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Step2247 [10]
4 years ago
10

For the following reaction, calculate how many moles of each product are formed when 0.356 moles of PbS completely react. Assume

there is an excess of oxygen.
2PbSs+3O2g→2PbOs+2SO2g
Chemistry
2 answers:
VMariaS [17]4 years ago
7 0

Answer:

0.356 moles of each products (PbO and SO2) are formed.

Explanation:

First step is to write out the equation and ensure it is balanced before proceeding.

2PbSs + 3O2g → 2PbOs + 2SO2g

2            3            2              2

The equation above is balanced because there are equal number of atoms of elemenents on both sides of the arrow.

By stating that oxygen is in excess, this tells that PbS is the limiting reactant (i.e the amount of products formed depends on it) and not on oxygen.

From the equation;

The products formed are PbO and SO2.

For PbO, it requires 2 moles of PbS to react completely in order to form 2 moles of PbO.

For SO2, it requres 2 moles of PbS to react completely in order to form 2 moles of SO2.

This is gotten from the equation.

PbO

2 moles = 2 moles

0.356 moles = x

upon cross multiplication, we are left with;

x = (0.356 * 2) / 2 = 0.356 moles

SO2

2 moles = 2 moles

0.356 moles = x

upon cross multiplication, we are left with;

x = (0.356 * 2) / 2 = 0.356 moles

galben [10]4 years ago
3 0

Answer:

When 0.356 moles of PbS completely react 0.356 moles of PbO and 0.356 moles of SO₂ are produced.

Explanation:

2PbS (s) + 3O₂ (g) → 2PbO (s) + 2SO₂ (g)

2 mol PbS ____________ 2 mol PbO

0.356 mol PbS ________         x

          x = 0.356 mol PbO

2 mol PbS ____________ 2 mol SO₂

0.356 mol PbS ________         x

          x = 0.356 mol SO₂

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Mrrafil [7]

Answer:

k = -0.006.

T₀ = 15 °C

Explanation:

Hola.

En este caso, considerando la gráfica mostrada en el archivo adjunto, podemos evidenciar que los datos dados se comportan de manera lineal, por lo que basado en la ecuación, T=k*h+To, podemos calcular la pendiente que basicamente es igual a k, tomando dos puntos en la gráfica:

k=\frac{12-13.5}{750-500}=-0.0006

Además, el valor de la temperatura inicial se puede extraer de la tabla, dado que esta es cuando la altura es 0 m, es decir 15 °C.

¡Saludos!

8 0
3 years ago
identify the reagents you would use to convert each of the following compounds into pentanoic acid: (a) 1-pentene (b) 1-bromobut
Morgarella [4.7K]

a)BH3.THF is used to convert 1-pentane to pentanoic acid and b)NaCN is used to convert Bromobutane to pentanoic acid.

a) The conversion of 1-pentane to pentanoic acid using BH3, also known as hydroboration-oxidation, is a two-step reaction involving the reaction of 1-pentane with borane (BH3), followed by oxidation of the resulting 1-pentylborane with hydrogen peroxide or other oxidizing agents.

In the first step, 1-pentane reacts with borane (BH3) to form 1-pentylborane, through a process known as hydroboration. This reaction is catalyzed by a Lewis acid, such as aluminum chloride, and proceeds via a hydride transfer from the borane to the 1-pentane.

In the second step, the 1-pentylborane is oxidized to pentanoic acid using hydrogen peroxide (H₂O₂) or other suitable oxidizing agents. The oxidation is catalyzed by an acid, such as hydrochloric acid (HCl), and proceeds via a proton transfer from the 1-pentylborane to the hydrogen peroxide. The end result is the conversion of 1-pentane to pentanoic acid.

The overall chemical reaction for the conversion of 1-pentane to pentanoic acid using borane (BH₃) and hydrogen peroxide (H₂O₂) is as follows:

1-pentane + BH₃ + H₂O₂ → pentanoic acid + H₂O + BH₂

b)The conversion of 1-Bromo butane to pentanoic acid using sodium cyanide (NaCN) proceeds via a nucleophilic substitution reaction. The reaction mechanism involves the following steps:

1. Attack of the nucleophile, NaCN, on the carbon atom of 1-Bromo butane to form a tetrahedral intermediate.

2. Loss of a proton from the tetrahedral intermediate to form a carbanion.

3. Protonation of the carbanion by water (or another proton source) to form pentanoic acid.

The overall reaction can be represented as follows:

1-Bromo butane + NaCN → Pentanoic Acid + NaBr

To know more about reagents, click below:

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3 0
2 years ago
Indicate the types of forces that are involved between the solute and solvent when forming a homogeneous solution between ch3ch2
Bess [88]
Answer:
            Dispersion Forces are found between n-Pentane (CH₃-CH₂-CH₂-CH₂-CH₃) and n-Hexane (CH₃-CH₂-CH₂-CH₂-CH₂-CH₃).

Explanation:
                   Dispersion Forces are present and developed by those compounds which are non-polar in nature. In given statement n-Pentane and n-Hexane both are non-polar in nature as the electronegativity difference between Hydrogen atoms and Carbon atoms is less than 0.4. 
                   When non-polar molecules approaches each other, a Dipole is induced in one of them, this step is known as Instantaneous Dipole, This generated Dipole on approaching another non-polar molecule induces dipole in it and the process propagates. Hence, creating intermolecular interactions.
8 0
3 years ago
Read 2 more answers
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On this reaction, 0.495 g = 0.495/78 moles =6.346 X 10⁻³ moles of Al(OH)₃.

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So, mass percentage of Al₂(SO₄)₃ is= (amount of Al₂(SO₄)₃/total amount of mixture)X100 = \frac{1.085 X 100}{1.45} =74.8 %.

7 0
3 years ago
I need an analysis and conclusion, for the lab, Types of reactions
Leya [2.2K]

Answer:

Explanation:

I am a bit confused is there instuctions on what to write on or anything like that

3 0
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