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1 year ago

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The interhalogen ClF₃ is prepared in a two-step fluorination of chlorine gas:Cl₂(g) + F₂(g) ⇄ CIF(g) CIF(g) + F₂(g) ⇄ CIF₃(g)(a)

Balance each step and write the overall equation.

1 answer:

bogdanovich [222]1 year ago

8 0

The interhalogen CIF3 is prepared in a two-step fluorination of chlorine gas : Cl₂(g) + F₂(g) ⇄ CIF(g) CIF(g) + F₂(g) ⇄ CIF₃(g). The balanced equation will be

Cl₂(g) + 2F₂(g) ⇌ ClF(g) + CIF₃(g)

Solution:

These are the two equation

Cl 2(g)+F 2(g) ⇌ ClF(g) (1)

ClF(g)+F 2 (g) ⇌ ClF 3 (g) (2)

Balance equation one

Cl 2(g)+F 2(g) ⇌ ClF(g)

This will be the result

Cl 2 (g)+F 2 (g)⇌2ClF(g)

Balance equation two

ClF(g)+F 2 (g)⇌ClF 3 (g)

The equation is already balanced

On adding both the equation the result will be

Cl 2 (g)+2F 2(g) ⇌ ClF(g)+ClF 3 (g)

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Using the van der Waals equation, the pressure in a 22.4 L vessel containing 1.00 mol of neon gas at 100 °C is ________ atm. (a

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Answer:

The answer to your question is P = 1.357 atm

Explanation:

Data

Volume = 22.4 L

1 mol

temperature = 100°C

a = 0.211 L² atm

b = 0.0171 L/mol

R = 0.082 atmL/mol°K

Convert temperature to °K

Temperature = 100 + 273

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Formula

$(P + \frac{a}{v^{2}} )(v - b) = RT$

Substitution

$(P + \frac{0.211}{22.4})(22.4 - 0.0171) = (0.082)(373)$

Simplify

(P + 0.0094)(22.3829) = 30.586

Solve for P

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P + 0.0094 = 1.366

P = 1.336 - 0.0094

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3 years ago

the temperature of a sample of water increases from 20celsius to 46.6celsius as it absorbs 5650 J of heat. what is the mass of t

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Answer:

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Explanation:

We have,

Initial temperature of water is 20 degrees Celsius

Final temperature of water is 46.6 degrees Celsius

Heat absorbed is 5650 J

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So,

$m=\dfrac{Q}{c\Delta T}\\\\m=\dfrac{5650}{4.186\times (46.6-20)}\\\\m=50.74\ kg$

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3 years ago

An equilibrium mixture of PCl 5 ( g ) , PCl 3 ( g ) , and Cl 2 ( g ) has partial pressures of 217.0 Torr, 13.2 Torr, and 13.2 To

antoniya [11.8K]

Answer: The new partial pressures of $PCl_5,PCl_3\text{ and }Cl_2$ when equilibrium is re-established are 223.4 torr, 6.82 torr and 26.4 torr respectively.

Explanation:

For the given chemical reaction:

$PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)$

The expression of $K_p$ for above reaction follows:

$K_p=\frac{P_{PCl_5}}{P_{PCl_3}\times P_{Cl_2}}$ ........(1)

We are given:

$P_{PCl_5}=217.0torr$

$P_{PCl_3}=13.2torr$

$P_{Cl_2}=13.2torr$

Putting values in above equation, we get:

$K_p=\frac{217.0}{13.2\times 13.2}\\\\K_p=1.24$

Now we have to calculate the new partial pressure of $Cl_2$ .

$P_{PCl_5}+P_{PCl_3}+P_{Cl_2}=P_{Total}$

$217.0torr+13.2torr+P_{Cl_2}=263.0torr$

$P_{Cl_2}=32.8torr$

The reaction is re-established and proceed to right direction by Le-Chatelier's principle to cancel the effect of addition of $Cl_2$ .

Now, the equilibrium is shifting to the reactant side. The equation follows:

$PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)$

Initial: 13.2 32.8 217.0

At eqm: 13.2-x 32.8-x 217.0+x

Putting values in expression 1, we get:

$1.24=\frac{(217.0+x)}{(13.2-x)(32.8-x)}\\\\x=40.4,6.38$

Neglecting the 40.4 value of 'x' because pressure can not be more than initial partial pressure.

Thus, the value of 'x' will be, 6.38 torr.

Now we have to calculate the new partial pressures after equilibrium is reestablished.

Partial pressure of $PCl_5$ = (217.0+x) = (217.0+6.38) = 223.4 torr

Partial pressure of $PCl_3$ = (13.2-x) = (13.2-6.38) = 6.82 torr

Partial pressure of $Cl_2$ = (32.8-x) = (32.8-6.38) = 26.4 torr

Hence, the new partial pressures of $PCl_5,PCl_3\text{ and }Cl_2$ when equilibrium is re-established are 223.4 torr, 6.82 torr and 26.4 torr respectively.

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3 years ago