3​,</u>

Empirical formula mass

= 88= 2 × Molar mass

<u>And the molecular formula is C6​H8​O6</u>

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Explanation: Hope this helps!

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A 3.452 g sample containing an unknown amount of a Ce(IV) salt is dissolved in 250.0-mL of 1 M H2SO4. A 25.00 mL aliquot is anal
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Answer:

1,812 wt%

Explanation:

The reactions for this titration are:

2Ce⁴⁺ + 3I⁻ → 2Ce³⁺ + I₃⁻

I₃⁻ + 2S₂O₃⁻ → 3I⁻ + S₄O₆²⁻

The moles in the end point of S₂O₃⁻ are:

0,01302L×0,03428M Na₂S₂O₃ = 4,463x10⁻⁴ moles of S₂O₃⁻. As 2 moles of S₂O₃⁻ react with 1 mole of I₃⁻, the moles of I₃⁻ are:

4,463x10⁻⁴ moles of S₂O₃⁻×\frac{1molI_{3}^-}{2molS_{2}O_{3}^-} = 2,2315x10⁻⁴ moles of I₃⁻

As 2 moles of Ce⁴⁺ produce 1 mole of I₃⁻, the moles of Ce⁴⁺ are:

2,2315x10⁻⁴ moles of I₃⁻×\frac{2molCe^{4+}}{1molI_{3}^-} = 4,463x10⁻⁴ moles of Ce(IV). These moles are:

4,463x10⁻⁴ moles of Ce(IV)×\frac{140,116g}{1mol} = <em>0,0625 g of Ce(IV)</em>

As the sample has a 3,452g, the weight percent is:

0,0625g of Ce(IV) / 3,452g × 100 = <em>1,812 wt%</em>

I hope it helps!

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Answer:

6.48 L

Explanation:

From the question,

Applying

PV/T = P'V'/T'......................... Equation 1

P = initial pressure of the helium balloon, V =  Initial volume of the balloon, T = Initial temperature of the balloon, P' = Final pressure of the balloon, T' = Final temperature of the balloon, V' = Final volume of the balloon.

make V' the subject of the equation

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Given: P = 1 atm, V = 4.5 L, T' = 253 K, T= 293 K, P' = 0.6 atm

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