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3 years ago

5

Oxidation is now defined in terms of change of oxidation number. Explore how earlier definitions of oxidation and reduction may

have led to conflicting answers for the conversion of P4 to H2PO2− and the way in which the use of oxidation numbers has resolved this.

1 answer:

Sav [38]3 years ago

5 0

Answer:

4

Explanation:

oxidation number resolved 4

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1. The gases SO2, O2 and SO3 are allowed to reach equilibrium at a constant temperature. The equilibrium constant for the reacti

irina [24]

Answer:

a)

$K_2=1.3x10^{2}atm^{-1/2}$

b)

$PCl_5\rightarrow PCl_3+Cl_2$

$[PCl_5]=0.0375M$

Explanation:

Hello!

a) In this case, since we can see that the second reaction is equal to the half of the first reaction, we can relate the equilibrium constants as shown below:

$K_2=K_1^{1/2}$

Thus, by plugging in the the equilibrium constant of the first reaction we obtain:

$K_2=(1.6x10^4atm^{-1})^{1/2}\\\\K_2=1.3x10^{2}atm^{-1/2}$

b) In this case, for the described reaction we can write:

$PCl_5\rightarrow PCl_3+Cl_2$

Thus, the corresponding equilibrium expression is:

$K=\frac{[PCl_3][Cl_2]}{[PCl_5]}$

In such a way, since we know the equilibrium constant and the concentrations of PCl3 and Cl2 at equilibrium, we can compute the concentration of PCl5 at equilibrium as follows:

$[PCl_5]=\frac{[PCl_3][Cl_2]}{K}\\$

$[PCl_5]=\frac{\frac{0.20mol}{4L} *\frac{0.12mol}{4L} }{0.04}$

$[PCl_5]=0.0375M$

Best regards!

7 0

3 years ago

The partial pressure of cartbon dioxide in the atmgsphere is 0239 toer Caiculate the partial pressure in mm He and atm Rpund eac

Verizon [17]

Answer :

The pressure of carbon dioxide in the atmosphere in mmHg is 0.239mmHg.

The pressure of carbon dioxide in the atmosphere in atm is $3.14\times 10^{-4}atm$ .

Explanation :

The conversion used for pressure from torr to mmHg is:

1 torr = 1 mmHg

The conversion used pressure from torr to atm is:

1 atm = 760 torr

or,

$1torr=\frac{1}{760}atm$

As we are given the pressure of carbon dioxide in the atmosphere 0.239 torr. Now we have to determine the pressure of carbon dioxide in the atmosphere in mmHg and atm.

<u>Pressure in mmHg :</u>

As, $1torr=1mmHg$

So, $0.239torr=\frac{0.239torr}{1torr}\times 1mmHg=0.239mmHg$

Thus, the pressure of carbon dioxide in the atmosphere in mmHg is 0.239mmHg.

<u>Pressure in atm:</u>

As, $1torr=\frac{1}{760}atm$

So, $0.239torr=\frac{0.239torr}{1torr}\times \frac{1}{760}atm=3.14\times 10^{-4}atm$

Thus, the pressure of carbon dioxide in the atmosphere in atm is $3.14\times 10^{-4}atm$ .

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3 years ago

What is the eccentricity of mercury

suter [353]

According to my astronomy book, it's about 0.206

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3 years ago

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ICI has a higher boiling point then Br. What is the best explanation for this?

Ahat [919]

Br2 experiences dipole-dipole interactions. ICl experiences dipole-dipole interactions. Br2 forms hydrogen bonds. ICl experiences induced dipole-induced dipole interactions.

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3 years ago

Compare and contrast the elements Potassium, K, and Neon, Ne. Use at least ten comparisons or descriptions.

vovikov84 [41]

Answer:

K is a group 1 element while Ne is a group 0 element

K is a good reducing agent while Ne is inert

K is metal while Ne is a gas

K has one valence electron while Ne has no valence electron

Explanation:

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3 years ago