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3 years ago

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An unknown compound contains only C, H, and O. Combustion of 6.10 g of this compound produced 14.9 g of CO₂ and 6.10 g of H₂O. W

hat is the empirical formula of the unknown compound?

1 answer:

Tanzania [10]3 years ago

4 0

Answer:

The answer to your question is C₃H₃O

Explanation:

Data

Combustion of a compound C, H, O

mass = 6.10 g

mass CO2 = 14.9 g

mass of water = 6.10 g

Reaction

Cx Hy Oz + O2 ⇒ CO2 + H2O

Process

1.- Calculate the moles of carbon

44 g of CO2 -------------- 12 g of carbon

14.9 g of CO2 ------------- x

x = (14.9 x 12) / 44

x = 4.06 g

12 g of C ------------------ 1 mol

4.06 g ------------------- x

x = (4.06 x 1) / 12

x = 0.34 moles

2.- Calculate the moles of hydrogen

18 g of water ------------- 1 g of hydrogen

6.10 g of water ---------- x

x = (6.10 x 1) / 18

x = 0.33 g

1 g of H ---------------- 1 mol of H

0.33 g ---------------- x

x = (0.33 x 1) / 1

x = 0.33 moles of H

3.- Calculate the mass of oxygen

mass of Oxygen = 6.10 - 4.06 - 0.33

= 1.71 g

16 g of O --------------- 1 mol of O

1.71 g of O ------------- x

x = (1.71 x 1) / 16

x = 0.11 moles

4.- Divide by the lowest number of moles

Carbon = 0.34 / 0.11 = 3

Hydrogen = 0.33 / 0.11 = 3

Oxygen = 0.11 /0.11 = 1

5.- Write the empirical formula

C₃H₃O

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Complete and balance the following redox equation using the set of smallest whole– number coefficients. Now sum the coefficients

Elena L [17]

Answer : The balanced chemical equation in a acidic solution is,

$BrO_3^-(aq)+6H^+(aq)+3Sb^{3+}(aq)\rightarrow Br^-(aq)+3H_2O(l)+3Sb^{5+}(aq)$

The sum of the coefficients is, 17

Explanation :

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

Rules for the balanced chemical equation in acidic solution are :

First we have to write into the two half-reactions.

Now balance the main atoms in the reaction.

Now balance the hydrogen and oxygen atoms on both the sides of the reaction.

If the oxygen atoms are not balanced on both the sides then adding water molecules at that side where the less number of oxygen are present.

If the hydrogen atoms are not balanced on both the sides then adding hydrogen ion $(H^+)$ at that side where the less number of hydrogen are present.

Now balance the charge.

The given chemical reaction is,

$BrO_3^-(aq)+Sb^{3+}(aq)\rightarrow Br^-(aq)+Sb^{5+}(aq)$

The oxidation-reduction half reaction will be :

Oxidation : $Sb^{3+}\rightarrow Sb^{5+}$

Reduction : $BrO_3^-\rightarrow Br^-$

First balance the main element in the reaction.

Oxidation : $Sb^{3+}\rightarrow Sb^{5+}$

Reduction : $BrO_3^-\rightarrow Br^-$

Now balance oxygen atom on both side.

Oxidation : $Sb^{3+}\rightarrow Sb^{5+}$

Reduction : $BrO_3^-\rightarrow Br^-+3H_2O$

Now balance hydrogen atom on both side.

Oxidation : $Sb^{3+}\rightarrow Sb^{5+}$

Reduction : $BrO_3^-+6H^+\rightarrow Br^-+3H_2O$

Now balance the charge.

Oxidation : $Sb^{3+}\rightarrow Sb^{5+}+2e^-$

Reduction : $BrO_3^-+6H^++6e^-\rightarrow Br^-+3H_2O$

The charges are not balanced. Now multiplying oxidation reaction by 3 and then adding both equation, we get the balanced redox reaction.

Oxidation : $3Sb^{3+}\rightarrow 3Sb^{5+}+6e^-$

Reduction : $BrO_3^-+6H^++6e^-\rightarrow Br^-+3H_2O$

The balanced chemical equation in acidic medium will be,

$BrO_3^-(aq)+6H^+(aq)+3Sb^{3+}(aq)\rightarrow Br^-(aq)+3H_2O(l)+3Sb^{5+}(aq)$

The sum of the coefficients = 1 + 6 + 3 + 1 + 3 + 3

The sum of the coefficients = 17

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Explanation : Given,

Vapor presume of 1‑Propanol $(P^o_1)$ = 20.9 torr

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Mole fraction of 2‑Propanol $(x_2)$ = 1-0.540 = 0.46

First we have to calculate the partial pressure of 1‑Propanol and 2‑Propanol.

$p_1=x_1\times p^o_1$

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$p_1$ = partial vapor pressure of 1‑Propanol

$p^o_1$ = vapor pressure of pure substance 1‑Propanol

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and,

$p_2=x_2\times p^o_2$

where,

$p_2$ = partial vapor pressure of 2‑Propanol

$p^o_2$ = vapor pressure of pure substance 2‑Propanol

$x_2$ = mole fraction of 2‑Propanol

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Now we have to calculate the mole fraction of vapor phase 1‑Propanol and 2‑Propanol.

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Thus, the mole fraction of vapor phase 1‑Propanol and 2‑Propanol is, 0.352 and 0.648 respectively.

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A typical example is our answer option, the mass of potassium required could be used to determine the mass of potassium chloride produced after a balanced reaction equation is written.

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