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Butoxors [25]
3 years ago
11

For the reaction COCl2(g)⇌CO(g)+Cl2(g), K= 2.19×10−10 at 373 K

Chemistry
1 answer:
vladimir1956 [14]3 years ago
6 0

Answer:

D) The equilibrium lies far to the left

Explanation:

According to the law of mass action, the equilibrium constant K for the reaction at 373K can be calculated as follows:

K = \frac{[CO][Cl_{2}]}{[COCl_{2}]} = 2.19×10^{-10}

([X] means = concentration of X)

This means that in the equilibrium the concentration of the reactant (that is in the denominator) will be much higher (around 10^{10} fold) than the concentrations of the products (that are in the numerator), and this means that the equilibrium lies far to the left (to the reactants side) as very small amount of product is being formed.

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The tarnish that forms on objects made of silver is solid silver sulphide. This can be removed by reacting it with aluminium met
steposvetlana [31]

Answer:

5.83 mol.

Explanation:

  • From the balanced reaction:

<em>2Al + 3Ag₂S → 6Ag + Al₂S₃,</em>

It is clear that 2 mol of Al react with 3 mol of Ag₂S to produce 1 mol of Ag and 1 mol of Al₂S₃.

Al reacts with Ag₂S with (2: 3) molar ratio.

<em>So, 2.27 mol of Al reacts completely with 3.4 mol of Ag₂S with (2: 3) molar ratio.</em>

<em />

  • The limiting reactant is Ag₂S.
  • The excess "left over" reactant is Al.

The reamining moles of excess reactant "Al" = 8.1 mol - 2.27 mol = 5.83 mol.

6 0
3 years ago
Use the information below to explain why the atomic radius decreases down a group.
notsponge [240]

Answer:

Detail is given below

Explanation:

Atomic radii trend along group:

As we move down the group atomic radii increased with increase of atomic number. The addition of electron in next level cause the atomic radii to increased. The hold of nucleus on valance shell become weaker because of shielding of electrons thus size of atom increased.

As the size of atom increases the ionization energy from top to bottom also  decreases because it becomes easier to remove the electron because of less nuclear attraction and as more electrons are added the outer electrons becomes more shielded and away from nucleus.

In A we can see that there is one positive charge and force of attraction is 2.30×10⁻⁸ N and distance is 0.10 nm

In B we can see that negative charge is further away from nucleus because of greater distance thus force of attraction will be less. 0.58×10⁻⁸ N

In C this distance further increases and force also goes in decreasing 0.26×10⁻⁸ N.

3 0
3 years ago
The equilibrium constant, Kc, for the following reaction is 5.10×10-6 at 548 K. NH4Cl(s) NH3(g) + HCl(g) Calculate the equilibri
levacccp [35]

<u>Answer:</u> The equilibrium concentration of HCl is 2.26\times 10^{-3}M

<u>Explanation:</u>

We are given:

Moles of NH_4Cl(s) = 0.564 moles

Volume of vessel = 1.00 L

Molarity is calculated by using the equation:

\text{Molarity}=\frac{\text{Number of moles}}{\text{Volume of solution (in L)}}

Molarity of NH_4Cl=\frac{0.564}{1}=0.564M

The given chemical equation follows:

                  NH_4Cl(s)\rightleftharpoons NH_3(g)+HCl(g)

<u>Initial:</u>         0.564

<u>At eqllm:</u>     0.564-x          x              x

The expression of K_c for above equation follows:

K_c=[NH_3][HCl]

The concentration of pure solid and pure liquid is taken as 1.

We are given:

K_c=5.10\times 10^{-6}

Putting values in above equation, we get:

5.10\times 10^{-6}=x\times x\\\\x=2.26\times 10^{-3}M,-2.26\times 10^{-3}M

Negative sign is neglected because concentration cannot be negative.

So, [HCl]=2.26\times 10^{-3}M

Hence, the equilibrium concentration of HCl is 2.26\times 10^{-3}M

5 0
3 years ago
A student wants to know why sandhill crane‘s fly south for the winter. The student uses a map of North America to make a model o
Ratling [72]

Answer:

a and c

Explanation:

im probably wrong  though

7 0
3 years ago
Read 2 more answers
Consider the formation of [Ni(en)3]2+ from [Ni(H2O)6]2+. The stepwise ΔG∘ values at 298 K are ΔG∘1 for first step=−42.9 kJ⋅mol−1
timurjin [86]

Answer:

kf = 1.16 x 10¹⁸

Explanation:

Step 1: [Ni(H₂O)₆]²⁺  + 1en → [Ni(H₂O)₄(en)]²⁺  ΔG°1 = -42.9 kJmol⁻¹

Step 2: [Ni(H₂O)₄(en)]²⁺  + 1en → [Ni(H₂O)₂(en)₂]²⁺  ΔG°2 = -35.8 kJmol⁻¹

Step 3: [Ni(H₂O)₂(en)₂]²⁺ + 1en →  [Ni(en)₃]²⁺  ΔG°3 = -24.3 kJmol⁻¹

________________________________________________________

Overall reaction: [Ni(H₂O)₆]²⁺  + 3en → [Ni(en)₃]²⁺  ΔG°r

ΔG°r = ΔG°1 + ΔG°2 + ΔG°3

ΔG°r = -42.9 - 35.8 - 24.3

ΔG°r = -103.0 kJmol⁻¹

ΔG°r = -RTlnKf

-103,000 Jmol⁻¹ =  - 8.31 J.K⁻¹mol⁻¹ x 298 K x lnKf

kf = e ^(-103,000/-8.31x298)

kf = e ^41.59

kf = 1.16 x 10¹⁸

7 0
3 years ago
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