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3 years ago

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List the following molecules in order from most reducedstate to most oxidizedstate: Formaldehyde (CH2O), Methane (CH4), Carbon D

ioxide (CO2), Formic Acid (HCOOH), and Methanol (CH3OH).

1 answer:

AleksAgata [21]3 years ago

3 0

Answer:

CH₄ < CH₃OH < CH₂O < HCOOH < CO₂

Explanation:

In order to know the oxidation number of C in these compounds, we will use the rule that states that the sum of the oxidation numbers of the atoms is equal to the charge of the compound, in these cases, zero because these are neutral molecules.

H is less electronegative than C, so it acts with the oxidation number +1.
O is more electronegative than C, so it acts with the oxidation number -1. When there is a double bond, it acts with the oxidation number -2.

<em>Formaldehyde (CH₂O)</em>

C + 2.H + O = 0

C + 2.(1) + (-2) = 0

C = 0

<em>Methane (CH₄)</em>

C + 4.H = 0

C + 4.(1) = 0

C = -4

<em>Carbon Dioxide (CO₂)</em>

C + 2.(O-double bond) = 0

C + 2.(-2) = 0

C = +4

<em>Formic Acid (HCOOH)</em>

In this case, we will just consider the H atom linked to the C atom.

C + 1.(H) + 1.(O-double bond) + 1.(O-single bond) = 0

C + 1.(1) + 1.(-2) + 1.(-1) = 0

C = 2

<em>Methanol (CH₃OH)</em>

In this case, we will just consider the H atoms linked to the C atom.

C + 3.(H) + 1.(O-single bond) = 0

C + 3.(1) + 1.(-1) = 0

C = -2

The order from most reduced to most oxidized is:

CH₄ < CH₃OH < CH₂O < HCOOH < CO₂

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The formula for the compound that contains Cu2+ actions and oxygen anions is

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Answer:

. CuO

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Since oxygen also has -2 the valency will cancel out the +2 on Cu leaving CuO

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Consider the following reaction: A(g)⇌2B(g). Find the equilibrium partial pressures of A and B for each of the following differe

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Answer:

a. Kp=1.4

$P_{A}=0.2215 atm$

$P_{B}=0.556 atm$

b.Kp=2.0 * 10^-4

$P_{A}=0.495atm$

$P_{B}=0.00995 atm$

c.Kp=2.0 * 10^5

$P_{A}=5*10^{-6}atm$

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Explanation:

For the reaction

A(g)⇌2B(g)

Kp is defined as:

$Kp=\frac{(P_{B})^{2}}{P_{A}}$

The conditions in the system are:

A B

initial 0 1 atm

equilibrium x 1atm-2x

At the beginning, we don’t have any A in the system, so B starts to react to produce A until the system reaches the equilibrium producing x amount of A. From the stoichiometric relationship in the reaction we get that to produce x amount of A we need to 2x amount of B so in the equilibrium we will have 1 atm – 2x of B, as it is showed in the table.

Replacing these values in the expression for Kp we get:

$Kp=\frac{(1-2x)^{2}}{x}$

Working with this equation:

$x*Kp=(1-2x)^{2} - -> x*Kp=4x^{2}-4x+1- - >4x^{2}-(4+Kp)*x+1=0$

This last expression is quadratic expression with a=4, b=-(4+Kp) and c=1

The general expression to solve these kinds of equations is:

$x=\frac{-b(+-)*\sqrt{(b^{2}-4ac)}}{2a}$ (equation 1)

We just take the positive values from the solution since negative partial pressures don´t make physical sense.

Kp = 1.4

$x_{1}=\frac{(1.4+4)+\sqrt{(-(1.4+4)^{2}-4*4*1)}}{2*4}=1.128$

$x_{1}=\frac{(1.4+4)-\sqrt{(-(1.4+4)^{2}-4*4*1)}}{2*4}=0.2215$

With x1 we get a partial pressure of:

$P_{A}=1.128 atm$

$P_{B}=1-2*1.128 = -1.256 atm$

Since negative partial pressure don´t make physical sense x1 is not the solution for the system.

With x2 we get:

$P_{A}=0.2215 atm$

$P_{B}=1-2*0.2215 = 0.556 atm$

These partial pressures make sense so x2 is the solution for the equation.

We follow the same analysis for the other values of Kp.

Kp=2*10^-4

X1=0.505

X2=0.495

With x1

$P_{A}=0.505atm$

$P_{B}=1-2*0.505 = -0.01005 atm$

Not sense.

With x2

$P_{A}=0.495atm$

$P_{B}=1-2*0.495 = 0.00995 atm$

X2 is the solution for this equation.

Kp=2*10^5

X1=50001

$X2=5*10^{-6}$

With x1

$P_{A}=50001atm$

$P_{B}=1-2*50001=-100001atm$

Not sense.

With x2

$P_{A}= 5*10^{-6}atm$

$P_{B}=1-2*5*10^{-6}= 0.9999 atm$

X2 is the solution for this equation.

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