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3 years ago

Which of the following is a product formed when K reacts with Cl2?

Chemistry

2 answers:

Lera25 [3.4K]3 years ago

6 0

Answer: KCl

Explanation: Potassium [K] has an atomic no of 19 and the electronic configuration is:

$K :1s^22s^22p^63s^23p^64s^1$ tends to get stable by losing one electron and forming $K^+$

Chlorine [Cl] has atomic no of 17 and thus the electronic configuration is :

$Cl :1s^22s^22p^63s^23p^5$ tends to get stable by gaining one electron and forming $Cl^-$

$K^+$ and $Cl^-$ attract each other and form an ionic bond. As we have a chlorine molecule that is two atoms of chlorine are present, 2 atoms of potassium will react to form 2 moles of KCl.

$2K+Cl_2\rightarrow 2KCl$

Stels [109]3 years ago

3 0

The answer is A.
Which is... KCI

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2 years ago

Consider the following equilibrium:

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Answer: The equation which is wrong is $K_p=K_c(RT)^{-5}$

Explanation:

For the given reaction:

$4Fe(s)+3O_2(g)\rightarrow 2Fe_2O_3(s)$

The expression for $K_c\text{ and }K_p$ is given by:

$K_c=\frac{1}{[O_2]^3}$

$K_p=\frac{1}{[O_2]^3}$

The concentration of solids are taken to be 1, only concentration of gases and liquid states are taken. The pressure of only gases are taken.

Relationship between $K_p\text{ and }K_c$ is given by the expression:

$K_p=K_c\times (RT)^{\Delta n_g}$

where,

$\Delta n_g$ = number of moles of gaseous products - number of moles of gaseous reactants

R = gas constant

T= temperature

For the above reaction,

$\Delta n_g$ = number of moles of gaseous products - number of moles of gaseous reactants = 0 - 3 = -3

Hence, the expression for $K_p$ is:

$K_p=K_c\times (RT)^{-3}$

Therefore, the equation which is wrong is $K_p=K_c(RT)^{-5}$

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2 years ago

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2 years ago

For the following reaction, KP = 0.455 at 945°C: At equilibrium, is 1.78 atm. What is the equilibrium partial pressure of CH4 in

Alborosie

Answer:

See explanation below

Explanation:

The question is incomplete. However, here's the missing part of the question:

"For the following reaction, Kp = 0.455 at 945 °C:

C(s) + 2H2(g) <--> CH4(g).

At equilibrium the partial pressure of H2 is 1.78 atm. What is the equilibrium partial pressure of CH4(g)?"

With these question, and knowing the value of equilibrium of this reaction we can calculate the partial pressure of CH4.

The expression of Kp for this reaction is:

Kp = PpCH4 / (PpH2)²

We know the value of Kp and pressure of hydrogen, so, let's solve for CH4:

PpCH4 = Kp * PpH2²

*: You should note that we don't use Carbon here, because it's solid, and solids and liquids do not contribute in the expression of equilibrium, mainly because their concentration is constant and near to 1.

Now solving for PpCH4:

PpCH4 = 0.455 * (1.78)²

PpCH4 = 1.44 atm

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2 years ago

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