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2 years ago

5

Which element is oxidized in the following reaction: 2FeCl₂ + Cl₂→2FeCl₃ if Fe goes from + 2 to +3 and Cl goes from 0 to −2?

1 answer:

Blizzard [7]2 years ago

3 0

Answer:

Fe.

Explanation:

The element which is oxidized is the element that losses electrons and its oxidation state be more positive.

The element which is reduced is the element that gain electrons and its oxidation state be more negative.

<em> Fe goes from + 2 to +3, so, it is the element that is oxidized.</em>

<em></em>

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Aclosed system contains an equimolar mixture of n-pentane and isopentane. Suppose the system is initially all liquid at 120°C an

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Explanation:

The given data is as follows.

T = $120^{o}C$ = (120 + 273.15)K = 393.15 K,

As it is given that it is an equimolar mixture of n-pentane and isopentane.

So, $x_{1}$ = 0.5 and $x_{2}$ = 0.5

According to the Antoine data, vapor pressure of two components at 393.15 K is as follows.

$p^{sat}_{1}$ (393.15 K) = 9.2 bar

$p^{sat}_{1}$ (393.15 K) = 10.5 bar

Hence, we will calculate the partial pressure of each component as follows.

$p_{1} = x_{1} \times p^{sat}_{1}$

= $0.5 \times 9.2 bar$

= 4.6 bar

and, $p_{2} = x_{2} \times p^{sat}_{2}$

= $0.5 \times 10.5 bar$

= 5.25 bar

Therefore, the bubble pressure will be as follows.

P = $p_{1} + p_{2}$

= 4.6 bar + 5.25 bar

= 9.85 bar

Now, we will calculate the vapor composition as follows.

$y_{1} = \frac{p_{1}}{p}$

= $\frac{4.6}{9.85}$

= 0.467

and, $y_{2} = \frac{p_{2}}{p}$

= $\frac{5.25}{9.85}$

= 0.527

Calculate the dew point as follows.

$y_{1}$ = 0.5, $y_{2}$ = 0.5

$\frac{1}{P} = \sum \frac{y_{1}}{p^{sat}_{1}}$

$\frac{1}{P} = \frac{0.5}{9.2} + \frac{0.5}{10.2}$

$\frac{1}{P}$ = 0.101966 $bar^{-1}$

P = 9.807

Composition of the liquid phase is $x_{i}$ and its formula is as follows.

$x_{i} = \frac{y_{i} \times P}{p^{sat}_{1}}$

= $\frac{0.5 \times 9.807}{9.2}$

= 0.5329

$x_{z} = \frac{y_{i} \times P}{p^{sat}_{1}}$

= $\frac{0.5 \times 9.807}{10.5}$

= 0.467

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3 years ago

A family is another name for a<br> on the periodic table of elements.

topjm [15]

Answer:

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Explanation:

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3 years ago

Hydrogen and iodine react to form hydrogen iodide, like this: H_2 (g) + I_2 (g) rightarrow 2 HI(g) Also, a chemist finds that at

telo118 [61]

This is an incomplete question, here is a complete question.

Hydrogen and iodine react to form hydrogen iodide, like this:

$H_2(g)+I_2(g)\rightarrow 2HI(g)$

Also, a chemist finds that at a certain temperature the equilibrium mixture of hydrogen, iodine, and hydrogen iodide has the following composition:

Compound Pressure at equilibrium

$H_2$ 61.8 atm

$I_2$ 46.5 atm

$HI$ 52.3 atm

Calculate the value of the equilibrium constant $K_p$ for this reaction. Round your answer to 2 significant digits.

Answer : The value of equilibrium constant $K_p$ for this reaction is, 0.952

Explanation :

The given chemical reaction :

$H_2(g)+I_2(g)\rightarrow 2HI(g)$

The expression of $K_p$ for above reaction follows:

$K_p=\frac{(P_{HI})^2}{P_{H_2}\times P_{I_2}}$

We are given:

$P_{H_2}=61.8atm$

$P_{I_2}=46.5atm$

$P_{HI}=52.3atm$

Putting values in above equation, we get:

$K_p=\frac{(52.3)^2}{61.8\times 46.5}\\\\K_p=0.952$

Therefore, the value of equilibrium constant $K_p$ for this reaction is, 0.952

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