A 3.4 g sample of an unknown monoprotic organic acid composed of C,H, and O is burned in air to produce 8.58 grams of carbon dioxide and 1.50 grams of water vapor. It was also determined that 279 mL of 0.1 M KOH was needed to titrate the acid completely.

Answers

Answer 1

Answer:

[tex]C_7H_6O_2[/tex]

Explanation:

Hello there!

In this case, we can divide the problem in three stages: (1) determine the empirical formula with the combustion analysis, (2) compute the molar mass of acid via the moles of the acid in the neutralization and (3) determine the molecular formula.

(1) In this case, since 8.58 g of carbon dioxide are released, we can first compute the moles of carbon in the compound:

[tex]n_C=8.58gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molC}{1molCO_2}=0.195molC[/tex]

And the moles of hydrogen due to the produced 1.50 grams of water:

[tex]n_H=1.50gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{2molH}{1molH_2O} =0.166molH[/tex]

Next, to compute the mass and moles of oxygen, we need to use the initial 3.4 g of the acid:

[tex]m_O=3.4g-0.195molC*\frac{12.01gC}{1molC}-0.166molH*\frac{1.01gH}{1molH} =0.89gO\\\\n_O=0.89gO*\frac{1molO}{16.0gO}=0.0556molO[/tex]

Thus, the subscripts in the empirical formula are:

[tex]C=\frac{0.195}{0.0556}=3.5 \\\\H=\frac{0.166}{0.0556}=3\\\\O=\frac{0.0556}{0.0556}=1\\\\C_7H_6O_2[/tex]

As they cannot be fractions.

(2) In this case, since the acid is monoprotic, we can compute the moles by multiplying the concentration and volume of KOH:

[tex]n_{KOH}=0.279L*0.1mol/L\\\\n_{KOH}=0.0279mol[/tex]

Which are equal to the moles of the acid:

[tex]n_{acid}=0.0279mol[/tex]

And the molar mass:

[tex]MM_{acid}=\frac{3.4g}{0.0279mol} =121.86g/mol[/tex]

(3) Finally, since the molar mass of the empirical formula is:

7*12.01 + 6*1.01 + 2*16.00 = 122.13 g/mol

Thus, since the ratio of molar masses is 122.86/122.13 = 1, we infer that the empirical formula equals the molecular one:

[tex]C_7H_6O_2[/tex]

Best regards!


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Answer:

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Answer:

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Explanation:

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Answer:

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Answer:

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What is the percent composition of water found in tin (IV) oxide octahydrate?

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Answer:

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Explanation:

Answer:

Look below

Explanation:

The following may be substituted cup for cup for vegetable oil in baked goods:

Applesauce, preferably unsweetened.

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Ghee.

Margarine, melted.

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Answers

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Explanation:

or something like that

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Answers

Answer:

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Explanation:

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Answer:

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Explanation:

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Answer:

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Answers

Answer:

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Answers

A: BE has more ionization energy than LI

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This question simply deals with ionization energy trends across the periodic table or down the group.

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Read more at: brainly.in/question/13610645

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Answers

Answer:

UF6

Explanation:

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Answers

Explanation:

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Please mark me has brainliest

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Answers

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Answers

Answer:

[tex]k=25.9M^{-2}s^{-1}[/tex]

Explanation:

Hello there!

In this case, since rate laws are written by considering the rate constant, concentration and specific orders of reaction of the concentration of the species contributing to the rate of reaction, we write:

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Best regards!

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Answers

Answer:

Pure substances are substances that are made up of only one kind of particles and has a fixed or constant structure. Pure substances are further classified as elements and compounds. An element is a substance that consists of only one type or kind of atom

Answer:

define the term pure substance​ :

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Answer:

C

Explanation:

 i hope i helped you and have a nice day...

__MgCl2 +
_Na3PO4 -->
NaCl
_Mg3(PO4)2 +
---
In the reaction above, how many moles of sodium chloride would be produced
if you started with 20 grams of magnesium chloride and excess sodium
phosphate? The equation is not balanced.
(If anyone can answer in depth if appreciate it )

Answers

Answer:

24.5 g of NaCl

Explanation:

We begin from the balanced reaction:

3MgCl₂ +   2Na₃PO₄  →    6NaCl   +  Mg₃(PO₄)₂

If the sodium phosphate is in excess, then the limting reagent is the magnessium chloride.

We convert mass to moles:

20 g . 1mol / 95.2g = 0.210 moles.

3 moles of MgCl₂ can produce 6 moles of NaCl

0.210 moles of salt, may produce (0.210 . 6) /3 = 0.420 moles

Ratio of reactant is twice the product

We convert the moles to mass:

0.420 mol .  58.45 g/mol = 24.5 g

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1.T

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